# An Intuitive Guide To Exponential Functions & e

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e has always bothered me — not the letter, but the mathematical constant. What does it really mean?

Math books and even my beloved Wikipedia describe e using obtuse jargon:

The mathematical constant e is the base of the natural logarithm.

And when you look up natural logarithm you get:

The natural logarithm, formerly known as the hyperbolic logarithm, is the logarithm to the base e, where e is an irrational constant approximately equal to 2.718281828459.

Nice circular reference there. It’s like a dictionary that defines labyrinthine with Byzantine: it’s correct but not helpful. What’s wrong with everyday words like “complicated”?

I’m not picking on Wikipedia — many math explanations are dry and formal in their quest for rigor. But this doesn’t help beginners trying to get a handle on a subject (and we were all a beginner at one point).

No more! Today I’m sharing my intuitive, high-level insights about what e is and why it rocks. Save your rigorous math book for another time. Here’s a quick video overview of the insights:

## e is NOT Just a Number

Describing e as “a constant approximately 2.71828…” is like calling pi “an irrational number, approximately equal to 3.1415…”. Sure, it’s true, but you completely missed the point.

Pi is the ratio between circumference and diameter shared by all circles. It is a fundamental ratio inherent in all circles and therefore impacts any calculation of circumference, area, volume, and surface area for circles, spheres, cylinders, and so on. Pi is important and shows all circles are related, not to mention the trigonometric functions derived from circles (sin, cos, tan).

e is the base rate of growth shared by all continually growing processes. e lets you take a simple growth rate (where all change happens at the end of the year) and find the impact of compound, continuous growth, where every nanosecond (or faster) you are growing just a little bit.

e shows up whenever systems grow exponentially and continuously: population, radioactive decay, interest calculations, and more. Even jagged systems that don’t grow smoothly can be approximated by e.

Just like every number can be considered a scaled version of 1 (the base unit), every circle can be considered a scaled version of the unit circle (radius 1), and every rate of growth can be considered a scaled version of e (the unit rate of growth).

So e is not an obscure, seemingly random number. e represents the idea that all continually growing systems are scaled versions of a common rate.

## Understanding Exponential Growth

Let start by looking at a basic system that doubles after an amount of time. For example,

• Bacteria can split and “doubles” every 24 hours
• We get twice as many noodles when we fold them in half.
• Your money doubles every year if you get 100% return (lucky!)

And it looks like this:

Splitting in two or doubling is a very common progression. Sure, we can triple or quadruple, but doubling is convenient, so hang with me here.

Mathematically, if we have x splits then we get 2x times as much stuff than when we started. With 1 split we have 21 or 2 times as much. With 4 splits we have 24 = 16 times as much. As a general formula:

$\displaystyle{ growth = 2^x }$

Said another way, doubling is 100% growth. We can rewrite our formula like this:

$\displaystyle{ growth = (1 + 100\%)^x}$

It’s the same equation, but we separate 2 into what it really is: the original value (1) plus 100%. Clever, eh?

Of course, we can substitute any number (50%, 25%, 200%) for 100% and get the growth formula for that new rate. So the general formula for x periods of return is:

$\displaystyle{growth = (1 + return)^x}$

This just means we use our rate of return, (1 + return), “x” times in a row.

## A Closer Look

Our formula assumes growth happens in discrete steps. Our bacteria are waiting, waiting, and then boom, they double at the very last minute. Our interest earnings magically appear at the 1 year mark. Based on the formula above, growth is punctuated and happens instantly. The green dots suddenly appear.

The world isn’t always like this. If we zoom in, we see that our bacterial friends split over time:

Mr. Green doesn’t just show up: he slowly grows out of Mr. Blue. After 1 unit of time (24 hours in our case), Mr. Green is complete. He then becomes a mature blue cell and can create new green cells of his own.

Does this information change our equation?

Nope. In the bacteria case, the half-formed green cells still can’t do anything until they are fully grown and separated from their blue parents. The equation still holds.

## Money Changes Everything

But money is different. As soon as we earn a penny of interest, that penny can start earning micro-pennies of its own. We don’t need to wait until we earn a complete dollar in interest — fresh money doesn’t need to mature.

Based on our old formula, interest growth looks like this:

But again, this isn’t quite right: all the interest appears on the last day. Let’s zoom in and split the year into two chunks. We earn 100% interest every year, or 50% every 6 months. So, we earn 50 cents the first 6 months and another 50 cents in the last half of the year:

But this still isn’t right! Sure, our original dollar (Mr. Blue) earns a dollar over the course of a year. But after 6 months we had a 50-cent piece, ready to go, that we neglected! That 50 cents could have earned money on its own:

Because our rate is 50% per half year, that 50 cents would have earned 25 cents (50% times 50 cents). At the end of 1 year we’d have

• Our original dollar (Mr. Blue)
• The dollar Mr. Blue made (Mr. Green)
• The 25 cents Mr. Green made (Mr. Red)

Giving us a total of $2.25. We gained$1.25 from our initial dollar, even better than doubling!

Let’s turn our return into a formula. The growth of two half-periods of 50% is:

$\displaystyle{growth = (1 + 100\%/2)^{2} = 2.25}$

## Diving into Compound Growth

It’s time to step it up a notch. Instead of splitting growth into two periods of 50% increase, let’s split it into 3 segments of 33% growth. Who says we have to wait for 6 months before we start getting interest? Let’s get more granular in our counting.

Charting our growth for 3 compounded periods gives a funny picture:

Think of each color as shoveling money upwards towards the other colors (its children), at 33% per period:

• Month 0: We start with Mr. Blue at $1. • Month 4: Mr. Blue has earned 1/3 dollar on himself, and creates Mr. Green, shoveling along 33 cents. • Month 8: Mr. Blue earns another 33 cents and gives it to Mr. Green, bringing Mr. Green up to 66 cents. Mr. Green has actually earned 33% on his previous value, creating 11 cents (33% * 33 cents). This 11 cents becomes Mr. Red. • Month 12: Things get a bit crazy. Mr. Blue earns another 33 cents and shovels it to Mr. Green, bringing Mr. Green to a full dollar. Mr. Green earns 33% return on his Month 8 value (66 cents), earning 22 cents. This 22 cents gets added to Mr. Red, who now totals 33 cents. And Mr. Red, who started at 11 cents, has earned 4 cents (33% * .11) on his own, creating Mr. Purple. Phew! The final value after 12 months is: 1 + 1 + .33 + .04 or about 2.37. Take some time to really understand what’s happening with this growth: • Each color earns interest on itself and hands it off to another color. The newly-created money can earn money of its own, and on the cycle goes. • I like to think of the original amount (Mr. Blue) as never changing. Mr. Blue shovels money to create Mr. Green, a steady 33 every 4 months since Mr. Blue does not change. In the diagram, Mr. Blue has a blue arrow showing how he feeds Mr. Green. • Mr. Green just happens to create and feed Mr. Red (green arrow), but Mr. Blue isn’t aware of this. • As Mr. Green grows over time (being constantly fed by Mr. Blue), he contributes more and more to Mr. Red. Between months 4-8 Mr. Green gives 11 cents to Mr. Red. Between months 8-12 Mr. Green gives 22 cents to Mr. Red, since Mr. Green was at 66 cents during Month 8. If we expanded the chart, Mr. Green would give 33 cents to Mr. Red, since Mr. Green reached a full dollar by Month 12. Make sense? It’s tough at first — I even confused myself a bit while putting the charts together. But see that each dollar creates little helpers, who in turn create helpers, and so on. We get a formula by using 3 periods in our growth equation: $\displaystyle{growth = (1 + 100\%/3)^3 = 2.37037...}$ We earned$1.37, even better than the $1.25 we got last time! ## Can We Get Infinite Money? Why not take even shorter time periods? How about every month, day, hour, or even nanosecond? Will our returns skyrocket? Our return gets better, but only to a point. Try using different numbers of n in our magic formula to see our total return: n (1 + 1/n)^n ------------------ 1 2 2 2.25 3 2.37 5 2.488 10 2.5937 100 2.7048 1,000 2.7169 10,000 2.71814 100,000 2.718268 1,000,000 2.7182804 ...  The numbers get bigger and converge around 2.718. Hey… wait a minute… that looks like e! Yowza. In geeky math terms, e is defined to be that rate of growth if we continually compound 100% return on smaller and smaller time periods: $\displaystyle{growth = e = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n}$ This limit appears to converge, and there are proofs to that effect. But as you can see, as we take finer time periods the total return stays around 2.718. ## But what does it all mean? The number e (2.718…) is the maximum possible result when compounding 100% growth for one time period. Sure, you started out expecting to grow from 1 to 2 (that’s a 100% increase, right?). But with each tiny step forward you create a little dividend that starts growing on its own. When all is said and done, you end up with e (2.718…) at the end of 1 time period, not 2. e is the maximum, what happens when we compound 100% as much as possible. So, if we start with$1.00 and compound continuously at 100% return we get 1e. If we start with $2.00, we get 2e. If we start with$11.79, we get 11.79e.

e is like a speed limit (like c, the speed of light) saying how fast you can possibly grow using a continuous process. You might not always reach the speed limit, but it’s a reference point: you can write every rate of growth in terms of this universal constant.

(Aside: Be careful about separating the increase from the final result. 1 becoming e (2.718…) is an increase (growth rate) of 171.8%. e, by itself, is the final result you observe after all growth is taken into account (original + increase)).

Good question. What if we grow at 50% annually, instead of 100%? Can we still use e?

Let’s see. The rate of 50% compound growth would look like this:

$\displaystyle{\lim_{n\to\infty} \left( 1 + \frac{.50}{n} \right)^n}$

Hrm. What can we do here? Remember, 50% is the total return, and n is the number of periods to split the growth into for compounding. If we pick n=50, we can split our growth into 50 chunks of 1% interest:

$\displaystyle{\left( 1 + \frac{.50}{50} \right)^{50} = \left( 1 + .01 \right)^{50}}$

Sure, it’s not infinity, but it’s pretty granular. Now imagine we also divided our regular rate of 100% into chunks of 1%:

$\displaystyle{e \approx \left( 1 + \frac{1.00}{100} \right)^{100} = \left( 1 + .01 \right)^{100}}$

Ah, something is emerging here. In our regular case, we have 100 cumulative changes of 1% each. In the 50% scenario, we have 50 cumulative changes of 1% each.

What is the difference between the two numbers? Well, it’s just half the number of changes:

$\displaystyle{\left( 1 + .01 \right)^{50} = \left( 1 + .01 \right)^{100/2} = \left( \left( 1 + .01 \right)^{100}\right)^{1/2} = e^{1/2} }$

This is pretty interesting. 50 / 100 = .5, which is the exponent we raise e to. This works in general: if we had a 300% growth rate, we could break it into 300 chunks of 1% growth. This would be triple the normal amount for a net rate of e3.

Even though growth can look like addition (+1%), we need to remember that it’s really a multiplication (x 1.01). This is why we use exponents (repeated multiplication) and square roots (e1/2 means “half” the number of changes, i.e. half the number of multiplications).

Although we picked 1%, we could have chosen any small unit of growth (.1%, .0001%, or even an infinitely small amount!). The key is that for any rate we pick, it’s just a new exponent on e:

$\displaystyle{growth = e^{rate}}$

Suppose we have 300% growth for 2 years. We’d multiply one year’s growth (e3) by itself:

$\displaystyle{growth = \left(e^{3}\right)^{2} = e^{6}}$

And in general:

$\displaystyle{growth = \left(e^{rate}\right)^{time} = e^{rate \cdot time}}$

Because of the magic of exponents, we can avoid having two powers and just multiply rate and time together in a single exponent.

## The big secret: e merges rate and time.

This is wild! ex can mean two things:

• x is the number of times we multiply a growth rate: 100% growth for 3 years is e3
• x is the growth rate itself: 300% growth for one year is e3

Won’t this overlap confuse things? Will our formulas break and the world come to an end?

It all works out. When we write:

$\displaystyle{e^x}$

the variable x is a combination of rate and time.

$\displaystyle{x = rate \cdot time}$

Let me explain. When dealing with continuous compound growth, 10 years of 3% growth has the same overall impact as 1 year of 30% growth (and no growth afterward).

• 10 years of 3% growth means 30 changes of 1%. These changes happen over 10 years, so you are growing continuously at 3% per year.
• 1 period of 30% growth means 30 changes of 1%, but happening in a single year. So you grow for 30% a year and stop.

The same “30 changes of 1%” happen in each case. The faster your rate (30%) the less time you need to grow for the same effect (1 year). The slower your rate (3%) the longer you need to grow (10 years).

But in both cases, the growth is e.30 = 1.35 in the end. We’re impatient and prefer large, fast growth to slow, long growth but e shows they have the same net effect.

So, our general formula becomes:

$\displaystyle{growth = e^x = e^{rt}}$

If we have a return of r for t time periods, our net compound growth is ert. This even works for negative and fractional returns, by the way.

## Example Time!

Examples make everything more fun. A quick note: We’re so used to formulas like 2x and regular, compound interest that it’s easy to get confused (myself included). Read more about simple, compound and continuous growth.

These examples focus on smooth, continuous growth, not the jumpy growth that happens at yearly intervals. There are ways to convert between them, but we’ll save that for another article.

Example 1: Growing crystals

Suppose I have 300kg of magic crystals. They’re magic because they grow throughout the day: I watch a single crystal, and in the course of 24 hours it sheds off its own weight in crystals. (The baby crystals start growing immediately at the same rate, but I can’t track that — I’m watching how much the original sheds). How much will I have after 10 days?

Well, since the crystals start growing immediately, we want continuous growth. Our rate is 100% every 24 hours, so after 10 days we get: 300 · e1 · 10 = 6.6 million kg of our magic gem.

This can be tricky: notice the difference between the input rate and the total output rate. The “input” rate is how much a single crystal changes: 100% in 24 hours. The net output rate is e (2.718x) because the baby crystals grow on their own.

In this case we have the input rate (how fast one crystal grows) and want the total result after compounding (how fast the entire group grows because of the baby crystals). If we have the total growth rate and want the rate of a single crystal, we work backwards and use the natural log.

Example 2: Maximum interest rates

Suppose I have $120 in an account with 5% interest. My bank is generous and gives me the maximum possible compounding. How much will I have after 10 years? Our rate is 5%, and we’re lucky enough to compound continuously. After 10 years, we get$120 · e.05 · 10 = $197.85. Of course, most banks aren’t nice enough to give you the best possible rate. The difference between your actual return and the continuous one is how much they don’t like you. Example 3: Radioactive decay I have 10kg of a radioactive material, which appears to continuously decay at a rate of 100% per year. How much will I have after 3 years? Zip? Zero? Nothing? Think again. Decaying continuously at 100% per year is the trajectory we start off with. Yes, we do begin with 10kg and expect to “lose it all” by the end of the year, since we’re decaying at 10 kg/year. We go a few months and get to 5kg. Half a year left? Nope! Now we’re losing at a rate of 5kg/year, so we have another full year from this moment! We wait a few more months, and get to 2kg. And of course, now we’re decaying at a rate of 2kg/year, so we have a full year (from this moment). We get 1 kg, have a full year, get to .5 kg, have a full year — see the pattern? As time goes on, we lose material, but our rate of decay slows down. This constantly changing growth is the essence of continuous growth & decay. After 3 years, we’ll have 10 · e-1 · 3 = .498 kg. We use a negative exponent for decay — we want a fraction (1/ert) vs a growth multiplier (ert). [Decay is commonly given in terms of "half life" -- we'll talk about converting these rates in a future article.] More Examples If you want fancier examples, try the Black-Scholes option formula (notice e used for exponential decay in value) or radioactive decay. The goal is to see ert in a formula and understand why it’s there: it’s modeling a type of growth or decay. And now you know why it’s “e”, and not pi or some other number: e raised to “r*t” gives you the growth impact of rate r and time t. ## There’s More To Learn! My goal was to: • Explain why e is important: It’s a fundamental constant, like pi, that shows up in growth rates. • Give an intuitive explanation: e lets you see the impact of any growth rate. Every new “piece” (Mr. Green, Mr. Red, etc.) helps add to the total growth. • Show how it’s used: ex lets you predict the impact of any growth rate and time period. • Get you hungry for more: In the upcoming articles, I’ll dive into other properties of e. This article is just the start — cramming everything into a single page would tire you and me both. Dust yourself off, take a break and learn about e’s evil twin, the natural logarithm. ## Other Posts In This Series Kalid Azad loves sharing Aha! moments. BetterExplained is dedicated to learning with intuition, not memorization, and is honored to serve 250k readers monthly. Enjoy this article? Try the site guide or join the newsletter: Math, Better Explained is a highly-regarded Amazon bestseller. This 12-part book explains math essentials in a friendly, intuitive manner. "If 6 stars were an option I'd give 6 stars." -- read more reviews ## 631 Comments 1. Anonymous says: Nice graphics! 2. Bob says: Fantastic Kalid! I consider myself a pretty bright guy and work with numbers every day but e has always been an opaque subject to me. Ever since high school it’s been hanging out with its friend the logarithm with nothing to do but mock me! I think this article is going to help me think about e in new ways and I may even put it to use. I’m really glad I found your site! 3. Kalid says: Great Bob, I’m glad you liked it! Yeah, e and its pesky friends like natural logarithms were a thorn in my side for a while. It really bugs me when I use a concept without *really* knowing what it meant. E got all this attention and I wanted to dig in and see what all the fuss was about I’ll be writing more on this topic as there are some really interesting ways of looking at “this constant, approximately equal to 2.71828…” 4. Jayson Waddell says: nice article, it’s so rare to find actual explanations when it comes to mathematical ideas, I was very happy to have found this one. 5. Kalid says: Thanks Jayson, the lack of intuitive explanations motivated me to start this site. I’m glad you are finding it useful 6. Stephen says: I have a question. you wrote that x=rate*time and that e^x=growth. using this information is it possible to find the smallest amount of time allowed, the gap between one time to the next. Does this give us the ability to give a value for that infinitesimally small value between each moment in time. Just a random ponderance. Nice guides specially enjoyed the vector calculus series. I was getting worried that i would have to take math as a second language. The words they use in calc. books are too big for me. 7. Kalid says: Hi Stephen, great question! Yes, if you take the unit of time smaller and smaller, you will get the instantaneous rate of growth at a point. And the surprising thing is that this is e^x! I want to write more about the calculus of it, but basically, when you have a certain amount of “stuff” (say, 10 units) then you are *growing* at 10 units per unit of time as well. Of course, once you grow just a little bit you have a new amount of “stuff” (10.1 units) and now you are growing at 10.1 per unit time. It’s a bit mind-boggling, but it’s the way e works — the current instantaneous rate of growth is equal to the current amount. I’ll be writing more on this as I get a good, intuitive understanding of it 8. joe says: Holy cow….somehow despite four semesters of calculus I forgot or failed to grasp that with any calculator I can do compound interest calculations as easily as circle areas. I’ve been dependent on financial calculators for twenty years. Now I’m going to read all your math-related posts. Keep writing! 9. That’s awesome Joe! I know what you mean, I had forgotten about being able to do compound interest as well — it’s funny how rarely we revisit old topics we’ve learned. E had always bothered me. I hope you enjoy the other posts, I’ll keep cranking them out 10. Dr Jani says: Thanks for the wonderful explanation. However, at places, confusion seems to arise due to wrong use of terminology. e.g. instead of ‘e is the fundamental rate of change shared by all continually growing processes.’ it should be ‘e is the fundamental net growth in all continually growing processes (in a unit time that would account for 100% simple growth)’, if I have understood correctly. Elsewhere, you do say growth=e^rt, where r is the rate of growth. So e is the total growth & not rate of growth. 11. Hi Dr. Jani, that’s a great point — the current description of e is confusing. As you say, e^1 = e = the amount of continuous growth after 1 unit of time (assuming growing at a rate of 100% simple growth per unit time). e^rt lets you compute the net growth for any rate and time. I’ll update the article to make this more clear, I appreciate the feedback! 12. Wow thank you so much for explaining e. The wikipedia article completely confused me about it 13. Kalid says: Thanks David. Yeah, it really bugs me when math topics are explained in a complicated way, I’m glad you liked it. 14. Sophie says: Kalid – thank you (and your contributors for their comments) for this site. I hope that your blog will be even more popular than Wiki and be the “goto” place for teachers and their students! 15. Thanks for the encouragement Sophie, you must have read my mind! I’m hoping this blog evolves into a place to think about new topics in a fresh, intuitive way, and share those “a-ha” moments from everyone. Wikipedia is a good reference, but encyclopedias tend to focus on facts vs. understanding. There’s a place for both 16. Jobie says: This is a great discussion. You should do the fundamental theorem of calculus, also. 17. Thanks Jobie, appreciate the comment. Yep, the Fundamental Theorem of Calculus is definitely on my list of upcoming topics 18. Dan H says: This was really cool – thanks for taking the time to writ it up. I realy have a grasp of e now! 19. Kalid says: Awesome Dan, I’m happy it helped! 20. Jut says: Very nice explanation. Even though I have a BS in Engineering, I still didn’t really know what e was. You brought back some memories and helped reinforce the ol’ e. Your intuitive explanation reminded me of this article on impedance: http://sci-toys.com/attention/2006/04/impedance-matching.html BTW, what is your profession and education? Thanks! 21. Hi Jut, I’m glad you liked it. I find there’s many subjects we just plow through without really understanding — that feeling has always bothered me :). I’ve got a BS in Computer Science and do programming/web development now, but enjoy a wide variety of subjects (there’s more in the about section). 22. Alice says: I think I get this whole expotental thing a little more. Thanks! In class, they kind of throw this thing at you. 23. Kalid says: Thanks Alice, that’s great! Yeah, things become much easier when you have an intuitive understanding, I’m happy it was able to help. 24. Saurabh says: Thanks man!!! I have a BE in CSE ,still could not get what ‘e’ is. passing exams and understanding the concepts is really totally different.Thanks again. Please keep it up. 25. Jayson Waddell says: Hello, what about exponential decay when the rate is negative? With positive rates it is easy to understand but for some reason I can’t see it with negative rates 26. Hi Jayson, negative exponents can be tough. Positive exponents imply “growing” and negative exponents mean “shrinking”. So, e^3 is “The amount of growth after 3 periods of time (assuming 100% growth rate, compounded).” What’s e^-3? Well, it’s a growth rate of -100%, but what does that mean? If you could shrink for 100% for one unit of time, you’d end up with zero. However, e is about continuous growth. So, as you shrink a little bit, your *rate* of shrinking slows down. Instead of shrinking at 100%, maybe you’re shrinking at 90%. As you shrink more, your rate may decrease to 80%. Then 70%. Eventually, your rate of shrinking is around 1% and it looks like you aren’t shrinking at all. Thinking about it more, negative exponents are a bit strange :). Another way is to consider negative exponents as negative *time*, rather than negative *growth*. e^3 * e^-3 = 1. So, e^-3 can mean “What value do I start with, and grow for 3 units, to get my current value?” e^-3 would be a really, really small number if it’s allowed to grow. When we look forward, we see growth. When we look backwards, we see decay. After all, 100 dollars in 2007 might look like “decay” when you look back at it from 2010 (when you have 1000 dollars). In some cases, an amount really is shrinking as time goes forward (and looking back in time looks like growth). A bit confusing, but hope this helps. 27. taz says: Can we look forward for more math explanation. The Articles you write can change the world (majority ) view on mathematics. We need such good teachers in our schools. Thanks 28. Thanks for the encouragement Taz — I have more posts in the works, but want to get them to high enough quality. I love helping people learn, and if I can help people enjoy “hated” subjects like math all the better 29. Nasir says: Awesome explanation Something I wanted to know for a long time. Please write more of the same kind of articles ! 30. Michael says: Well..! Finding good teachers is like finding a treasure. I got to your explanation by.. serendipity. Thank you very much! Internet allows finding people who “just” do extraordinary pieces of work. All the best! 31. Hi Nasir and Michael, thanks for the kind words! I have more articles in the works, so I hope you enjoy them as well. I completely agree about the Internet in terms of bringing people together and discovering new things — I’m so happy I was born in this “modern” time period :). 32. Dan says: The best explanation of e and ln and have ever seen, and I have looked before. Absolutely awesome. 33. Wow, thanks Dan, happy you liked it. I try to explain things the way I wish I were taught. 34. sophia says: What about -e to the x power? My daughter’s pre-cal teacher gave this as homework and test question to graph. I believe the base should be positive, else you will get a hilly graph. Please explain 35. Hi Sophia, good question. Did the teacher mean -(e^x) or (-e)^x? If it’s the first, it looks like the reverse of the e^x and isn’t too hard to graph. If it’s the second (negative base), things get tricky. For whole numbers like (-e)^1, (-e)^2, (-e)^3 you can plot it out, and it will be hilly, jumping from negative to positive. The tricky thing comes with intermediate values like (-e)^.5. This means the square root of -e, which is an imaginary number! Any non-integer exponent will have this problem. So, the only real values you can plot are for the integers 1,2,3,4,5, etc. I’d ask the teacher to double-check for a typo, but otherwise you can only plot it out for the integer values. 36. sophia says: Thanks a bunch Kalid. You confirmed what I thought. I intend to ask the teacher about this again. My objective is for my daughter and the rest of the class to fully understand the concepts as they are introduced. Thanks so much. 37. No problem, glad it was useful for you! 38. jim says: Now I have clue on e! Thanks! I want to be an actuary now! 39. Kalid says: Thanks Jim 40. Charanjeev Singh says: I just “e”njoyed my a-ha moment! 41. Awesome, very nice! 42. Icegirl says: WOW! Amazing how i was able to condense 16 years of learning into an hour long window! All those years spent into oblivion when teachers would come to class, scribble something in greek (literally)on the board, while students hoped helplessly that they’d someday understand all this! With your tutorial, the world has hope again! I wish teachers would focus on smaller syllabus, but more pertinent, practial, day-to-day stuff! Thank you! You are my official math messiah! 43. Hi Icegirl, thanks for the wonderful comment! I’m happy the article was able to help — yes, the goal is to focus on practical, “this makes sense” insights. 44. Kalid, Thanks for the great explanation. My comment is looking for a comment from you. A new money system. Every one gets 100 units a day for a year. At the end of the year there are 36,500 units of currency in circulation for each person. The 100 units have diminished as a proportion of the whole to the point of 1/365. On the first day of the second year the daily allotment is increased by 1/365. This keeps the proportion of the daily allotment constant at 1/365. Over year this approximates e, and if the growth is continuous, it is e. I call the type of money salmoney, and this particular version excalibrator. I guess what I’m looking for is does this resonate with you as an apt or accurate use of e. Thanks for your time. My website is jaspersbox.com It is the attempt to create a new currency with a built in universal dividend. Thanks for listening. S. Clark 45. Hi Steven, thanks for writing, glad you liked the article. I’m not quite sure what you meant by “increasing by 1/365″ — do you want to keep the proportion of money handed out the same? (Even in the first case, giving out 100 per day, the percent changes day after day). You *can* use e to convert between simple and compound interest — I’m planning on a follow-up article on this topic, which may do a better job of answering your question than I’m doing now 46. todd says: This is a great explanation…. It has been twenty years since I have studied this area of mathematics and then I just memorized it and went on about my business. I am now working on a wind turbine design and (to confess) having to look a few things that I have forgotten up. I came upon a speed gradient and surface roughness equation as applied to wind velocity that utilized the natural log of the height / surface roughness length. I did an internet search on the concept of the natural log to brush up on it and found this little jewel. This is one of the best explanations I’ve yet to come across. 47. Hi Todd, thanks for the wonderful comment! I’m really glad it was useful to you, and hope you figured out that turbine design 48. Hi Kalid, thanks a lot for the intuitive explanation. I’ve wondered a lot about “e”. I linked to this article from my lj above. Keep your good work going! 49. Hi Deepak, thanks for the comment — glad you liked it! 50. Fantastic website. I love the explanation. Well definitely be back to read up some more. 51. Kalid says: Thanks Shaw, glad you like it! 52. barry says: Nice job, Kalid. Now I’m going to try to explain it to my wife using your approach. 53. Kalid says: Thanks Barry, hope she enjoys it 54. Denis says: Hi Kalid, I have a question here. You have given an example to us as follows: “I have$120 in an account bearing 5% interest. I keep it for 10 years. Assuming compound growth, I’d have 120 * e^(.05 * 10) = 197.85 after the 10 years.”

