Understanding Exponents (Why does 0^0 = 1?)

We’re taught that exponents are repeated multiplication. This is a good introduction, but it breaks down on 3^1.5 and the brain-twisting 0^0. How do you repeat zero zero times and get 1?

You can’t, not while exponents are repeated multiplication. Today our mental model is due for an upgrade.

Viewing arithmetic as transformations

Let’s step back — how do we learn arithmetic? We’re taught that numbers are counts of something (fingers), addition is combining counts (3 + 4 = 7) and multiplication is repeated addition (2 times 3 = 2 + 2 + 2 = 6).

Repeated addition works when multiplying by nice round numbers like 2 and 10, but not when using numbers like -1 and sqrt(2). Why?

Our model was incomplete. Numbers aren’t just a count; a better viewpoint is a position on a line. This position can be negative (-1), between other numbers (sqrt(2)), or in another dimension (i).

Arithmetic became a way to transform a number: Addition was sliding (+3 means slide 3 units to the right), and multiplication was scaling (times 3 means scale it up 3x).

So what are exponents?

Enter the Expand-o-tron(TM)

Let me introduce the Expand-o-tron 3000.

Yes, this device looks like a shoddy microwave — but instead of heating food, it grows numbers. Put a number in and a new one comes out. Here’s how:

• Set the growth to the desired change after one second (2x, 3x, 10.3x)
• Set the time to the number of seconds
• Push the button

And shazam! The bell rings and we pull out our shiny new number. Suppose we want to change 1.0 into 9:

• Put 1.0 in the expand-o-tron
• Set the change for “3x” growth, and the time for 2 seconds
• Push the button

The number starts transforming as soon as we begin: We see 1.0, 1.1, 1.2… and just as finish the first second, we’re at 3.0. But it keeps going: 3.1, 3.5, 4.0, 6.0, 7.5. As just as we finish the 2nd second we’re at 9.0. Behold our shiny new number!

Mathematically, the expand-o-tron (exponent function) does this:

$\displaystyle{original \cdot growth^{duration} = new}$

or

$\displaystyle{growth^{duration} = \frac{new}{original}}$

For example, 3^2 = 9/1. The base is the amount to grow each unit (3x), and the exponent is the amount of time (2). A formula like 2^n means “Use the expand-o-tron at 2x growth for n seconds”.

We always start with 1.0 in the expand-o-tron to see how it changes a single unit. If we want to see what would happen if we started with 3.0 in the expand-o-tron, we just scale up the final result. For example:

• “Start with 1 and double 3 times” means 1 * 2^3 = 1 * 2 * 2 * 2 = 8
• “Start with 3 and double 3 times” means 3 * 2^3 = 3 * 2 * 2 * 2 = 24

Whenever you see an plain exponent by itself (like 2^3), we’re starting with 1.0.

Understanding the Exponential Scaling Factor

When multiplying, we can just state the final scaling factor. Want it 8 times larger? Multiply by 8. Done.

Exponents are a bit… finicky:

You: I’d like to grow this number.

Expand-o-tron: Ok, stick it in.

You: How big will it get?

Expand-o-tron: Gee, I dunno. Let’s find out…

You: Find out? I was hoping you’d kn-

Expand-o-tron: Shh!!! It’s growing! It’s growing!

You:

Expand-o-tron: It’s done! My masterpiece is alive!

You: Can I go now?

The expand-o-tron is indirect. Just looking at it, you’re not sure what it’ll do: What does 3^10 mean to you? How does it make you feel? Instead of a nice tidy scaling factor, exponents want us to feel, relive, even smell the growing process. Whatever you end with is your scaling factor.

It sounds roundabout and annoying. You know why? Most things in nature don’t know where they’ll end up!

Do you think bacteria plans on doubling every 14 hours? No — it just eats the moldy bread you forgot about in the fridge as fast as it can, and as it gets more it starts growing even faster. To predict the behavior, we use how fast they’re growing (current rate) and how long they’ll be changing (time) to figure out their final value.

The answer has to be worked out — exponents are a way of saying “Begin with these conditions, start changing, and see where you end up”. The expand-o-tron (or our calculator) does the work by crunching the numbers to get the final scaling factor. But someone has to do it.

Understanding Fractional Powers

Let’s see if the expand-o-tron can help us understand exponents. First up: what does at 2^1.5 mean?

It’s confusing when we think of repeated multiplication. But the expand-o-tron makes it simple: 1.5 is just the amount of time in the machine.

• 2^1 means 1 second in the machine (2x growth)
• 2^2 means 2 seconds in the machine (4x growth)

2^1.5 means 1.5 seconds in the machine, so somewhere between 2x and 4x growth (more later). The idea of “repeated counting” had us stuck using whole numbers, but fractional seconds are completely fine.

Multiplying exponents

What if we want to two growth cycles back-to-back? Let’s say we use the machine for 2 seconds, and then use it for 3 seconds at the exact same power:

$\displaystyle{x^2 \cdot x^3 = ?}$

Think about your regular microwave — isn’t this the same as one continuous cycle of 5 seconds? It sure is. As long as the power setting (base) stayed the same, we can just add the time:

$\displaystyle{x^y \cdot x^z = x^{y + z}}$

Again, the expand-o-tron gives us a scaling factor to change our number. To get the total effect from two consecutive uses, we just multiply the scaling factors together.

Square roots

Let’s keep going. Let’s say we’re at power level a and grow for 3 seconds:

$\displaystyle{a^3}$

Not too bad. Now what would growing for half that time look like? It’d be 1.5 seconds:

$\displaystyle{a^{1.5}}$

Now what would happen if we did that twice?

$\displaystyle{a^{1.5} \cdot a^{1.5} = a^3}$

$\displaystyle{partial \; growth \cdot partial \; growth = full \; growth}$

Looking at this equation, we see “partial growth” is the square root of full growth! If we divide the time in half we get the square root scaling factor. And if we divide the time in thirds?

$\displaystyle{a^1 \cdot a^1 \cdot a^1 = a^3}$

$\displaystyle{partial \; growth \cdot partial \; growth \cdot partial \; growth = full \; growth}$

And we get the cube root! For me, this is an intuitive reason why dividing the exponents gives roots: we split the time into equal amounts, so each “partial growth” period must have the same effect. If three identical effects are multiplied together, it means they’re each a cube root.

Negative exponents

Now we’re on a roll — what does a negative exponent mean? Negative seconds means going back in time! If going forward grows by a scaling factor, going backwards should shrink by it.

$\displaystyle{2^{-1} = \frac{1}{2^1}}$

The sentence means “1 second ago, we were at half our current amount (1/2^1)”. In fact, this is a neat part of any exponential graph, like 2^x:

Pick a point like 3.5 seconds (2^3.5 = 11.3). One second in the future we’ll be at double our current amount (2^4.5 = 22.5). One second ago we were at half our amount (2^2.5 = 5.65).

This works for any number! Wherever 1 million is, we were at 500,000 one second before it. Try it below:

Taking the zeroth power

Now let’s try the tricky stuff: what does 3^0 mean? Well, we set the machine for 3x growth, and use it for… zero seconds. Zero seconds means we don’t even use the machine!

Our new and old values are the same (new = old), so the scaling factor is 1. Using 0 as the time (power) means there’s no change at all. The scaling factor is always 1.

Taking zero as a base

How do we interpret 0^x? Well, our growth amount is “0x” — after a second, the expand-o-tron obliterates the number and turns it to zero. But if we’ve obliterated the number after 1 second, it really means any amount of time will destroy the number:

0^(1/n) = nth root of 0^1 = nth root of 0 = 0

No matter the tiny power we raise it to, it will be some root of 0.

Zero to the zeroth power

At last, the dreaded 0^0. What does it mean?

The expand-o-tron to the rescue: 0^0 means a 0x growth for 0 seconds!

Although we planned on obliterating the number, we never used the machine. No usage means new = old, and the scaling factor is 1. 0^0 = 1 * 0^0 = 1 * 1 = 1 — it doesn’t change our original number. Mystery solved!

(For the math geeks: Defining 0^0 as 1 makes many theorems work smoothly. In reality, 0^0 depends on the scenario (continuous or discrete) and is under debate. The microwave analogy isn’t about rigor — it helps me see why it could be 1, in a way that “repeated counting” does not.)

