Multiplying Triangles

It should not be a surprise that the first hints of the meaning of imaginary numbers came from the ancient Greeks. Diophantus discovered the following equation:

(a² + b²)(c² + d²) = (ac – bd)² + (ad + bc)²

Just to foreshadow a little bit, this is like the formula for multiplying complex numbers:

(a + bi)(c + di) = (ac – bd) + (ad + bc)i

But Diophantus’s equation does a better job of revealing the mystery of imaginary numbers. We know from the Pythagorean formula that when you have a right triangle, a² + b² = c². So the left of Diophantus’s equation represents two right triangles being multiplied together. The right side is also a triangle. So Diophantus found that two right triangles multiplied together equals another right triangle. That is a pretty amazing discovery!

Lets try it out. Our first triangle has sides of 3 and 4. The second triangle has sides of 5 and 12. Since they are right triangles we know by the Pythagorean theorem that the hypotenuses are 5 and 13 respectively. Plug that in and we get:

(3² + 4²)(5² + 12²) = 4225 = (3·5 – 4·12)² + (3·12 + 4·5)²

Using the Pythagorean theorem we get:

5²·13² = 4225 = 33² + 56²
5·13 = 65

So the length of the hypotenuse of the new triangle is equal to the product of the hypotenuses of the original triangles. Pretty neat! But here is something even neater. Lets find the arc sin (AKA sin^-1) for the triangles:

arc sin (³⁄₅) = 36.87°
arc sin (⁵⁄₁₃) = 22.62°

And we know that the legs of the product triangle are 33 and 56, and the hypotenuse is 65. So

arc sin ((⁵⁶⁄₆₅) ) = 59.49°

If you add up the angles of the original triangles, you get 59.49 degrees! Amazing! When you multiply two right triangles together, you get a new right triangle. The length of the hypotenuse is equal to the product of the lengths of the hypotenuses of the original triangles. And the main angle of the product angle is equal to the sum of the main angles of the original triangles! You can check this approximately on the figure above which is shown to scale (minus a small amount of error introduced making the figures – a limitations of Microsoft Paint).

Introducing The Complex Plane

Of course, we don’t really multiply triangles; we multiply vectors. And a vector is really just a point in the coordinate plane. Any point in the coordinate plane can be written as the pair (x,y), which determines a right triangle. One leg x, and the other leg is y. The hypotenuse is the line segment from the origin to the point (x,y). When you multiply two vectors together you multiply their lengths – which is given by the hypotenuse – and add their angles.

So what happens if you multiply the vector (0,1) by itself? The length is 1·1 = 1. And the angle is 90 degrees. So 90 + 90 = 180. So the product vector is of length 1 at 180 degrees – which means that the product vector is (-1,0). So this vector is the square root of negative one! We just found i!

We could write complex numbers as vectors, but we stick with the i notation. So we use i instead of (0,1). If we had the point (1,1) we would write it as 1 + i. More generally, we represent the y axis with i, so (a,b) becomes a + bi. The formula for multiplying two complex numbers is given by our Diophantus-like:

(a + bi)(c + di) = (ac – bd) + (ad + bc)i

This view also explains why a negative number squared is a positive. The angle of a negative number is 180 degrees. Square it and you get 180 + 180 = 360. We’ve come full circle and back to the positive direction. Similarly, the angle for a positive number is 0 degrees. So if you square a positive number the result is still zero degrees – it is still a positive number.

Let’s test this by using the formula for complex numbers to multipy the triangles (or vectors) in the example above: multiplying a triangle (or vector) of sides 3,4, and 5 by a triangle with sides of 5,12, and 13.

(4 + 3i)(12 + 5i) = 48 + 36i + 20i – 15 = 33 + 56i

Notice that I had to rearrange the terms compared to Diophantus’s equation. I had to make the real part 4 for the first triangle and 12 for the second. This is actually an advantage of complex numbers. They tell us which length belongs to which leg of the triangle. Diophantus’s equation does not do that.