If I calculate it in the compound interest formula, the results are different:
$120 * (1 + 0.05)^10 =$195

Could you explain why? Thanks

55. Kalid says:

Hi Denis, great question — clarifying the meaning between simple and continuous growth rates is something I’d like to write about more.

At a high level, “simple” interest is the *final* amount of growth, i.e. what you get at the end of the year. So 5% simple interest for 10 years would be (1.05)^10, as you wrote.

Continuous interest is the *current* rate of growth, i.e. how fast are we growing at this instant. In that case, you use e^(.05 * 10) to find the impact of 5% current growth for 10 time periods.

The reason continuous growth is more is because the “interest you earn, earns interest”. That is, after 1/2 a year you have earned 2.5%. After another half year, that 2.5% interest has earned 2.5% on itself. You can cut the time slice smaller and smaller (like we did for 1/3 of a year above) and see how small chunks of interest can earn more. With simple interest, you just say “I’ll give you 5% at the end of the year… and no compounding before then. If you’re lucky, I’ll spread out these payments each month.” Most banks list *simple interest* as their annual percentage rate, as it’s sometimes easier to think about.

However, continuous growth is more common in science and the real world (radioactive decay), as most things have a current growth rate. We don’t figure out the final amount and work backwards, as banks may. I’d like to write more about this, thanks for bringing it up.

56. Denis says:

Thank Kalid. So I guess it is more appropriate to use e for scientific calculation but not bank transaction calculation as bank’s compound growth rate is usually annual but not continuous.

57. Kalid says:

Hi Denis, that’s right — banks usually express their interest rates in terms of simple interest over the year.

58. Mark says:

It seems trite to say, after so many comments of the same; however, I have to echo: this is explanation was fantastic!!! When I read about mathematic principles that are so clearly laid out, it reminds me of how much math is such a direct in-road to philosophy. Thanks for helping reveal the mystery. I’m looking forward to you writing more of the same. Very best regards, Mark

59. Kalid says:

Hi Mark, thanks for the wonderful comment! It’s great to hear that explanations are working, it helps me tailor upcoming articles to do more of the same.

Yes, I think any math principle can be explained simply — it’s us humans who tend to complicate things :). The irony is that it takes a lot of thinking and effort to reveal the “mystery” which is simple after all. I’ll do my best to keep cranking out articles, thanks for dropping by.

60. Anonymous says:

I have one for you, How can you explain Euler’s formula re^j(theta) in a understandable way.
Read r multiplied by e to the power of imaginary unit j x angle theta

61. Kalid says:

Euler’s formula is on the list — but first I need to do an explanations of complex numbers, possibly sine and cosine as well

62. Very nice explanation indeed!

63. Kalid says:

Thanks Garry!

64. Hi Kalid,

As others have already said, that’s a really nice explanation of the constant e.

I think you can explain Euler’s identity eiπ + 1 = 0 using not much more machinery than you’ve developed for the real-valued case. In fact, knowing the limit as n tends to infinity of (1 + x/n)n is ex is almost enough.

First, interpret the (1 + ix/n) term geometrically as a complex number with real part 1 and imaginary part x/n. For large n this is a number very close to 1 but rotated slightly towards towards the positive imaginary axis. The angle of that rotation will be approximately x/n.

Multiplying two complex numbers results in a new number with the sum of their angles and product of their moduli. So taking the nth power of (1 + ix/n) will result in a new complex number with angle approximately x for large n since the nth power is just n rotations by x/n.

Substituting x = π (a rotation of π radians) gets you to -1 and so the identity holds.

I realise that this is not really a proof, more a sketch really. Also, it takes advantage of a lot of facts about the polar representation of complex numbers which in turn depend on e. That said, in terms of giving me an intuition for Euler’s identity this is the most satisfying explanation I’ve heard so far.

65. Hmmm… it seems the sup HTML tags didn’t work in that last comment of mine. Also, “π” (pi) looks a lot like “n” (en) which is unfortunate.

