Calculus – BetterExplained https://betterexplained.com Math lessons that click Wed, 04 May 2022 22:30:39 +0000 en-US hourly 1 Integral of Sin(x): Geometric Intuition https://betterexplained.com/articles/integral-sinx/ https://betterexplained.com/articles/integral-sinx/#disqus_thread Wed, 20 Nov 2019 00:01:09 +0000 https://betterexplained.com/?p=12244 You're minding your own business when some punk asks what the integral of $\sin(x)$ means. Your options:

  • Pretend to be asleep (except not in the engineering library again)
  • Canned response: "As with any function, the integral of sine is the area under its curve."
  • Geometric intuition: "The integral of sine is the horizontal distance along a circular path."

Option 1 is tempting, but let's take a look at the others.

Why "Area Under the Curve" is Unsatisfying

Describing an integral as "area under the curve" is like describing a book as a list of words. Technically correct, but misses the message and I suspect you haven't done the assigned reading.

Integral of Sin(x): Geometric Intuition

Unless you're trapped in LegoLand, integrals mean something besides rectangles.

Decoding the Integral

My calculus conundrum was not having an intuition for all the mechanics.

When we see:

$\int \sin(x) dx$

We can call on a few insights:

  • The integral is just fancy multiplication. Multiplication accumulates numbers that don't change (3 + 3 + 3 + 3). Integrals add up numbers that might change, based on a pattern (1 + 2 + 3 + 4). But if we squint our eyes and pretend items are identical we have a multiplication.

  • $\sin(x)$ just a percentage. Yes, it's also fancy curve with nice properties. But at any point (like 45 degrees), it's a single percentage from -100% to +100%. Just regular numbers.

  • $dx$ is a tiny, infinitesimal part of the path we're taking. 0 to $x$ is the full path, so $dx$ is (intuitively) a nanometer wide.

Ok. With those 3 intuitions, our rough (rough!) conversion to Plain English is:

The integral of sin(x) multiplies our intended path length (from 0 to x) by a percentage

We intend to travel a simple path from 0 to x, but we end up with a smaller percentage instead. (Why? Because $\sin(x)$ is usually less than 100%). So we'd expect something like 0.75x.

In fact, if $\sin(x)$ did have a fixed value of 0.75, our integral would be:

$\int \text{fixedsin}(x) \ dx = \int 0.75 \ dx = 0.75 \int dx = 0.75x$

But the real $\sin(x)$, that rascal, changes as we go. Let's see what fraction of our path we really get.

Visualize The Change in Sin(x)

Now let's visualize $\sin(x)$ and its changes:

Integral of Sin(x): Geometric Intuition

Here's the decoder key:

  • $x$ is our current angle in radians. On the unit circle (radius=1), the angle is the distance along the circumference.

  • $dx$ is a tiny change in our angle, which becomes the same change along the circumference (moving 0.01 units in our angle moves 0.01 along the circumference).

  • At our tiny scale, a circle is a a polygon with many sides, so we're moving along a line segment of length $dx$. This puts us at a new position.

With me? With trigonometry, we can find the exact change in height/width as we slide along the circle by $dx$.

By similar triangles, our change just just our original triangle, rotated and scaled.

  • Original triangle (hypotenuse = 1): height = $\sin(x)$, width = $\cos(x)$
  • Change triangle (hypotenuse = dx): height = $\sin(x) dx$, width = $\cos(x) dx$

Now, remember that sine and cosine are functions that return percentages. (A number like 0.75 doesn't have its orientation. It shows up and makes things 75% of their size in whatever direction they're facing.)

So, given how we've drawn our Triangle of Change, $\sin(x) dx$ is our horizontal change. Our plain-English intuition is:

The integral of sin(x) adds up the horizontal change along our path

Visualize The Integral Intuition

Ok. Let's graph this bad boy to see what's happening. With our "$\sin(x) dx$ = tiny horizontal change" insight we have:

Integral of Sin(x): Geometric Intuition

As we circle around, we have a bunch of $dx$ line segments (in red). When sine is small (around x=0) we barely get any horizontal motion. As sine gets larger (top of circle), we are moving up to 100% horizontally.