Impossible Solutions

Suppose you are a city dweller on your way to catch the bus. As you approach the bus stop you see your bus just starting to pull away, so you run after it. You are running at six yards per second. The bus has a 30 yard head start, and is not moving. But it is accelerating at one yard per second per second. Will you catch the bus?

The answer is no. After six seconds you will have gone 36 yards, but the bus will have gone 18 yards (if you remember your high school physics, you know that distance = ½ acceleration × time²). Add in the 30 yard head start and the bus is still ahead of you. And after six seconds the bus has accelerated to a faster speed than you. So you will never catch the bus.

More formally, let t be the time, starting at 0.

Your position = 6t
Bus’s position = ½t² + 30

If you catch the bus then your positions are the same at the same time. So set the equations equal to each other and solve for t.

6t = ½ t² + 30
0 = ½ t² – 6t + 30

Using the quadratic formula, we get:

t = 6 ± √(36 – 4·½·30)
t = 6 ± √(-24)

As you may remember from high school math and physics, you discard all answers that involve the square roots of negative numbers. So there is no solution. We can see that graphically. The line is the graph of your position, and the parabola is the graph of the bus. The graphs never meet because there is no solution.

But what if there is an imaginary solution? As you probably know, i is √-1. We were stumped because we had √-24, which is then √24i². Let’s see if we can catch that bus after all.

t = 6 ± √-24
t = 6 ± √24i²
t = 6 ± 2i√6

It doesn’t seem like we’ve made much progress – what does it mean to say that we caught the bus at time 6 ± 2i√6? Does that mean we made it? Perhaps in some alternate universe? You can see why the great mathematician and philosopher Rene Descartes rejected imaginary numbers – they were impossible solutions to real problems.

But imaginary numbers refused to go away. In fact, there were clues to their power and utility going back to the ancient Greeks.

Taylor Series

One of the basic facts about calculus is that finding integrals is tough. The fundamental theorem helps a lot, but as the expressions get more and more complicated it gets tougher and tougher to find the integral. It makes you long for the good old days when everything you integrated was a polynomial in the form of axⁿ. That is undoubtedly what drove people to invent the Taylor series. The central concept is to find a way to express a non-polynomial as a polynomial. For example, could you write sin x as a simple polynomial?

How would you do this? Ponder this for a minute (no, really – take a minute and think about how you would do it). Here is one hint: you can’t use a simple polynomial. Instead use a power series: a₀ + a₁x + a₂x² + … + a_nxⁿ.

The secret is to start at the point x=0 (which makes the math really easy). and set the the various derivatives of the power series to the derivatives of sin x.

Step One: Find the derivatives of sin x

Lets start by finding the first five derivatives of sin x at the point x = 0:

sin 0 = 0
sin’0 = cos 0 = 1
sin”0 = -sin 0 = 0
sin⁽³⁾0 = -cos 0 = -1
sin⁽⁴⁾0 = sin 0 = 0
sin⁽⁵⁾0 = cos 0 = 1

Step Two: Find the derivatives of the power series

Now lets find the first five derivatives of the power series at x = 0. Lets call the power series a(x).

a(0) = a₀ + a₁x + a₂x² + … + a_nxⁿ = a₀
a’(0) = a₁ + 2a₂x + 3a₃x² + … + na_nx^n-1 = a₁
a”(0) = 2a₂ + 2·3a₃x + 3·4a₄x² + … + (n-1)*na_nx^n-2 = 2a₂
a⁽³⁾(0) = 2·3a₃ + 2·3·4a₄x + 3·4·5a₅x² … + (n-2)·(n-1)·na_nx^n-3 = 2·3a₃
a⁽⁴⁾(0) = 2·3·4a₄
a⁽⁵⁾(0) = 2·3·4·5a₅
a⁽ⁿ⁾(0) = n!a_n

Now you see why it makes sense to find the derivatives at x = 0. All the higher degree terms drop out and you are left with a constant term of n!a_n. Note that step two is always exactly the same, even if you are finding an approximation for cos x or e^x or any other complicated function.