So my second paragraph should read (in LaTeX) “… Euler’s identity $e^{i\pi} + 1 = 0 …” and “… of$(1 + x/n)^n$is$e^x$…”. Hope that’s not too confusing. 66. Kalid says: Hi Mark, thanks for the wonderful comment! I really, really like that insight, I’ll have to use it when explaining Euler’s formula. I’d first like to tackle complex numbers to give people (including myself) an intuition for it. Also, I’d like to cover the series expansion of e^x as well [1 + x + x^2/2 + ...]. One of the great things about e is that it turns a “mind-bending” operation like imaginary exponent into a “understandable” operation like a series of multiplications/rotations. Again, thanks for stopping by, that comment helped me a lot. 67. Kalid says: Thanks Siva :). 68. mclaren says: This is really beautiful. Congrats on an elegant explanation. There’s an alternative formula for e that might tickle you: e = lim as [delta x] -> 0 of (1 + [delta x])^(1/[delta x]) The simplest explanation I ever heard for the importance of exponential grwoth in phenomena like population growth is that with an exponetial function, the growth rate is proportional to the size. I.e., the bigger the population, the faster the growth. It works in reverse, viz., radioactive decay: the less radioisotpe remains, the slower the decay. 69. Anonymous says: awesome i never thought of e as a base like that. When you compared it to the radius of a unit circle it just clicked. No one ever (or has yet to) said that to me in calc or any other of the engineering courses i’ve taken. 70. Anonymous says: You didn’t understand what makes e so important. It’s because it’s the ‘unity’ of diferentiation/integration d(e^x)/dx = e^x 71. Anonymous says: Any positive number different than 1 could be the e as you have been using in this article… 72. Kalid says: @mclaren: Thanks, that’s another way to think about it [It's the same limit but restated]. Yes, the application of “growing as much as you currently have” is part of what makes e great. @Anon1: Thanks, I’m glad to see it helped! Everyone has a different “a ha!” moment. @Anon2: Being reversible is one great property of e^x and will be covered eventually, but I find it isn’t what helps people “get it” intuitively. A lot of the time you see e in formulas related to growth rates (heat transfer, radioactive decay), and you want to know why it’s there. 73. Kalid says: @Anon3: e is the unique number you get when you compound 100% (unit growth) continuously for a unit time period. Yes, any number can be the base of an exponent, but only 2.71828 (e) emerges from the unit rate and time period. 74. Richard says: Hi Kalid, I can follow your explanation of e but I can’t relate (1.33)^3 = e^x = 2.37037. If I take the natural log of 2.37037 it equals 0.863046. In other words for this particular case how do I determine what rate*time is? Thanks, Richard 75. william says: i feel smarter 76. @Richard: That’s a great question — I think I need to clarify simple vs. compound interest in an upcoming article. (1.33)^3 means “3 periods of 33% simple interest”. In this case, it happens to equal 2.37. ln(2.37) = 0.86 = rate * time, which can be interpreted as: One period (time = 1) of 86% continuous growth (rate = .86) would be 2.37 total. .86 periods (time = .86) of 100% growth (rate = 1.0) would be 2.37 total. Regular exponents (1.33^3) measure simple interest. e^x measures compound interest. Natural log helps convert between the two. 77. Your site is wonderful. And this entry is fantabulous. 78. Kalid says: Hi Asad, thanks for the support! 79. Ivo says: Hello, I found this posting recently and would like to express my admirations. I also have a question: how could add an unexpected conditions influencing on the growth? Let me use your Bacteria sample: During their growth, at different time, some of them dies because are too old, other – because of illness, still others – because have been killed somehow… So the numbers are 1 -> 2 -> 4 -> 3 -> 9 -> 8 -> 7… how could implement this (let’s name it “unknown factor”) to the growth? x = e^(r*t) Thank you 80. Hi Ivo, great question. I’m not an expert on this but one way to do it is to subtract a negative term that represents the unknown factor. For example, you could try growth = (2 – sin(x)) * e^x and as sin(x) varies between 1 and -1, your growth will dip up and down (it will generally trend upward). A sigmoid curve is another way to represent “exponential growth that levels off” as populations compete for resources: http://en.wikipedia.org/wiki/Logistic_function Hope this helps. 81. Anonymous says: That was awesome. Good job! 82. wow says: I wish I had read this when I was in class eleven. 83. Kalid says: Thanks, glad you liked it! Yeah, I wish I could go back in time and tutor myself, but the closest I can do is write it for other people to read 84. Adam says: Fantastic article! Thanks for the great overview in layman’s terms. 85. Kalid says: Thanks Adam! I find that people who feel a need to explain things in “mathematician’s terms” don’t really understand the topic at a deep level — it’s a good litmus test 86. Jon says: A fine explanation of a complicated subject — well done! You might want to revisit your examples, though. The first and third make use of rates which most people already think of in “compound interest” terms, rather than needed to introduce e by moving from simple interest to compound interest. If I’m told that bacteria increase by 100% in 24 hours, then after 24 hours I expect to have twice as many. Similarly, if I’m told that a radioactive sample has a half-life of three months, I expect to have 50% remaining after three months, 25% after six months, and so on. 87. D.jef says: Excellent work, and you’re damn right about Wikipedia maths. A comment on exponential rates of decay: Why is it we can write decay rates as e^(-RT)? Just to expand your own observations, suppose we put some small deposit X in the bank, which grows exponentially over some time T at some rate R to the sum of Xe^RT. Now suppose we fritter all our gains away at some exponential rate r so that after that same period T we’re back to our original deposit of X. If both processes work ‘exponentially’, then just as we could multiply X by e^RT to find the value of the larger sum Xe^RT, multiplying this large sum by some other value e^rT will bring us back to our original deposit of X. So X.e^RT.e^rT = X, which gives e^rT = 1/e^RT, which is written e^(-RT), and which also gives us r = -R D.jef 88. John Smith says: I have to agree. The best explanation this far! 89. Kalid says: @Jon: Thanks, glad you liked the article. Yes, you have a great point — the terminology is a bit confusing, I’m planning on going back to rephrase. The problem is we often don’t speak about continuous growth, so interest is assumed to be simple. I’m planning on writing more about this to clarify. @D.jef: Thanks for the details! Yes, another way to think about negative rates is time going in “reverse” (aka you look backwards and you’re shrinking) or the “reverse” rate (shrinking instead of growing). @John Smith: Thanks! 90. Anonymous says: please, e not E 91. Robert R says: Fantastic article. Great explanation of e. I use e all of the time in my work, and have never really understood it at the intuitive level. This article makes it very clear. Nice work. 92. @Anon: I like e, but sometimes lowercase can look strange in a title. @Robert: Thanks! I’m really happy the intuitive meaning shined through. I used e throughout physics in college without a good grasp of what it meant, and why it was “really” there (and not some other exponent). 93. Sarnath, from India says: It was really a great amazing tutorial! Simple and Efficient! I recommended to many of my colleagues too. Thank you, 94. Kalid says: Thanks Sarnath, glad you found it useful 95. Connie says: Thanks for the article. I’m teaching Algebra II for the first time and your article will help me give a clear explanation about e and natural logs. 96. Kalid says: You’re welcome Connie, I hope your students find it helpful! 97. Matthew B. says: Hey, this is a pretty sweet site. All I ever wanted to know about e! Plz send emails. Btw, I had to read this for a math class. It was worth it. Ok bye. 98. Ken says: Nice article. Can you explain how this view of “e” relates to other conceptions of “e” mathematically (perhaps in a second article)? For example, see http://en.wikipedia.org/wiki/E_(mathematical_constant) Alternative characterizations. It would be nice to intuitively understand how these different ways of representing “e” are really part of the same mathematical concept. Thanks! 99. Kalid says: Hi Ken, great suggestion. Yes, I was planning on doing a follow-up to show how the alternate characterizations of e can actually mean the same thing :). I like the interest definition to start (since it’s the most “tangible”) but the infinite series expansion (1 + x + x^2/2 + x^3/3!…) has an intuitive explanation as well. I’ll definitely do a follow-up on this. 100. sudarshan says: hi kalid, what a great attempt.I am glad there is some one out there in the world to make mathematics interesting.I was so happy to read your article after having loads of frustration tryin to understand e and natural log.why dont you write a comprehensive mathematics course.I am sure the professors will learn from you!!!!!!!!!!!!!!! 101. marcus says: thank you very much 102. Bagish says: Really really informative and good use use of graphics 103. Kalid says: @Sudarshan: Thanks for the kind words! Yes, I think math (or any subject) can be interesting if presented in the right light. I’d love to create a math course — right now, I’m taking it one article at a time :). @Marcus: You’re more than welcome. @Bagish: Appreciate the comment — I find diagrams really help when approaching complicated topics. 104. tchuk says: cheers, nice one really nice one. I’m an engineering masters student, yet e keeps eluding me 105. Kalid says: Thanks, glad you found it useful. Yes, it’s funny how often we use an idea without understanding it deep down. It took me a long time to “get” e also. 106. sujatha says: I never understood the significance of e.But I am glad that I understood a very important part of maths atleast today 107. Kalid says: Thanks Sujatha, glad it worked for you. 108. Louis says: Kalid, Thank you so much for this great explanation! I have been searching for an explanation like this for a long time, as I never fully understood the concept of e, but now I understand it much better. I do have one point of confusion, so I hope you don’t mind me asking: I understand why why y=e^x describes Compound Interest, because the growth is continuous. However, as you stated above, bacteria cannot function until they are fully born (not while they are still growing out of their mother cell), so that growth is not really continuous (new bacteria born every 24 hrs in your example above). So wouldn’t a better equation for bacteria growth be y=2^x (where x = 1 period or 24 hours), rather than y=e^x (continuous growth), even though in my college biology class we learned that the equation was y=e^x? Thank you! 109. Kalid says: Hi Louis, thanks for the comment! You bring up a great point, I should have made the bacteria example more clear. I’m planning on doing a follow-up article on simple vs compound vs continuous interest, as it’s a point I’ve even confused myself about :). When studying a single cell, the equation 2^x probably fits best, as the cell can’t do anything until it has split. When studying bacteria cultures (millions of cells), the individual cells aren’t likely on the same phase. At any moment, some are 1/4 done, some are just getting started, some are 99% done, etc. So if you start at noon, there’s some growth at 1pm, even though a given bacteria takes 24 hours to mature. This is one reason e^x makes sense for bacteria — the growth appears continuous on a macro scale. The other reason e^x makes sense is that it can model *any* growth rate (even 2^x) by using e^(ln(2) * x). In this case, ln(2) becomes a “scaling factor” that transforms e^x into the rate you need. I’ll be doing a follow up on this as well. Hope this helps — I need to change the last bacteria example because it’s confusing, given what I said above. By the way, this is what I love about blogs — getting immediate feedback on what’s working, and what isn’t :). Thanks again for the comment. 110. Nicola says: Thanks for writing this page. It really helped me a lot. I’ve been looking everywhere for a concrete definition of e and I’ve finally found it here. I now understand why it is natural despite the fact that it has always sounds random! 111. Kalid says: Thanks Nicola, glad you found it useful! 112. Spodbod says: Thanks for that, I am finishing a PhD and e appears often in the equations i use, it is nice to know why it is there rather than just accept it. Brill explination 113. Kalid says: Awesome, glad it helped. It’s amazing how many ideas there are out there that we only realize the meaning of much later. 114. Thomaz says: That’s a great explanation but not sure if using the money example is a good thing because in real life, it doesn’t work like that, it happens in discrete steps, like monthly, it’s still compound but not continuous(with steps tending to infinite). 115. Bill says: Great article, I really enjoyed the explanations! I am having a head scratcher moment though .. it’s been a long time since high school, but with the given equation FV=PV*e^(r*t), how would one solve for r, knowing the other variables? 116. Kalid says: @Thomaz: Thanks for the comment. I agree, I may have to change the money example since it is most commonly *not* compounded continuously — mostly monthly or daily, as you say. e was discovered by the theoretical maximum return of infinitely compounding interest, so money does have its role (even if banks aren’t so generous as to give you that maximum rate). @Bill: Thanks, glad you liked it! Great question — to solve for r, you need to “cancel” e by using the natural log. So you’d do: FV/PV = e^(r*t) and take the natural log of both sides: ln(FV/PV) = ln( e^(r*t) ) and since ln and e “cancel”, you get ln(FV/PV) = r * t You can then divide by r or t to find one or the other variable. 117. Brendan says: I absolutely love your intuitive approach. I’m studying physics right now and this is soooo much more understandable and fundamental then “e (or ln) cuts the y-axis (or x) at 45 degrees”. Thanks e^(a million)! 118. Kalid says: Thanks Brendan! Yes, many people are taught e by examining its properties, like having a slope of 1 at x=0. But that’s like saying a circle is the set of all points x^2 + y^2 = r^2. It’s an interesting property, but doesn’t help a beginner! Just say a circle is “round” (so you can visualize it) and then start diving into the details. Glad you liked it. 119. Harisha says: I know people who know about this feels this simple, but i highly appreciate for the way the author has given explanation. even for the first time people shoould understand without any further micro doubt also 120. Kalid says: Thanks Harisha — yes, there’s many things we think we “know” but don’t actually understand until we take a deeper look. Glad you liked it. 121. jbaty says: Louis (#110) and your answer #111 pose a similar concern I have with your radioactive example. You stated that 20 lb would be 2.706 lb after a 3 half-life period(12month/4month). The conventional answer to this is 2.5 lb (20lb*1/2*1/2*1/2). I have never seen this approached any other way for this decay. (After 3 half-life time 1/8 remains is all books on decay I have seen) Are we again getting confused between simple, compound and continuous. Seems similar to the scaling issue in your answer and it should be written as 20lb * e^(ln(2) * x). Where x = -3. 122. Hi jbaty, that’s a great question. I need to revise the example to be more clear, since 50% after 3 months was not meant to be the half life, but the “decay constant”. Looking back, I should have used different numbers. From what I understand, there are a few ways to describe decay (http://en.wikipedia.org/wiki/Exponential_decay): Use “decay constant”: This is the continuous decay rate used in e^rt. Half life: The time it takes to lose 50% of your value. This is used with a “compound interest” formula (2^(x*t)), rather than a continuous one. As you mention, half life = ln(2)/decay constant So we can use either 2^-(half life * time) = e^-(ln(2) * decay constant * time) The negative signs since we really want the inverse. I’ll try to clarify these examples. 123. Scott says: Kalid, Thanks for the article. I definatley have a better grasp on e. I do have one question. Coming from engineering, I am having trouble understanding more of the mathematics of e. Why is it that the derivative or integral of e^x is itself, e^x? Thanks 124. Hi Scott, glad you liked the article. Great question — I’m planning on doing a follow-up on the calculus of e. There are several explanations for e’s derivative, but one approach I like is this: e represents growth of 100% per time period (we found it by continuously compounding 100% growth). That is, if you have 1 unit of “stuff”, you are growing at 100% per time period, or 1 per time period. If you have 2 units of “stuff”, then you are growing at 2 units per time period. If you have 17.8 units of stuff, you are growing at 17.8 units per time period. If you are at some number e^x (if x = 3, then you are at 20.08), you are growing at 20.08 per time period. That is, your current amount if e^x (20.08) and your current rate of change is 20.08. e is that special number where your rate of change exactly equals your current value. For some numbers, like 2^x, your rate of change is less than your current value. For other numbers, like 3^x, your rate of change is greater than your current value. But e is “just right”. [Sometimes people define "e" to be the number e^x where dy/dx = 1 at (0,1). I don't like this definition because it misses the intuition about growth]. Hope this helps — I’ll be getting into this more. 125. djamwal says: Thanks Kalid. You explained in one post what many a math teachers failed me in all their years. Needless to say, Keep up the great work. 126. gbuch says: I made a bigger post, but it seems I’ve lost it… Anyway, great explanation, but I can’t understand how can you work with both simple and compound interest, at the same time. I mean, first you divide the yearly interest rate as if it were simple (let’s say, 100% a year became 50% every 6 months). Then you compounded these 2 periods of 6 months (and 50% for .5 year became 1.5^2-1=1.25=125%). However, I don’t get it how can you use both. I mean: A)Had you used simple interest all the way: 1 year interest rate: 100% .5 year interest rate: 100%/2=50% going back: 1 year interest rate: 50%*2=100% B)Had you used compound interest: 1 year interest rate: 100% .5 year interest rate: (1+100%)^(1/2)-1 = 2^(1/2)-1 =~ 1.414-1 = 41.4% Going back: 1 year interest rate: (1+41.4%)^2-1=100% 127. @djamwal: Many thanks, glad the explanation was useful for you. @gbuch: Great question — take a look at http://betterexplained.com/articles/a-visual-guide-to-simple-compound-and-continuous-interest-rates/ for more details. Basically, continuous growth happens when you compound simple interest. It’s important to realize what type of growth a “rate” like 100% refers to — whether it’s simple (base price) or includes compounding effects (price after tax & shipping, if you get me). 128. Anonymous says: thanks so much! all this time, i thought e was just a number and i’d been completely missing the point! 129. Kalid says: Awesome, glad it become more clear for you! 130. pooman says: u rok 131. gbuch says: Still, when we say that Uranium 232 has a half live of ~70 years, it means that it loses 50% of it’s weight in 70 years, or (1+.5)^(1/2)-2 = 22% of it’s weight in 35 years. If it had lost 25% in 35 years, then it wouldn’t have a half-life of 70 years. 132. @pooman: Thanks. @gbuch: I know what you mean — the terms get mixed up enough since it’s more common for us to think about “final values” (50% less) than the instantaneous rate of decrease. e is 100% instantaneous growth, which turns into 271.8% growth when all is said and done. If something actually grew continuously at 100%, we’d call it 271.8%. Similarly, if something has 50% less after 1 year, we say it decays at “50% per year” though in reality its daily decay is not (50%/365). But it’s a great point, I think I need to write a follow up on this :). 133. Mayilvaganan says: Thanks for demystifying it. For years, I wondered where it came from 134. Chidu says: Found it very informative! Cool explation! 135. Kalid says: @Mayilvaganan & Chidu: Glad you liked it! 136. Anonymous says: What is the value of e^(tau)(pi)? thanks 137. Hi there, not sure I understand the question. If you have a value for tau, you can just plug it in to the equation. 138. zaphod says: This seems to breakdown for very large values of n(1,000,000,000,000,000), well short of the limit at infinity. Can you explain what this means, in layman’s terms? I mean, it’s obviously a limitation, but how to interpret/explain it? Thanks! 139. Hi Zaphod, thanks for the comment but I’m not quite sure I understand the question. As n gets large (millions, billions, trillions) the value (1 + 1/n)^n gets closer to the real value of “e” (which is an infinite, non-repeating decimal). It’s a bit like getting closer and closer approximations to pi: you can start with a square, then a hexagon, then an octogon and keep adding sides to approach the shape of a circle. As n gets larger, it approaches the shape of continuous growth. Hope this helps! 140. wow! you are a rare person.good teacher.the best thing i like about your articles is that they are so simply written,and still (rather so) they bring out exactly why the concepts you explain came to be. Nice work. you have got one more fan. 141. Kalid says: Thanks Tejesh! Yes, I find complex language can get in the way of conveying an idea, so I’m glad you’re enjoying the simpler style. Appreciate the note. 142. xatrrz says: your site is great! keep the good work up, perhaps someday we will buy schoolbooks from you 143. Glad you enjoyed it! One day I’d love to turn this all into a book :). 144. Hendrik Jan says: Nice article, although I agree with comment 128 (and others) that mixing simple and compound interest without explicitly mentioning the (fundamental) distinction is a tad confusing. It is precisely this confusion which always causes my students to mess up: they often think 20% interest in 10 years means 2% interest per year. Another very natural way to get to the number ‘e’ is this (someone might find it enlightening). Suppose you’re in a car on a long straight road and there are distance-markers alongside the road, e.g. every km (or mile, if you prefer :-). Now you start driving at marker “1 km” and make sure your speedometer says you’re driving 1 km/h (yes, that’s slow :). Very gradually you increase your speed, such that once you reach the “2 km” marker (after less than an hour) your speed is 2 km/h. You continue speeding up this way, always making sure your speed is equal to the distance-marker. It does take some time to reach marker “200″, but once you’re there you’re driving 200 km/h and reach the 400 km marker within the next hour (driving 400 km/h by then). If you now plot the graph of time vs distance, you’ve plotted the function distance=e^time. I’ve found that for my students this is one of the most intuitive / accessible introductions to the “meaning” of the number ‘e’. 145. Kalid says: Hi Hendrik, thanks for the comment! Yes, I agree the distinction between simple and compound interest can be confusing. I’m still thinking about the best way to make that distinction, I can see how it’s strange that we take 100% and split it into 50%, then do multiplication (1.5)^2 vs 50% * 2. I’ll be thinking more about how to introduce this simply. I like that “driving” explanation of e as well, it helps intuitively introduce some of its interesting properties for calculus :). 146. Kalid, Just read my comment from a long time ago. Your response was fine, it was my explanation that was lacking. This is the scenario. Imagine that the monetary authority decides to double the amount of money in circulation to the benefit of the general public at large. But instead of doing it all at once they do it one day at a time. Instead of creating an amount of money equal to the sum of all existing money, 1/365 is created and distributed. On day two, money has increased by 1/365, so the amount of money created will increase by 1/365 as well. A full year of this will yield the algorythm (1 + 1/365)^365 = e You have a great site. Check out my book “Mathematical Measures, Mathematical Pleasures, Mathematical Treasures” on Lulu.com. It is available for free download or purchase of a hard copy. I do not deal with e, but I do deal with pi and phi, in what I think are unique ways. Thanks for your site. 147. Hendrik Jan says: Re:147. Maybe an approach like the following: Your bank agrees to give you 100% interest per year. You’re free however to distribute those 100% over a period of one year whichever way you want. So you could have them pay you 100% at the end of the year. Or they could give you 50% after half a year and another 50% after the second half. In that case your total interest would obviously be more than 100%: 1.5 x 1.5 = 2.25 so you would have gotten 125% interest in total. Since you’re clever, you realise that getting 25% every 3 months would yield even more money: 1.25^4 = approx 2.44, giving 144% interest. You now start wondering if there’s a maximum to the amount of money you would be able to collect by dividing the year in every smaller and smaller fragments. Maybe you could amass an enormous amount of money in just one year! [etc.] Just an idea… BTW, I do like the base-idea — phrasing the classical limit in terms of interest. I’ll remember that the next time this limit comes up in the classes I teach. 148. Cassie says: I don’t get it! I am trying to figure out the growth in Tokyo-Yokohama over a 10 year period. THe average annuyal growth rate was .86%. In 1991, the population was 27.245 million. What was the population in 2001? I am trying to use the formula f(x) = ab^10. So, in my mind, 27.245 million * .86 ^ 10, but the answer does not make sense. It should be more not less than the original amount. Help! 149. Hi Cassie, great question. It’s important to remember that you need to use 1 + rate, not rate by itself. That is, if you double every year (100% growth) you’d multiply by 2 (1 + 100%). In this case, your growth is .86% so you multiply by (1 + .86%) = 1.0086. So the formula would be f(x) = original * (1 + rate)^10 = 27.245 million * (1.0086)^10 The reason is that ‘growth’ implies more than the original. So when we have 50% growth, we mean we have 50% more than the orignal (1), or 1 + .50% = 1.50 after the growing is done. 150. Hi kalid, I’m from Perú. I never understood ‘e’, I just use it. I love math when I understand the logic and the reality behind the concept, when you really know what something means, you get the feeling that math isn’t as abstract as you thought. Thanks! 151. Kalid says: Hi Rodrigo, thanks for the comment! You captured the feeling exactly, I love getting that gut feel and understanding for a topic, vs. just knowing the formula. I’m happy it was useful for you! 152. Falco says: You should write a book on math topics dude! 153. Kalid says: Thanks Falco — I’d love to turn this into an ebook (or regular book) one day :). 154. Viru says: Nice article Kalid. I will be going through your other stuff too now. I wonder if it makes sense for you to explain the fourier transform also in a intuitive way like this if possible; I can always wish can’t I :-). 155. Chiming in with my compliments. I understand you. That is, I understand the concepts you aim to explain. Nicely done, keep it up! Benjamin 156. Kalid says: Thank you Benjamin! Glad it worked for you. 157. Kalid says: @Viru: Thanks! Yes, I think the Fourier Transform would be a great topic as well, it’s on the list :). 158. Sharath Bhat says: This is the kind of stuff that the text books should be made of. A superb article. Keep them coming. Thanks a lot. 159. Kalid says: Thanks Sharath, glad you enjoyed it :). 160. anac7ronox says: amazing like the other articles here. take care and God bless! 161. Kalid says: Thanks :). 162. sassieston says: Thanks alot explaining that (wish I had you as my math teacher) 163. Kalid says: Thanks sassieston :). 164. Daitoshi says: Whew! This really helped on my exams. Thanks! ^_^ Stupid definitions and complicated explanations elsewhere were not helping me any. Thanks for the charts and stuff. They were kewl. 165. Kalid says: Awesome, glad it was helpful! 166. eli says: clear, intelligent and enlightening explanation. 167. Hi Eli, thanks for the kind words! 168. Raghavendra says: Hi Kalid, great explanation… I was trying to teach my niece about e / natural logs etc. and now I have found a wonderful way – thanx to your work. I am going to come back to this site more often and not going to miss your posts… I also wish this explanation was available when I was in school … Once again very good stuff… 169. md says: Just curious. When it comes to exponential growth, aren’t we talking about a special class of “continuously growing processes?” What about linear continually growing processes? Snow and rain accumulation are two simple examples. Is there need to more precisely define the domain of continuously growing processes that e applies? I’m not a math guy, so I don’t know. Thanks! 170. I want to thank you for your excellent explanation of E and ln. I consider myself a big math guy, and E and ln had always bugged me because they were never explained, just used and thrown about. You’ve made me realize WHY they are so important and have probably advanced my “mathematical mind” years ahead of where it previously sat. 171. @Raghavendra: Thanks for the comment! Your niece is lucky to have someone who’s interested in math teaching her :). Appreciate the kind words. @md: Great question, I may go back and make that distinction clearer. Linear processes grow the same amount, no matter how much “stuff” is already there. For example, you could accumulate snow at 1 inch per hour. After 12 hours, you’d have 12 inches. The new snow doesn’t care how much old snow is on the ground. For exponentially growing things, the amount of growth depends on how much “stuff” is already there. For populations, the number of new people being born depends on how many people are already there (a village of 100 people may have 1 birth a year; a village of 1000 people may have 10 births a year). Basically, exponential growth happens when you can measure change as a percentage (rate) vs a fixed static amount. Great question! @Ryan: Thanks! I’m the same way — I really enjoy math, but felt very uncomfortable that e and ln weren’t intuitive in the same way addition, multiplication and even complex numbers (rotations) were. I hate using “magical” formulas, and wanted to know why e was 2.718 and not some other number. Glad it was useful. 172. Rebecca says: This is a great explanation and makes the concepts surrounding e so much more intuitive. You deserve to be paid for making a complicated topic so accessible. 173. anonymous says: Thank you, thank you, thank you! 174. @Rebecca: Thanks, I’m glad it was helpful for you! There may be an ebook in the works :). @Anonymous: You’re welcome! 175. Awesome!!! I had tried to grasp the concept for ages and now it is finally clear. (Mostly from dark obscure math textbooks long after finishing college) *Med student in college. Thanks. 176. Thanks verde! Glad it was helpful. Same here, the real meaning of e didn’t hit me until long after I studied it in school. 177. Anonymous says: Well Explained..!!! Excellent 178. Kalid says: Thanks, glad you liked it. 179. AJ says: Very nicely done; I wish my high school math teachers had been this clear on these subjects. As it was, I had to figure it out myself. 180. Thanks AJ, glad it was helpful! I know what you mean, a few words could have saved me years of anguish :). There’s something to be said for figuring things out yourself, but concepts like this shouldn’t be kept in the dark. Thanks for dropping in. 181. Typo: “10 periods of 3% growth = 30 1% changes, happening over 10 years” should say “3 1%” not “30 1%”. 182. M says: I never really understood e, and now I do. Thanks for a superb writeup. 183. Imran Fanaswala says: Good article, small correction: “Clearly having 30% growth for 1 year is better than 3% growth for 10 years” 184. Ravi says: Khalid,you are “da man”. 185. Anonymous says: Could you update the Wikipedia article? 186. Kalid says: @Ian, Imran: Appreciate the comments, I need to clarify that example. 30% continuous growth for 1 year (and stopping) has the same net impact as 3% growth for 10 years, after all is said and done. It probably wasn’t clear that years 2-10 had no growth in the 30% case. Thanks for the feedback. @M, Ravi: Thanks! @Anonymous: That’s a good idea. Unfortunately I don’t really have much (any) authority on Wikipedia so might be tough to add in my fairly drastic changes :). I’m happy that people can find me from the wiki article though. 187. Lisa Coffey says: Greetings. I just found your site and really am enjoying reading all the comments. Thank you for taking the time to post all of this neat information. I have a question. Please don’t laugh; I’m a novice! I just tried calculating “e” with Excel 2003 (I know… my abacus was broken… had to make do). Note the interesting results, which prompted an “eeeeee!” instead of an “aha!”: Input for n is left justified; output for (1+1/n)^n is indented: At relatively low values of “n” (100 million), we’re happily moving in toward our target of 2.718281828459: 1.000000000000000000000000000000E+08 2.718281786395800000000000000000E+00 But as n further increases, e jumps beyond our target (2.7182818…) by just a bit: 1.000000000000000000000000000000E+09 2.718282030814510000000000000000E+00 As n increases, it gets worse: 1.000000000000000000000000000000E+10 2.718282053234790000000000000000E+00 1.000000000000000000000000000000E+11 2.718282053357110000000000000000E+00 1.000000000000000000000000000000E+12 2.718523496037240000000000000000E+00 1.000000000000000000000000000000E+13 2.716110034086900000000000000000E+00 1.000000000000000000000000000000E+14 2.716110034087020000000000000000E+00 1.000000000000000000000000000000E+15 3.035035206549260000000000000000E+00 Creeps, now look: 1.200000000000000000000000000000E+15 2.903201529183850000000000000000E+00 1.400000000000000000000000000000E+15 2.541075307073900000000000000000E+00 1.600000000000000000000000000000E+15 2.903201529183850000000000000000E+00 1.800000000000000000000000000000E+15 3.316934014348890000000000000000E+00 Even if Excel just couldn’t handle the multi-decimal place calculations, it still seems odd that the output is so strange. Instead of 1.0000000 or 0.0000000, the output jumps around like a kangaroo on steroids. Ok, not that bad, but still. Thanks for your thoughts. I might just have to find my abacus. Wonder if Excel 2007 solves the problem. Sincerely, Lisa Coffey 188. javed Khanzada says: Hi Kalid, Your explanation is the best I have come across on the whole internet that I have searched! We use e all the time in engineering and were always taught to use it and not understand it. However , lack of understanding always lead to a stone wall and lots of things never became clear. Thanks to you I can now revisit them in a new light and really appreciate your work. A big thank you. Javed 189. @Lisa: Thanks for the comment! I love these “novice” questions since they often reveal things that the “experts” didn’t consider. In fact, this very article is a novice question about “Hey, what does e really mean?”. Anyway, this is a great question. I think you’ve run into Excel’s limit of precision and things are starting to break down :). I tried with the Excel 2007, and it seems to jump around the same way! Goes to show that sometimes the abacus can beat the computer. Once you crank n really high (like 1e100) then Excel just gives 1.0000 (since 1/n is essentially 0). I don’t have many details on how Excel does its calculations, but yep, this is definitely a bug/limitation you’ve found. @javed: Thanks for the wonderful comment! Yes, I totally agree — by just memorizing and moving on to the next lesson, we set ourselves up for more confusion down the road. Glad it was useful! 190. Lisa Coffey says: Kalid, thank you very much for your thoughts on calculating e with Excel and the unruly spreadsheet’s likely precision-calculation limitations. Still not sure why the output flip-flops for n slightly greater than 1.0-E15 rather than iterates 0′s or simply plain old 1.0000000000 – in fact, (1+1/n)^n yields 2.90320152918385 for both n=1.2-E15 and n=1.6-E15. Perhaps Microsoft can explain how Excel does its calculations. If they can explain this seeming calculation meltdown in more detail, I’ll surely pass along the info. Meanwhile, thank you for your kindness in running the numbers. Looking forward to your future discussions of e^x = d(e^x)/dx, friend of the cloning scientist. Meanwhile, any thoughts about the Riemann Hypothesis and where to find the clearest explanation for how to calculate the non-trivial zeroes along the ley line x=1/2? I’m probably the last math lover in world to find out about this problem. Would like to explore it further, but the books I’ve read so far on the subject (by Marcus du Sautoy, e.g.), while fascinating and relevant, seem to be more historical and philosophical. I know that this site is dedicated to your friend e, but e seems to play a key role in predicting the distribution of the primes and thus is sure to play a role in proving or disproving the RH. Thank you again for your gracious spirit and great postings by you and your guests. Lisa 191. december says: Please provide me with an example of a javascript that calculates the principle using the formula for calculating principle 192. vaibhav says: e always troubled me much.But no longer.THANKS A LOT! 193. amin elsersawi says: 120 * e^(.05 * 10) does not equal 120 * (1 + 0.05)^10 because e = 2.718 for almost less than infinity, not for 10. The article is excellent. 194. Hi Kalid Like many others I stumbled over you looking about for an “internet teacher”. Logs are like calculus in two ways it seems to me: 1) they are forever; 2) they are the first gone because most easily dissipated without constant use. Well, anyway, my work is metaphysics and I always thought Pythagoras had it right, that number truly represents reality, and I set about to better or more fully explicate just that. Along the way I stumbled upon a derivation for 1/7. Because I am bipolar and on lifetime disability and didn’t finish college, I need to make something of and for myself to get out of poverty, and as a nobody, nobody listens. So I need to turn anything I do into something, and this derivation might perhaps do the trick. But I can’t discover if anyone else has done a successful derivation. I am asking you because the solution seems tailor-made to elaboration by “e”. Would you happen to know who if anyone has a clear-cult derivation? If you can help me with this I will email you in private the derivation I have, which, may I so say, is just plain cool as h_ _ _. I look at the damn thing and wonder how I sis it! Go figure. Thanks, Kalid, so much for your work. In philosophy the concepts are often easier than expressing them, so I really, really appreciate what you have mastered and so ably delivered. Bravo, kudos. Charles 195. Anonymous says: this was absolutely fantastic and my friend skoda absolutley adored it. 196. Having flunked Maths at school due to never turning up, I had always been baffled by the e^x key on the calculator. Even when working with Orders of Growth with Algorithms it never came up. Now doing the Maths part of my physics degree it comes up in the prep work and like you I needed a complete explanation. Thank you for producing this page, it’s the perfect walkthrough of how to arrive at e and therefore at e^x. Mandy 197. vaibhav says: hi,i appreciate your article but how i reconcile that ,as in your 3rd example,(e^rt) also hold for negative rates 198. vaibhav says: hi, i appreciate your article but how do i reconcile that (e^rt) also holds for negative rates,as in example 3.please tell. 199. Bazza says: Kalid mate, I just spent most of my night examining population growth rate formulas that I did not understand until I read through your page here, describing e and its magical qualities. Thanks a million for your great explanation! To be honest, I am still wrestling with some of the concepts involved – it’s been years since I picked up a math text and I’m just now returning to university and diving head first into all the sciences; but you’ve helped me immeasurably. Thanks!!! -Vaibhav, I was just contemplating the same question you have asked about how (e^rt) holds for negative rates; I think the reason is due to the idea that raising e to a negative power is the same as writing the reciprocal of e: (e^-rt)=(1/e^rt). When you use the reciprocal of e, the rate of growth becomes a rate of decay because it is now negative. When Kalid writes the decay of 10kg as 10 * e^(-1 * 3) he is assigning a negative value to the growth rate, 100%, and multiplying it by 3 years, which becomes e^-3 or a -300% growth rate. I double-checked all the calculations and it’s working out for me; but please correct me if I’m wrong! 200. Tyler says: Kalid, thanks a lot for your article which is both intriguing and intellectually stimulating. Since I am reading a lot of books on the quantitative finance, getting a solid understanding of e is so critical. Both John Hull and Paul Wilmott, two hot writter in the area of quant finance, are unable to demystify the e though they are quite comfortable with it due to their concrete maths background. From my perspective, it is better for you to make an attempt,as an aid,to explain the concept in a more calculus oriented but accessible way. A bit of Taylor Series, a bit of basic series concept will be enough and perfect for everyone. James Stewart’s Magnum opus, Caculus, has an intuitive and heurist chapter explaining the concept of Taylor Series with the property of e as an example. 201. Kalid says: @Lisa: Thanks for the wonderful comment (and apologies for the delayed reply)! Unfortunately I don’t know much about Excel’s workings, I feel it has to do with limits of floating-point numbers as they are stored in computers. Regarding the Riemann Hypothesis, I’ve only read the books (Marcus du Sautoy) but don’t know much beyond that. My pure math knowledge is very sparse, so it’s a “bit” out of my league right now :). But good luck in your search! @vaibhav: Thank you! @amin elsersawi: You’re right, e approximates continuous growth, and only roughly approximates yearly compound growth. For small interest rates these appear similar. @Charles: Thanks for the kind words! I’m unclear what you mean about the derivation of 1/7 — the decimal expansion, or some other meaning? @Anonymous: Glad you and your friend liked it @Mandy: Thanks Mandy! It’s funny how such a “basic” idea can be understood so many years later. I first learned about e in high school and only now do I really “get” it. @vaibhav: Bazza nailed it, check out his explanation. Another way to see it is that a negative rate really means negative time. (After all, e^x = e^rate*time). So if you grow by 2x when moving forward in time, you grow by 1/2 when moving backwards (i.e. you shrink). @Bazza: Thanks for the kind words, I’m so happy it was useful! I know the frustration of an idea not clicking all too well :). Great job on the explanation, you got it. The reciprocal is indeed a negative exponent, and one way to think about that is as negative time. @Tyler: Thanks for the comment and pointers! Yes, I’d love to do a more detailed treatment of e as it relates to a Taylor series (I think this can be understood intuitively by looking at e in a slightly different manner, as a “factory” that produces more factories). I’ll have to check out Stewart’s book as a reference. Thanks again for the note. 202. Rebecca Tyson says: Thank you very much for the clear and intuitive explanation above. I will use it in my precalculus class! Regards, Rebecca 203. Doyin says: Kalid, not only are you intelligent, but you seem to have a very humble demeanor. Not only do you write a very excellent article for the benefit of others, but you have taken time to acknowledge all those who have complimented you work! It is not very often you find people as smart as yourself being very courteous and selfless, willing giving back in form of tutorials!! inspiring. Thank you. 204. Kalid says: @Rebecca: You’re welcome — I hope your students find it helpful! @Doyin: I really appreciate the comment, thanks for stopping by. I try to approach learning with a curious attitude of discovering new things, and am just excited to share what I’ve learned :). Often times people leave comments and feedback helping me understand an idea better, so I’m always happy to have a conversation. Thanks again for the kind words. 205. Carlos says: You’re all wrong! “e” is that short guy that plays on Entorage. 206. Juan says: Kalid, Thanks for the excellent explanation. My regards for spending the time to explain it in an understandable fashion appended with some witty humor. Juan 207. Kalid says: @Carlos: I don’t watch the show, but I like that name. @Juan: Thanks for the kind words — I try to add a bit of sugar to make the math taste better :). 208. delfeld says: In response to #171, isn’t linear growth simply e^0? If snow is falling, the fallen snow is not producing snow, so the rate of growth is not changing. The continuous linear growth rate would be e^0, or 1. Linear growth is not a “growth” at all; the snow does not produce snow. However, interest on money is money producing money, and population growth is actually people producing people. If nothing is produced then the rate of production is 0, and e^0=1. e only talks significantly about production. Likewise, in bacteria cultures where growth depends not only on production of bacteria, but also on proximity and quantity of food, and proximity and quantity of poisons (to bacteria), and since growth of bacteria has a decay-like effect on food and growth-like effect on toxins (from waste produced), the population eventually levels off, or else dies almost instantaneously. Though toxins and food do not themselves “grow” (or produce themselves, assuming a controlled environment), they do *act* like they have growth due to the effects of actual growth of bacteria. So food and toxins are dependent on bacteria producing bacteria, and, in turn, are dependent on e! 209. @delfeld: Thanks for dropping by, and for the bacteria explanation! The term “growth” can be confusing, as when snow is falling it means (accumulation / addition of snow) whereas for bacteria cultures or money it implies exponential growth. At a high level, the main difference is whether your growth rate *changes*. If you always change at 1 inch per hour (snow accumulation), then it’s linear growth and e^x doesn’t really apply — just multiply rate * time! If the amount you change varies over time (like interest on money; the more money you have, the more interest you earn) then you can use some version of e^(rate * time) to determine the impact. 210. delfeld says: Kalid, I agree that the term “growth” is confusing: it refers to both addition from an outside source (e.g., more snow falling), and also addition by self-replication. English is an imperfect language! But I am guilty of not helping the matter, I guess. As you suggested, (self-replicating) growth can also be represented as partially linear. Say you add a fixed amount of money to your interest-bearing account every day. Then the equation would be: {initial value}*(e^(r*t))*(r2*t2) With closed systems, like a bank account where no deposits are made, (r2*t2) is one instance of rate at a time interval of 1 (i.e., the addition of the principle), (1*1)=1. So: {initial value}*(e^(r*t))*(1) = {initial value}*(e^(r*t)) Likewise, linear accumulation (or non-self-replicating growth) can be represented expotentially: {initial value}*(e^(r*t))*(r2*t2) However (as long as snow is non-reproductive!), there is zero rate times any time interval you pick, so (r*t)=0 and (e^0)=1. So: {initial value}*e^0*(r2*t2) = {initial value}*(r2*t2) 211. J says: This is awesome… Great job…Wikipedia pales in comparison to your explanatory power…please write my next math book! 212. @delfeld: Thanks for the follow-up! Yes, I like that hybrid model with linear (r2*t2) and exponential (r1*t1) growth rates & times. @J: Glad you enjoyed it 213. John Maxwell says: I had a hard time understanding the stuff under the “What about different rates?” heading. What you seem to be trying to say is that if you halve the interest rate (e.g. by reducing 100% interest to 50% interest) then the result is the same as halving the number of compounding periods. But I’m not sure this is quite correct. Let’s say that you are getting 100% interest per year and your interest is being compounded twice yearly. In this case, you will be getting 50% interest at the end of every six months. In one scenario, you halve the number of compounding periods while keeping the interest rate constant. In this case, you will be awarded 50% interest annually. In the other scenario, you half the amount of interest while keeping the number of compounding periods constant. In this case, you will be awarded 25% interest at the end of every 6 months. These scenarios aren’t equal, because in the latter scenario, you are making more money. I could believe that the difference between 2n compounding periods and i% interest and n compounding periods and 2i% interest decreases as n goes to infinity. But you don’t mention anything about this in the section. 214. Hi John, thanks for the comment. You got it — that relationship only works if the number of time periods goes to infinity — that is, the growth is continuous. But luckily in this case, we’re dealing with e, which is defined to be continuous growth. That relationship doesn’t directly work when you are dealing with yearly intervals. (There are ways to convert a yearly compounded growth rate into its equivalent continuous rate so you can use these tricks — that’s enough for another article) 215. anon says: Amazing article 216. @anon: Glad you enjoyed it. 217. jee says: It was brilliant…………. 218. Kalid says: Thanks Jee, glad you enjoyed it! 219. jz says: nice work. one thing tho, i think it would be beneficially to add the idea of going from a discreete presentation of the growth function and then derive the continuous one. what i had in mind is something like: n \ e = / (1 + 1/n) i which becomes e = lim n -> inf (1/n)^n as n approaches infinity. in case the retarded ascii above doesnt parse, i mean a very similiar approach to how you define the integral when going from discreete riemans sums to the integral. thinking in discreete small entites and adding those are generally very natural and so this apporach would explain the exponential n you end up with… 220. kelly says: Thank you so much! Your tone is very conversational, and you talk plain English! I am bookmarking this page and coming back again. I have been searching for three days to figure out the definition of e and the uses of natural logs. Ugh. I finally had my Eureka moment when I got to your explanation. 221. @jz: Thanks for the comment! I think the approach is pretty similar to that, in that we start with 1 discrete period, then go to 2, then to 3, etc. A deeper discussion about the difference between discrete and continuous definitions may be great for a later article, thanks for the suggestion. @kelly: Glad you enjoyed it! Yes, I think math (and most subjects) are more easily understood and enjoyed when they’re talked about. Happy you were able to have your “aha”. 222. spirit says: Hi Kalid, very clever: the ingenious are always fanciful, and the truly imaginative never otherwise than analytic (E.A.Poe) Many thanks for your articles. Spirit 223. Paul says: Your explanation of the material is by far the best. Just amazing! Thanks! 224. Kalid says: @Spirit: Glad you enjoyed it! @Paul: Thanks for the kind words :). 225. Marcey says: Kalid I was stumped by the question, “What is e?” … even my text couldn’t provide the answer for me! It stated that e=2.71… but there was no explanation as to why or HOW! Thank you for this amazing explanation! I think you just helped me understand this week’s Algebra assignment! ~marcey 226. @Marcey: Thanks for the comment! Really glad you were able to get an intuition for it :). 227. This is a great explanation! I am 49 years old and have never known what e is all about. It is thanks to your article that I get it and now can explain it to my son who is 13 years old. Thanks .. I wish there are more teachers like you in this world .. Divakar New Delhi 228. @Divakar: Thanks for the note! I’m really glad you were able to share the ideas :). 229. WerK says: Greetz from Czech Republic. Excellent article, I stumbled onto it when I found that I don’t understand logarithms at all. Your explanations are very easily understandable, I’m definitely going to read the rest of the site as well. Thanks ! 230. Kartheek says: Simply brilliant! I am so glad I came across your website, thanks for bringing e from the realms of mystery into practical reality. 231. @Werk: You’re welcome, glad you enjoyed it! @Kartheek: Appreciate the comment! Yes, e was a mysterious concept for me for such a long time also. Glad it was helpful. 232. Sreenath Chary says: Another brilliant post…..I have made your site as a favourite on my Chrome browser. I will read all of them. Wish these were available when I was growing up! 233. chris says: Excellent explanation – thank you so much. I couldn’t get anywhere with the wiki description. No more e for esoteric now it’s e for easy. 234. @Sreenath: Thanks for the support! @Chris: Glad you enjoyed it — happy it took on a new meaning for you :). 235. Christian says: Love it! Thanks very much – I wish everything was taught like this. 236. Arya says: The whole Mr Blue/Red/Green thing got a bit confusing. I find it easier to just think: 33% of$1 = 0.33, so $1.33 overall. Then you go on to 33% of 1.33 = 1.78 and then 33% of that = 2.37. But altogher very well written and quite easy to understand. 237. @Christian: Thanks! @Arya: Thanks for the note. Yes, it does get tricky once there are multiple levels of interest (earning interest, which earns interest, etc.). The advantage of discussing it is that things later down the road become more clear when you break interest into components (more details here: http://betterexplained.com/articles/developing-your-intuition-for-math/). Thanks for the comment! 238. ticha says: thank you thank you thank you! this was the BEST explanation of e ever! I followed everything. thank you!!!!! 239. @ticha: You’re welcome! Really glad you found it helpful :). 240. Jon says: erm kalid is e part of a logarithmic system? 241. bb says: Thanks for your insightful guide. It’s great. Hope that you can provide more useful articles like this Cheers 242. Kalid says: @Jon: e (or any number) can be used as the base for a logarithm. e is an especially good choice because it shows up so frequently in natural systems — and it’s logarithm is called the natural log. @bb: Thanks! 243. If nothing else, this article at least explains why I’m broke… Fantastic explanation! 244. Kalid says: @Steve: Thanks 245. Golgoon says: Absolutely fab article and clear explanations, if only my 3 different maths teacher at highschool taught like this, am sure I would have ended up studying Maths. I love it but found it hard to understand how “e” is the base for compound growth and interest calculations. Lightened up my whole week. Thank you! Please post more of this difficult to grasp math ideas. Merci again 246. Brilliant explanation Kalid! Its amazing how we learn all the tools and techniques of math without really knowing the insight behind them. Thank you for making this fabulous site 247. @Golgoon: Thanks! I was the same way, e confounded me for such a long time. I’ll try to keep the articles coming :). @Yahya: You’re welcome, happy you were able to find it useful. 248. Anonymous says: I LOVE YOU MAN!!!! 249. biz-niggety says: Holeee-Shhiit! You blew my mind with this, Kalid. Absolute kudos. I revel in the fact that my stumbling upon this site did, in fact, immediately follow my visit to wikipedia and the “obtuse” definition. What a tremendous relief. Thanks a bunch! 250. saw says: Hi Sir, Truly amazing explanation of “e”. which i had used during my college days many years ago.And which i had forgotten about! Out of renewed interest in the subject, I stumbled upon this site while browsing the net. Thanks saw 251. Thanks for the article.. One small suggestion, the comments in this page is growing rapidly. Some new comer will think he has to read such a long page to understand the concept of “e” and run away. why don’t you display only few comments in initial view and make complete comments listing as optional 252. AlyssaJo says: Thank you so much. e was a completely foreign concept to me until I came across your website. Thank you!!!!!! 253. jaiwant says: You have just cleared up 15 years of confusion in a few minutes.I cant tell you how much this article means to me ! 254. @Anonymous, biz, shaw, AlyssaJo, Jaiwant: Thank you! @Jaiwant: Thanks for the suggestion, though I think I’ll keep the comments as is. Most blogs are like this, so I think people will be able to separate the content from the comment section. 255. Eric says: Thank you very much for the concise and layman’s explanation! I am older student enrolled in calculus and many core concepts (such as e) are a bit rusty.. Your refresher was very helpful!! 256. @Eric: You’re welcome! Glad it was helpful :). 257. Wow, you realy made me hungry for me. Since today I realy hated this stupid e because in school you have to use it but wont ever know what it actually is! This article brought some light in the darkness of the mathpart of my brain. 258. @GrandMaster-D: Awesome, glad you’re enjoying e again :). 259. John says: Yowzers! This is a gem, I wish I had this resource taught to my A-Level maths teacher (who was a v. good teacher). We were told just to accept e for what is was, which became clearer over time but you have now closed the open case for me. Great job. 260. Jeremy says: Hey Kalid, Thanks for the explanation–it’s very clear, and I have a much better intuitive understanding of e now. However, I think there’s a bit of confusion when you talk about continuous rate of growth versus growth over a given time period. You seem to treat them as completely separate, when in fact it’s perfect valid to specify a continuous rate of growth by saying how much overall growth you get after a certain period of time. Take Example 1. You say that you have a magic crystal that doubles in size every day, and you say that because the crystals grow continuously, we need to use e to compute the rate of growth. But this is incorrect! They may be growing continuously, but they still grow by a factor of 2, not e, each day. To clarify, let’s try using your formula. You say that if I have 300 kg of crystals (let’s just call it 300 crystals), and we let them grow for 10 days, we have 300 * e^(1 * 10) = 6.6 million crystals. But then if we let 1 crystal grow for 1 day, we’d have 1 * e^(1 * 1) = e crystals, which is wrong! We know (by assumption) that letting one crystal grow for one day yields two crystals, not e crystals. Now, that doesn’t mean that the crystals are not growing continuously. It just means that the growth curve is a little bit shallower than you’d expect, so that when you compound continuously, you wind up with two crystals at the end of the day. For example, let’s start a crystal growing, and look at it after 12 hours. You might think that you’d have 1.5 crystals, since it has been growing for half the time. But then, as you point out, those 1.5 crystals would grow for another 12 hours, leaving 2.25 crystals at the end. So what’s going on? The answer is not that we’re going to have e crystals after a day, since we’ve already assumed we’ll have 2 crystals after a day. The real answer is that we have *less* than 1.5 crystals at 12 hours. In fact, we’ll have sqrt(2) ~= 1.4 crystals after 12 hours, so that when those 1.4 crystals grow continuously for another 12 hours, we have exactly 2 crystals at the end. The same thing is true with rates of interest from a bank. If a bank says they’re giving you 5% interest annually, that doesn’t mean they’re cheating you out of the interest that accumulates continuously. The bank may very well be compounding interest continuously, but doing so in a way that adds up to a 5% annual interest rate. If so,$120 will grow over 10 years to become $120 * 1.05^10 =$195.47, not $197.85. 261. Jeremy says: Hey Kalid, Scratch that, I’m totally off base. I learned even more than I thought from your article. In Example 1, it looks like your first crystal has produced its weight in smaller crystals, but you’re not counting the additional growth of those smaller crystals when you say that the crystal has doubled. In that case, I agree that your example is correct! And you’re right about Example 2 as well. Say the bank specifies that they’ll give me 5% interest each year. They can compound interest at whatever rate they’d like (yearly, monthly, daily, instantaneously) in order to achieve that overall 5% rate. However, if they compound instantaneously (using e to calculate the return), then I’ll have slightly more after 10 years than I would if they compounded yearly. Alright, got it. Thanks again! 262. Jeremy says: Wait, hmm, I’m still a bit confused about Example 2, and I think it has to do with APR vs. APY, which you discuss in another article. In your example, it looks like you have an APR of 5%. Since this is compounded continuously, your APY is e^.05 = 5.13%, and after 10 years, you’ll have$120 * 1.0513^10 = $120 * e^(.05 * 10) =$197.85. Once I’ve figured out the continuous APY, I can compute interest year-by-year.

