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Filed under: The Complex Plane — Justin @ 7:59 pm

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.

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Filed under: Calculus — Justin @ 3:23 pm

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

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Filed under: Uncategorized — Justin @ 6:17 pm

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.

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