Ultimately, the various $\sin(x) dx$ segments move us horizontally from one side of the circle to the other.

A more technical description:

$\int_0^x \sin(x) dx = \text{horizontal distance traveled on arc from 0 to x}$

Aha! That's the meaning. Let's eyeball it. When moving from $x=0$ to $x=\pi$ we move exactly 2 units horizontally. It makes complete sense in the diagram.

The Official Calculation

Using the Official Calculus Fact that $\int \sin(x) dx = -\cos(x)$ we would calculate:

$\int_0^\pi \sin(x) dx = -\cos(x) \Big|_0^\pi = -\cos(\pi) - -\cos(0) = -(-1) -(-1) = 1 + 1 = 2$

Yowza. See how awkward it is, those double negations? Why was the visual intuition so much simpler?

Our path along the circle ($x=0$ to $x=\pi$) moves from right-to-left. But the x-axis goes positive from left-to-right. When convert distance along our path into Standard Area™, we have to flip our axes:

Integral of Sin(x): Geometric Intuition

Our excitement to put things in the official format stamped out the intuition of what was happening.

Fundamental Theorem of Calculus

We don't really talk about the Fundamental Theorem of Calculus anymore. (Is it something I did?)

Instead of adding up all the tiny segments, just do: end point - start point.

The intuition was staring us in the face: $\cos(x)$ is the anti-derivative, and tracks the horizontal position, so we're just taking a difference between horizontal positions! (With awkward negatives to swap the axes.)

That's the power of the Fundamental Theorem of Calculus. Skip the intermediate steps and just subtract endpoints.

Onward and Upward

Why did I write this? Because I couldn't instantly figure out:

$\int_0^\pi \sin(x) dx = 2$

This isn't an exotic function with strange parameters. It's like asking someone to figure out $2^3$ without a calculator. If you claim to understand exponents, it should be possible, right?

Now, we can't always visualize things. But for the most common functions we owe ourselves a visual intuition. I certainly can't eyeball the 2 units of area from 0 to $\pi$ under a sine curve.

Happy math.

Appendix: Average Efficiency

As a fun fact, the "average" efficiency of motion around the top of a circle (0 to $\pi$) is: $\frac{2}{\pi} = .6366$

So on average, 63.66% of your path's length is converted to horizontal motion.

Appendix: Height controls width?

It seems weird that height controls the width, and vice-versa, right?

If height controlled height, we'd have runaway exponential growth. But a circle needs to regulate itself.

$e^x$ is the kid who eats candy, grows bigger, and can therefore eat more candy.

$\sin(x)$ is the kid who eats candy, gets sick, waits for an appetite, and eats more candy.

Appendix: Area isn't literal

The "area" in our integral isn't literal area, it's a percentage of our length. We visualized the multiplication as a 2d rectangle in our generic integral, but it can be confusing. If you earn money and are taxed, do you visualize 2d area (income * (1 - tax))? Or just a quantity helplessly shrinking?

Area primarily indicates a multiplication happened. Don't let team Integrals Are Literal Area win every battle!

]]> https://betterexplained.com/articles/integral-sinx/feed/ 21 Intuition for Taylor Series (DNA Analogy) https://betterexplained.com/articles/taylor-series/ https://betterexplained.com/articles/taylor-series/#disqus_thread Thu, 24 Jan 2019 06:14:43 +0000 https://betterexplained.com/?p=11904 Your body has a strange property: you can learn information about the entire organism from a single cell. Pick a cell, dive into the nucleus, and extract the DNA. You can now regrow the entire creature from that tiny sample.

There's a math analogy here. Take a function, pick a specific point, and dive in. You can pull out enough data from a single point to rebuild the entire function. Whoa. It's like remaking a movie from a single frame.

The Taylor Series discovers the "math DNA" behind a function and lets us rebuild it from a single data point. Let's see how it works.