I think I was confused in my first post because I assumed you were talking about a 5% APY, in which case it doesn’t matter how it’s compounded, right? (Effectively, the compounding is already included in the APY.) If you’re compounding continuously, the APR in this case would be ln .05 = 0.0488.

Anyway, thanks again for the article, and apologies for filling your comments with my nonsense.

263. Aaron says:

Kalid!!!

Wow-wow-e-wow…

…wow. I’ve been studying for my CFA, and this is brethtakingly helpful!

Thank you!

(wow!)

264. Aaron says:

(breathtakingly)

265. @John: Thanks! Yes, I hear you — e was just a concept to memorize for a long time for me also. Glad it helped you.

@Jeremy: No worries, I appreciate the discussion! Yes, I think I need to clarify the meaning of APR vs APY in those examples. For the crystal, I meant to say that the main crystal is doubling (growing its own weight in 1st generation baby crystals) but those baby crystals are growing too (making 2nd, 3rd, 4th generation and so on).

For the money example, I meant to say the bank was giving us a 5% APR (current rate of growth), but if we wanted to maximize return we’d ask for our interest to be calculated every instant (leading to an e^.05 APY).

But it looks like you’ve clarified these ideas in your head, so understand this well enough! I’ll see if I can go through and make the wording more clear — APR vs APY can be very confusing since “interest rate” can imply either (and APR/APY sounds so formal).

@Aaron: Thanks, really glad it helped you!

266. Joanne says:

Thanks for the great explanation. I must teach a lesson this next week, and have been laboring to put the explanation into words that my students will truly understand. Your explanation will help a lot.

