The Formula for Taylor Series

We have computed power series representations for some functions, including the following.

$\begin{eqnarray} \frac{1}{1-x}&=& \sum_{n=0}^\infty x^n = 1+x+x^2+x^3+x^4+\cdots\\ \frac{1}{1+x} &=&  \sum_{n=0}^\infty (-x)^n = 1-x+x^2+\cdots\\ \frac{1}{(1-x)^2}&=&\sum_{n=1}^\infty n\, x^{n-1}=1+2x+3x^2+4x^3+\cdots\\ \ln(1+x)&=&\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}=x - \frac{x^2}{2} + \frac{x^3}{3} -\frac{x^4}{4}+ \cdots\\ \tan^{-1}(x)&=& \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}=x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\\  \end{eqnarray}$

All of these have radius of convergence $R=1$, which is a result of their geometric series origins.

You may be wondering if there are other series that can be represented by power series.  The answer is yes.  (A function is called analytic if it can be expressed as an infinite power series around some point $a$.) 

Taylor's theorem tells us how to find the coefficients of the power series expansion of a function. 

This says that if a function can be represented by a power series, its coefficients must be those in Taylor's Theorem.  This formula works both ways: if we know the $n$-th derivative evaluated at $a$, we can figure out $c_n$; if we know $c_n$, we can figure out the $n$-th derivative evaluated at $a$.  To use this theorem, we have the conventions that we define $f^{(0)}(x)$ to be $f(x)$, and $0!$ to be $1$.

(Notice that this series is centered around $a=0$.)  By Taylor's Theorem, $c_nx^n=x^n$ since $x^n$ are the terms of our series.  So $c_n=1$ for all $n$.  On the other hand, also by Taylor's Theorem, $c_n=\frac{f^{(n)}(0)}{n!}$, so we must have $f^{(n)}(0)=n!$ for all $n$ here.  Let's see if this is true.

We let $f(x)=\frac{1}{1-x}=(1-x)^{-1}$.  DO:  Find $f'(x),f''(x),f^{(3)}(x),f^{(4)}(x)$, etc., until you see a pattern.

$\begin{array}{lllll} f(x)&=&(1-x)^{-1}&&=\frac{1}{1-x}\\ f'(x)&=&-(1-x)^{-2}(-1)&=(1-x)^{-2}&=\frac{1}{(1-x)^2}\\ f''(x)&=&-2(1-x)^{-3}(-1)&=2(1-x)^{-3}&=\frac{2}{(1-x)^3}\\ f^{(3)}(x)&=&2(-3)(1-x)^{-4}(-1)&=2\cdot 3(1-x)^{-4}&=\frac{2\cdot3}{(1-x)^4}\\ f^{(4)}(x)&=&2\cdot 3(-4)(1-x)^{-4}&=2\cdot 3\cdot 4(1-x)^{-5}&=\frac{2\cdot3\cdot4}{(1-x)^5}\\ &\vdots&&&\vdots\\ f^{(n)}(x)=&&&&=\frac{n!}{(1-x)^{n+1}}\\ \end{array}$

We have found $f^{(n)}(x)$ and from this we will find $f^{(n)}(0)$.  Plugging in $x=0$ to our derivatives above, we see that $f^{(n)}(0)=n!$.  Then $$c_n=\frac{f^{(n)}(0)}{n!}=\frac{n!}{n!}=1, \text{ so that }c_nx^n=1\cdot x^n=x^n.$$ These are indeed the coefficients of our series, and Taylor's Theorem holds for that series.