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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:

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:

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:

Exercises

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

June 7, 2007

Making Sense of Logarithms

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.

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

multiply 3 by 100 using natural logs.

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June 6, 2007

The Product Rule Really is Intuitive

Filed under: Calculus — Justin @ 8:28 pm

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}

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The Fundamental Theorem of Calculus

Filed under: Calculus — Justin @ 5:22 pm

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.

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