Intuition for Taylor Series (DNA Analogy)

Pulling information from a point

Given a function like $f(x) = x^2$, what can we discover at a single location?

Normally we'd expect to calculate a single value, like $f(4) = 16$. But there's much more beneath the surface:

  • $f(x)$ = Value of function at point $x$
  • $f'(x)$ = First derivative, or how fast the function is changing (the velocity)
  • $f''(x)$ = Second derivative, or how fast the changes are changing (the acceleration)
  • $f'''(x)$ = Third derivative, or how fast the changes in the changes are changing (acceleration of the acceleration)
  • And so on

Investigating a single point reveals multiple, possibly infinite, bits of information about the behavior. (Some functions have an endless amount of data (derivatives) at a single point).

So, given all this information, what should we do? Regrow the organism from a single cell, of course! (Maniacal cackle here.)

Growing a Function from a point

Our plan is to grow a function from a single starting point. But how can we describe any function in a generic way?

The big aha moment: imagine any function, at its core, is a polynomial (with possibly infinite terms):

\displaystyle{f(x) = c_0 + c_1 x + c_2 x^2 + c_3x^3 + \cdots}

To rebuild our function, we start at a fixed point ($c_0$) and add in a bunch of other terms based on the value we feed it (like $c_1x$). The "DNA" is the values $c_0, c_1, c_2, c_3$ that describe our function exactly.

Ok, we have a generic "function format". But how do we find the coefficients for a specific function like sin(x) (height of angle x on the unit circle)? How do we pull out its DNA?

Time for the magic of 0.

Let's start by plugging in the function value at $x=0$. Doing this, we get:

\displaystyle{f(0) = c_0 + 0 + 0 + 0 + \cdots = c_0}

Every term vanishes except $c_0$, which makes sense: the starting point of our blueprint should be $f(0)$. For $f(x) = \sin(x)$, we can work out $c_0 = \sin(0) = 0$. We have our first bit of DNA!

Getting More DNA

Now that we know $c_0$, how do we isolate $c_1$ in this equation?

\displaystyle{f(x) = c_0 + c_1 x + c_2 x^2 + c_3x^3 + \cdots}

Hrm. A few ideas:

  • Can we set $x = 1$? That gives $f(1) = c_0 + c_1(1) + c_2(1^2) + c_3(1^3) + \cdots$ . Although we know $c_0$, the other constants are summed together. We can't pull out $c_1$ by itself.

  • What if we divide by $x$? This gives:

\displaystyle{\frac{f(x)}{x} = \frac{c_0}{x} + c_1 + c_2 x + c_3x^2 + \cdots}

Then we can set $x=0$ to make the other terms disappear... right? It's a nice idea, except we're now dividing by zero.

Hrm. This approach is really close. How can we almost divide by zero? Using the derivative!

If we take the derivative of the blueprint of $f(x)$, we get:

\displaystyle{f'(x) = (c_0)' + (c_1 x)' + (c_2 x^2)' + (c_3x^3)' + \cdots}

\displaystyle{f'(x) = 0 + c_1 + (2\cdot c_2 x) + (3\cdot c_3x^2) + \cdots}

Every power gets reduced by 1 and the $c_0$, a constant value, becomes zero. It's almost too convenient.

Now we can isolate $c_1$ using our $x=0$ trick:

\displaystyle{f'(0) = 0 + c_1 + (0) + (0) + \cdots = c_1}

In our example, $\sin'(x) = \cos(x)$ so we compute: $f'(0) = \sin'(0) = \cos(0) = 1 = c_1$

Yay, one more bit of DNA! This is the magic of the Taylor series: by repeatedly applying the derivative and setting $x = 0$, we can pull out the polynomial DNA.

Let's try another round:

\displaystyle{f''(x) = 0 + 0 + (2\cdot c_2) + (3\cdot 2 c_3x^1) + \cdots}

After taking the second derivative, the powers are reduced again. The first two terms ($c_0$ and $c_1x$) disappear, and we can again isolate $c_2$ by setting $x=0$:

\displaystyle{f''(0) = 0 + 0 + 2\cdot c_2 + 0 + \cdots}

For our sine example, $\sin'' = -\sin$, so:

\displaystyle{f''(0) = \sin''(0) = -\sin(0) = 0 = 2\cdot c_2}

or $c_2 = 0$.

