One of the things that you may recall is that the log of 1 plus x is equal to the

integral of 1 over 1 plus x, dx. If you remember that, great.

If you don't, don't worry about it. Accept it as a given and let's move on

from there. What could we do with that integrand, 1

over 1 plus x? That is really the geometric series

evaluated at negative x, and so, we could integrate the sum k goes from 0 to

infinity of negative x to the k with respect to x.

By rearranging the terms, a thing that we are allowed to do as long as we are

within the convergence domain, and pulling out the negative 1 to the k, then

we get the sum, as k goes from 0 to infinity, of negative 1 to the k, times

the integral of x to the k. That is x to the k plus 1 or k plus 1 if

we add an arbitrary constant out in front, well, that's actually not going to

be so bad. Why?

When x equals 0 we obtain the log of 1 which is 0.

So forget that constant. And now, we have an answer with a bit of

re-indexing. As the sum k goes from one to infinity, a

negative 1 to the k plus 1 times x to the k over k.

That is perhaps more easily remember as x minus x squared over 2 plus X cubed over

3, et cetera. Notice there are no factorials here.

This is x to the k over k. Now are there any problems with this?

Well, since we used the geometric series in this derivation, we must be careful to

remain within the convergence domain for this.

x must be less than 1 in absolute value. This fits well with the principle that

Taylor polynomials approximate well only on the convergence domain.

If we look at the terms of this series as an approximation to log of 1 plus x and

we get increasingly something that seems to fit what the curve is doing.

But only on the domain of convergence as x goes to negative 1 to 1, that is, we're

looking at values of log from 0 to 2. Outside of this region these polynomial

curves do not become better and better approximations.

They in fact become worse and worse. Let us consider another example, this

time the arctan function, one of the things that you may recall is that arctan

has as its derivative 1 over 1 plus x squared.

So, we can integrate this to obtain arctan.

You will notice also that 1 over 1 plus x squared can be represented as the

geometric series of negative x squared. So, that we can integrate the sum of

negative x squared to the k. k goes from 0 to infinity.

Again, by rearranging terms doing a little bit of work here.

We see that we get the sum, k goes from 0 to infinity, negative 1 to the k times

the integral of x of 2k. That is x to the 2k plus 1 over 2k plus

1. We have to deal with the constant of

integration. But again, since arctan of 0 equals 0 it

is irrelevant. We therefore obtain the Taylor series, x

minus x cubed over 3. Plus x to the 5th over 5, minus x to the

7th over 7, et cetera. Once again, no factorials in those

denominators. And once again, it is important that x

remain less than 1 in absolute value in order for this series to converge,

because we used the geometric series as its basis.

There is one last series that will be very helpful to us.

This is the binomial series. It is the Taylor series for 1 plus x to

the alpha, where alpha is a constant. The series is 1 plus alpha x plus 1 over

2 factorial times alpha times alpha minus 1 times x squared.

Well, there are lot of other terms here. Let me say simply that the kth order term

is alpha choose k times x to the k, where by alpha choose k, I mean a binomial

notation. This is the number alpha, times alpha

minus 1, times alpha minus 2, all the way down to alpha minus k plus 1.

Take that product and divide by k factorial.

For example, if alpha equals 2, then what does the binomial series give you?

1 plus alpha times x plus 1 half alpha times alpha minus 1 times x squared and

that simplifies to x squared. All other terms involve an alpha minus 2

hence the coefficients are 0. And we obtain, of course, what the Taylor

series for this polynomial must be. When alpha equals negative 1, we're

really looking at a variation of the geometric series for 1 over 1 plus x.

This however, holds for all values of alpha.

For example, if we take alpha equal to 1 half.

So that we're looking at the square root of 1 plus x.

Then one can evaluate these binomial coefficients and obtain a Taylor series

for that. In this case, in particular this is only

going to remain valid for x less than 1 in absolute value.