267. @Joanne: Thank you, and good luck with your lesson!

268. Rajakumar says:

Is there any difference between exponential ‘e’ and ‘E’.

269. @Rajakumar: Nope, they are the same thing. I believe e [lowercase] is the technical way of writing it (though in some cases it may be capitalized in a title, as in the wikipedia article and the heading for this one, which I’m changing now).

270. Ritu says:

I must say…i have no words to appreciate your understanding of different topics. Keep it up!!! you have helped me love maths again

271. Kalid says:

272. John says:

Superb work, Kalid. Just to clarify one of your goals:
“e^x lets you predict the impact of any growth rate and time period”. This is only true for a “continuous” growth process, as opposed to a discrete process like simple interest, correct?

273. @John: Thanks. Yes, e^x models continuous growth. However — items like simple interest can be converted to their equivalent continuous rate. So you could say “I’m earning x% simple interest per year, modeled by (100+x)^n, which is equivalent to y% continuous interest, modeled by e^(y*n).” Both graphs will take the same value at the desired time interval (the continuous one will interpolate between points).

Take a look here for more details: http://betterexplained.com/articles/a-visual-guide-to-simple-compound-and-continuous-interest-rates/

274. Mario Panzieri says:

Yours’ possibily the clearest, most interesting explanation of maths I have ever heard, 5 years of college notwithstanding. Thank you very much for helping us understand, it’s priceless.

275. @Mario: Thanks for the kind words — you’re more than welcome! That really means a lot, as my goal was to help people understand and enjoy math [especially people who suffered through it in college like I did ] :).

276. Quwackers says:

WOW thnx alot I have to a Maths B assignment and all the other web sites I have viewed have only confused me as to what e really does now I understand I dont know what I would have done without this thanks alot. : )

277. @Quwackers: You’re welcome!

278. Art Malm says:

For many years I’ve wondered what the heck “e” was and some day I would find out. Not that I ever tried real hard, but it was always in the back of my mind, popping to the front whenever I ran into an expression with an e or ln.

Today my search ended. Thank you very much for an extraordinary piece of teaching. Unfortunately, the question which led me to your site raised simply another question: ‘Why the hell would somebody use the natural log of the hydrogen ion concentration in an expression setting the standard for lead concentration in drinking water? C’est la Guere!

My now more educated guess at the answer? Either out of ignorance or to obscure the answer, given that a far simpler expression would use the negative log(10) of the hydrogen ion concentration, the otherwise very familiar “pH”.

279. Nice article! You said:

“e is like a speed limit (like c, the speed of light)…”

Maybe you have a nice explanation on the constant ‘c’ too?

280. @Mohamad: Thanks! I think that’d be a fun article, I don’t have an intuitive understanding of that concept yet but am excited to learn about it. Appreciate the suggestion!

281. kalai says:

Good. Really so thanks for your detailed ans. it will help me to know lot and lot.

282. Said Jazaerli says:

thanks for the nice explanation.
would you please explain to me the folowing expresion in an intutively way:
e^jwt=cos(wt)+jsin(wt), where j is the complex number: j=sqr(-1).

283. @Kalai: You’re welcome!

@Said: Great question — I plan on covering Euler’s formula in a later post.

284. Jia Liu says:

There are some very good explanations about e and log in an old and small book by W.W. Sawyer: Mathematician’s Delight. It is really worth reading.

285. Kalid says:

@Jia: Thanks for the suggestion!

286. Artek says:

Hello Kalid!

I really like your way of explaining mathematical concepts. With the idea of bacteria growth I finally got it why it is called natural ;). I have had only one tougher time with radioactive decay – didn’t know why you used the negative exponent of e (it did not seem 100% intuitive to me to just flip e upside down and this is it. As I came to understand it, the idea of decay could be put in this way: (1 – 1/n)^n (we lose a bit every 1/n of the 100% time – or we lose n-bits over the time, where bit- is a fraction of the actual state of the whole). This is similar to e, but needs some adjustment: ((n-1)/n)^n then we turn it upside down: (n/(n-1))^-n which is ((n-1+1)/(n-1))^-n and there emerges e:
(1 + 1/(n-1))^-n which is (1 + 1/(n-1)) * (1 + 1/(n-1))^-n and therefore (1 + 1/(n-1)) * e^-1. Of course when n gets unlimited the whole thing turns out to be e^-1. I don’t know if this thinking is in agreement with actual decay process – I don’t know if it’s correct. However I needed more step-by-step explanation to understand why you just proposed that 1/e is the rate of decay. Maybe you could provide some intuitive one, as you did in other examples..
Regards and praise for what you’ve done!

287. Dan says:

Hello,
I have a question that relates to e that I was wondering if you could explain to me…

I am trying to understand this study in which someone was trying to calculate the average rate of change in human skull size over the past 1.5 million years. starting off, he had a data set consisting of roughly 100 skull samples, which had all been radiometrically dated.
Therefore, he started off with a graph of skull size vs. time. He then proceeded to take the natural log of both axes, claiming that this avoids interdependency issues within the data set.
How exactly does taking the natural log of this relationship do this?
Thank you!

288. Glen says:

Great explanation! You’ve made everything so clear to me. I only have one problem, and it probably comes down to my negligble understanding of limits.
When n approaches infinity for (1+1/n)^n
my basic knowledge tells me that 1/n approaches 0, so the entire expression approaches 1, rather than e. On the other hand I have tried and tested it myself and it definitely approaches e. Can you explain?

Cheers.

289. robin says:

I don’t know whether to be sad that you weren’t my math teacher when I was a kid, or happy that I found this article. THANKS.

290. Landy says:

True mastery is demonstrated by an ability to make the complicated seem mundane; as you have.

291. DerekSmith says:

Hi Kalid,

Either I have misunderstood your Magic Crystals problem or you have mis-stated the problem.

You define the problem as 300kg of crystals growing to double in a 24 hour period. For continuous growth, the rate would have to be Ln(2) i.e. 0.693… , not 1 as you state in the example.

Ten days then would be E^10*0.693… which is 1024kg, not 6.6 million kg as you calculated.

Another way of looking at it is if in 24 hours the crystal weight doubled, then in ten days the weight would have doubled ten times or 2^10 which equals 1024.

http://instacalc.com/embedpreview.html?d=&c=bG4oMil8RV4oMC42OTMxNDcxODA1NioxMCl8Ml4xMHx8fHw&s=sssssss&v=0.9

292. Kalid says:

@Derek: Thanks for the note! I’ve updated the example to be more clear about what I mean.

I was trying to say that a single crystal grows at a rate of 100% per day — that is, if you look at the baby crystals flying off, it appears to generate its own weight. Of course, the baby crystals start growing on their own, but we don’t track that — only the amount the parent generates.

So, in this case we’re going from the input rate/growth rate for a single item (“APR”) to the total compounded rate (APY). In your correction, we’re going from the total compounded rate back to the input rate, and indeed want to use natural log to reverse it out.

I’ve updated the article to make this more clear, thanks for the feedback!

293. Kalid says:

@Landy: Thanks! I don’t think I’ve actually mastered the topic yet, there’s so much more I’d like to learn :).

@robin: Thank you!

@Artek: Great discussion! Yes, I should go in and make it more clear why a negative rate would mean decay.

One intuitive way to think about it: if going forward in time is getting larger, then going backward in time is getting smaller. So you can consider e^(rate * time) for e^(-1) to be “100% growth, but 1 second ago”. I’ll think about other ways to make this clear!

@Glen: Great question. I think the key is that for a limit, you can never replace 1/n with exactly 0, even though it “approaches” zero. 1/n becomes a tiny, microscopically small amount, but never actually, truly zero (it’s a small number outside of our range of “error”, but not zero).

Therefore there’s always a little amount of growth it can contribute, which is taken to a giant exponent. I plan on doing an article on limits (or infinitesimals) in the future :). I personally don’t like limits that much and am starting to prefer the hyperreal number system which makes infinitely large and small numbers easier to work with.

@Dan: Interesting question — I’m not that well-versed in statistics so I can only hazard a guess. I couldn’t actually find a good definition of what interdependence means! (It’s not correlation, otherwise they would have used that word?).

From quick research, it seems log-log graphs can show interesting distributions (like Zipf for word frequencies).

Intuitively, one way to see the log is to “correct” for growth. If you have something doubling (2^x) you can take the log (ln(2^x)) and see that it’s a straight line, or a constant growth rate [leading to exponential growth].

If you do a log-log, you are correcting for growth on both the output and time axis. So perhaps this cancels out the “interdependence”, whatever that may be?

294. Albert Stifter says:

I am seriously interested in maths,but I am one of the slowly understanding and I never got satisfied with my very poor results in past..
I think u are “Star number one “and yor way should be somehow implemented to to school systems around the globe…One day I want to study maths and I imagine if you would be my teacher I would be no retard anymore…

295. Anonymous says:

My teacher had asked the students to find out what really ‘e’ is, everybody found the same definitions but didnt build up any logic..i really liked your site
hats off!

296. Kalid says:

@Albert: Hi Albert, glad it was helpful! A lot of things, math included, are just a matter of looking at things from the right perspective. Reading and writing were once thought to be the work of scribes, and now everyone does it. Math was once thought the work of mathematicians, but everyone can do it!

297. Marie says:

Do you know chemistry too? If so, can you help me relate this equation to your explanation of natural logs? Someone mentioned it earler: ” dG = -RT ln K, which relates the reaction rate constant K to Gibbs Free Energy”

Also, is the independent variable always time? Could it be growth with respect to concentration or temperature or something like that?

298. Kalid says:

@Marie: Great questions! The independent variable doesn’t need to be time actually, I intuitively see it as something related to the “rate” of growth (in fact, in e^x, x is a combination of “rate * time” — the net growth after 1 unit of time). I can imagine how the temperature would determine the rate or the total growth after that unit of time.

My chemistry is very rusty, but from looking up the equation it seems that you compute K from the identity:

K = e^(-dG/RT)

One way of seeing the natural log is determining the “rate” to get a total rate of growth. So ln(K) = -dG/RT might be seen as a “rate” … and when you multiply by -RT you get the dG back out. Hope this helps!

299. Marie says:

Thank you SO much! That’s tremendously helpful to me.

300. @Marie: You’re welcome!

301. misty says:

thank you so much..your explanation about this e really helps me..honestly..i really don’t know what does this e means..but after i have read these article..it gives me a lot of idea about that constant e.

tHANKS..!!!

302. JR says:

Wow that was a great way of explaining e. Now when I do my calculus homework I know what e is about. All I can say is… Thank You So Very Much!!!!

303. Kalid says:

@misty: You’re welcome!

I’m a CE student and I’m thinking about double-majoring with pure mathematics..

thanks a lot.

305. Kalid says:

306. Rodrigo says:

Nice stuff, pretty good explanations.

I think there is a market to good math books with this kind of straightforward approach.

Congratulations!

307. Kalid says:

@Rodrigo: Thanks! Yes, I’m working on my ebook now actually :).

308. Emily says:

Kalid,

I am a freshman in highschool and am writing a report on natural logarithms and the number e. Your explanation was by far the easiest to understand out of all the articles I’ve read. Thank you!! (Why does my teacher want this report? We’re only in 9th grade!!)

-Emily

309. @Emily: Thanks for the note — really glad it was useful! It’s awesome that you’re getting to learn about e so early :). Good luck on the report!

310. AK says:

THANKS A LOT!
it really helped me with finance stuff…

311. Kalid says:

@AK: You’re welcome!

312. Evans says:

I had been wallowing in a miasma of deceit and confusion over this little e until today!!! I tried many sources with patience but with no success in my confidence of its comprehension until I landed at betterexplained.com; that’s after being taken in circles in Wikipedia etc. Receive my thanks compounded to e.

eVans

313. AWesome site Khalid, thanks so much for your insightful, “”intuitive”” explanations. Jazakallah, you’ve spent quite a lot of time and effort in creating such an effective site. After searching all over the web for explanations on e, and ln, i have to say, ur site is the best there is! Keep up the good work, may you be rewarded here, and hereafter.

314. Kalid says:

315. ali says:

Great article. I completed my graduation but never knew the real “e”. now i am know its importance and glad that i happened to come across this site.

next thing i gonna do is forwarding this article to my geeky friends who, i m sure, dont know the real “e”

now i m in mood of exploring this whole site

Regards,
ali

316. Kalid says:

@Ali: Awesome, glad you figured it out! Yes, I didn’t learn about the real meaning of e until after graduation as well :). Hope your friends enjoy the site too, thanks for sharing it.

317. Hi Kalid,

Nice article. Question – if a population grows at variable rates between 1 and 2 percent annually, is it growing exponentially? See my article “What Is Exponential?” for my own views.

You may also like to read my “Scales of e” article which extends the heuristic tool the Rule of 70 (or the Rule of 72) to include both positive and negative rates of growth, and also allows for variable rates of growth (not just constant rates of growth).

Regards,

David

318. Alessandro Orefice (from Brazil) says:

That´s amazing explanation. My dream is that all the Math teachers will be able one day to explain Mathematics like you.
Great job!

319. Kalid says:

@Alessandro: Thanks for the kind words!

320. This is the first time I’ve really “gotten” e. thank you!

321. Kalid says:

322. Kalid says:

@Carl: Awesome!

323. AJ 88 says:

Thanks dude for doing this. I’m a 3rd year degree student in mathematics and never fully understood ‘e’ until reading all this. It’s astonishing that schools don’t teach this from the beginning!

324. Dimiter says:

Thank you very much for explanation!
I loved math when I was kid but after entering into Calculus and other maths I got scared because nobody explained me how we obtain those constants. We just had to learn them and not too look for a logic behind them.
Many thanks again!

325. Anonymous says:

Though most of this is correct, e is actually defined as the number that, when raised to the x power, is equal to both its derivative and its integral.

326. @AJ: Awesome, glad it helped! Yes, I think schools feel the need to use the “Calculus” version of e and miss out on its more humble origins.

@Dimiter: You’re welcome! Yep, I was in the same boat… at some point it struck me that I didn’t really _understand_ this number we’d been using for so many years…

@Anon: e has several equivalent definitions, just like a circle; I don’t know if one is more official than the other.

The integral/derivative is useful in many contexts, but unfortunately it implies you need to understand calculus to use e.

An analogy: a circle can be defined as all points the same distance from a given point (the center), or x^2 + y^2 = r^2. Both convey the same idea, but one requires you to understand algebra.

327. Kaido says:

Nice job kalid. May I use this in my blog to explain this for estonian student ?

328. Kalid says:

@Kaido: Sure!

329. Hi Kalid

Similar to others, I am a math/science guy (working in IT) and consider myself intelligent/educated, but I gained a great deal from reading your explanations of e and ln.

Your explanations are concise but lucid, and bring an intuitive understanding of what used to murky concepts to me (and many others). Quite clearly you have a deep understanding which enables you to simplify the complex. Thank you for sharing – much appreciated.

Regards
Pete.

330. Marisano says:

Cool site – thanks!

Here’s a minor correction to make this page even cooler: I think that, “Suppose I have $120 in a count with 5% interest.” from the “Example 2: Maximum interest rates” section should read, “Suppose I have$120 in an account with 5% interest.”

That is, that “a count” should be replaced with “an account”.

Cheers!

331. Marisano says:

Another minor error, this time in “Example 3: Radioactive decay”:

Yes, we do begin with 10kg and expect to “lost it all”

Yes, we do begin with 10kg and expect to “lose it all”

That is, “lost” should read “lose”.

332. Marisano says:

Also from “Example 3…”

[Decay is commonly given in terms of "half life", or non-continuous growth. We'll talk about converting these rates in a future article.]

“for non-continuous growth”??

Thanks.

333. Kalid says:

@Marisano: Thanks for all the corrections! I just updated the post :).

334. I am a math/science guy and consider myself intelligent/educated, but I gained a great deal from reading your explanations of e.Thank you very much.

335. I am a math/science guy (working in IT) and consider myself intelligent/educated, but I gained a great deal from reading your explanations of e.Thank you very much.

336. Matt says:

Just wanted to say thanks for the great explanations. Really helped me out. Thanks.

337. Great post! Thank you for taking the time to explain…

338. Me says:

Hello All,

The latest Crop Circle actually is communicating in binary code…..e^(hi)pi+1=0…..can anyone shed light on what this means….thanks

339. jack says:

“I even confused myself a bit while putting the charts together”

Finally! A teacher who will admit something like that. It shows students that it’s OK to get confused and it’s normal, rather than pretending to be perfect. Thanks for being honest!

340. jack says:

Many comments here by people who need to tell others they are intelligent. Great, that’s fine. Being born with intelligent genes might be nice and all and good for showing off. Winning the genetic lottery is great and I would congratulate anyone who wins any sort of lottery.

But really, it’s those underdogs who are NOT bright but are ultra-motivated and persistent the ones who should be showing off their ability to keep learning despite difficulties. Hopefully this determined attitude will help these people become successful.

They just need a great teacher, and Kalid is awesome.

341. Kalid says:

@jack: Thanks for the comment! I agree, sometimes teaching seems more about demonstrating your own knowledge vs. improving what other people know. I really think the vast majority of topics are within anyone’s grasp if explained correctly.

342. Kalid says:

@jack: You’re more than welcome! I find it’s dishonest to pretend an idea is simple and obvious when it took some of the greatest minds centuries to find it. Getting confused is part of learning.

343. Pat Eblen says:

This may be covered somewhere on this page but I offer it anyway:

The number e is unique in that it is the only number that can define a function such that the value of the function is proportional the slope of the function at every point. That is, the rate of change of the function is proportional to the value of the function at every point: f(x) = e to the x power, AND (the slope of f(x)) = e to the x . There is no other number that this is true for. So e shows up anywhere that something is changing at a rate that is proportional to the amount of the “something” that is current.

This can also be viewed as such: What if one wanted to define a function such that the value of the function is proportional to the slope of the function at every point? Since the slope has to increase as the function increases, it is reasonable to speculate that the function is an exponential of the form, f(x) = a to the x. Now, roughly sketch the curve. If x is o, f(x) = 1, so there is one point on the curve, that is, it crosses the y axis at y = 1. If x = – infinity, f(x) = 0, so the curve asymptotically approaches the x axis as x becomes more negative. If x = 1, f(x) = a, another point on the curve. Now the curve can be sketched. The ONLY number that satisfies the requirement is a = e = 2.718+.

344. Leerick says:

Exactly what I was looking for!!!

345. SS says:

Just found this. Always wanted to revisit “e”, and understand the rt. This was a big step in the right direction. Now, I am primed to read about “i”, and rediscover Calculus.

You’re a true scholar, spreading info. I hope you find success chasing money also, though I doubt that’s your motivation in life. And that, you must embrace! A rare breed.

346. Another Jack says:

Great page!!!

I completely forgot all of this from school. Reading this page, I think I might understand it better than ever.

347. Kalid says:

@Another Jack: Awesome, hope it’s clicking for you.

348. Kalid says:

@SS: Thank you for the support and kind words — I just want to share the insights that helped me get over those frustrations when learning. I’m all about rediscovering calculus :). Yep, I’m not about chasing money really — I’d like to support myself and give myself the freedom to pursue my passions, like this site.

349. Kalid says:

@Leerick: Great!

350. Anonymous says:

Thank you soooooooo much for this. I’ve been trying for weeks to get a clear explanation on exponential growth and I didn’t understand it until I saw your page. Please keep up the amazing work on this and I hope this website never goes down so that others may learn for years to come.

351. Jerry Perfetti says:

Beautiful great. Best explanation I have seen on
topic.

352. Kalid says:

@Jerry: Thanks, glad you enjoyed it!

353. Kalid says:

@Anonymous: You’re welcome! It took me a long time to really understand exponential growth also, and thanks for the encouragement.

354. I finally get “e” 10 years after calculus in college:) Brilliant job e(xplaining) it!

355. Jenna says:

Finally! After all this time, with no one able to explain to me when to use ln, I get it! Thank you so so so so so much!!!! When I get an A on my precalculus exam tomorrow, it will be thanks to you!

356. Roger says:

I have just started working toward my Master’s degree in Fisheries Resource Management. I have just downloaded your book…I can see it will be a great resource. Thanks so much…

Roger

357. Kalid says:

@Roger: You’re welcome, and thanks for the support.

358. Loui says:

Oh my gosh, thank you! I’ve been struggling with this for months! You’re a Godsend.

359. Kalid says:

360. Diego says:

Gee, and I thought I was the thick one. This article brought to mind the Pink Floyd song where the cruel teacher goes home to be beaten by his psychopathic wife.

In my case, that would be Mr. Shaw. See, in my case, I sat in math class thinking I was too dumb to follow, and Mr. Shaw I think took pleasure in perpetrating this line of thinking.

So, 20 years later, on a Friday evening, I come across a logical, precisely and concisely presented version of the facts surrounding e and get it. At first reading. No doubt in my mind. It’s understood. Cold.

Ah, I sure hope Mr. Shaw didn’t get knocked around too much when he went home…

361. Kalid says:

@Diego: Thanks for the story :). I think we all have demons we’re overcoming, math seems to evoke more than other subjects (mine are around a calculus class I took in college, the pain of which inspired me to start writing). I firmly believe that anything can be explained simply, and if an explanation doesn’t make us say “Aha! Now I get it!” then it’s broken.

362. Huckleberry says:

Just wanted to be among the many to say thanks. My appreciation for ‘e’ has grown dramatically. Cheers. (I will sleep more contentedly this evening.)

363. David Hughes says:

This web page has given me a dose of pure joy, I have always had trouble with e, and lost marks in exams over decay functions, now they seem so transparent… thank you

364. Kalid says:

@David: Awesome, really glad it helped! I was giddy with excitement when e started coming together also :).

365. Joy says:

hi khalid,
i did not understand the part about taking different rates
i am in high school,
i came across your blog when i was doing permutation and combination and was hooked, i recommend this site to anyone with difficulty in the topics covered. the cool thing is you made maths like reading a newspaper or novel.