As we keep taking derivatives, we're performing more multiplications and growing a factorial in front of each term (1!, 2!, 3!).

The Taylor Series for a function around point x=0 is:

\displaystyle{f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots}

(Formally, the Taylor series around the point $x=0$ is called the MacLaurin series.)

The generalized Taylor series, extracted from any point a is:

\displaystyle{f(x) = f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+\cdots }

The idea is the same. Instead of our regular blueprint, we use:

\displaystyle{f(x) = c_0 + c_1 (x - a) + c_2 (x-a)^2 + c_3(x-a)^3 + \cdots}

Since we're growing from $f(a)$, we can see that $f(a) = c_0 + 0 + 0 + \dots = c_0$. The other coefficients can be extracted by taking derivatives and setting $x = a$ (instead of $x =0$).

Example: Taylor Series of sin(x)

Plugging in derivatives into the formula above, here's the Taylor series of $\sin(x)$ around $x = 0$:

\displaystyle{\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots}

And here's what that looks like:

Intuition for Taylor Series (DNA Analogy)

A few notes:

1) Sine has infinite terms

Sine is an infinite wave, and as you can guess, needs an infinite number of terms to keep it going. Simpler functions (like $f(x) = x^2 + 3$) are already in their "polynomial format" and don't have infinite derivatives to keep the DNA going.

2) Sine is missing every other term

If we repeatedly take the derivative of sine at x = 0 we get:

\displaystyle{\sin(0) \xrightarrow{\text{derive}} \cos(0) \xrightarrow{\text{derive}} -\sin(0) \xrightarrow{\text{derive}} -\cos(0) \xrightarrow{\text{derive}} \sin(0) \dots}

with values:

\displaystyle{0, 1, 0, -1, \dots}

Ignoring the division by the factorial, we get the pattern:

\displaystyle{(0) + (1)x^1 + (0)x^2 + (-1)x^3 + (0)x^4 + (-1)x^5+ \dots }

So the DNA of sine is something like [0, 1, 0, -1] repeating.

3) Different starting positions have different DNA

For fun, here's the Taylor series of $\sin(x)$ starting at $x =\pi$ (link):

taylor series sin x at x=pi - Wolfram|Alpha 2019-01-18 16-45-04

A few notes:

  • The DNA is now something like [0, -1, 0, 1]. The cycle is similar, but the starting value has changed since we're starting at $x=\pi$.

  • Written as calculated numbers, the denominators 1, 6, 120, 5040 look strange. But they're just every other factorial: 1! = 1, 3! = 6, 5! =120, 7! = 5040. In general, the Taylor series can have gnarly denominators.

  • The $O(x^{12})$ term means there are other components of order (power) $x^{12}$ and higher. Because $\sin(x)$ has infinite derivatives, we have infinite terms and the computer has to cut us off somewhere. (You've had enough Tayloring for today, buddy.)

Application: Function Approximations

A popular use of Taylor series is getting a quick approximation for a function. If you want a tadpole, do you need the DNA for the entire frog?

Intuition for Taylor Series (DNA Analogy)

The Taylor series has a bunch of terms, typically ordered by importance:

\displaystyle{f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots}

  • $c_0 = f(0)$, the constant term, is the exact value at the point
  • $c_1 = f'(0)x$, the linear term, tells us what speed to move from our point
  • $c_2= \frac{f''(0)}{2!}x^2$, the quadratic term, tells us how much to accelerate away from our point
  • and so on

If we only need a prediction for a few instants around our point, the initial position & velocity may be good enough:

\displaystyle{\text{Linear model} = \text{initial point} + \text{velocity effect} = f(0) + f'(0)x}

If we're tracking for longer, then acceleration becomes important:

\displaystyle{\text{Quadratic model} = \text{initial point} + \text{velocity effect} + \text{acceleration effect}} \displaystyle{ = f(0) + f'(0)x + \frac{1}{2}f''(0)x^2}

As we get further from our starting point, we need more terms to keep our prediction accurate. For example, the linear model $\sin(x) = x$ is a good prediction around $x=0$. As we get further out, we need to account for more terms.