Step Three: Solve for the values of the a_n’s

sin 0 = 0 = a₀
sin’0 = 1 = a₁
sin”0 = 0 = 2a₂
sin⁽³⁾0 = -1 = 3!a₃. So a₃ = -1/3!
sin⁽⁴⁾0 = 0= 4!a₄. So a₄ = 0
sin⁽⁵⁾0 = 1 = 5!a₅. So a₅ = 1/5!

At this point the pattern emerges. Plug the a_n’s back into the power series to get:

sin x = x – x³⁄_3! + x⁵⁄_5! – x⁷⁄_7! + …

This is a pretty accurate approximation. The sin of 1 (in radians of course) is .841470985. And plugging 1 into the power series – just going up the x⁷⁄_7! term – gives .841468254. Not bad.

Uses of Taylor Series

The nice thing about Taylor series are that they are easy to integrate. Now we can finally integrate everything.

Limitations

There are some limitations to this technique. The biggest is that we are simply assuming that the power series converges, but that may not be the case. That is why tests for series convergence are so useful. A second, and more practical, limitation is that we built our approximation around the point x = 0. The approximation is pretty good near the point x = 0, but it gets worse and worse the farther you get from that point. You would have to rebuild the power series at a different point.

That is easy enough to do. If you want to find values around the point 3, then rebuild your power series using (x-3) instead of x. Otherwise the process is identical.

Tips and Tricks

What if you wanted to find the Taylor series for sin(x²)? You could start taking derivatives:

sin’(x²) = 2x*cos(x²)
sin”(x²) = 2cos(x²) – 4x²sin(x²)

As you can see, it gets ugly in a hurry! The thought of slogging through the process of taking the derivative a few more times is not pleasant. Luckily there is an easier way. Just substitute x² for x into the Taylor series for sin x, to get:

sin(x²) = x² – x²³⁄_3! + x²⁵⁄_5! – x²⁷⁄_7! + …

Rewriting we get:

sin(x²) = x² – x⁶⁄_3! + x¹⁰⁄_5! – x¹⁴⁄_7! + …

Another easy technique is to solve something like ^{sin x}⁄_(x+1) by taking the Taylor series for sin x and then dividing each tearm by x + 1. So:

^{sin x}⁄_(x+1) = ¹⁄_x+1[ x - x³⁄_3! + x⁵⁄_5! - x⁷⁄_7! + ... ]

Exercises

There are a few Taylor series that you will repeatedly encounter, so they are worth memorizing. As an aid, calculate the following:

1. e^x
2. sin x
3. cos x
4. ln(1 + x)
5. (1 + x)^p

Making Sense of Logarithms

Necessity may be the mother of invention, but it also takes that eureka! moment. By the way, the original eureka moment came from the ancient Greek mathematical genius Archimedes. He was struggling with finding a way to measure the volume of solids. This is easy for nice geometric shapes like cubes and spheres, but very difficult for irregular shapes like rocks. There is no formula for the volume of a rock! Legend has it that Archimedes figured it out while taking a bath. He noticed that the water level rose when he entered the tub, and then realized that he could find the volume of a rock by submerging it and seeing how much water was displaced. He shouted eureka! and ran naked through the streets of Syracuse in joy.

The necessity that drove the discover of logarithms was due to the fact that multiplying and dividing large numbers by hand is extremely difficult. It was so difficult that it held back Kepler’s work in astronomy for years. Here is the eureka part of the discovery of logarithms.

Look at the following infinite series:

1, x, x², x³, x⁴, x⁵, x⁶, …

In order to multiply any two terms in the series, you simply add the exponents. This turns multiplication into addition, and division into subtraction. That is a huge increase in the speed of calculation. There is only one catch, which is that you have to find a way to represent numbers using the same base.

But even using a number as small as 2, this is not possible. What if you have to multiply 3 by 100? You need to do one of two things:

Method One: Express numbers as fractional exponents. Thus

3 × 100 = 2^1.5849 + 2^6.6438
= 2^8.2287
= 300

This is the approach we use now, although we use the natural logarithm, e, as our base rather than 2. But people did not know how to use fractional exponents back when logarithms were first developed. So they used the second approach.

