# Representations of functions as power series

## Differentiation and integration of power series

• Theorem. Let $$f(x)=\sum_{n=0}^\infty c_nx^n$$ be a power series with radius of convergence $$R$$. Then the following are true.
• For $$a-R<x<a+R$$, we have $f'(x)=\sum_{n=0}^\infty c_n(x^n)'=\sum_{n=1}^\infty c_n nx^{n-1}$
• For $$a-R<x<a+R$$, we have $\int f(x)\,\mathrm dx=\sum_{n=0}^\infty c_n\int x^n\,\mathrm dx=C+\sum_{n=0}^\infty c_n\frac{n^{n+1}}{n+1}.$
• In other words, for $$x$$ for which a power series is absolute convergent, you can derivate or integrate into the infinite sum.
• Warning 1 Checking for absolute convergence is a must. Otherwise, the formula might not work. Mathematicians like to make fun of Physicists for not checking for absolute convergence here.
• Warning 2 The formula only holds in the open interval $$(a-R,a+R)$$. Even if a boundary point is in the domain, the formula might not hold there.
• This is the formula we were waiting for.
• It shows that if a function is given by a series, then in the interior of its domain (that is, excluding the boundary points) its antiderivative is also going to be a power series.
• Now we will see how to write elementary functions as power series.
• In this section, we will look at special cases, at which we can use the convergent geometric series formula.
• In the next section, we will see how to convert an infinitely differentiable function to a power series in general.

## Using the convergent geometric series formula to convert some elementary functions into power series.

• Recall that the power series $$f(x)=\sum_{n=0}^\infty x^n$$ has domain $$(-1,1)$$, and in its domain, we have $$f(x)=\frac{1}{1-x}$$.
• Consider now the function $$\frac{1}{1-x^3}$$.
• We can write $\frac{1}{1-x^3}=\sum_{n=0}^\infty x^{3n}.$
• We get a power series which is also a geometric series. Its radius of convergence is 1.
• Consider the function $$\frac{1}{x-3}$$.
• We can write $\frac{1}{x-3}=-\frac{1}{3-x}=-\frac13\frac{1}{1-\frac{x}{3}}=-\frac13\sum_{n=0}^\infty\frac{x^n}{3^n}=-\sum_{n=0}^\infty\frac{x^n}{3^{n+1}}.$
• This time, the radius of convergence is 3.
• Consider the function $$\frac{x^2}{x^2-1}$$.
• We can write $\frac{1}{x^2-1}=-\frac{1}{1-x^2}=-\sum_{n=0}^\infty x^{2n}.$ The radius of convergence is 1.
• Therefore, we get $\frac{x^2}{x^2-1}=x^2\frac{1}{x^2-1}=-x^2\sum_{n=0}^\infty x^{2n}=-\sum_{n=0}^\infty x^{2n+2}.$
• Consider the function $$\frac{1}{(x+3)^2}$$.
• We have $$\frac{1}{(x+3)^2}=\left(-\frac{1}{x+3}\right)'$$, so we first express the latter as a power series, and then derivate into it: $-\frac{1}{x+3}=-\frac13\frac{1}{1-\left(-\frac{x}{3}\right)}=\sum_{n=0}^\infty(-1)^{n+1}\frac{x^n}{3^{n+1}}.$ The radius of convergence is 3.
• This gives: $\frac{1}{(x+3)^2}=\left(-\frac{1}{x+3}\right)'=\sum_{n=0}^\infty\left((-1)^{n+1}\frac{x^n}{3^{n+1}}\right)'=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^{n-1}}{3^{n+1}n}.$
• Consider the function $$\ln(2+x)$$.
• Since $$\int\frac{1}{2+x}\,\mathrm dx=\ln|2+x|+c$$, let's first convert the rational function, then integrate it.
• We have $\frac{1}{2+x}=\frac12\frac{1-\left(-\frac{x}{2}\right)}=\sum_{n=0}^\infty(-1)^{n+1}\frac{x^n}{2^{n+1}}.$
• The radius of convergence is 2. Since if $$-2<x<2$$ then $$0<2+x<4$$, on this interval we have $$|2+x|=2+x$$.
• Therefore, we get $\ln(2+x)=\int\frac{1}{2+x}\,\mathrm dx=\sum_{n=0}^\infty\int(-1)^{n+1}\frac{x^n}{2^{n+1}}\,\mathrm dx=\sum_{n=0}^\infty(-1)^{n+1}\frac{x^{n+1}}{2^{n+1}(n+1)}.$
• The radius of convergence is the same as the original: 2.
• Exercises. 11.9: 1, 2, 5, 7, 11, 13, 15, 17, 24, 25, 28