Similarly, $e^x \sim 1 + x$ works well for small interest rates: 1% discrete interest is 1.01 after one time period, 1% continuous interest is a tad higher than 1.01. As time goes on, the linear model falls behind because it ignores the compounding effects.

Application: Comparing Functions

What's a common application of DNA? Paternity tests.

If we have a few functions, we can compare their Taylor series to see if they're related.

Here's the expansions of $\sin(x)$, $\cos(x)$, and $e^x$:

\displaystyle{ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \xrightarrow{DNA} [0, 1, 0 -1, \dots] }

\displaystyle{ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \xrightarrow{DNA} [1, 0, -1, 0, \dots] }

\displaystyle{ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \xrightarrow{DNA} [1, 1, 1, 1, \dots] }

There's a family resemblance in the sequences, right? Clean powers of $x$ divided by a factorial?

One problem is the sequence for $e^x$ has positive terms, while sine and cosine alternate signs. How can we link these together?

Euler's great insight was realizing an imaginary number could swap the sign from positive to negative:

{\begin{aligned}e^{ix}&=1+ix+{\frac {(ix)^{2}}{2!}}+{\frac {(ix)^{3}}{3!}}+{\frac {(ix)^{4}}{4!}}+{\frac {(ix)^{5}}{5!}}+{\frac {(ix)^{6}}{6!}}+{\frac {(ix)^{7}}{7!}}+{\frac {(ix)^{8}}{8!}}+\cdots \\[8pt]&=1+ix-{\frac {x^{2}}{2!}}-{\frac {ix^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {ix^{5}}{5!}}-{\frac {x^{6}}{6!}}-{\frac {ix^{7}}{7!}}+{\frac {x^{8}}{8!}}+\cdots \\[8pt]&=\left(1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+{\frac {x^{8}}{8!}}-\cdots \right)+i\left(x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots \right)\\[8pt]&=\cos x+i\sin x.\end{aligned}}

Whoa. Using an imaginary exponent and separating into odd/even powers reveals that sine and cosine are hiding inside the exponential function. Amazing.

Although this proof of Euler's Formula doesn't show why the imaginary number makes sense, it reveals the baby daddy hiding backstage.

Appendix: Assorted Aha! Moments

Relationship to Fourier Series

The Taylor Series extracts the "polynomial DNA" and the Fourier Series/Transform extracts the "circular DNA" of a function. Both see functions as built from smaller parts (polynomials or exponential paths).

Does the Taylor Series always work?

This gets into mathematical analysis beyond my depth, but certain functions aren't easily (or ever) approximated with polynomials.

Notice that powers like $x^2, x^3$ explode as $x$ grows. In order to have a slow, gradual curve, you need an army of polynomial terms fighting it out, with one winner barely emerging. If you stop the train too early, the approximation explodes again.

For example, here's the Taylor Series for $\ln(1 + x)$. The black line is the curve we want, and adding more terms, even dozens, barely gets us accuracy beyond $x=1.0$. It's just too hard to maintain a gentle slope with terms that want to run hog wild.

Intuition for Taylor Series (DNA Analogy)

source

In this case, we only have a radius of convergence where the approximation stays accurate (such as around $|x| < 1$).

Turning geometric to algebraic definitions

Sine is often defined geometrically: the height of a line on a circular figure.

Intuition for Taylor Series (DNA Analogy)

Turning this into an equation seems really hard. The Taylor Series gives us a process: If we know a single value and how it changes (the derivative), we can reverse-engineer the DNA.

Similarly, the description of $e^x$ as "the function with its derivative equal to the current value" yields the DNA [1, 1, 1, 1], and polynomial $f(x) = 1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots$. We went from a verbal description to an equation.