Advanced: Repeated Exponents (a to the b to the c)

Repeated exponents are tricky. What does

$\displaystyle{\Large (2^a)^b}$

mean? It’s “repeated multiplication, repeated” — another way of saying “do that exponent thing once, and do it again”. Let’s dissect it:

$\displaystyle{(2^3)^4}$

• First, I want to grow by doubling each second: do that for 3 seconds (2^3)
• Then, whatever my number is (8x), I want to grow by that new amount for 4 seconds (8^4)

The first exponent (^3) just knows to take “2″ and grow it by itself 3 times. The next exponent (^4) just knows to take the previous amount (8) and grow it by itself 4 times. Each time unit in “Phase II” is the same as repeating all of Phase I:

$\displaystyle{(2^3)^4 = 2^3 \cdot 2^3 \cdot 2^3 \cdot 2^3 = 2^{3+3+3+3} = 2^{12} }$

This is where the repeated counting interpretation helps get our bearings. But then we bring out the expand-o-tron: we grow for 3 seconds in Phase I, and redo that for 4 more seconds. It works for fractional powers — for example,

$\displaystyle{(2^{3.1})^{4.2}}$

means “Grow for 3.1 seconds, and use that new growth rate for 4.2 seconds”. We can smush together the time (3.1 × 4.2) like this:

$\displaystyle{(a^b)^c = a^{b\cdot c} = (a^c)^b}$

It’s different, so try some examples:

• (2^1)^x means “Grow at 2 for 1 second, and ‘do that growth’ for x more seconds”.
• 7 = (7^0.5)^2 means “We can jump to 7 all at once. Or, we can plan on growing to 7 but only use half the time (sqrt(7)). But we can do that process for 2 seconds, which gives us the full amount (sqrt(7) squared = 7).”

We’re like kids learning that 3 times 7 = 7 times 3. (Or that a% of b = b% of a — it’s true!).

Advanced: Rewriting Exponents For The Grower

The expand-o-tron is a bit strange: numbers start growing the instant they’re inside, but we specify the desired growth at the end of each second.

We say we want 2x growth at the end of the first second. But how do we know what rate to start off with? How fast should we be growing at 0.5 seconds? It can’t be the full amount, or else we’ll overshoot our goal as our interest compounds.

Here’s the key: Growth curves written like 2^x are from the observer’s viewpoint, not the grower.

The value “2″ is measured at the end of the interval and we work backwards to create the exponent. This is convenient for us, but not the growing quantity — bacteria, radioactive elements and money don’t care about lining up with our ending intervals!

No, these critters know their current, instantaneous growth rate, and don’t try to line it up with our boundaries. It’s just like understanding radians vs. degrees — radians are “natural” because they are measured from the mover’s viewpoint.

To get into the grower’s viewpoint, we use the magical number e. There’s much more to say, but we can convert any “observer-focused” formula like 2^x into a “grower-focused” one:

$\displaystyle{2^x = (e^{ln(2)})^x = e^{ln(2)x} }$

In this case, ln(2) = .693 = 69.3% is the instantaneous growth rate needed to look like 2^x to an observer. When you enter “2x growth at the end of each period”, the expand-o-tron knows to grow the number at a rate of 69.3%.

We’ll save these details for another day — just remember the difference between the grower’s instantaneous growth rate (which the bacteria controls) and the observer’s chart that’s measured at the end of each interval. Underneath it all, every exponential curve is a scaled version of e^x:

$\displaystyle{a^x = (e^{ln(a)})^x = e^{ln(a)x} }$

Every exponent is a variation of e, just like every number is a scaled version of 1.

Why use this analogy?

Does the expand-o-tron exist? Do numbers really gather up in a line? Nope — they’re ways of looking at the world.

The expand-o-tron removes the mental hiccups when seeing 2^1.5 or even 0^0: it’s just 0x growth for 0 seconds, which doesn’t change the number. Everything from slide rules to Euler’s formula begins to click once we recognize the core theme of growth — even beasts like i^i can be tamed.

Friends don’t let friends think of exponents as repeated multiplication. Happy math.

Other Posts In This Series

1. Karen says:

2. Kalid says:

Great question. At an intuitive level, I see an irrational exponent as a different amount of time, in between known rational numbers, and ultimately approximated by a rational number.

When we write “3 * sqrt(2) = 4.24″ we “know” that we’re really taking an approximation, and that 3 * sqrt(2) goes on forever.

Similarly, “3^sqrt(2) = 4.72″ is just an approximation and the real decimal result goes on forever.

3. Steven says:

There is a small problem with your 2^3^4 = 2^12 example: exponentiation is (usually?) understood to be right-associative (ie top-down), therefore: 2^3^4 = 2^(3^4) = 2^81

4. I’m pretty sure your “repeated exponentiation” section is not quite right – when there are repeated exponents like that, in the standard notation, we work from right-to-left. Thus, 2^3^4 means 2^(3^4) which is neither 2^(3*4) nor 2^(4^3)…

2^3^4 is in fact 2417851639229258349412352 while 2^(3*4) is 4096 and even 2^(4^3) is only 18446744073709551616.

5. Kalid says:

@Steven, sabik: Thanks for the comments! My mistake, I didn’t realize it was right-associative. I’ll add in the parentheses to make this more clear.

6. Paul says:

Amazing work once more Kalid. You make learning fun again.

7. Kalid says:

@Paul: Thank you — I find math enjoyable when I’m working with analogies that make sense to me.

8. Todd says:

Forgive me if I ask a dumb question. It’s been awhile since I’ve been in any sort of math class.

If I put 3 in the expand-o-tron for 0 seconds, I get 3 because there was no time expanding. If I put it in for 1 second, shouldn’t I get 9 since it spent one second expanding? But I though 3^2 was 9. Maybe I’m reading it wrong.

9. @Todd: Great question — I should make the operation more clear. Normally, we start off with 1.0 and see how it grows, so with a setting of 3x power for 2.0 seconds, we’d get 1.0 * 3^2.0 = 9.0.

If we started with 3.0 and ran it for 1 second, we’d get: 3.0 * 3.0^1 = 9.0.

Most of the time we want to see what happens to a “unit” amount.

10. Nobody says:

It’s great to see another article. As with many of your articles, it helped me realize that I have not understood something I’ve been using for so long.

About evaluating i^i: it appears that I have the problem half-way done.
So, I put 1 in with i times growth for i seconds. Meaning, each second I rotate 1 counter-clockwise 90 degrees in the complex plane. However, how do I look at i seconds intuitively and thus complete the computation?

11. Tomer says:

Nobody: Unfortunately the “^i” operation is a little confusing. You should start by understanding e^i and e^(a*i) where a is a real number. Then remember that x = e^(log x), and i = e^(log i)

so i^i = (e^(log i))^i = e^(i*log(i))

Analogies only go so far!

12. Kalid says:

@Nobody: Great question! (Tomer, thanks for the details).

It’s hard to think about “i” seconds, just like it’s hard to think about -1 seconds — it’s easier to think about what the transformation does. Negative numbers flip the direction, and i brings items into the complex plane. When you take i as an exponent, it changes the direction in which you are growing (instead of growing in the real dimension, you start growing in the imaginary dimension). There’s a lot more to say, but that’s the intuitive approach I take with it :).

13. I really appreciated this article, and just want to suggest that you make it more clear at the top that you ALWAYS start with one.

I really struggled with your explanation of 0^0 until I got down into the comments and saw that that was the case. Thanks.

14. Kalid says:

@Andrew: Thanks for the comment! I’ll make that more clear, really appreciate the feedback.

15. When you take i as an exponent, it changes the direction in which you are growing (instead of growing in the real dimension, you start growing in the imaginary dimension).

Umm, what? When you have i as an exponent, you go in circles, getting neither larger nor smaller.

16. NF says:

Is there a Nobel prize for explaining math

17. @Sabik: It’s tricky — consider i^i. It “seems” like it should stay on the unit circle (that’s where it started, magnitude of 1), but it doesn’t. I should be more clear: i as an exponent changes your _instantaneous_ rate of change by 90 degrees.