Method Two: Choose a really, really small base. One of the first bases chosen was 1.0001. Using this approach we get

3 × 100 = 1.0001¹⁰⁹⁸⁶ + 1.0001⁴⁶⁰⁵⁴
= 1.0001⁵⁷⁰⁴⁰
= 300

I left off one really important step, which is finding the logarithms of 3 and 100 to the base of 2 or 1.0001. But there is no magic here. In olden days you simply looked these values up on a chart. Of course, working out the values on those charts was a time consuming process that took years. Log tables were worth more than their weight in gold. Eventually slide rules replaced charts. Nowadays we simply use a calculator.

Exercises

1. multiply 3 by 100 using natural logs.

The Product Rule Really is Intuitive

One of the great geniuses of mathematics is Gottfried Liebniz. He discovered calculus along with Newton, and created the superior form of notation that we use today. He also flushed out the basics of calculus. And when he wasn’t doing calculus, he was working on philosophy. Liebniz was one of the greatest philosophers of the Enlightenment. But even the greatest geniuses stumble. Einstien divided by zero in his theory of relativity and missed out on predicting the Big Bang. And Liebniz got stuck for years because he expected that the derivative of u(x)v(x) should be u’(x)v’(x). Unfortunately the real answer is a lot more complicated:

derivative of u(x)v(x) = u’(x)v(x) + u(x)v’(x)

Apparently Liebniz got the correct answer but didn’t believe it was right. So he spent years trying to redo the problem before he accepted the truth. However, here we will learn why the real answer makes all the intuitive sense in the world.

When you multiply two lengths together you get a rectangle. So don’t think about u(x)v(x) as one value, think of it as a rectangle. Now consider what happens to the change of the area of the rectangle as u(x) and v(x) change, which is shown in the picture below.

The change in area = u’(x)v(x) + u(x)v’(x) + u’(x)v’(x)

However, when the change of x is small, which is the case when you are taking a derivative, then the far diagonal corner of the rectangle effectively disappears. So the last term of u’(x)v’(x) goes to zero. So you are left with the product rule:

derivative of u(x)v(x) = u’(x)v(x) + u(x)v’(x)

A Simple Test

Lets try a simple test of the product rule. Find the derivative of (2x + 1)(3x + 4) using the product rule, then by expanding and taking the power rule of the terms.

Product rule = 2(3x + 4) + (2x + 1)3 = 12x + 11

Now using the power rule. First we expand:

(2x + 1)(3x + 4) = 6x² + 11x + 4

Then apply the power rule to get:

12x + 11

Both answers are the same. So we can use the the product rule.

The Quotient Rule

The Product Rule also leads directly to the quotient rule.

v(x)÷u(x) = v(x)u(x)^-1

Then apply the product rule and the chain rule:

v’(x)u(x)^-1 + v(x)u’(x)(-1)u(x)^-2

     v’(x)u(x) – v(x)u’(x)
= ——————————-
        u(x)²

Fractional Exponents

If you are comfortable with implicit differentiation then you can now handle fractional exponents.

y = xⁿ = x^p/q
y^p = x^q

This is the essential insight – rewrite xⁿ as x^p/q and then take the qth power of each side. At that point it is just a matter of taking the derivative of both sides (using implicit differentiation) and then collecting the terms. Ultimately you get the expected answer, which that the derivative of x^p/q is ^p⁄_qx^{p/q – 1}

The Fundamental Theorem of Calculus

Calculus has given us two powerful techniques. Integrals let us find the area under a graph, and derivatives find its slope, or rate of change. You might wonder if it is possible to combine these two concepts, and find the rate of change of the area under a graph.

The method of doing this is almost identical to finding the derivative. The secret to finding derivatives was to find the rate of change of two points, and let those points get closer and closer together until they were, more or less, identical. To find the rate of change of the integral we will find the area under the region defined by two points on the graph. Then let those points get closer and closer together until they are, more or less, the same point.

Look at the above graph of a generic function. Think of what happens if we add the interval from x to x + h. As h gets smaller and small the interval we add gets skinnier and skinnier. When it becomes infinitely thin the area we add is simply equal to the height of the graph. And that value is f(x). The derivative of an integral is the original function!