366. Rafay says:

I effing love you for this!!!!!

367. Kalid says:

@Joy: Thanks, glad you’re enjoying the site! Yes, the explanation on the different rates isn’t the best — I’d like to revisit it.

Here’s another way to see it, maybe it will help:

e is about taking growth, breaking it into tiny chunks, and apply each chunk. So, if we say we have 100% growth, we mean:

* Break 100% into tiny increments (of 1%, say)
* Apply one increment at a time, and take the total result

So instead of 100% growth (going from 1 to 2), we want

1 + 1% + 1% + 1% … (100 times) = 1 * (1.01)^100 ~ e [not quite e, but very close]

e is a shortcut for “100% growth, as fast as possible”.

Cool. Now what about 50% growth? What does that mean?

Well, we do the same thing: take 50%, divide it into small chunks (1%), and apply them individually:

1 + 1% + 1% + 1% … (50 times) = 1 * (1.01)^50 = ?

Well, what is that question mark?

Well, if we say (1.01)^100 = e, then taking only half the multiplications must be the square root of e! (Think about it this way: 2 * 2 * 2 * 2 = 16. If we take half the multiplications, or 2 * 2, we must be at the square root of 16).

So, (1.01)^50 = 50 multiplications = half the multiplications of e = square root of e = e^(1/2)

In general, a new interest rate (like 75% per year) is just a different number of multiplications to use for e: and 75% growth would be e^(75/100) = e^(3/4).

e is really e^(100%) = e^(1), and assumes we’re talking about 100% growth. We can modify the number of multiplications to get the impact for any growth amount we need.

Hope this helps!

368. Ehab says:

thanks very much for this amazing explaination
and excuse me if this comment is specified in questions only!!!

369. Vinay says:

Khalid – I knew exactly what e is and how it evolved, but was always had difficulty using it for exp. growth/decay.. your explanation helped me a lot. I am an engineering student and would visit your site regularly.

vinay

370. Anonymous says:

You rock! thanks a bunch!

371. Ashwin says:

A wonderful concise and an effective explanation. Thanks a ton. The examples make this arcticle the best. Thanks -Who ever you are.

372. @Ehab, @Vinay, @Anonymous, @Ashwin: Thank you!

373. Befuddled says:

Awesome article!
Question: Is the fact that e, the base rate of growth, is an irrational number just by chance, or is there an intuitive understanding as to why e ought be irrational?

374. Kalid says:

@Befuddled: Great question. There’s a formal proof here: http://en.wikipedia.org/wiki/Proof_that_e_is_irrational but I haven’t got an intuitive feel for it yet. My gut tells me that “natural” constants (e, pi, phi) shouldn’t line up perfectly with the integers (or a ratio of integers) — it’s just too much of a coincidence! :). After all, there are infinitely more irrationals than rationals.

375. Clement says:

if e^2x+1 = a …. i need you to help me make x the subject of the formula

376. Raza says:

I just want to say thank you for doing this. You won’t believe how many years I have thought about this and never getting a satisfying answer. I am in 3rd year Computer Science and have taken quite a few Math courses but they all avoid explaining rudimentary subjects like this.

377. Kalid says:

@Raza: You’re more than welcome, I appreciate the kind words! I was exactly in your situation — going through my courses without a real clue for what it all _meant_. Glad it helped :).

378. anonymous says:

Hi Kalid,

As many have said before, wonderful explanations.

Your explanation speaks to a constant rate of growth (linear) but what if the term in the exponent is quadratic, exp(-x^2)? This would be a Gaussian function, wouldn’t it?

I’m wondering how the derivation would look.

Thanks!
Anon

379. Kalid says:

@Anon: Thanks, and great question. Yes, this leads to a Gaussian function, but I don’t have a good intuition for the meaning behind it yet. Definitely something to dive into further :).

380. Kevin says:

wow you are so cool! i loved your explaination

381. Martin says:

Great explanations. Thanks. From what you explain, it seems to me that e is not so much the natural rate of growth or maximum rate of growth as the most efficient rate of growth. The more divisions you perform, 2, 12, 100, 1000, etc., the more efficient your growth is becoming. In the real world, I guess that this level of efficiency is impossible to obtain but it’s easy to see why from your explanations it is a solidly based estimate.

382. Kalid says:

383. Kalid says:

@Martin: Thanks for the comment! Yes, in practical terms we just hope to approximate e and use it in calculations (similar to how we approximate pi or sqrt(2) and use them). It’s nice to know the theoretical ideal (max efficiency, as you say) that we’re reaching for.

Great article. Best explanation I’ve ever read

385. Kalid says:

386. Dude, Thank you so much. I was planning on spending a lot of time until I’d finally understand natural logarithms and the constant e. Then I came across your article…

I can’t say it more clearly than this:
You just saved me a ton of time! Thank you!

387. Kalid says:

@Rowan: You’re welcome! e and ln bothered me for a long time, glad they were able to save you some

388. Samir says:

Kalid,

I disliked math in school because teachers explained it so dull and unengaging.

Now it happens that I love physics and plan to study it, but I need to get up to date on my math. I’ve got several books and are studying them, but well, I get stuck.

Everytime I get stuck, I come here. Everytime I try to learn a new concept and I somehow “feel” the solution lingering below my threshold of complete understanding, I know that there must be an intuitive approach.

Then I come here and most of the time, there were you, kind enough to share insights and relate concepts, and my mind = blown about how intuitive math actually is, and I get that silly euphoric rushy feeling in my head!

Thank you so much for that!!!

389. Kalid says:

@Samir: Thanks for the heartfelt message! That euphoric feeling is exactly what I strive for too — and I think it’s there, available to us for any topic if we’re able to find the right explanation :). You are more than welcome and I’m so glad it helped!

390. Anonymous says:

Great article! I always wondered about e, and now I am starting to get it! Its found everywhere it seems, and without knowing what it is its very strange. If only I understood how it relates to imaginary exponents…

391. Dileep says:

Hello Kalid

Thanks a lot for explaining the stuff in an intuitive way.. I have studied Electrical engineering and seen ‘e’ everywhere….
I didnt understand the real meaning of ‘e’ until I read this article today.. Keep up the great work bro!

392. Kalid says:

@Dileep: Thanks, understanding ‘e’ was a long time coming for me too :). Appreciate the support!

393. Anonymous says:

Amazingly explained

394. Narayan says:

Thank you very much….

395. talespin kit says:

Its awesome, was really, really helpful.

396. Joanna says:

Amazing site! You have explained this math concept so simply, yet deeply. Even my math teacher can’t explain as good as you.

397. Kalid says:

@Joanna: Thank you!

398. Simple Simon says:

This is a lot easier to understand. Brilliant! The Wikipedia is uni level and i am 13 phew

399. yonggook says:

beautiful explain. thanks alot

400. Kalid says:

@Simple Simon: Awesome, glad it helped!

401. filio623 says:

Wow I’m in my senior year of engineering school and have used e in countless equations and problems but I don’t think I’ve ever really understood it till now. Great explanation!

402. Kalid says:

@filio623: Thanks! Glad it clicked for you :).

403. Rickard says:

Kalid, I’ve been struggling a lot with this material. Your methods and insights are simply priceless. Thank you so much.

I’m having some minor problems with the infinite series definition of e. Each term can be thought of as “interest” on the term that precedes it. But why are the terms determined by “1/n!”?

I’ve gathered that the definition is a special case of the taylor expansion of e^x. Is there a more simple way of understanding why each term is that particular size? If not, will I have to learn about taylor series to make it click for me?

404. Matti_Meikäläinen says:

Unbelievable

405. joely says:

Have you considered making this available as an e-book?

406. WCW says:

You are a revolutionary.

407. Syd Polk says:

Outstanding explanation. Wish I had this in high school.

408. Olya says:

Thank you very much for this article. Math seems so complicated to me. And it is the subject I need most because I’ve decided to major in economics. Your articles answer the questions I’ve been asking for a long time now.

409. Dave Neary says:

e is also e^1 where e^x is the function satisfying the identity d/dx (f(x)) = f(x). A function whose slope is the same as its value at every point on the curve.

I do like your parallel to compounding interest over smaller time periods (which is basically the equivalent to deriving the Maclauren series for e^x with x=1), and showing e as a universal constant in a certain type of calculation is really good. Impressed!

Dave.

410. OSC says:

Very good and clear !

411. grillermo says:

Awesome, why can’t more mathematicians do this? it just take a little bit of humility and being able to actually talk like a human being.

412. Arun says:

I wish you were my teacher in college; I really do!

413. Kalid says:

@Rickard: More than welcome. Check out http://betterexplained.com/articles/developing-your-intuition-for-math/ which describes the infinite series for e in terms of compound interest. Hope it helps!

@joely: Hi, I have en ebook at http://betterexplained/ebook =). I’d like to make some follow-ups to it as well.

@Matti_Meikäläinen, WCW, Syd, OSC, tumutanzi: Thanks!

@Olya: Thank you — I really need explanations to click deep down before I feel comfortable. I hate plug and chug!

@Dave: Thanks! I like thinking about the different ways to define e… I shy away from the Calculus definition at first because it introduces yet another thing to learn. But once you know it, another tool in the belt :).

@grillermo: Exactly. Let’s not pretend a concept that took humanity thousands of years to develop is “obvious”.

@Arun Thanks!

414. NANDEESH says:

Hi,
I knew permutations and combinations. But I came across the concept of derangements recently. I found that the total number of permutations is ‘e’ times the total number of derangements. To explain this in terms of growth, is it correct to say the set of derangements grows to the set of all permutations as we loosen the conditions for derangement gradually?
Let me give an example.
Consider three letters A,B &C. If we mix them up such that no letter is in its original position we get 2 arrangements (B,C,A) and (C,A,B).
There are however a total of 6 permutations:(A,B,C,),(A,C,B),(B,A,C),(B,C,A),(C,A,B)&(C,B,A).
The ratio of all permutations to derangements is 6/2=3.
If we take 4 items the permutations are 24 and the derangements are 9. The ratio is 2.7.
For 5 items there are 120 permutations and 44 derangements. The ratio is 2.7.
As the number of items increases the ratio gets closer to ‘e’=2.718281828…

It seems ‘e’ pops ups where you least expect.
Can this be explained in terms of growth???

415. Kalid says:

@NANDEESH: Wow, the relationship to e is really interesting. I know e shows up in other statistics puzzles, it would be fun to examine this one too.

416. ewan says:

Just a quick thanks for an excellent explanation.

417. Kalid says:

@ewan: Thanks, I appreciate it!

418. Ron Nimmo says:

Kalid: I really think the “intuitive” approach is long overdue; among other things, it engages the power of the subconscious mind into the understanding process. Ultimately, this kind of approach will lead to clearer and more elegant formal explanations and proofs.
However, I think that your definition of e as the continuous rate of growth is not exactly correct. e is the original quantity plus the rate of growth. In other words e = 1 + 1.718 Therefore 1.1718 is the continuous rate of growth, or actually the amount of growth resulting from compounding a 100% rate continuously.

I also think the following passage is ambiguous:

Although we picked 1%, we could have chosen any small unit of growth (.1%, .0001%, or even an infinitely small amount!). The key is that for any rate we pick, it’s just a new exponent on e:

{growth = e^{rate}}

The size of the unit of subdivision that you pick (i.e. .1, .01, .001) is not dependent upon the rate of growth, but is a reflection of the number of times it is compounded. The actual rate of growth per time period at “simple interest” is independent of the number of compunding periods.
The compounding frequency and rate become interchangeable at the point at which you “normalize” any given growth rate to (1 + 1/100)^100.

419. patrick says:

You have done what was not done for me in 4 yrs of college!!! I have an engineering degree and you have made this so easy to understand, thank you very much and may God bless you…

420. Kalid says:

@patrick: Thanks for the note! I was the same way — I didn’t get e until way after college ;).

421. Jose says:

Hey Kalid, you are an excellent teacher! I had the same problems when I first saw the number e at high school. Now I cannot wait to read your other articles. Thanks a lot!

422. Kalid says:

@Jose: Thank you, really glad it helped!

423. Fitzer says:

Thanks mate, that was very helpful

424. jim says:

Hello Kalid,
Thank you for your excellent insight of “e”. I used your explanation to teach my 11 year old son the meaning of “e”. He got it, thanks to your explanation. Did you ever follow through on the calculus of e (aka post 126)? I am looking for insight why “e” (whose origins are from continuously compounding interest) establishes the base for the only function whose derviative equals itself. I hope you had some time to give it some thought. Thanks Jim

425. Kalid says:

@Fitzer: You’re welcome!

@Jim: Awesome, glad it helped your son! I love knowing when the content works for a younger audience.

Yep, there’s a post discussing the calculus of e here: http://betterexplained.com/articles/developing-your-intuition-for-math/. It walks through several definitions (such as why the derivative equals itself) and how it relates to the growth analogy. Hope you enjoy it!

Wow, Kalid! That really put it together for me! Equations with Pi were connected to the circle in some way. But “e” never appeared to have a connection to a fundamental concept until now where you tied it together. When I was explaining “e” to my son, he made the connection but asked the question “so why does a 100% continous growth give rise to a curve whose rate of change equals it value?” I had looked in my old calculus books and the internet but the best I found was that “e has the magical property” discussion (i.e. the authors probably didn’t know). Your explanation nailed it exactly! And the Euler expansion and Ln now makes sense to me. I really wish I had this insight when I was in college. It would have allowed me to have much better insight into the equations that used “e” and how to apply them intuitively instead of mechanically. At least my son will have the benefit of your insight, intuition, and wisdom. This is really a great resource for students learning math. Any chance of monthly newsletter containing the latest insight? thanks so much, I can not tell you how much this means to me.

427. J says:

I didn’t really get e until I read this. The comparison with circles and pi was simple and intuitive. This is a fantastic explanation on what it is in normal language as well as how to derive it mathematically. Great talent.

428. Ron N. says:

I was looking at Post 300, your response to Marie’s chemistry question, in which you clarified Gibb’s equation:

“My chemistry is very rusty, but from looking up the equation it seems that you compute K from the identity:

K = e^(-dG/RT)

One way of seeing the natural log is determining the “rate” to get a total rate of growth. So -ln(K) = -dG/RT might be seen as a “rate” … and when you multiply by -RT you get the dG back out. Hope this helps!”

It did indeed help Marie but from K = e^(-dG/RT) you derived -ln(K) = -dG/RT.
However,I am pretty sure that +ln(K) = -dg/RT. Then if you multiplied by -RT you would get -RT ln(K) = dG, which is what you wanted to demonstrate. In other words there is no minus sign in front of ln(K).

Ron N.

429. Kalid says:

@Ron (#423): Thanks! Yep, I find the intuitive explanations are the ones that actually sink in.

Good point on e, I need to clarify “rate”, “yield” vs. “total” — e is clearly the total amount (2.718) and not the yield (1.718) and not the instant rate (1.0). As you say, 1.718 is the yield you get when compounding the 100% rate (I’m using the banking terms for APY and APR). I’ll need to go back and clarify this in the post.

On the compounding: yep, that’s exactly it — you need to normalize to some rate (like units of 1/100). I’m not sure of the ambiguity, but would like to correct it if there — is it the implication that the unit of subdivision must be based on your current rate of growth?

@Jim: Awesome — I’m really happy it helped when explaining it to your son! There’s so many explanations that just repeat the static definitions without getting into the meat of why.

I’d love to do a newsletter, it’s something I’ve been kicking around but hope to launch something soon :).

@J: Thanks! I strive to explain things in language I would use when talking to “myself” (why be needlessly formal?).

@Ron (#433): Whoops! Yes, that’s right — there’s an extra negative sign. I’ll tweak the comment!

430. Ronald Nimmo says:

Response to Pat Eblen Post # 345:

Pat, you made a very interesting comment, but what is the proof of this statement that you made?

If x = 1, f(x) = a, another point on the curve. Now the curve can be sketched. The ONLY number that satisfies the requirement is a = e = 2.718+.

431. Ribbles02 says:

Wow! This is one of the most lucid and enjoyable explanations I’ve found on math. Very, very good writing with a very, very good understanding. I’m definitely coming to this site more.

432. Kalid says:

@Ronald: Thanks for chiming in!

@Ribbles02: Really appreciate it! My goal is to make math as understandable as I can :).

433. James Fisher says:

> We’d multiply one year’s growth (e^3) by itself two times

“Multiplying x by itself twice” is a really ambiguous phrase; it could be x^2, as you use it, but I would interpret it as x^3; it could even be x^4 ((x^2)^2).

I would change it to simply

> We’d multiply one year’s growth (e^3) by itself

434. Kalid says:

@James: Great feedback — yes, that could be extremely confusing. I’ve just fixed the article, appreciate it!

435. Nice explanations and anologies.

436. Pat Eblen says:

@Ronald Nimmo, @Kalid, Ron – Great question. No explanation of e is complete without saying that it is the only number that can generate a function of the form a^x, whose slope is proportional to the function value at every point, and this is the fact that I thought would be helpful. This is the essence of why e shows up so often. Of course when x=1, f(x)=a since a^1=a, but then how to prove that a=2.718… is doable, but I do not know how. If you find out how to derive this magic number from the slope/function value rule, other than trial and error, please let me know. There must be an elegant proof. Thanks for the comment. Pat

437. Frank says:

This is awesome

438. D Rajendra says:

Explained very nicely.

439. Kalid you say in the above “When dealing with compound growth, 10 years of 3% growth has the same overall impact as 1 year of 30% growth (and no growth afterward).”

I think it would be more clear if ‘continuous compound growth” replaces “compound growth” in above, which I am sure is what you meant to say. The above sentence is not valid for regular discrete compounding, though it is valid for simple interest situation i.e. 10 years of 3% simple interest (where principal is not reinvested) is same as 1 year of 30% interest but definitely not if yearly interest is compounded.

BTW, It’s a great website. You explain so well.

440. kalid says:

@D: Thank you.

441. kalid says:

@Kumar: Great point! I’m amending the article. Yes, it’s important to separate continuous compound growth from the regular case, where you can’t simply exchange the rates. Appreciate the kind words!

442. waqas says:

Nice way writing . completely understandable

443. Anonymous says:

thnks alot i hve been working on exponential growth and decay few days ago but not been able even to understand the concept bt this page really clears alot of my confusions

444. Emilie says:

AMAZING! Forget textbooks and wiki, this is my new favourite go to website! Incredible!

445. kalid says:

@Emilie: Wow, thank you! I hope to keep writing for a long time to come :).

Amazing, thank you so much. I’ve been trying to get a better understanding of what it is that i’m learning rather than memorizing formulae, your detailed explanations are a real help. Kudos!

447. kalid says:

@Adrian: Awesome, glad it helped! e bothered me for a long, long time too :).

448. Selim says:

This is an awesome article. Many many thanks for making the topic this clear and this easy to follow and understand.

449. kalid says:

@Selim: You’re welcome, happy it helped!

450. kelly vanessa says:

hi, i was looking for articles which could really help me understand exponential functions and ln for our exams tomorrow. im really happy i found your website. thank you so much!

451. kalid says:

@kelly: Thank you, and thanks for the support!

452. Deya says:

thanks Kalid! it was great!

453. kalid says:

@Deya: You’re welcome, glad you enjoyed it!

Bless you and thanks. I am not even a math student, only a curious reader. How many explanations have left me dizzy and frustrated!? I don’t understand it all, of course, but I did get enough through my head to congratulate you for your humane explanations. I’ll be studying and looking forward to more.

455. kalid says:

@desiree: Thanks so much for the note! Making explanations “humane”, as you said well, is something I really hope to achieve . I appreciate it!

456. kukku says:

WOW kalid really gr8
Actually I was doing a project on the topic e .
But the thing is I just want something new about it .
I was searching for an integer approximation for e (like 22/7 for pi).
once again thank you for the article

457. Husam says:

fantastic job bro, u helped me a lot explaining one biggest questions in my mind,please keep your work going and are you from india?

458. Sonali says:

Khalid, Great article. But I am confusing myself. I thought i understood it but as I thought more and more about it I have confused myself.

Take following example -
If I have $1 that grows at a rate of 200% in 1 year with TWO time period (i.e. 100% per 1/2 year). by e ( rate * time) this will be e^2. But if I use your Mr Blue/Mr Green theory this gives me an answer of$4 ( if I have done it correct). I think I am missing something here. Please can you help?

459. kalid says:

@Husam: Thanks, glad it helped! Nope, I’m actually Sri-Lankan by heritage.

460. kalid says:

@Sonali: Great question. It’s important to remember that e^x refers to continuous growth, where you always generating some return and getting growth on that return.