@NF: Heh, I think the rumor is Alfred Nobel hated mathematicians so there’s no prize for pure math :).

18. CL says:

Thanks for the great visualization of how exponents work. I’m a huge fan of your articles.

The step I can’t intuitively grasp is in the multiplying exponents section: “What if we want to two growth cycles back-to-back? Let’s say we use the machine for 2 seconds, and then use it for 3 seconds at the exact same power”.

To me, if I put something in the microwave for 2 seconds, then again for 3 seconds, intuitively this would be like adding together two doses of microwaving:

x^2 + x^3 instead of x^2 * x^3 (i.e. running the microwave for 2 min, and then again for 3 min is 2 + 3 = 5 min, not 2 * 3)

I’m not sure how to picture what x^2 * x^3 means in microwave terms…maybe something like increasing the wattage by 3 times.

Anyway, keep the great articles coming!

19. mike says:

yet another “a-ha” moment when reading this…=)

20. This certainly explains exponents better then I’ve ever encountered explanations before.
(In essence it doesn’t differ much from the repeated multiplication explanation but introducing “time” in the Expand-o-tron 3000 simplifies imagining non-integer powers and their effect on outcomes.)

21. @CL: Great question! I had the same thoughts myself, I may need to go back and make that section more clear.

Exponents act like multiplication, but the amount you multiply by has 2 inputs: growth rate and time (vs regular multiplication, which is just growth rate).

You can combine regular multiplication: Multiplying by 3, and then multiplying by 15 is the same as multiplying by 15 (3×5) all at once.

With exponents, using x^2 and then x^3 is the same as multiplying by x^(2+3) = x^5 all at once.

The tricky thing is to remember that the time is being combined, but it’s happening in the exponent you raise it to.

Hope this helps!

@Theo: Thanks! That was the goal — to help expand our insight so things like fractional (or even negative) exponents can make sense. Negative counting is confounding.

22. Thank you Kalid. Another reminder that math concepts are easy to teach and very hard to explain. Your analogies and patient decomposition of the problems are wonderful to read.

23. CJ says:

Excellent article! You are good. You are on my RSS feed.

24. @Stuart: You’re welcome — I agree, I think the difficulties in understanding math is more to due with complex explanations, not complex ideas. Thanks for the kind words :).

25. Apples says:

Here’s my own little explanation for powers to the i, though it’s not nearly as well-written as yours, of course :3

It’s called Euler’s Identity, and is presented by purists as e^i*pi+1=0. We can prove it’s true, but it doesn’t seem to make any sense. After all, what the hell do i, e, and pi have to do with each other? Let’s see if we can boil this down a bit.

For this example, we’ll work with 2^i, which is equal to about 0.769+0.639i, if you plug it into a calculator.

First, let’s remember that, as Kalid put it above, we can turn an exponent into a grower-oriented one by changing it. So 2^i = e^ln(2)i.

Euler’s formula tells us that e^xi = cos(x)+sin(x)i

If we plug in ln(2) to this, we get cos(ln(2))+sin(ln(2))i, which is about 0.769+0.639i, the answer the calculator gave us before!

However, this doesn’t really help us much unless we already have a really good understanding, intuitive of what Cos and Sin are (and I’m not going to explain all of that here). Instead, we can look at this another way.

e^x is another way of saying e^(rate*time), correct? Well, rather than say we grow for i seconds, how about we use i as our rate of growth, since switching the order in which we multiple doesn’t change the result?

Multiplying by i is simply doing a 90 degree turn. So as Kalid previously said, we just start at 1 and start growing in a circle. So when we say “we grow for ln(2) seconds”, we mean traveling around the unit circle on the complex plane a length of ln(2) radians! (Read Kalid’s article on radians for more info) If you already know what Cos and Sin are, this should be making sense by now. We can check our work with this as well.

The circumference of a circle is 2pir, where r is the radius. Since we start at 1 before we start growing in a circle, the radius is 1, which means the circumference is 2pi. This means moving ln(2) radians around the circle is about the same as rotating 39 degrees.

Let’s return to Euler’s Identity. With what we’ve just established above, e^pi*i means moving pi radians around a unit circle with a radius of 1, right? Moving pi radians around the circle is the same as doing a 180 degree turn, since it’s half the circumference of the whole thing (2pi), right? Which means we go from 1 to -1.

Which brings us to e^i*pi=-1.

Hope this cleared up some confusion…

26. Jeff says:

I would LOVE to hear Kalin do an article on Euler’s Identity or really on Euler’s formula. I have forever wanted to understand the relationship between e, i, pi, and rotational position. I appreciate your comments Apple but I’m just not quite there yet.

It would also give Kalin an opportunity to bring together many of his articles into one giant mathematical relationship that has many practical applications.

27. Jeff says:

So here is the heart of all my Euler’s Identity questions:

Lets just look at e^pi for a moment:
1*e^pi means you start at one and then move in the positive direction at an increasing “unit rate” for pi seconds. You will end up at a seemingly arbitrary location (23.1407).

However e^i*pi means you start at 1 and ROTATE at an increasing “unit rate” for pi seconds. You end up making exactly 1/2 circle! This means there is some relationship between e and pi. 2^i*pi or 5^i*pi will put you at some random spot on the circle, but a growth rate of e takes you exactly pi radians in pi seconds. Doesn’t that seem strange?

28. Kalid says:

@Apples, Jeff: Thank you both for the wonderful comments! Yes, I’d love to do a follow-up on Euler’s formula, and seeing i as “rotating” your rate of growth is how I’m learning to make intuitive sense of it :).

29. Jeff says:

Thanks Kalin I am looking forward to it! At the very least can you help me understand this one point? I feel like the answer is staring me right in the face but I can’t wrap my mind around it.

1*e^pi means you start at 1, “grow” by a factor of e for pi seconds and you end up at 23.141.

Now if we throw a little “i” in the mix then it gives us 1*e^(i*pi). If this means our growth takes a “left turn” so to speak and we rotate our growth, then why don’t we travel 23.141 radians in pi seconds?

Thanks again Kalin for all your work.

30. I find thinking about exponents as “how many patterns can I make with some objects if the objects come in x varieties of color and I have y objects” for x^y.

For instance, let’s say we’re lining up a bunch of eggs.
Our eggs come in 2 colors: brown and white.
x represents the variety. x = 2.
y represents the number of objects (eggs in this case). y = 3

If we line up 3 eggs in a row, how many different patterns could we make with white eggs and brown eggs?
white, white, white. white, white, brown. white, brown, white. white, brown, brown. brown, white, white. brown, white, brown. brown, brown, white. brown, brown, brown.

That’s 8 patterns. 2^3 = 8.
If we had one egg, we’d only have to possible patterns: white. brown. 2^1 = 2.

If we didn’t have any eggs, we’d have exactly one pattern: no eggs. (the one and only pattern is the absence of eggs).

If we had -1 eggs, then how many patterns would we have? So -1 eggs means we take one egg away from however many we have. If we started out with 10 eggs, we have 1024 patterns (2^10). If we took one egg away, we’d have 512 patterns (2^9). So taking one egg away gives us exactly half as many eggs if we hadn’t taken that last egg away. 2^-1 = 0.5.

Now lets say we have (2^3)^4.
Inside the parenthesis, we have 3 eggs, each with 2 possible colors. We have 8 possible patterns. If we laid 4 groups of eggs side by side, how many patterns could we have? Now we will use a group of 3 eggs as our new object instead of using just using one egg as our object.
We know each group of 3 eggs has 8 patterns (2^3). So each group of 3 eggs has 8 possible variations. So our new x is 8. Our new y is 4.
(2^3)^4 = 8^4.
But instead of thinking in terms of groups, we can just count the number of eggs. If we have 4 groups of 3 eggs, then we have 12 eggs in a row.
Each egg has two colors. Thus, (2^3)^4 = 2^12.

I find this pretty intuitive. What do you think?

31. Kalid says:

@Jeff: Great question. I’d like to do a more complete follow-up on Euler’s theorem, but consider this:

e^(i*pi) means we are growing at 100% (rotated) for pi seconds. Growing at a 90 degree angle does not change your *magnitude*, it only changes your *direction*. So we always end up growing at 100% (without the compounding interest) and end up going only 100*pi or 3.14 radians around the circle. We aren’t “growing” in the correct direction to increase speed; instead, we are growing to change our velocity.