Lets see if we can use this insight to find the integral. Suppose we take the function f(x) = x². We now know that the derivative of the integral is x². So what function has a derivative of x²? Try and guess (really, grab a pencil and write it down).

The answer is ¹⁄₃x³. Apply the power rule and you’ll see that it works out.

If we apply differentiation in reverse we find the following.

Let f(x) = axⁿ
F(x) = integral of f(x) = ¹⁄_n+1axⁿ⁺¹

The Fundamental theorem of calculus: the derivative of the integral of f(x) is f(x). Similarly, the integral of the derivative of f(x) is also f(x).

The fundamental theorem of calculus says that we’ll never need to take a Riemann sum again! Instead well simply reverse the process of differentiation. This is called the anti-derivative.

Applications

Lets apply this process to the problem we did with Riemann sums. One of them was finding the area under f(x)=x² from 0 to 10.

The integral F(x) = ¹⁄₃x³.

Area from 0 to 10 = F(10) – F(0) = 1000/3 – 0 = 333.333

That is the exact answer we got before. And the formula clears up a minor mystery, which is why it seemed suspicious that the answer was exactly one-third of 1000.

There is another detail that may seem mysterious, which is taking f(10) – f(0). But F(x) gives the area under the graph throughout its entire graph up to point x. f(x) could be thought of as taking the area under the graph at precisely the point x. But we don’t want the area under the entire graph, we only want the area from x = 0 to x = 10. So you subtract one value from the other to get the region in which you are interested.

Now that we have the fundamental theorem of calculus, we can find the integrals of all types of functions, such as f(x) = x⁴⁸⁴. That is a heck of a lot easier than Riemann sums!

The Ease of Derivatives

Suppose you are worried that your speedometer is incorrectly calibrated. So you drive at a constant speed and count mile markers on the freeway. After three hours you count 180 mile markers. How fast are you going? This is an easy problem because your speed is constant, but it can easily get more difficult. Once you start adding in acceleration it gets more difficult. And that is where derivatives help.

Integrals tell us how far we have travelled. Derivatives tell us how fast we are travelling. If you look at a graph, the derivative corresponds to the rate of change of the function. It is a measure for how steep the graph is at a given point.

Finding the Derivative

Lets start with y = 3x. You probably already know the rate of change, or slope, of this graph. It is 3. But you could easily find out by picking two arbitrary points and calculating the difference in values of the function by the values of the points themselves. Lets pick 1 and 5 as our two points.

slope = (3×5 – 3×1) ÷ (5 – 1) = 12 ÷ 4 = 3

We run into an immediate problem if we try this on the graph of a parabola such as y = x². The rate of change is different depending on which points you choose. However, there is an solution. Just pick to points that are very close together. We want to find the limit at point x, so lets make the other point x + h, with h being the distance between the two points.

slope = [ (x + h)² - x² ] ÷ h
= [ x² + 2hx + h² - x² ] ÷ h
= [ 2hx + h² ] ÷ h
= 2x + h
= 2x

Since we want the two points to be as close together as possible, we let h = 0 and the answer is 2x. (Yes, this blithely ignores all the rigor of limits and continuity, but the goal is to understand the concept, not to get bogged down with limits and epsilon-deltas. But rest assured that it does all work out).

The Power Rule

Now lets try y = x³

slope = [ (x + h)³ - x³ ] ÷ h
= [ x³ + 3hx² + 3xh³ + h³ - x³ ] ÷ h
= [ 3hx² + 3xh² + h³] ÷ h
= 3x² + 3xh + h²
= 3x²

Once again, we are relying on letting h = 0 while ignoring the issue of limits and continuity.

Now lets try y = x⁴

slope = [ (x + h)⁴ - x⁴ ] ÷ h
= [ x⁴ + 4hx³ + 6x²h² 4xh³ + h⁴ - x⁴ ] ÷ h
= [ 4hx³ + 6x²h² + 3xh³ + h⁴] ÷ h
= 4x³ + 6x²h + 4xh² + h³
= 4x³

Do you see the pattern? Here are what we have found. If the slop

slope( x² ) = 2x
slope( x³ ) = 3x²
slope( x⁴ ) = 4x³

Generally, the slope of xⁿ = nx^n-1. This is called the power rule. This can be proven formally using mathematical induction.