Because you said “two time periods (100% per 1/2 year)”, you are implying that growth happens in “chunks”, and the Mr. Blue/Mr. Green model will work for you (and $4 is the correct answer: you go from 1 to 2 in the first 6 months, and 2 to 4 in the next 6 months). If you were growing continuously, and didn’t have to wait to compute your interest, then you could say your final amount would be e^2 (about 7.4). 461. Anonymous says: This is genius! What i’ve been trying to work out for a week now in one page. Thank you so much; you should teach 462. sonali says: Thanks Khalid. I am reading all your articles and planning to buy a book on kindle as well. Your articles have made me interested in maths. Otherwise I have always thought that I don’t like maths and am poor in maths! 463. kalid says: @Anon: Thanks! I would like to ramp up the article production / teaching this year. 464. kalid says: @Sonali: You’re welcome — I’m really happy to hear about the enthusiasm! A lot of the time we think “I’m not good at math” but it’s really just the way the material was presented. 465. Jason says: Fantastic article that I am very happy I stumbled on. Very, very well done. I took calculus in college and still didn’t have any idea what e represented until today. Thanks! 466. kalid says: @Jason: Awesome, glad you enjoyed it! 467. nitharsun says: really nice .please publish a book containing this type of matters .it will helpful to the learners like me.awesome explanation.all my doubts cleared.thank you 468. kalid says: @nitharsun: Awesome, glad you enjoyed it! I do have a book but it might be good to do one focusing more on exponents also. Thanks for the suggestion. 469. Lawrence says: Awesome article. Thanks for simplifying such a potentially muddy subject/idea. Working with economic growth models in my graduate program sent me scrambling, and this find has been a goldmine! Thanks! 470. kalid says: @Lawrence: Awesome, glad it helped! 471. jeff says: Hi Kalid, thanks god, at last i can see a glimmer of light at the end of the long tunnel that is Acturaial CT1. I’ve been studying for the exam thinking that just my determination and drive would get me through it and this has been puzzling me since. Thanks so much for all you do for the wannabe maths wizzards that some of us are. Thanks. PS it would be great if you could put a word on the Force of Interest. 472. @at@atat@@at@@@ says: how would i find the approximate equation of a line that is similar to a decaying exponential function? i have a line that does decrease over time at a slowing rate, though instead of constant decay it sometimes repeats numbers as y approaches 0, and actually does reach 0 as there cant be half of the unit that im using. hopefully im making some sense here…. instead of constant decay, there is a chance of decay. so starting with one hundred, we lose alot, then some of those, some of the remaining until say, 2 are left. then those have a chance of going but dont for a few t units, then 1 left for a few times until it completely dissapears. cheers! 473. doug says: Thanks for the video. I’ve never really understood e but now it makes more sense 474. kalid says: @doug: Glad it helped! 475. Giridhari says: splendid! you made so understandable 476. kalid says: @Giridhari: Awesome, glad it helped! 477. GoneMoon says: Hi to everyone, this is my first post, because I REALLY need help. Everywhere i see know this advertisements about Forex thing, got attracted, started to read about it, but still really confused about What is forex . All the explanations i find on the internet are very complicated. Too many smart words. Can someone explain what is forex by his own words ? Tired of websites. Thanks. 478. Nick says: Wow I am in 7th grade and I understood that, and no they do not teach e in middle school I just wanted to know what e was Awesome description that was really simple thanks 479. kalid says: @Nick: Awesome! You’re off to an early start on exponents :). 480. Nazar says: Hi Kalid, Thanks for such a clear intuition. Before I understood e only from analytical point of view. e was just an magical irrational number such that exponential function (a^x) with base e would have tangent line with a slope equal to 1, hence these nice properties and ubiquitous usage. I wonder if there any intuition on “opposite” sequence that leads to e: (1+x)^(1/x), x->0 ? Thanks, Nazar 481. alkmini says: thamk you for your intuitive explanation kalid it helped a lot 482. Nelson says: Thanks very much for writing this page. It really helped me a lot. I’ve been looking everywhere for a definition of e which I could understand and I’ve finally found it here. Your explanation was the best. Could you please make this point clear to me?: I have read in Weakipedia that the importance of e is due to the fact that the exponential function e^x is equal its derivative. Besides, “e is the unique real number such that the value of the derivative (slope of the tangent line) of the function f(x) = ex at the point x = 0 is equal to 1.” as written there. Could you explain in simple words this concept? What does for a function mean to have the same derivative as its original function? why is it important? Someone (Anonymous, Nov 22) have first post here “e is the unity of diferentiation/integration. What is the concept? My language is not English, so I am not sure to make me understand. Thanks a lot!! 483. kalid says: @Nelson: Great question, glad you enjoyed the article. Check out this article: http://betterexplained.com/articles/how-to-develop-a-mindset-for-math/ The 2nd half covers different definitions of e, including “the derivative being the same as the origina function”. In a sentence, it’s another way to say “The rate of growth is exactly equal to your current amount: you are always growing at 100%”. e^x has several definitions depending on your point of view, just like a circle — hopefully the article above helps! 484. Nelson says: Kalid, I appreciate your help !! Thanks very much !! 485. kalid says: @Nelson: You’re welcome! 486. Martha says: After reading your explanation, I understand (I hope!) that e is a constant, because all rates of growth are equal to the growth at 100%, but they have been squeezed or stretched, lengthened or shortened to accommodate the variations. (Is that right?) Your comparison to pi made it clear to me- all circles have the same curve, whether they are the edge of a dime or the edge of the sun. Thanks very much! 487. kalid says: @Martha: That’s pretty much it. e^x assumes 100% growth rate for time “x”. If you adjust the rate (50% growth, 200% growth) it’s like scalene e up and down (e^0.5x or e^2x), which is like changing the radius of a circle (to shrink it to a dime, or expand it to the sun). The core pattern is the same. 488. Tim says: Hey Kalid– I had a minor aha moment when I realized that simple growth could be viewed as compound growth compounded only once. Not a big deal, but insight sometimes comes in increments. Gratefully yours, –Tim 489. kalid says: @Tim: Awesome insight, I love it! Usually insights come in small steps and build on each other, so I’ll take whatever I can get :). 490. Thanks Kalid, for explaining the practical use of e in an easy understandable fashion, I’m passed 50, math was useless, dry stuff brought to me by uninterested teachers. Your type of explaining gives me the courage to have a go at understanding the world of math in a proper way. Marc 491. kalid says: Hi Marc, I really appreciate the comment. I’m with you — math can be so boring and dry unless presented properly, I’m happy if this was able to help :). 492. Moisés says: Hi Kalid! I loved the article. I am a biologist but I am interested in maths. I have a question about the e number. Why this expression, e^-rt, can be simplified to this other (1-r)^t ? Thanks for this so comprehensive article! 493. kalid says: Hi Moisés, great question. Your formulas have the rate as “-r” so we are shrinking. e^(-rt) is the formula for continuous, compound shrinking and (1-r)^t is the formula for compound but *not* continuous shrinking. With continuous shrinking, we are shrinking at every possible instant. With non-continuous shrinking, the “shrinks” only happen at set intervals (every 3 months, for example). Check out this article for the difference: http://betterexplained.com/articles/a-visual-guide-to-simple-compound-and-continuous-interest-rates/ 494. Idan says: I’ve been looking for such an explanation for a long time now, much appriciated imma go check some more articles out many thanks for putting up this website 495. Gautama's friend says: Thanks a lot!!! your explanations are helping me to﻿ develop new mathematical models 496. kalid says: @Gautama’s friend: Awesome, glad it helped :). 497. Dewei says: Hi, Kalid I am from China, and I found your amazing article through a translated version. I was confused on e for a long long time until spending 2 hours on this page, thank you very much! I have a further question, maybe you answered before, that is “What is the essence of the a in a^x?” When a=e, you explained here, but what if a=2, a=10? Is there some intuitive explanation? I will keep reading your other articles:) Best Regards! Dewei 498. kalid says: Hi Dewei, so glad it helped! The base “a” is how much you’ve grown at the *end* of each time period. So if you double at the end of a year, your base is 2. There’s more here: http://betterexplained.com/articles/understanding-exponents-why-does-00-1/ Separately, it might seem natural to have 2 or 10 as a base (things double every year!). This is true, but only for human systems we can control, which only compound at the end of each interval. Natural phenomena pay no attention to our calendars and are always growing at an instantaneous rate (such as 100% growth, compounded continuously). e^x is the best way to work with instantaneous rates. 499. Ganesh. G. S says: Its a nice explanation and helps me to understand what is e 500. kalid says: Thanks Ganesh! 501. yousra says: thannnks 502. rob hughes says: thank you so much, ive been researching this and ive been finding it difficult to understand on other sites :L 503. kalid says: @Rob: Awesome, very glad it helped! 504. Tc Madu says: Hey Kalid, You should have one of those icons for sharing your articles on facebook. I think every math student in the world should see these. Great work! 505. Lin says: This was fantastic. I remember asking a college professor “where e comes from?” – many years ago. He was enthusiastic and anxious to explain. After about 30 blackboards and endless computations (lots of limits) – he had numerous blank stares and opened mouths. Yet proudly he stood, confident he cleared it all up. But we (students) looked at each other and nodded as we told him “we get it now”. Now I can actually say – I do! Awesome post. 506. kalid says: @Lin: Glad you liked it! Ack, that situation is all too common. Even if the teacher is excited and enthusiastic, they can still not explain it in a clear way. Unfortunately, we’re afraid to say the emperor has no clothes (“Nope, still not seeing it!”). 507. Riley says: Thank you so much! I am in a college calc class and had never got to the point were I understood e. I knew how to work it, but it never made sence were it came from. you explained it in an excelently simplisitc fashion! The biggest thing is that I never had understood e is 100% and it’s what you do to it that changes it. I just knew if u had e you went by what power it was at. 508. Bella says: Hi! you probably already read my comment about my essay paper! i was wondering if you could help me organize all my ideas to make it a little more simplistic! Thanks so much your website is great! 509. Peter says: I’d suggest you’ve maybe missed something right at the beginning. Although you use the word ‘compound’ in your opening paragraphs, I don’t think you have emphasised enough that e is the signature of a process whose rate is determined by the process itself; it’s that feeding-back that is the key. Not sure that I put that terribly well myself, but anyway for me the clearest example is the flow of water out of a full barrel with a hole at the bottom. The distance the jet spurts out is dependant upon the pressure, but the pressure is contantly decreasing because of the jet. Similarly in electronics the voltage across a charged capacitor discharging through a resistor; e is inolved because as the cap discharges, the voltage decreases so there is less electrical pressure behind the discharge. Sure that is implicit in the rest of your explanation, but for me the aha moment about e was just this feeding-back quality, and I think it’d be even better if this was up front. 510. JavierLF says: Please, tell me, As it applies to myself, “Get you hungry for more: In the upcoming articles, I’ll dive into other properties of e.” When will you “dive into”? Thanks. Javier. 511. kalid says: Hi Peter, thanks for the note. Great point — I think the key is “your change depends on how much you have”. With a barrel and pressure, the pressure depends on the remaining water. Or with a population of deer, the more there are, the higher the (absolute) number of deer each year (even if it’s just 10% more or whatnot — the actual number each year keeps increasing). Appreciate the note about what works, I’ll have to think about how to work it in. 512. Elliot Robinson says: Hi Kalid: thank for doing this. I’m a 73 year old geezer but still interested in learning. I took 3 semesters of calculus (didn’t get good, or even fair, grades) and I don’t think I ever learned where ‘e’ came from! Fortunately I made my living working for the telephone company, mostly finding the broken wire or bad connection or low electron tube or all the other things that went wrong with the system until we made it all bits and fiber and they didn’t need me anymore. how the technical world has changed!. Just love your articles, wish I had been able to learn a lot of this stuff when I was a lot younger but now it’s mostly “entertainment” but I will try to “inflict” the knowledge on my grandchildren. thanks! elliot 513. Marc says: Finally I have an intuition! Thanks! Greetz from Switzerland 514. Anonymous says: Khalid, glad to see you still here. Below is my question/comment from 5 years ago. Your comments were lucid and spot on. I’m still on the transcendental number e, so here goes. I actually think that I put too much prelude last time, and ended up being not clear, so here goes. Imagine a money in which every one is given and original allocation of 365 units of currency. On the the first day everyone is given one unit of currency. This represents 1/365 of the original allotment. The next day they receive a dividend that is 1/385 higher than the first day. This keeps its proportion to the total allotment constant at 1/365. If repeated over a year, the daily allotment will have grown to 2.74.. (1 + 1/365)^365 a close approximation of e. The money supply will grow at the rate of the exponential function e^x. Money will have a half life(doubles) of about 8 months. In about 7 years it will have increased to 1,000 times in original amount. So the money is summed up into currency divisions that are 1,000 time the original. You cut three zeros off as Mexico did in 1993. Another version is the 100% solution. Everyone is given an original allotment of 100 units. You may do with this as you please. Every account is taxed at the rate of 1% per day. This tax is distributed on a per capita basis.If you keep your account at 100 you receive exactly what you are taxed. If you have less you receive a net subsidy, if you have more you pay a net tax. This means that currency units deteriorate to the point of 1/e at the end.of 100 days. I call this the e period. I did not understand the math until I read you explanation of radioactive decay. The rate of decay is 100% in 100 days, but it is negatively compound as the elements are. Your site is wonderful. I learned a lot here in a few minutes. I will be back. My math site is mathfortheages.com and you mat find in interesting. My focus is on squares, square roots, the pythagorean theorem, the regular polygons and polyhedra. My money is stuff is at different sites, mainly at usbig a group supporting a basic income or citizens dividend. excalibrator is designed to do that. google usbig and exclaibrator and that has the best article on that. I will be looking over your whole website, and I think your explanation of e is the best I have seen. I have been negligent in not returning sooner. stephen clark on October 17, 2007 at 4:14 pm said: Kalid, Thanks for the great explanation. My comment is looking for a comment from you. A new money system. Every one gets 100 units a day for a year. At the end of the year there are 36,500 units of currency in circulation for each person. The 100 units have diminished as a proportion of the whole to the point of 1/365. On the first day of the second year the daily allotment is increased by 1/365. This keeps the proportion of the daily allotment constant at 1/365. Over year this approximates e, and if the growth is continuous, it is e. I call the type of money salmoney, and this particular version excalibrator. I guess what I’m looking for is does this resonate with you as an apt or accurate use of e. Thanks for your time. My website is jaspersbox.com It is the attempt to create a new currency with a built in universal dividend. Thanks for listening. S. Clark Kalid on October 17, 2007 at 11:26 pm said: Hi Steven, thanks for writing, glad you liked the article. I’m not quite sure what you meant by “increasing by 1/365″ — do you want to keep the proportion of money handed out the same? (Even in the first case, giving out 100 per day, the percent changes day after day). You *can* use e to convert between simple and compound interest — I’m planning on a follow-up article on this topic, which may do a better job of answering your question than I’m doing now 515. I call the dividends of excalibrator, compound disinterest, it is money paying a rent to humanity at large. 516. Isabella :) says: Thank you sooooo much!! You are such a great teacher! 517. Jeremy Weiss says: I was not satisfied with this artcle. I also read most of the comments. I remain unsatisfied. Here is why. This is not intuition. This is scence. Intuition means feeling. If the transcendental number “e” is from God, then how can I feel the meaning of this? Khalid, I commend your efforts but you & the commentators have merely rehashed what I am already well aware of.Therer is neither deep insight nor intuitition on this page. Interestingly there was no mention of the gamma function, Sterling’s formula or the relationship to permutations. I felt nothing in the end. 518. Ahmed says: Perfectman. Thanks!! 519. Anonymous says: @ Jeremy Weiss I think (in fact I’m nearly certain) you are mistaking a pre-calc supplement for a deep concept in a third year mathematics course in college. Unless you were lucky enough in high school you might have been given a fictional story about how a kid in your class uses pert shampoo to remember the formula A=Pe^rt. I wouldn’t take such rudimentary, although illusive, insights into e for granted. 520. Robin Allison says: @ Jeremy Weiss I think (in fact I’m nearly certain) you are mistaking a pre-calc supplement for a deep concept in a third year mathematics course in college. Unless you were lucky enough in high school you might have been given a fictional story about how a kid in your class uses pert shampoo to remember the formula A=Pe^rt. I wouldn’t take such rudimentary, although illusive, insights into e for granted. 521. Robin Allison says: ^^Clarification: it should be “an article for a deep concept…”. By no means do I think you mistook the two topics haha, just the audience this article was for. 522. Jeremy Weiss says: Robin A, Yes your assessment is probably accurate. Yet I am not dogmatic. Here is my take on this. Permutation means rearrangement which implies change and (possibly) growth. Growth is built upon passion which is the excess of desire over will. Passion multiplied by drive equals knowledge( drive is defined as desire fixated to will). Putting all of this together, we conclude that the number “e” is derived from primal passion in the monad. 523. Coi says: This has been really useful for me! Thanks very much for your time in making this and the rest of the site. “e” has always been a tricky subject for me…after going through this material I am starting to wonder why it was so difficult in the first place, it’s all so clear now! 524. kalid says: Hi Coi, really glad it’s helping! 525. Keerthi Narayana says: Excellent work Kalid to explain the concepts in simple pedagogic way. We need articles like this to make all the concepts simpler and appealing to everyone. 526. kalid says: Thanks Keerthi! 527. Jerry says: Thanks for the explanation!! I am a statistics major and I am embarrassed to report I do not know the explanation until today! Thanks! 528. kalid says: Thanks Jerry, glad it helped! 529. Simon says: This is so great Kalid thanks. I went through engineering school never really understanding what the number e really meant. I mean, I know how and when to use it but I didn’t know where it came from. It gave me a feeling of euphoria truly understanding what it means. And that at an age of 31. The world should have more teachers like you 530. kalid says: Thanks Simon, really appreciate it. I had that same euphoric feeling when it started to click as well. 531. Mikala says: Wow, thanks so much. My 12 year old son, having mastered pi, has been asking me to explain e. Of course it’s made me realize how little I understood it back in math class. This is a totally graspable explanation, which I can now share with him. You rock. 532. kalid says: Hi Mikala, that’s awesome! I love it when people can share their knowledge with others :). 533. Ioan says: The key thing is that $\displaystyle{y(x) = e^kx}$ represents continuous growth in proportion to the current value of the function y(x). That’s why it satisfies $\displaystyle{y' = ky}$. 534. alemayehu kassa says: it was interesting. 535. Victor Batorsky says: Misstatement (I think): “Of course, we can substitute 100% for any number (50%, 25%, 200%) and get the growth formula for that new rate.” I think you meant to say: Of course, we can substitute any number (50%, 25%, 200%) for 100% and get the growth formula for that new rate. Or did I miss something. 536. kalid says: Thanks Victor, that’s a typo, fixing it up now. 537. great what is e^ infinity, e^ – infinity, same with -e pl 538. kalid says: @skghosh: e^infinity is the same as any number to infinity (2^x), and will grow without bounds. e^-infinity will shrink to 0. -e is just -2.7818… 539. Harmony says: Thank u for putting together such a thorough and clear explanation of e!!! I can’t believe teachers and textbooks just sorta throw it in without explaining it hardly at all, especially considering how important it is!!! Thank you! 540. This is a pretty amazing. I especially like this part: “Phew! The final value after 12 months is: 1 + 1 + .33 + .04 or about 2.37.” But say we are talking about 5 different periods (n = 5). How do we determine the contribution from each Mr.? n = 1 1 + 1 + .25 n = 2 1 + 1 + .33 + .04 n = 5 1 + 1 + ??? 541. “Phew! The final value after 12 months is: 1 + 1 + .33 + .04 or about 2.37.” How do you write a polynomial like this for any value of n? I cannot figure it out. 542. I found your website while searching for Fourier series tutorials. I like your style and impressed by your selection of examples. I think example (especially from everyday life) is the best way to explain or create intuition in someone about the subject. On a separate note, which statement is correct? In article “An Intuitive Introduction to Limits” you declared “e” as your favourite number but in the article ” An Intuitive Guide To Exponential Functions & e” you said you were always bothered by number “e”. Regards, Shahid Khilji 543. kalid says: Glad you liked it! When I wrote the e article (2007), it had always bothered me, but I’ve since grown to enjoy it :). 544. kalid says: Hi Harmony, really glad it helped! 545. kalid says: Hi Cody, the key to split interest into “n” parts and multiply it out. For earning interest every 6 months (i.e., 2 times a year) you get: (1 + .5)^2 = (1 + .5) * (1 + .5) = 1 (original amount) + [.5 + .5] (the interest the original earned) + [.5 * .5] (interest the interest earned) = 1 + 1 + .25 You can “expand out” the multiplication for any n; for n = 5 you have (1 + 1/5)^5. However, it becomes cumbersome to track it like that when you can just compute the exponent. If you want to draw it out, divide the growth into 5 steps, and each step add a new interest element (Start with blue. Step 1: Blue earns Green; Step 2: Blue earns Green, Green earns Red; Step 3: Blue earns Green, Green earns Red, Red earns Orange; Step 4: Blue earns Green, Green earns Red, Red earns Orange, Orange earns Pink; Step 5: Blue earns Green, Green earns Red, Red earns Orange, Orange earns Pink; Pink earns Black). 546. Anonymous says: You’re the man…now all that is left for me is figuring out why e^x is its own derivative 547. Khalid — this is truly amazing. First, your explanation of e is wonderful. Second, you take the time and energy to reply to everyone’s comments and follow-up. It is a testament to your dedication that more than 5 years after you wrote an article, it is still being commented on. Thank you! 548. thanks for your time in explaining e! It’s very clear! 549. gautham says: Hi kHlaid, In your statement, “Mathematically, if we have x splits then we get 2^x times more “stuff” than when we started. With 1 split we have 2^1 or 2 times more. With 4 splits we have 2^4 = 16 times more.” I understand that you say 2^x because we are doubling ? Assuming we are doubling, 1 becomes 2 i.e 1 split gives 2^1 = 2 (2 times more or 100% growth) 2 becomes 4 i.e 2 split gives 2^2 = 4 (4 times more or 300% growth) this ‘x’ times more .. is it computed with respect to the original value of 1? because when 4 becomes 8 i.e 4 splits gives 8 => 2^4 = 16 (16 times more or 1500% growth) Assuming even this is computed w.r.t the original value of 1. When 1 becomes 8, its just 700% growth and doesn’t tally ! Am I missing something ? Regards, Joe 550. NikoBellic says: At least bank don’t continuously compound the loans we take FROM them either, right? 551. kalid says: @Niko: That’s right :). But, they’ll often advertise the better rate (i.e., the lower uncompounded APR when showing the interest you’ll have to pay, and the higher, compounded APY when you are earning interest). 552. Kalid says: @gautham: Whoops! I was probably a little sloppy with the 100% growth vs 200% of the previous value. I meant a doubling progression, so you gain your entire value (100%) and end up at 200% (2.0) of your original amount. 553. Ravi Shankar says: This is simply great! I have always wanted to understand “e” and its origins better and this was the first time I came across something like this. Now I can explain it to my son as well. Thanks for the knowledge share. 554. kalid says: Thanks Ravi, glad you were able to share it with your son! 555. I like very much math. This explonation is realy simple and very nice to undestading. Thanks for sending to me thise e-mail Kazik 556. manigo says: Excellent explanation. Remarkable how one can complete years of higher education without encountering the like. Thank you! 557. Kumar Kakumanu says: Hi Kalid, There is a minor typo I just wanted to let you know. In the sentence “1 period of 30% growth means 30 changes of 1%, but happening in a single year. So you grow for 30% a year and stop.” I think it should say “1 period of 30% growth means 1 change of 30%, and happening in a single year. So you grow for 30% a year and stop.” -kumar 558. Pete says: Kalid: Thanks for the best explanation I`ve seen. I think I am starting to grasp some of it now. How could I explain it to someone else by using the graphs of 2^x, e^x, 3^x and their derivatives, like this: http://geogebratube.org/student/m29170 ? “- Hey, look here! The graph of the derivative of e^x is the same as that of e^x. If we imagine that the function is a plot of bacterial population, we see that the growth is proportional to current population-size.” Would not 2^x and 3^x be that as well? 559. kalid says: Comment from Dr. J which was eaten by the spam filter: Posting again to get LaTeX to display the equation: I like this website, but I’m questioning your use of the word “growth.” Doesn’t growth represent a change or gain, i.e., the final amount less the initial amount? It seems that when you use the word “growth” in your equations, you really mean “total amount” or “balance.” Consider the example of semiannual compounding at 100% annual interest in the “Money Changes Everything” section. After one year, a$1 investment will be worth $2.25. The balance is$2.25, but the growth is $1.25 =$2.25 (final amount) – $1.00 (initial investment). So $\displaystyle{\text{total amount} = (1 + 100\%/2)^2 = 2.25}$ Please tell me if I am misunderstanding your terminology. Thanks. Great point, I need to clarify. e is indeed the total, not the change. I was thinking of starting at 1 and multiplying in the “growth factor” of e (so going from 1.0 to 2.718) but that may not be clear. I’ll have to revisit the terminology used to clarify. 560. Doug Bennett says: Thank you so much for posting this material! I’m getting back into math for my engineering degree and it’s hard to get momentum on the new concepts of calculus if you don’t have an intuitive understanding of the fundamental numbers and concepts you’re using. I will definitely be back to learn more to get a better understanding of how these concepts all make sense. A+ on the explanation and great use of diagrams! 561. Jim says: The best explanation of what e actually is that I’ve ever heard. Thanks! 562. Mike says: Superb work! I’m going to read the rest of your series now! I’ll certainly recommend this to anyone who’s confused by e! Have you thought abut writing a book? 563. David says: Kalid, This site has got to be one of the best of all time! How you have the patience and dedication astounds me. You even respond to all feedback! 564. kalid says: @Doug: Thanks — glad you’re enjoying the diagrams, I find them sorely missing in most textbook explanations! @Jim: Glad to hear it! @Mike: Awesome, check out the rest of the articles when you have time. I have a book on Amazon (Math, Better Explained) but I guess I need to make the ads for it more prominent! @David: Thanks, I really appreciate it. I just try to answer comments a little bit every day, and in batches :). 565. hakizimana says: i don’t understand the reason why we always assign “e” with 2,718 yet you told us above that it’s not a number. 566. kalid says: @hakizimana: We don’t really assign e a value, we discovered that all growth had a common internal rate, and called it “e” (which happens to be 2.718). This is like discovering that all light moves at the same speed, and calling that speed c (which happens to be 186,000 miles/sec). 567. Joshua Marler says: It would be so fantastic if you could cover another article on Exponential Decay. Such as the decay of amplitude of sound etc. Would go brilliantly along with your Sine Wave intuition. This article is brilliant by the way. Helped me have a small mental breakthrough in regards to Audio Programming. Big Thanks 568. wes says: where were you when i was struggling in math in high school… not born yet… shit ! no wonder i didn’t do so well… you are so helping me make up for it now !!! 569. kalid says: Thanks wes! 570. kalid says: Thanks Joshua! I’d like to do a follow-up to discuss more applications of the exponential function, how we model various decay rates/growth rates with it. For a long time I just used e as an abstract tool, but it’s become more real to me over the years. I almost see it like a little machine that can be dialed in to follow a desired path (“Need 50% decay? Ok, dial that in, and let’s chart out that path…”). 571. Biplob says: I still can’t grasp one thing. 1. Why at 100% rate if 1 unit yields 1 unit [net 2 units] at the end of the year, then it should yield, x = .5 unit in six months? It should be the case of simple rate. I’m asking this because you’ve said, instead of a year, if we count six months, 1 unit gives .5 unit- this is the core point because Your derivation of e starts from here as you’ve come to the decision that this .5 unit also yields some units for next six months. Then you divide the year into more & more sections and using same principle have come to e. But when this is not the case of simple rate, how can you say it’s .5 after six months? As this is the case of compounding, Isn’t it more logical to think that after six months 1 unit would yield x units which together with 1 unit [net (1+x) units ] will yield 1 unit [net 1+ 1] units when compounded for next six months ?! 572. kalid says: Hi Biplob, thanks for the comment. I’m not sure I understand the question exactly, but let me see if I can explain. Simple interest is a type of artificial, man-made growth which looks something like this: You give me$100 on January 1st, and I promise 100% “simple interest” for the year. That means on Dec 31st I’ll give you $100 (original) +$100 (interest) = $200. Ok. Now, what should you have in your account on June 30, halfway through the year? You’d think you should have$100 (original) + $50 (interest for half year). And then in the next six months, you earn the remaining$50 of interest.