@Jeremy: Thanks for the explanation! I think the set interpretation has some merit for whole numbers, but is too far removed from the traditional “repeated multiplication” explanation to help with repeated fractions (what does 0.5^2.3 mean in regard to sets with a number of elements?).

But, this may be because I’m not well-versed with it (and I think that may be the ‘modern’ definition of exponents). I always welcome other mental models as they may map better to different problems :).

32. Mj DeYoung says:

Don’t get a big head (-: but I think you would make a very gifted math teacher – putting many so called “math teachers” in the US to shame, shame, shame.

Will you be writing any books?

33. SHAUNA says:

If you have something like SQRT(8) + SQRT(2), the numbers underneath the square roots are not the same, but you can combine the terms. EXPLAIN?

34. Anonymous says:

Dear Sir,

I am sorry that someone had not brought this to your attention sooner, but you are plain wrong. 0^0 is an indeterminate form. Defining 0^0 = 1 is not only misleading, it is a mistake so deep that when mathematicians finally realized how much damage it was causing, our understanding of mathematics was greatly changed for the better.

Consider the following.

Choose ANY positive number ‘c,’ and for x>0, let

f(x) = x^(-c/log(x))

Naively “plugging in” x=0 looks like we should have

0^(-c/(-infinity)) = 0^0

But we need to be more careful. Taking the log base e (or “ln” if you like) of f(x) and using properties of logs gives

log(f(x)) = (-c/log x)*(log x) =-c

so that f(x) –> e^(-c) as x –> 0 (from the right). A similar argument works for negative c. The point is, c can be any number you want. Thus,

0^0 = any number you want.

Sometimes defining something which does not have a definition to begin with is OK, as in the case of a PARTICULAR function, such as g(x) = (x-1)/(x^2-1), which has a singularity at x=1. It is perfectly fine to let g(x) = 1/2 (or anything else, if you don’t mind violating continuity). However, to define 0/0, 0^0, 1^infinity, infinity – infinity, or any other indeterminate form will lead to many, many problems in mathematics. Historically, this has happened quite a lot, and it took us mathematicians quite awhile to realize the source of the problem, which was that people kept trying to define these things.

35. Kalid says:

It seems the debate about 0^0 is still not complete; the goal for the article is to present an intuitive (non-rigorous) interpretation which explains why 0^0 = 1 “makes sense” in many circumstances.

36. Tim says:

Typo: At the start you describe addition as “combing numbers”, you meant “combining”. Great site! Thank you so much- your take on maths is so intuitive and interesting.

37. Tim says:

Also, the URL “http://betterexplained.com/articles/intuitive-guide-to-angles-degrees-and-radians/” is not an actual link (Under subheading “Advanced: Rewriting Exponents For The Grower”). Probably because of the preceding colon.

Happy Math!

38. Kalid says:

@Tim: Glad you enjoyed it, and thanks for the fixes! I just fixed them now.

39. That’s a great explanation.
But this make me wonder of how other people brain operate. For me most of the time I understand math as rule of the game and I don’t need interpretation to digest this. How about others?

40. x to the nth says:

Have to re-read this but when i entered 0^0 into my calculator and indeed “1” came-up i know i mis-comprehended a critical point in high school math class…

41. DJ says:

Excellent explanation. I teach a high school Algebra 2/Trig course and I will definitely be using this approach when presenting exponents to my students this fall.

42. Kalid says:

@watchmath: Thanks — I guess it depends on your personality. I need math to make sense at a deep intuitive level, it just doesn’t feel right if the rules seem “arbitrary” to me :).

@x: Yes, I think many calculators may be programmed with that result.

@DJ: Great, glad it was useful!

43. thios is a realy good website about exponents

44. Kalid says:

45. Sully says:

This material is incredibly easy to digest and apply. Do you have any material on logarithms in general?

46. Here is a possibly more simple understanding of why 0^0 = 1 (depending on your point of view).

Take a total map to be a function that is defined on every point in the domain.

Then,
We can think of n^m as the number of total maps (ahem functions) from the m element set to the n element set. To see this draw 2 circles, but n dots in one and m dots in the other (it is advisable to pick small n and m!). Then very methodically draw all possible total maps. Maybe make a tree.

Now, the empty set is a subset of every set, thus there is always a map from the empty set to any set. Moreover, this map is unique. It follows that there is a map from the empty set to the empty set, and this map is unique. Then the number of maps from the 0 element set to the 0 element set is 1. Thus

0^0 = 1.

47. Kalid says:

@Jonathan: Thanks for the comment! Yes, that’s another, more detailed/combinatorial way to think about it. Depending on the context of the problem, growth or combinations may make sense. Often times people see e^x and are trying to figure out the meaning of a 0 exponent; other times, you have n^m (n choices, m decision) and this interpretation helps too.

48. lampan says:

Hi.. Is your book available anywhere in India? Thanks.

49. Kalid says:

@Lampan: Hi, you can get the ebook anywhere that supports PayPal, Google Checkout, or credit card. Thanks!

50. Alley says:

Hi Kalid,

My math teacher brough up the 0!=1 parodox today but wasn’t able to explain it so I thought I’d research it myself. It makes sense to me now except for I cant see why the expand-o-meter always starts out at one. If I accept that fact everything else is clear I just dont see why it should be and one. Thanks!

51. Kalid says:

@Alley: Great question! Here’s how I see it — exponential growth is about, well, growing stuff with multiplication! So we need to start with something to grow and compare. The reference point is 1, our original amount, and we see what we transform it into.

If we don’t do any changes, then our input = output, so we finish with 1.

Another way to think about it “times 3″ really means “1 times 3″, that is, start with your original amount and then scale it up. We often drop the implicit “1”, but when doing exponents it can help to explicitly talk about it. Hope this helps!

52. Lee says:

What will happen if I place a base number of 1 to the expandotron, set it to zero growth and turn back time for 1 second?

53. Kalid says:

@Lee: Let me see if I get the question. With the base number 1 in the expandotron, we’re saying that after a unit of time, we have the same result. In fact, with 1 as the base, any amount of time will have the same result (no change). If we use imaginary numbers, strange things can occur, but that’s another article :).

Turning back time for 1 second is asking “1 second ago, given this rate of growth, what was our number?”. Since the rate of growth doesn’t have any effect, the number is the same as the original. When you aren’t changing, 1 second ago you are still at your present amount.

In math terms, 1^x = 1 for any real x. Hope this helps!

54. ducnhuandoan says:

Will you see:
30. Asked Is 0 to the power of 0 equals 0 a final answer 27 Oct 2010 05:54
31. First answer to Is 0 to the power of 0 equals 0 a final answer 31 Oct 2010 12:29
as my comment on your proof 0^0 = 1.

55. Don says:

Thanks that really help though i still have a question in mind

by the way i do A-Level maths curently

i saw in a cartoon an equation that read

46x*87y over the sq root of 90?

i mean is that even close to possible cos i know you said that ‘once you work out the impossible, anything can happen’.

56. JJeigers says:

0 to the power 0 doesnt equal anything
it equals infinite or negative infnite

i will explain everything

if you have zero, its a value, aint positive nd aint negative
when you power that ‘nothing’ by ‘nothing’, as a result, you dont actually get a number
0^0 does not equal 1 or 0
it equals infinity OR negative infinity

57. Don says:

0^0 wont equal 1, i can believe that
when 2 rules clash, neither would be valid
e.g. if you type 0^0, the calculator would be saying ‘syntax error’ ( if you have a scientific calc) when what its trying to say is that it cant have an answer because well the only time you can have 2 or 4 answers is in quadratic or sim quad equ

and calc cannot display 2 answers because that we cannot then perceive a calc as a correct instrument of mathematics.

and by the way jjeigers, infinity isnt a number
its a CONCEPT, its a way that we think of number
or bacially a limitless thing. It cease to exist as a number. However, google and googleplex are, google is 10^100 and google plex is 10^google(this is a huge number. probably close to infinity? no.)