Phew! A few items to ponder.

Happy math.

]]> https://betterexplained.com/articles/taylor-series/feed/ 17 How to Add 1 through 100 using Calculus https://betterexplained.com/articles/how-to-add-1-to-100-using-calculus/ https://betterexplained.com/articles/how-to-add-1-to-100-using-calculus/#disqus_thread Fri, 17 Feb 2017 20:52:31 +0000 https://betterexplained.com/?p=10127 Earlier we saw a few ways to add a set of numbers (1 to 10).

adding 1 to 100 strategies

And the formula we found was:

\displaystyle{\text{Sum from 1 to n} = \frac{n(n+1)}{2}}

\displaystyle{\text{Sum from 1 to 100} = \frac{100(100+1)}{2} = (50)(101) = 5050}

It seems that regular arithmetic, algebra, geometry, or even statistics could help work out the equation.

But how about Calculus? Is this bringing a nuclear missile to a gun fight?

Let's find out.

The sequence to add (1 2 3 4 5 6 7 8 9 10...) looks a lot like $f(x) = x$. At every position on the x-axis, we put in a number and get the same one out.

Intuitively, the integral is "repeatedly adding a bunch of stuff" -- it seems like we could put it to work. From the rules of Calculus (or using Wolfram Alpha) we get this:

\displaystyle{ \int x = \frac{1}{2} x^2 }

Intuitively: Add up things following the $f(x) = x$ pattern and you end up with $\frac{1}{2} x^2$.

Well, let's see: the actual sum from 1 to 100 is 5050. But using the Calculus equation we get:

\displaystyle{ \frac{1}{2} x^2 = \frac{1}{2}100^2 = \frac{10,000}{2} = 5000 }

Uh oh: there's a difference. What's going on?

Calculus works with continuous patterns, and we used a discrete one.

Here's what's happening:

calculus add area under x

Calculus was built to measure smoothly changing functions, like a line, parabola, circle, etc. The pattern we have is a jumpy staircase (going from 1 to 2 without ever passing through 1.5, or 1.1, or 1.0001). In math class, books harp on analyzing whether a function is "continuous", aka changes smoothly enough for Calculus to work.

So when a pattern changes smoothly, Calculus works great. If a pattern changes suddenly, Calculus can only give an approximate answer. So what's the plan?

Use Calculus where possible, on the smooth part, and adjust for errors in the jumpy part.

calculus integral of x discrete

The area under the line is the integral. We a bunch of triangles above the line we need to include.

  • How many of them? 1 for each item (x)
  • How big are they? They're half a of a 1x1 square, so they have area 1/2.
  • What's the total area to add back in? $x \frac{1}{2} = \frac{x}{2}$

So our final formula should be

\displaystyle{\text{Integral} + \text{Adjustment} = \frac{1}{2} x^2 + \frac{x}{2}}

Aha! Learning Calculus doesn't mean we hunt around for Official Calculus Problems.

Nope. Take your scenario (adding 1 to 100) and realize what Calculus brings to the table: finding patterns in smoothly changing functions. Use Calculus on the smooth parts and adjust (or ignore) the other parts.

(Ironically, Calculus works by making jumpy approximations for smooth functions, and is in fact "jumpy" under the hood. If you are planning on working with jumpy patterns, use Discrete Calculus.)

Example: Adding the first n squares

Let's take this further: what's your guess for the sum of the first 100 square numbers?

\displaystyle{1^2 + 2^2 + 3^2 + 4^2 + ... + 100^2}

Hrm. Getting the exact formula is tricky. But maybe we don't need the exact count, just an estimate.

With Calculus, we'd say: The pattern isn't continuous, but it looks like $f(x) = x^2$. Let's integrate x^2 from 0 to 100.