This sounds nice — except it’s not fair! From Jan 1 – June 30, you had $100 which earned$50 in interest. From July 1 to Dec 31, you had $150 which earned$50 in interest. How could the larger amount earn the same interest as the smaller?

Well, banks love doing this (with simple interest they can pay you less), but nature doesn’t. Generally speaking, nature doesn’t have a “gap” between when interest is earned and when it’s used. Once the growth happens, its impact is accounted for immediately.

The goal was to show the difference between artificial, “staircase-like” simple interest (represented by (1 + r)^n) and the smooth accumulation of interest represented by e^rt.

573. Anonymous says:

Hey Kalid, thanks for the kind reply. I’ve read this article innumerable times & I was exactly with the same problem as others’ have fallen when understanding what the hell “e” is. Wiki deteriorates the whole thing more. But your site is the first things that seems like a heavenly place for me who always seeks intuitive meaning of an equation.

I’ve grasped the intuition of “e”…I also have understood that what “e^rt” stands for…all your logical statements from the first to the last….I can feel it! Continuous Compounded growth! It’s happening naturally everyday. And “e” is the one hero to exactly express them. It’s significance is obviously clear. And your insight is one hell of a thing I admit.

But still… still except one thing about the intuitive process you have come to “e” through- I can’t get it, as you’ve said,

“You’d think you should have $100 (original) +$50 (interest for half year).”

You’d “think”? Why I’d think so? What’s the intuitive/mathematical reason for “thinking” so? Just for the sake of simple dividing policy? Interest for one is $100, hence for half year it’s$50?

At first glance, I also thought so but later I can’t take this so easily.

Is this should be illogical if I think that I should get not ” $50″ but ”$x” and the value of the “x” should be such that this “$100 +$x” would yield net $200 after a year when compounded for next half a year! So, that the equation (1 + r)^n) still holds true?!! Have I clarified myself now? 574. raj says: you are earth, i mean i can dig you(blog) so i can get more resources. very intuitive. 575. Joe says: I’m 61, and I guess I first heard about e when I was 16. I got a degree in math since then. But I never quite got the purpose of e until now. Bad texts and bad teachers are an extraordinary burden on math education. 576. kalid says: @Anonymous: More than happy to help — very glad the insights are working. For the sentence “You’d think you should have$100 (original) + $50 (interest for half year).” I was trying to show the problem with naive reasoning, which might be our guess for how interest should be divided up. If I invest$100 and expect to double my money at the end of the year, shouldn’t I have $150 halfway through? Nope! Because if you did, it means you earned$50 in the first 6 months (starting with $100), and then$50 in the last 6 months (even though you started with $150). As you write, you need to work out the amount so (1 + r)^n holds true :). So in the case of interest being paid every 6 months, if you were to double you’d have (1 + r)^2 = 2 which becomes r = 41%. That means at 6 months you should have$141, so after another 6 months you get

$141 + 41% ~$200

Of course, we could say we get interest every 3 months, or 1 month, or every day, and we end up realizing we need to pay interest based on e :).

@raj: Hah, glad you enjoyed it.

@Joe: Appreciate the comment, thanks! I agree, a poor explanation can turn us off from a subject entirely. It’s really important to be honest with ourselves about what explanations are working (or not) and always try to improve them.

577. Tim says:

Brilliant! Thanks so much. I love the analogy of pi to e. I did electrical engineering in grad school and never really got an intuitive handle on e.
Thanks.

578. Great work Khalid! I am mathematics advisor in secondary education and I can say that I have never found such an original and creative contemplation of the number e. I’ll use some of your ideas in a future presentation with due reference to your name.
Please go on, you are an acute mathematical mind and you, so to say, make mathematics land from the heavens in a very tangible, intuitive way. By the way, I do not know your mastering of higher mathematics but it would thrill me to read your intuitive approach to general topology ideas.

579. kalid says:

Wow, thank you Stathis, I really appreciate the kind words. I’d love to keep going as long as I can, and something like Topology would be a fun topic to explore. My formal math knowledge isn’t that deep (beyond a regular engineering education) and I’d like to get into more theoretical topics down the line. Thanks for the note!

580. anupam says:

Back then…it was all rote learning we did.
Atleast,I wish we had teachers like you.

581. kalid says:

Thanks Anupam, glad you enjoyed it! Many of us had rote learning experiences, but we can always share what helped move beyond it :).

582. Anonymous says:

Thanks for taking your valuable time to enlighten the rest of us. 18 years of education and thousands in tuition couldn’t explain e better than you did.

I read somewhere that math is the language of the nature. I think I get that a little more after reading this article.

583. Aliya says:

Kalid! how are you??
I need help! can I get your email? or rather, I want to know the different ways of approximating e, I know the Compound interest one… and the 1/n!.. I was wondering If your could help me with the continued fraction and showing how e is irrational and trascedental..

584. leotrim says:

well nice job kalid, but don’t you think that the right formula on the top should
be (1+return)^return,and not (1+return)^x.Just for correctsy,anyway nice explained

585. kalid says:

Hi leotrim, thanks for the note. I might be misunderstanding your comment, but if you have return=50% and have it for x=3 periods of time, you’d get (1 + 50%)^3 = (1.5)^3 = 3.375. You don’t want to do (1 + 50%)^50%!

586. michael bersudsky says:

I would like to thank you so much for your wonderfull insights. realy I am speachless

587. Kalid says:

Thanks Michael, glad you enjoyed it!

588. cody says:

clear + intuitive. thank you!

589. kalid says:

Thanks Cody!

590. Angel says:

WOW! Thank you Kalid! This article was out of this world! AWESOME

591. kalid says:

Thanks Angel glad you enjoyed it!

592. RK says:

Excellent Khalid, clear, precise and fussfree!! Can you care to explain how this e concept is applied in mortgage calculations? Amortization of a home loan?

593. hedi says:

This one page explained ‘e’ better than my 4 years in engineering school

594. jasper says:

I’m 49 years old and never understood e until now.

Of course I could have paid attention in class way back when…Better late than never though.

Thanks !

595. Chandra says:

I googled “what does e to the power of x mean” and went across 3 sites before encountering your explanation. I can easily say Google needs changes in their search engine, as this site should result right at the top. Great work!!

596. Kevin says:

Love this site, Kalid!

I’m in year 11 and the typical textbook explanations just don’t cover it in enough depth for me, I like to build a deep intuition of the concepts, which is precisely why I love your site!

e is a fascinating number…

I was a little stuck at your explanation on how to use ‘e’ with different rates, but now I think I get it.

Basically, you can imagine a year of 50% growth as the same thing as half a year, with 100% growth per year. This is because per half year, the growth is 50%.
Also an intuitive explanation for why the derivative of e^x is e^x, is that e^x represents 100% growth, continuously. This means that at each instant, the gradient will be equal to the amount that there is currently.

But of course, I can’t explain it as well as you, Kalid, so I’ll leave it to those who do it best!

597. kalid says:

Thanks Heidi, Jasper and Chandra!

Kevin, really glad you’re enjoying it! You got it, from an instantaneous and continuously compounding perspective, 1 year of 50% growth = half a year of 100% growth.

I think of it like this: An individual “cell” intends to grow from 1 to 1.5 during the period in question. That’s the battle plan. Of course, with the compounding, the colony grows more (children get children) but each cell is unaware of that and just grows according to its own goal.

You got it, having your gradient equal to yourself every instant is 100% growth (more on this here: http://betterexplained.com/articles/developing-your-intuition-for-math/).

Btw, everyone has a different and useful take on a concept, please don’t feel like you can’t get your explanations out there! I write what clicked for me, but another analogy might resonate better for someone.

598. surinder singh says:

Thanks

Surinder

599. Hi Surinder, thanks for the note! Happy to clarify the doubt. Understanding the difference between a “nominal” rate (what is listed) and the “actual” rate (what you get) is the key to e.

Imagine a bank that promises 100% return at the end of the year (a magical bank, I know).

You deposit $1000 on Jan 1st — how much will you have on Dec 31st?$2000. After all, they promise 100% return.

Ok. But let’s say you’re not very patient, and want to peek at your account after 6 months (end of June). How much should you have?

You might say $1500 — after all, after halfway through you should have earned half the interest, right? This sounds good, but there’s a problem. If the bank shows$1500 at the end of June, it means:

* From Jan 1st to June 30, you earned $500 on your$1000 initial balance
* From July 1st to Dec 31, you earned $500 on your$1500 initial balance

That isn’t good. Why should both periods return the same interest when you started with different amounts?

The only fair way to compute interest is to include all the effects of compounding.

Now, there is a concept of “doubling every year”. If you have an investment that goes $1000 (starting),$2000 (year 1), \$4000 (year 2)… it is growing at 100% per year.

But, the continuous interest rate that causes this doubling is actually only 69.3%.

69.3% continuous growth = doubling at end of year (100% increase)

100% continuous growth = 2.718 at end of year (171.8% increase)

The terminology does get confusing though — we need to ask whether 100% means the final result “after all compounding effects are included” or whether 100% is the instantaneous rate of growth (and therefore becomes 2.718 after all compounding is included).

Try out the bank example and see what would make it “fair”. If you wanted to get 100% return at the end of the year, how much should each 6-month period pay? How about every 3 month period?

600. Chris Davis says:

Thanks, I still can’t calculate e, but now I know how it is used with interest, which is a lot more than I figured out after reading Wikipedia.

601. Deepan Datta says:

First I want to thank you for this Better Explanation which surely put the physical concept of ‘e’ and ‘ln’ deeper into my brain.Now I have a query, can you please explain me the physical interpretation of (-e^x) and (-e^-x)? As you wonderfully explained it for e^x and e^-x.

602. James says:

For sure, I have to start maths afresh. I am doing an MSc. in signal processing and find this articles very useful. Do you have such notes in statistics and probability especially on stochastic processes, mutual information and Entropy and how they link up with “e”. I also think you are doing more that enough and I feel like donating to your site, so that you continue doing the great job. i will definitely be donating once I am through will my studies.

Thank you. Finally I found a place that shows what is e. But can you give an example using the population growth dynamics in terms of survival rate?

Wow. Just wow. Such beginner friendly explanations are hard to come by (espescially in maths ). Got it on my first attempt.

605. Bhashit Parikh says:

Hi Kalid,

really great explanations. Wish all my math teachers were as ingenious and interesting as you.

606. That video was so perfect! Why weren’t you my math teacher? Many thanks!

607. dg says:

I was playing around with (1 + 1/n)^n
and inputted large values of n in excel.
This may be an excel issue but when I input the following values for n
1. 10^12 – result 2.718523496
2. 10^13 – result 2.716110034 – why is it moving lower from 2.718?
3. 10^14 – result 2.7161100
4. 10^15 – result 3.0350352 – now it is greater than 2.718?
5. 10^16 and greater- result 1.000? why 1.00?

I thought as it approaches infinity, the value would be 2.718.
Is this an excel computational bug issue?

608. kalid says:

Hi dg, that’s exactly it — Excel has computation limits on the precision. There are many floating-point issues which can happen, many scientists avoid using Excel for large computations.

Putting 1 + 1/(10^16) is likely to result in “1.0″ (no remainder), so when you take it to the large power it just stays at 1. [This happens on Google's Calculator too: https://www.google.com/?gws_rd=ssl#q=1+%2B+1%2F(10%5E16) ]

609. Actually dg — I tested this out on both Excel and Wolfram Alpha, and it is working fine. There is no issue with Excel — it is computing the values properly and it shows that the result starts to approach e.

610. Thank you very much. This post is very informative. I was looking for a simple explanation of e and I found it here.

611. Richard Warren says:

e was being discussed on the radio in the UK the other day (BBC Radio 4, In our time, 25 September), and they used the 100% interest example at the beginning of the show, but it was too quick for me and I spent the rest of the programme wishing I had grasped the vital explanation. Now I feel I have an understanding of e comparable to my understanding of pi. Thanks very much.

612. Jac says:

Hi great article

i have a question. In the article you say this:

What if we grow at 50% annually, instead of 100%?
If we pick n=50, we can split our growth into 50 chunks of 1% interest:

(1+.50/50)^50=(1+0.01)^100/2=e^1/2

I understand this, but why did you use n= 50, could you use n=10 in this case would it still work ? thanks

613. Anonymous says:

hi great article,

you say at the end of the video and in the article that e can approximate jagged systems that don’t grow smoothly. Can you give an example of this because I don’t see it.
Thanks

614. dg says:

In the above you compare 30% in one year vs 10 years of 3% growth by saying:

10 years of 3% growth means 30 changes of 1%. These changes happen over 10 years, so you are growing continuously at 3% per year.
1 period of 30% growth means 30 changes of 1%, but happening in a single year. So you grow for 30% a year and stop.
The same “30 changes of 1%” happen in each case. The faster your rate (30%) the less time you need to grow for the same effect (1 year).

Question:
How do you break it down to 30 changes of 1%? Isn’t there an infinite number of changes happening when considering continuous growth?
Unless you are saying that the number of continuous changes is 30?
Or am I missing something in your explanation?
I would think that the number of continuous changes over 10 year exceeds that over a 1 year period. I am having trouble understanding the merger of time and rate explanation – rate x time.

615. vivek says:

Hi Khalid,

Very informative post. But I think the concepts tend to evaporate over time hence there must be some place to apply these concepts so that it sticks there for long time. Do you know something that may help us to achieve this thing.

PS: I am a programmer to so, its very difficult to go back to school books.

Thanks!

616. chris says:

Thankyou for the great article – i’m learning about cell survival curves , the survival fraction of cells after a radiation dose. It is describes as e^-(ax + bx^2) . Trying to get an intuitive feel to this is there a way to get the head around e^-(x^2) ? or how about the general e^(x^n) ??

617. kalid says:

@dg: Great question. When dealing with continuous growth, we can mix and match rate and time as we please. Here’s why.

Let’s say we have 1 year of 30% continuous growth. That means each individual dollar intends create interest of 30% when all is said and done. Along the way, its interest will start growing, but the original dollar doesn’t know! The 30% growth (from a continuous case) means “An individual dollar, Mr. Blue, knows he will create a Mr. Green that is 30% as large at the end of the period. What Mr. Green happened to be doing along the way is none of Mr. Blue’s business.”

So, with this model, it doesn’t matter if Mr. Blue builds up Mr. Green in one year of 30%, or in 30 years of 1% progress each year. At the end, Mr. Blue has made a Mr. Green who is 30% as large as he is.

Clearly, it’s better for us, the investors, to get the growth in 1 year vs. 30, but either approach results in the same total amount (e^{30% * 1} = e^{1% * 30} = e^{.30} = 1.35. Either way, have a total growth from 1 to 1.35. The extra .05 is because our interest did work along the way. (We’re assuming the growth happens in whatever time is necessary, then stops afterwards.)

@vivek: Practice problems are key, there’s some at Khan Academy or http://www.exampleproblems.com/.

@chris: Great question. When there’s multiple terms it helps to break them apart:

$\displaystyle{e^{-(ax + bx^2)} = e^{-ax} \cdot e^{-bx^2}}$

Basically, we have two decay factors acting simultaneously (they are decay factors, not growth factors, since the exponent is negative, and will be shrinking our value).

The first is a regular decaying exponential function, e^{-ax}. The second is an exponential function which is shrinking extremely quickly — with the *square* of time.

The general term e^{-bx^2} is a Gaussian function (bell curve) [http://en.wikipedia.org/wiki/Gaussian_function] and is basically a decaying exponential on steroids. Instead of a nice sloping curve it will be shrinking extremely quickly. What a regular exponential would see in 100 time periods, it sees in 10. I’m not that familiar with this function (or stats in general) but would be a fun thing to investigate later on.

618. chris says:

thanks so much Kalid, the tip off that it is a bell curve is great . and the description of it as a decaying exponential on steroids really helps i will dig further in to the function e^(-bx^2).

hmm, I find it a little odd that it is used to describe a physical phenomenon, killing of cells with radiation, and that the good old regular decay is not sufficient.

619. kalid says:

Awesome, glad it helped. Yep, I’m not sure why it needs both: maybe there is a bell curve of possibilities, each of which is diminishing exponentially?

620. Lars says:

Thank you! Finally I can begin to grasp the weird e thing! I have always enjoyed mathematics, but everytime I encountered e I became frustrated. Your e-xplanation is exactly what I needed!

621. Hello Kalid,

Came here after finishing Barbara Oakley’s “Learning How to Learn” course on coursera. I am thankful to her that not only she has designed a perfect course but she has introduced many learning gems such as your website in the course.

I was always struggling to understand mathematical terms and its use in real life. From this first article I am very sure that I will finally find peace with Maths.