0^0 is one of those questions that prove we haven’t found out everything about maths. we may know a heck of a lot, but not everything

58. souvik das says:

0^0=1 by
=(1-1)^0
=(i-0*1) {by applying binomial theorem}
=(1-0)
=1
hence proved

59. souvik das says:

0^0=1 by
=(1-1)^0
=(i-0*1) {by applying binomial theorem}
=(1-0)
=1
hence proved
is it right?

60. Befuddled says:

Hi. Just discovered your website; great stuff!
With all due respect, I don’t think you gave fractional exponents enough attention.
First of all you state that
2^2 means 2 seconds in the machine (4x growth).
According to your explanation, 2^2 should really mean a 2x growth followed by another 2x growth; 4^1 should mean a single 4x growth. But if this is the case, I have no understanding of what 2^(1/2) means: what does half of a 2x growth mean? It obviously can’t mean that the original amount only has a .5x growth because in that case 2^(1/2) would have to be 1.5 when, in fact, it is irrational.
Again, I think your approach to this topic and to knowledge, in general, is awesome, but I wish you can elaborate further on this.
Thank you.

61. Kalid says:

@Befuddled: Great question, happy to help clarify. I think the key is realizing we’re talking about continuous growth, which happens instant by instant (see the article on e: http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/).

When we see “2^2″ and think “2x growth for 2 seconds” we need to clarify what “2x” really means. Sure, we grew from 1 to 2 during the span of 1 second. But how did we get there? Did all our growth happen in the final .0001 second? Probably not.

Halfway through, we had *some* amount of growth. But was it exactly half? Let’s see. If at 0.5 seconds we are at 1.5x our initial amount, it means

* From 0.0 to 0.5 we started at 1 and grew .5, ending at 1.5
* From 0.5 to 1.0 we started at 1.5 and grew .5, ending at 2.0

Hrm — there’s a problem here. Our starting amount was *different* for each interval yet we grew the same amount! Something’s not right.

e to the rescue! It lets us figure out our *instantaneous growth rates* which work for any starting amount. In fact, the natural log (ln) can tell us our instant growth rate needed to get to any growth amount.

So, we can say ln(2) ~ .693 which means we need to grow instantly at about 69.3%, for one second, to end up at 2.0 at the end of the second.

We’re partway there — what does 2^(1/2) mean? Well, it says “Grow at that instant rate of 69.3%, but only for half a second”. This is where e comes in — it lets us figure out the net effect. So we can say

e^(.693 * 1/2) ~ 1.414…

which means “If you grow at 69.3% growth for only half a second, you will end up at 1.414…”, which is the square root of 2!

The key points:

* 2x means “continuous growth”, so halfway through (in time) doesn’t mean halfway through (in amount grown)
* We can use e and natural log (ln) to figure out where we are after x seconds of growth, or what our amount is after exactly half a second

Hope this helps! Feel free to ask any questions if this didn’t make sense.

Another example: Let’s say you are growing at 4x each second. You start off at 1.0, and after 2 seconds you are at 4^2 = 16. What is your amount after 1 second of growth? Should it be 8? (This is a similar question to what you asked — but instead of going in-between 1 and 2 for 1.5 seconds, we’re going in-between 1 and 3 to get to 2 seconds. Either way, the halfway amount in time is not the halfway amount in growth. The halfway amount in time is actually the square root of the final amount, since repeating that growth (multiplying) again gives the full amount).

62. Befuddled says:

Finally finished the e and ln articles and just came back to read your response. Beautiful and satisfying response. Thank you!

63. @Befuddled: Awesome, glad it helped!

64. Alex says:

The scenario gets more interesting when you consider 4 doors (1 car, 3 goats). In that situation contestant A has 1/4 chance of winning.

Now consider contestant B. He cannot choose the door that contestant B picked. But he can only win if contestant A didn’t pick the car, which gives him 3 winning scenarios. But contestant B has 9 scenarios in which he can loose. So 3 winning scenarios out of 12 possible scenarios gives a chance of 3/12 = 1/4 There is no change!

Now consider the switching scenario. Contestant A picked a door at random. The remaining 3 doors form a set. When Monty open a door showing a goat, he filters that set. The set represents a total amount of chance = 3/4. Because Monty opened one door, that amount of chance gets split between the 2 remaining doors = 3/4 * 1/2 = 3/8.

Switching from the original door with chance = 1/4, to one of the doors from the filtered set improves your chances from 1/4 to 3/8

Thanks.

65. Kalid says:

@Alex: Whoa, really neat scenario — I hadn’t thought of the 2-person game!

66. RB says:

Just discovered your site. As a math teaching specialist I appreciate the visual model. Niggling over precise math concepts, though interesting and entertaining, does, in my estimation undermines the purpose of models in math (at least from a teaching perspective). This model makes fuzzy concepts much easier to handle and explain. I am curious; it seems that most (all?) responders to this blog have a substantial understanding of mathematics. Has anyone attempted to use this model to explain the concepts of exponents to learners new to this area of math? I would be greatly interested in its efficacy.

Now for my niggle:)

On Jan 10 Don was responding to the previous comment and, correctly, stating that infinity was NOT a number. However, his example of google is not a number either, it is an internet company (corporation). The number he was looking for was googol (10^100), which is stil nowhere near infinity!

67. Kalid says:

@RB: Thanks for the note! I’m actually not sure how effective this analogy has been in the classroom, but I know it’s helped un-fuddle exponents for me :). Good point re: google vs. googol, we’ve forgotten the spelling of the original word!

68. Nkateko says:

Hi,isn’t 0^0 not defined?. i’m confused. How come when i put 0^0 in my calculator,i get “Math error” instead of 1?

69. Kalid says:

@Nkateko: Technically, 0^0 is indeterminate (i.e. has no singular value, it can equal different things depending on how you compute it) but many mathematicians choose to define it to be 1 because it makes many equations much more convenient. The calculator may not have this assumption programmed in however.

70. thanks EXPAND-A-TRON! you really worked!!;-)

71. kalid says:

@fayal: Thanks, I love the expand-o-tron!

72. Scott Robertson says:

I found this by googling what I thought everyone would want to know, and as usual a question that keeps me up nights isn’t even a part of the known universe…here it is. I had googled “why the exponent goes down one” and I found this excellent intuition which I plan to study more carefully. My application is the power rule of differentiation, used to find the derivative. They say that the power rule is just a trick, I say if it is equal to something and the something it is equal to is differentiation, then the power rule is not just a trick but an identity of the facts behind differentiation. If that is so, then there should be an intuition behind the power rule that gives an explanation of why, for example, 2x^3 diff-ates to 6^2. I don’t believe in coincidences. Is there a way you can use the expandotron to develop an intuition for the power rule that reveals the heart of differentiation to a beginner like me? I believe it involves considering dx the UNIT, and dy/dx the RATIO or proportionality factor, and y the function or dependent variable, and of course x as the independent variable. My guess about the heart of calculus is that it is the linear equation but I have to go over this page and try to see what I think you would say and see if we come up with the same answers. Thanks!

73. Laura says:

So when you say that the scaling factor is always 1, we can also think of this as the original number we are putting into the expand-o-tron that will be affected by both the growth and the duration. This is why when the exponent is 0(the duration=0) and no changes are made, we get an answer of 1?

74. kalid says:

@Scott: Thanks for the suggestion, I’d like to do an article on the power rule. The main intuition for 2x^3 going to 6x^2 is the number of combinations that arise.

Intuitively, think of a cube (3 by 3 by 3). When you make it bigger in all dimensions by 1 ( to a 4 by 4 by 4) you can consider this “adding a layer” to the 3 outer sides. That’s why x^3 => 3x^2. There is a tiny piece in the corner which is “ignored” in the Calculus sense (it becomes negligible as your changes get smaller).

For x^2 => 2x, it’s similar (to make a square larger, add a strip to 2 sides). I’d like to write more about this!

@Laura: Yes, I always consider 1.0 going into the expand-o-tron to start. I see exponents as a “scaling factor” that have to operate on something else. So even

3^2

can be considered

1 * 3^2

That is, we start with something (1) and make it 3x bigger for “2 seconds”. When the duration is 0, it’s like we never used the machine, and we get the original amount out (scaling factor of 1). That’s the intuition that clicks for me ;).