\displaystyle{ \int x^2 = \frac{1}{3} x^3 }

The indefinite integral is $\frac{1}{3} x^3$, the running total for how much we have. From 0 to 100 it would be

\displaystyle{ \frac{1}{3}100^3 = \frac{1,000,000}{3} \sim 333,333 }

That's our guess, without a calculator. And the actual answer? 338350.

calculus area under x squared

How close were we? 99.9%. Not bad for something we worked out by hand in a minute!

Truly internalizing Calculus means it helps other elements of your math understanding, even regular addition problems.

Happy math.

PS. To keep building your intuition, check out the Calculus Guide.

]]> https://betterexplained.com/articles/how-to-add-1-to-100-using-calculus/feed/ 12 Quick Insight: Easier Arithmetic With Calculus https://betterexplained.com/articles/calculus-arithmetic/ https://betterexplained.com/articles/calculus-arithmetic/#disqus_thread Thu, 28 Jul 2016 17:39:19 +0000 https://betterexplained.com/?p=6304 Let's say you had to estimate 1 + 2 + 3 + 4 + 5 + 6 + 7 + ... + 100.

What would you do? Well, you could work out the exact formula:

\displaystyle{\frac{(n + 1)(n)}{2}}

and plug in n=100 to get 5050.

But we just want a rough answer. You have a list of numbers, they follow a simple pattern, and want a quick estimate. What to do?

The "easy" way (well, the Calculus way) is to realize 1 + 2 + 3 + 4 is about the same as $f(x) = x$. The first element is f(1) = 1, the second is f(2) = 2, and so on.

From here, we can take the integral:

\displaystyle{ \int x = \frac{1}{2} x^2 }

We usually see the integral as a formal, elegant operation, which artfully accumulates one function and returns another. Informally, we're squashing everything together in that bad mamma-jamma and seeing how much there is.

The result $\frac{1}{2} x^2$ should be pretty close to what we want.

calculus add area under x

The exact total is our staircase-like pattern, which accumulates to 5050.

The approximate answer is the area of that triangle, $\frac{1}{2} base * height = \frac{1}{2} 100 * 100 = 5000$. The difference is because of the corners in the staircase which overhang. $\frac{x}{2}$ is one-half, x times (the size of overhang (1/2) times the number of pieces (x)).

The net result is using a smooth, easy-to-measure shape to approximate a jagged, tedious-to-measure one. (This is a bit of Calculus inception, since we usually use rectangles to approximate smooth shapes.)

More Estimates

This tactic works for other sequences:

What's the sum of the first 10 square numbers? 1 + 4 + 9 + 16 + 25 + ... + 100 = ?

Hrm. The formula is probably tricky to work out. But without our Calculus-infused Arithmetic, a quick guess would be:

calculus add square numbers x squared

\displaystyle{\int x^2 = \frac{1}{3} x^3}

Our first hunch should be "one third of 10^3" or 333. But as we saw before, there's an "overhang" that we missed. Let's call it 10%, for an estimate of 330 + 10% ~ 370.

The exact answer is 385. Not bad! The actual formula is:

\displaystyle{P_{n}=\sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {2n^{3}+3n^{2}+n}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}} }

I'd say $\frac{x^3}{3}$ isn't bad for a few seconds of work.

Data doubles every year. What does lifetime usage look like?

The integral (squashed-together total) of an exponential is an exponential. In Calculus terms,

\displaystyle{\int e^x = e^x}

The key insight is that all exponential growth is just a variation of $e^x$. If $e^x$ accumulates exponentially, so will $2^x$.

So the total usage to date will also follow an exponential pattern, doubling every year also. Contrast this with a usage pattern of "1 + 2 + 3 + 4 ..." -- we grow linearly ($f(x) = x$), but total usage accumulates quadratically ($\frac{1}{2}x^2$).

My goal is to incorporate math thinking into everyday scenarios. We start with an arithmetic question, convert it to a geometry puzzle (how big is the staircase?), and then use calculus to approximate it.

I know a concept is clicking when I can switch between a few styles of thought. Imagine the problem as a script: how would Spielberg, Tarantino, or Scorsese direct it? Each field takes a different look. (To learn how to think with Calculus, check out the Calculus Guide.)

Happy math.

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