Lets find the derivative for 7x².

slope = [ 7(x + h)² - 7x² ] ÷ h
= 7[ x² + 2hx + h² - x² ] ÷ h
= 7[ 2hx + h² ] ÷ h
= 7 [2x + h]
= 14x

This leads to a more general version of the power rule.

slope( axⁿn) = n·ax^n-1

This more general version should technically be proven using mathematical induction, but hopefully you see from how we were immediately able to factor the 7 in the previous example that the pattern should hold.

Exercises

1. Calculate the derivate of 14x⁹
2. Calculate the derivate of 4x²⁹
3. Go through the method of finding a derivative for y = 3x³. Then compare your answer to the power rule.

The Power of Riemann Sums

In the last post we learned the secret behind the technique of Riemann sums. But in order to keep the math simple, we only applied it to easy problems such as finding the area under y = x. Now we will show off the power of this new technique by finding the area under y = x² from the interval from 0 to 10.

The first approximation uses 5 rectangles. The width of each rectangle is 10÷5. The height of rectangle one is (10÷5)². The height of rectangle two is (2·10÷5)², and so on. We’ve already factored the width in the formula below. Make sure that you stare at it long enough to get the idea.

area = (10÷5)[(1·10÷5)² + (2·10÷5)² + ... + (5·10÷5)²]

Now let’s do it for n rectangles. The width of each rectangle is 10÷n. Notice how each n replaces each 5 in the previous formula.

area = (10÷n)[1·10÷n)² + (2·10÷n)² + ... + (n·10÷n)²]
= (10÷n)[1·(100÷n²) + 2²·(100÷n²) + ... + n²·(100÷n²)]
= (1000÷n³)[1 + 2² + 3² + ... + n²]

Now we are at crux of the problem. In the easier cases we looked at in the last post we had to sum up the first n integers. Now we have to sum up the first n integers squared. This is not quite as intuitive (although there is a fairly intuitive geometric way to visualize this using blocks) so I’ll just give the formula, which is n(n+1)(2n+1)÷6. Put that in our equation:

area = (1000÷n³)n(n+1)(2n+1)÷6

Grab an pencil and work out the algebra and you get:

area = 166.666 ( 2 + ³⁄_n + ¹⁄_n²)

since n is infinite, the answer is 166.666 × 2 = 333.333

Our new technique has born fruit. We have now advanced our capabilities and can solve problems that we could not solve before.

Foreshadowing the Fundamental Theorem of Calculus

The answer we got is suspicious: 333.333. We were looking at a graph inside a 10×100 box and we got an answer that is exactly one third the total area. That was with the function y=x². When we work with y = x the area under the graph is exactly half the area of the box. Hmmmmmm. If the formula were y = x³ would the area by one fourth? And would it be one-fifth for y = x⁴? Hold that thought, because we will revisit it with the fundamental theorem of calculus.

Let me close with a final promise. The technique of Riemann sums is a complicated and roundabout. But I hope that you worked out the arithmetic enough to understand the theory; it will deepen your understanding in the long run. After this we will never need to take a Riemann sum again. Instead we will leverage the our knowledge of differentiation by employing the fundamental theorem of calculus and get our answers much more easily.

Exercises

1. Find the area under the graph y = 3x² from 4 to 8.
2. Find the area under the graph y = x³. Hint: the formula for the first n cubes is (n(n+1)/2)².

The Brute Force Method of Integration

The best way to appreciate the power of integration is to begin with what you already know how to do. Suppose you were driving in your car at a steady rate of 60 miles per hour for 5 hours. How far did you travel? The answer is easy, multiply the speed by the time to get 300 miles. But let’s think about what is going on geometrically. If you graph the speed of the car you have the graph y = 60. When you multiplied 60 by 5, you found the area of the rectangle with a height of 60 and a width of 5. You found the area under the graph in the interval of (0,5) on the x-axis.