75. Patrick Robichaud says:

From what I understood from this article, and correct me if I am wrong, when you calculate 0^0 and get a result of 1, the “1” is actually the scale factor, or the number by which the 0 is multiplied. so basically you multiply 0 by 1, and still get 0…. so why is it that you would write 1 as the answer, and not just the answer itself of 0?

76. kalid says:

@Patrick: Great question, something I need to clarify. My analogy for exponents is:

original * growth^duration = new

so if we have 0^0 we get

original * 0^0 = new
original * 1 = new
original = new

I.e., we didn’t change at all. The scale factor is multiplied into the original amount, so 0^0 = 1 is another way of saying “we didn’t change at all”.

For something like 3^2, we write

original * 3^2 = new
original * 9 = new

which is to say, our new value will be 9x our original one. Normally we don’t think about the original amount and write “3^2 = 9″ but then it becomes harder to see why 3^0 = 1. So in my head, I even see 3^2 as “1 * 3^2″, i.e. we have a starting point and begin to change it.

77. Tom says:

Hi Kalid,

Do you have any insightful ways of thinking about why (a^x).(b^x)=(ab)^x ? I can see how to extend from natural numbers etc but can’t think of anything that makes sense on a deeper level. I thought about picturing the whole surface z=x^y (which I think is helpful for example for linking the generally rather separated school topics of indices and surds) but seemed to get caught up in rather a lot of dimensions if trying then to take a product between z=x^y (for a^x) and z1=x1^y1 (for b^x). So basically confusion… Hope what I’m trying to say makes sense…

Thank you (and thank you also for all the fantastic material on this site!)

Tom

78. kalid says:

@Tom: Great question, and thanks for the kind words! For

(a^x) * (b^x) = (ab)^x

I’d think about it like this: Imagine 2 expand-o-trons going at the same time, side by side:

* The first takes x = 3 seconds to turn 1 into a^3
* The second takes x = 3 seconds to turn 1 into b^3

Ok. Now, when we write (a^3) * (b^3) I think “I want to apply 2 separate growth processes, one after another”.

What if we wanted to apply them at the same time? What would the equivalent be? Well, every second you’d have to grow by a AND b, i.e a*b. (Growing by 2x and 5x simultaneously would be growing by 10x).

So, the result is (ab)^3 => each second you grow “ab” which takes each individual growth rate into account. Hope this helps!

79. mohamed says:

amazing explanation

80. John Jordan says:

Hey Khalid,

I’m interested in hearing your explanation of this section of the article: There’s much more to say, but we can convert any “observer-focused” formula like 2^x into a “grower-focused” one.

I’ve been playing about with complex exponents the last two days and am struggling to find a consistent way of describing what they are; imagine I had to describe it to a child.

Equations of the form, a.i^n, are easy deal with; simply rotate your unit, a, by an angle, with the angle specified by n, i.e. where n=1, angle =90°, n=0.5, angle = 45° etc. This corresponds to the view of the observer.

Equations of the form, a^bi, are what I’m struggling with, e.g. e^i covers a distance of one radian in one unit of time, …but why??? What does a^bi refer to? Is it the particle/traveller’s rate?

My difficulty is in trying to explain equations of the form a^bi, as I see some inconsistencies in the explanations provided here (maybe it is just me), e.g. your (excellent!) explanation of i^i, in the Euler’s Formula article, is not quite the same as trying to explain 2^i, as in, you describe arriving at i after one unit of time for i^i..but for something like 2^i, you do not arrive at 2, but travel a distance of 0.693 radians.

It seems like e is assumed to be the ”unit speed” about the unit circle; must this simply be taken as given? Or is it related to the continuous perpendicular ”growth” described by imaginary exponents?

Is it just me being stupid??/?

81. kalid says:

@John: Great question. The articles were written at different time periods (and different levels of my understanding!) so there may be some inconsistencies in the explanations. Let me try to clarify.

1) e means “Start at 1.0, grow continuously at 100% interest for 1 unit of time”. So it’s a given, in the sense that e is the simplest type of continuous growth we can describe (100% growth for 1 period).

2) e^x means “modify your growth by x”. What does this mean? If x=3 we can interpret this as “grow for 3 times as long”, and if x = i, this can mean “Have your growth be perpendicular, not in the same direction.” In this case, e^i means “have 100% perpendicular growth, applied continuously.”. Perpendicular growth moves us in a circle, so 100% perpendicular growth moves us our *current amount* (1.0) along the circle. We start on the number line at 1.0, and move 1.0 along the unit circle (that was the interest we earned).

3) Equations like a^bi are variations of this growth. When we take a as the base (not e), we’re saying “I want to get to a first, then have my exponential growth applied”. a = e^ln(a), so we can always rewrite it in terms of e (instead of jumping to a directly, start at e, grow by ln(a) to get to a, and then keep going). Rewriting in terms of “e” lets us see growth in its simplest terms. The “^bi” part is like before: rotate our growth (i) but also go for longer (b). so a^(bi) = [e^ln(a)]^bi = e^(ln(a) * b * i), which means “Figure out how much extra growth we need to get to a [which is ln(a)], make this amount even larger [b], and then apply it on a circle [i].

In the case of 2^i, we see that 2 is less than e, so we have a *shrink* to get to 2 [ln(2) = .693]. So instead of 100% interest, we only get 69.3% interest and only go .693 units around the circle: 2^i = e^[ln(2)*i] = e^[.693 * i]. So exactly, we only travel 0.693 radians around.

Hope this helps!

82. Dan says:

I really like how this article got me thinking. I read it a long time ago, and recently had an AHA moment. Example: 9^1/2 = 3. As you grow from 1 to 9, half way there you are at 3. This is obvious with a base, or growth factor of 3x. 3^0=1, 3^1=3 and 3^2=9. What if we have a growth factor of 2, or 4? Does this still hold true? We always start at one and grow, grow, grow. 2^0=1, 2^1=2, 2^2=4, 2^3=8, and 2^4=16. 2^x = 9 so x = ln(9) / ln(2). so x=3.16993, or 2^3.16993 = 9. Does 2^(3.16993/2) = 3? You bet!!

83. can zero have an exponent of zero?

84. kalid says:

@phil: Yep! But its meaning has to be “defined” (i.e., chosen to be consistent with other rules we want). In many cases, we define 0^0 to be 1.

85. jj says:

0/0 is indeterminate because there are an infinite number of correct answers through the definition of division. 0/0=1 is correct because 1×0=0. 0/0=2 because 2×0=0 is correct. Since there is no unique answer, 0/0 is indeterminate. 0³/0³= 0° = 0/0.
Therefore 0° is indeterminate machine or no machine. Remember figures don’t lie, just liers figure.

86. kalid says:

@jj: Yep, 0^0 is indeterminate but it’s useful to pick a value for it, to simplify many equations.

87. confused man says:

wtf why did it say x^3 * x^4? why did it jump to multiplication for “two growth cycles back to back”? what is the logic of that? You do this again when you go a^1.5 then a^1.5. why is it multiplication all of a sudden?

also, why does 2 growing for 1.5 amount of time give 2.82842712475?

88. Anonymous says:

If you take a block of cheese, and you cut it zero times from all 6 faces of the block of cheese, you still have 1 block of cheese.

89. Anonymous says:

Instead of thinking about growing think about dividing it that about of times. If you have 1 block of cheese, and you cut it in half you have 2 blocks of cheese. If you keep those pieces jammed together, turn it sideways, and cut it in half again, you have 4 blocks of cheese, if you keep it together and flip it onto the Z plane (XYZ) and cut it in half again you have 8 peices, but if you break it apart and XY cut the peices you get and you “quarter” them again you get 2^2 x 2^2 16 blocks of cheese

90. Anonymous says:

I should also add a note on the “to the 1.5th” power, or 3/2 power. If your intention is to halve the block of cheese, you can choose to approach that “halving” process from any angle possible on the block so long as you are cutting it “symmetrically”. If you cut something in half from one angle, and then halfway through the process of cutting it from another angle (it doesn’t have to be “perpendicular on cut 2 so long as the resulting cut effectively quarters the volume upon completion, but you stop halfway through the cut……sometimes you will only have 2 blocks of cheese equal in volume with gouges cut into them, and other times you will have 3 peices because one of the halves was effectively “halved” when you were halfway through the cut. The chance that you will have of making 3 instead of 2 is approximately 82.84271………% of the time, so that is why it equals 2.8282871…….

