# 11.9: Calculus with Power Series

Now we know that some functions can be expressed as power series, which look like infinite polynomials. Since calculus, that is, computation of derivatives and antiderivatives, is easy for polynomials, the obvious question is whether the same is true for infinite series. The answer is yes:

## Theorem 11.9.1 |

Suppose the power series \[f(x)=\sum_{n=0}^\infty a_n(x-a)^n\] has radius of convergence \(R\). Then \[\eqalign{ f'(x)&=\sum_{n=0}^\infty na_n(x-a)^{n-1},\cr \int f(x)\,dx &= C+\sum_{n=0}^\infty {a_n\over n+1}(x-a)^{n+1},\cr }\] and these two series have radius of convergence \(R\) as well. |

**Example 11.9.2 **Starting with the geometric series:

\[\eqalign{ {1\over 1-x} &= \sum_{n=0}^\infty x^n\cr \int{1\over 1-x}\,dx &= -\ln|1-x| = \sum_{n=0}^\infty {1\over n+1}x^{n+1}\cr \ln|1-x| &= \sum_{n=0}^\infty -{1\over n+1}x^{n+1}\cr }\]

when \(|x| < 1\). The series does not converge when \(x=1\) but does converge when \(x=-1\) or \(1-x=2\). The interval of convergence is \([-1,1)\), or \(0 < 1-x\le2\), so we can use the series to represent \(\ln(x)\) when \(0 < x\le2\).

For example

\[ \ln(3/2)=\ln(1--1/2)= \sum_{n=0}^\infty (-1)^n{1\over n+1}{1\over 2^{n+1}} \]

and so

\[ \ln(3/2)\approx {1\over 2}-{1\over 8}+{1\over 24}-{1\over 64} +{1\over 160}-{1\over 384}+{1\over 896} ={909\over 2240}\approx 0.406 .\]

Because this is an alternating series with decreasing terms, we know that the true value is between \(909/2240\) and \(909/2240-1/2048=29053/71680\approx .4053\), so correct to two decimal places the value is \(0.41\).

What about \(\ln(9/4)\)? Since \(9/4\) is larger than 2 we cannot use the series directly, but \(\ln(9/4)=\ln((3/2)^2)=2\ln(3/2)\approx 0.82,\) so in fact we get a lot more from this one calculation than first meets the eye. To estimate the true value accurately we actually need to be a bit more careful. When we multiply by two we know that the true value is between \(0.8106\) and \(0.812\), so rounded to two decimal places the true value is \(0.81\).