That was an easy integral; we can make this more difficult. Suppose you are in a spaceship and you are accelerating. So your speed is not a constant of 60 mph, but it starts at zero and accelerates by 40 mph every hour. How far does the spaceship travel in 5 hours? Your speed is given by the formula y = 40x. The graph is still a straight line, but now the area under the graph makes a triangle rather than a rectangle. This is still easy to solve because we have a nice formula for the area of a triangle, which is one half the base times the height. The base is five hours, and the height is your maximum speed after five hours, which is 5 × 40 = 200. So the total distance traveled is ½ × 5 × 200 = 500 miles.

But now lets find a really difficult integral. Suppose the speed of your spaceship is given by y = 20x². Now how far do you travel in five hours? The answer is the area under the graph. But there is no nice geometric shape that matches the area under a parabola. So how do you solve this problem? That is why we need calculus. In particular, we will use a technique called Riemann sums.

The Secret of Riemann Sums

Look at the graph of y = x³ above. Clearly the area under the graph does not make a nice geometric shape like a triangle or a rectangle. But we can approximate the area using rectangles. This is called the method of Riemann sums. The more rectangles we use, the better the approximation. And here is the secret: we can get the exact answer by using an infinite number of rectangles. And believe it or not, we do have the mathematical tools to sum up the area of an infinite number of rectangles, at least in some cases.

A Simple Beginning: y = x

Lets start really simple to make the math easy. Lets find the area under y=x from 0 to 10. We already know the answer using the formula for the area under a triangle (one half base times height); it is 50. Buts let’s find it using the method of Riemann sums. We’ll start with just a few rectangles, and work our way up.

• Let’s start with 5 rectangles. Each rectangle has a width of 10÷5 = 2. The height increases with each successive rectangle:
  area = 2·2 + 2·4 + 2·6 + 2·8 + 2·10 = 60
• If we use ten rectangles of width 10÷10 = 1 we get:
  area = 1·1 + 1·2 + … * 1·10 = 55.
• If we use n rectangles of width 10÷n, then we get:
  area = (10÷n)·1·(10÷n) + (10÷n)·2·(10÷n) + … + (10÷n)·n·(10÷n)

You may want to stare at this last example for a bit until you appreciate the pattern. The first (10÷n) is the width of each rectangle. The next two terms are the heigaht, which is given by y=x. Rectangle 2 is twice as high as rectangle 1. Rectangle 3 is three times as high. Rectangle n is n times as high as rectangle 1. Grab a pencile and work out the algebra. It works out to the following:

area = (100÷n²)(1 + 2 + 3 + 4 + … + n)

Now we are at the secret of Riemann sums. In order to find the area, we need to find a way to add up the first n integers. The solution is straightforward. Suppose we have to add up the first 100 integers. Do this by pairing them up. Add 1 and 100 to get 101. Add 2 and 99 to get 101. Add 3 and 98 to get 101. There are 50 pairs that total up to 101. So the answer is 50(101). The formula for the sum of the first n integers is ½ n (n + 1). Lets use that and rewrite the formula.

area = (100 ÷ n²) ⁿ⁄2 (n + 1)

Do the algebra (grab a pencil and work it out) and you get:

area = 50 + ⁵⁰⁄n

Since we have an infinite number of rectangles, the area is 50. Exactly what we got before! We have a new technique for finding the area under the graph.

Exercises

1. Use the method of Riemann sums to find the area under the graph of y = 4x from 0 to 20. Compare to the answer given by the formula for area under a triangle.
2. Use the method of Riemann sums to find the area under the graph of y = 4x from 10 to 25 (hint: this does not make a triangle, but you can solve it by calculating the area for a large triangle with a base of 25 minus a small triangle with a base of 10). Compare to the answer given by the direct geometric method.
3. Use the method of Riemann sums to find the area under the graph of y = 7 from 0 to 15 (hint: this is actually an easier problem since the area under the graph is a rectangle rather than a triangle. It requires a different, but easier, formula for summing up an infinite series)