91. Anonymous says:

a block (cube for arguments sake) has 6 faces, 8 corners, 12 edges. be mindful of the operator you’re using in any given application. it has 6 faces, because it has 3×2 faces per plane (XYZ, 3D), it has 8 corners because it has 2^3(XYZ) corners, it has 12 edges because it has 3×4 sides per (XYZ) plane

92. Anonymous says:

I should correct myself, 2 posts up (wish you could edit on this thing) you take the inverse of the chance that youre approaching the second cut in a fashion that will halve one of pieces before you stop

93. Anonymous says:

Faces (6) x corners (8) x edges (12) = 576
any unique combination for multplying 2 of those “aspects” of the cube = 6×6, 8×8, 12×12, 6×8, 6×12, 8×12 the sum is 480

480 possibilities out of 576 = 1/6, the inverse 5/6ths multiply by the 3 planes (XZY) = 2.5 or 1 + your 1.5 cuts.

I haven’t got far enough in math to fully analyze and integrate pi and e, but I am assuming somewhere in the mix they come into play since you can still create assymetrical lines that would divide the cube in “half” as far as volume/mass was concerned without the end result of the quartering process being identical “looking” peices.

94. Anonymous says:

(2^1.5 + 1/6(1/pi^pi))/(2+5/6) = accuracy of 99.9811% any chance this is a calculator abbreviation? Do we live in a pi-dimensional world

95. Wesley says:

Great insight and explanation! I learned a lot from posts in this series.

96. kalid says:

Thanks Wesley!

I have this small doubt that needs clarification. suppose i rephrase the first example shown i.e 1 * (3^2) by saying that we are growing 1 three times per second. so after two seconds, won’t the equation be 1* (3 times/ second * 2 seconds) = 6.
I’m not able to understand why the multiplication by the time period would be wrong!

98. kalid says:

@Ninad: Excellent question. The microwave grows whatever *is inside* by 3x per second. So we start with 1.0, grow for one second (to 3.0), then grow for another second (to 9.0). After the first second, what’s “inside the microwave” has already changed! The growth is always relative to your current amount, not your original amount.

Put another way, imagine you had a bank account that was earning 200% interest (a fantasy, I know). After the first year you go from $100 to$300. How much would you expect to have at the end of year 2? You’d be angry if the bank still computed interest off your original \$100 from last year, right?

However, there may be situations where you have a static rate, and want to do 3 * 2 and not 3^2 [i.e., always compute growth off the original amount]. Check out http://betterexplained.com/articles/a-visual-guide-to-simple-compound-and-continuous-interest-rates/ for a few different examples.

Ah! now i get it.. the difference between simple growth and compunded growth!
thanks a ton for the immediate reply!
sadly i stumbled upon this site in 2013! hope to see more new explanatory articles!
all the best!

100. Ediz Ersoy says:

IT is NOT true that 0^0 =1 This is false.

For some real number “a” not equal to zero:

a^n = a^n*a^0

Thus we should be able to divide both sides by a^n and get: 1=a^0

IF a=0 or a^n=0 then we run into problems.

a^0 is defined only if a does not equal zero.

101. kalid says:

@Ediz: 0^0 is technically undefined [you get conflicting answers for x^0 and 0^x as x -> 0], but 1 is a very reasonable (and useful) definition, if we had to pick a value.

102. Pablo says:

Kalid, would you please expand this analogy for negative numbers? What does it mean in this context (-2)^2 is a negative number and (-2)^3 is a positive number?
Thanks for your explanations. They are brilliant.

103. Kalid says:

@Pablo: Glad the explanations are helping! For negatives, I imagine a flip (http://betterexplained.com/articles/rethinking-arithmetic-a-visual-guide/).

So, with whole numbers, (-2)^2 = -2 * -2 = 4. Basically, we started at 1.0, flipped backward and grew 2x (1 * -2), then flipped forward again and grew 2x (-2 * -2 = 4).

If you have fractional powers (like -2^(0.5)), that is a “partial flip” which requires imaginary numbers to work out :).

104. Barton says:

I’ve read many of your articles and loved them all. Thinking of the base of an exponent as a growth rate and the exponent as a time is a nice way to extend beyond repeated multiplication. I’ve been trying to apply it to other formulas, though, and have become a bit flummoxed. For instance, take the formula for kinetic energy: K=(1/2)mv^2. Translating that to your analogy would be: “put an appropriate amount of mass in the oven, turn the power on so that the mass will grow so that it is v-times larger each second, and cook it for to seconds.” The output, obviously, represents energy. I don’t have too many qualms imagining the first part of the analogy applying (should I?), but I don’t really no how to grasp turning the power to a level representing velocity and cooking for two seconds. Why would I do that?

Can you think of a way that intuitively applies your analogy to kinetic energy?

I also tried my hand at explaining the formula for electric field intensity from a point charge: E=kq/r^2=kqr^(-2). So: “put in a lump representing units of charge into the oven, shrink it at the rate of the distance from the charge, for two seconds.” With that one I make a little more progress: one way to think of an e-field is field lines per unit area. The number of field lines is static, while an increasing radius causes growth in area for a spherical shell in two dimensions. So it seems a little plausible to say that you should shrink at a rate of the radius, for two seconds–each second representing a dimension. Maybe what you call time represents dimensions, sometimes.

Trying to apply that thinking to kinetic energy didn’t get far though.

A lot of the problems come about when considering equations with variables that are not the exponent when those variables explicitly or implicitly have something to do with time, such as t, v, a, i, etc. d=(1/2)a*t^2–two timey variables but neither represent time in our analogy.

I’m not posting these ideas to criticize your approach. I love your approach and it seems like if I were too understand these equations in a different way than I currently understand them, I might make a few other intuitive insights along the way.

Wanna take a shot at helping me out?

105. kalid says:

Hi Barton, thanks for the note, happy to help (btw, constructive criticism is always welcomed, I want to constantly evolve the explanations here).

In the case of physics formulas with whole-number exponents (1/2 mv^2, etc.) the purpose of the “square” isn’t really growth applied to v, but “v*v”. That is, we are mainly using exponents for the shorthand of repeated multiplication. In a similar way, we aren’t trying to “repeatedly add” m “v^2″ times, the multiplication “m * v^2″ represents a scaling of energy by both amounts. In formulas, I try to figure out which sense the operations are taking on (scaling, repeating, applying, growing).

When we see e^x in a formula, where x is the variable that changes, it’s a near guarantee we’re talking about growth (and applying growth rates). But when we have x^2, x^3, etc., it’s more likely we are just trying to apply x in two multiplications, use the 2nd or 3rd dimension, etc. [vs modifying a rate of growth]. Hope that helps!

106. One who knows Trig in 9th Grade says:

Okay okay I have heard of this debate on 0^0 and it is like this:

Middle School teacher:

Any number to the power of 0 is 1 and that includes 0

Clever student:

Um 0^0 is undefined. Let me show you.

Okay you know that 0 to any positive exponent is 0 and to any negative exponent is undefined.

You also know that 0^1 = 0 and that decreasing the exponent is the same as dividing by the base.

so 0^0 = 0/0 and 0/0 is undefined.

High School teacher:

But that disagrees with the law that x^0 = 1 so it must be 1

Cleverer student: 0/0 = 0

Calculus teacher: Lim x and y as both get closer and closer to 0 is 1 so 0^0 = 1

Even Cleverer student:

That doesn’t prove it for 0^0 only for negatives and positives

and it all comes down to

Everyone: 0^0 = 0/0 and 0/0 is undefined

We have heard your side and even your proofs and we agree that 0^0 is an exception to x^0 = 1

I am on that side.

107. @One who knows Trig in 9th Grade:

0^0 depends on usage in Calculus or in Discrete Mathematics.

Calculus, in general, is an example of Continuous Mathematics and 0^0 deals with Continuous Exponents, thus, 0^0 in General Calculus is Undefined.

Discrete Mathematics, on the other hand, deals with distinct objects; something which has a boundary, like a continent, you know what belongs to Asia and what belongs to Africa. 0^0 in Discrete Mathematics is simply 1.

Therefore, 0^0 depends on what Context (Whether it is Calculus or Discrete Math).

108. One who knows Trig in 9th Grade says:

@Took Engineering For 5 Years & Computing Science for 10 Years:

Even following that as you decrease the exponent by 1 you divide by the base you get 0^0 = 0/0 and that doesn’t have a bit of calculus in it, just simple algebra 1.

109. Eric V says:

@Barton, post 105
I would respectfully depart from Kalid’s take on this. For the case of the equation for kinetic energy, KE = 1/2 mv^2, the power of 2 doesn’t really come from the idea of v*v. It comes from the fact that we started with a ‘something’ with v in it, we did a process and ended up with a 1/2 in front and a 2 in an exponent. This should look suspiciously like integration, if we remember that lovely power rule our instructors made us memorize. I can explain this in a basic idea in a moment, but first my take on exponents in general.
I appreciate Kalid’s example to explain it as growth in the Expand-O-Tron. It directly relates to a lot of real world examples, and may well be easier than the way I first looked at it, so kudos Kalid. As you (Barton) point out the time-growth analogy doesn’t really click here if ‘velocity’ is our time under which KE is growing. So here’s how I’ve always viewed exponents. Start with the view as a shorthand for multiplication. First observe that we have a progression, from addition, to multiplication, to exponents. They do all seem related as a family of processes. They do all use a second number to modify the first number, each ‘more powerful’ process just does ‘more’ of a modification. If we draw a 2D graph with just a few entries we may be able to look beyond one pair of input/output, and see the graph as a whole. We only see a few points corresponding to integers, but the fact they appear to lie in a curve seems to indicate something should be in-between the integer data points we’ve graphed. Now I just need to find a way to describe those ‘in-between’ numbers. 3^1, 3^2, and 3^3 all make sense. What about 3^1.5? When I look at the graph I get a clue, but not a full answer. 3^1.5 should be, and is, between 3^1 and 3^2, but not half way. There must be something about the way exponents ‘boost’ or grow numbers that makes it more potent as the graph goes to the right. For me, I approach it the same way I think of multiplying by a fraction. If I only know how to multiply by integers AND if I have the compliment, division by integers, then I don’t need to multiply by 1.5, I multiply by 15 and divide by 10. I can now multiply by any rational. But there are some numbers in-between the rationals, hmm…
Instead of * by 1.55 I could * by 155 and divide by 1000. Instead of * by 3.14159 I could * by 314159 and divide by 100000. It’s messy but it gets the job done, and all in terms of integers. Now if I can get the idea of taking a limit of something as a number approaches a target I can at least see how a computer can go a lot closer to the target than I could with paper and pen, and at least form a loose idea that the idea of * by an irrational is just the end result of an infinitely long process of multiplying and dividing by integers. It’s all made of integers (and they’re made of primes, but that’s another dragon to slay!).

// programmer geekiness to follow, skip ahead a bit if this hurts your brain
And ,oh by the way, for the computer programming geeks among us, it should make sense that all other types of numbers are based on integers, and all other process are based on add and subtract. If we wanted to make our own user defined data type called Multiple Precision so that it is essentially a floating point decimal not limited to the ability of the processor but to any number of digits we want, we start by creating a class called MPint (multiple precision integer). We teach the computer that an MPint is just a linked list of nodes where each node is one digit (char data type is one byte and this is a base 256 number system, yowza!) and a pointer to the next node. Once we teach the computer what this new integer is we then teach it how to add, then how to subtract… then how to do cosine… it’s a lot of fun! It may seem we’re re-inventing the wheel but what if you wanted to generate a graphic of a Mandelbrot set and the computer quit on you just because you can’t zoom in any more once you’ve gotten as fine detail as you can with the floating point decimal, of what if you really wanted the googol-plex’th digit of π?
// programing geekiness done; return void

Likewise with exponents. I know how to do the task with integers, all I need is a compliment/inverse function, the root, and that should do the trick. Instead of x^1.5 I raise x^15 and take the 10th root. For x ^ irrational, again consider a limit of the process as I take a larger and larger root.

Now this is pretty heavily a ‘math’ way of thinking and not a friendly, cuddly little kitty for the first time learner to exponents. This kitty has teeth, and gas too! You don’t want to stay around him too long!

I does let you see exponents firstly through the lens of ‘shorthand for multiplication’, then later through another lens: take a number and modify it by a greater process than multiplication.

Now back to your question, what is “v^2″ in KE = 1/2mv^2. It’s hard to think of putting mass in the Expand-O-Tron for some velocity-wise amount of time, or that mass grows in the domain of velocity instead of time. Rather it means that velocity was somehow acted upon in the way that only exponents can. To see how this really happened remember my little spoiler from earlier, it comes from an integral:
KE = ∫ (momentum) dv, but what does that mean?
It’s area under a curve, o.k., but what does it mean?
It ‘counts up’ (integrates) all the little momentum changes needed to bring the mass from being still (ground state, let’s leave absolute motion and relativity for later) up to the velocity it has right now. More simply, the KE that it has now, is the same as all the energy needed to bring it from motionless to its current state of motion. Sometimes I think of this a little bass-ackwardly ’cause I’m a physics kind of guy, and tell myself:
The particle doesn’t really HAVE any KE at all, never did. No energy was added to the PARTICLE, energy was added to the system to change the motion of the particle (remember the relativity thing I said I wouldn’t get into, never trust a smiling physicist!)The power of 2 popped up because of the integral. v^2 acts on v in a bit more of a way than v*2 would.
//B.T.W. the ∫ can be typed with b, I spelled momentum as there is no lower case rho unless I could change the font to Symbol for one letter.//

Happy Math and Happy Gillmore!

110. will says:

cool

111. abc says:

can you explain this-
0^0=0^(m-m)=0^m/0^m=0/0 which is not defined

112. Farseen Abdul Salam says:

Now I have a confusion. If I cook number 1 in the expand-o-tron with base 2 for 1 secs, won’t the number 1 be doubling constantly so that the number will converge to 2.718281828459045, which equals e?
Or does the expand-o-tron double or triple or do the magic on the number at the very last moment? If this was the case, what about the fractional powers?

And my biggest ‘Huh?’ is what does it mean to raise a number to a complex power?
For example: e^(i*pi) = -1

113. kalid says:

Hi Farseen, great question. The expand-o-tron just guarantees that at the *end* of the interval, you will have reached your desired growth rate.

Under the hood, it’s using e^x to have smooth growth. So when you type “2” as the growth factor, it knows to run e^x for ln(2) units of time. In other words, you specify the final result you want, not the instant-by-instant growth factor. [Inside the machine, we use ln to figure out the corresponding instant-by-instant factor for some amount of growth].

Check out this article for more on this final result to instant-by-instant conversion:

http://betterexplained.com/articles/think-with-exponents/

and

http://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/

for e^(i*pi)

114. Srinivasan says:

sir,

I have gone through ur book on better explained math book.Excellent book.

but i am not clear with advanced topic on chapter 10 rewriting exponents for the grower.

can u please mail me in simple statements about what that particular topic

expecting a positive reply from ur side.

regards

Srinivasan.K

Asst.Professor

Sri Venkateswara College of engg

Chennai

ph:9842092575

115. Dominic says:

Your part on negative exponents is completely wrong…
”This works for any number! Wherever 1 million is, we were at 500,000 one second before it.”
10^6 = 1 000 000
10^5 = 100 000
That’s not 500 000?!

What you say works only for 2 and any exponent you give it.
It’s a very misleading part, as I’m writing this; I reread to make sense of what you said. In reality, by your explanation, your microwave working on 45^-1 would have to go back in time not 1 second but 2 seconds (0.022 = 45/45/45). That could be explained saying that 45 is the same as 45^1 and that you have to go through 45^0 before getting to 45^-1… but that kills the progression of an article that’s made to culminate to x^0. Anyhow, that flaw helped me understand more than the article itself, so anyway. (Hidden exercice? =O )