*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.

- It shows that if a function is given by a series, then in the

- Recall that the power series \(f(x)=\sum_{n=0}^\infty x^n\) has domain \((-1,1)\), and in its domain, we have \(f(x)=\frac{1}{1-x}\).
- Recall also that the partial sums give approximations for \(f(x)\). Click here for an illustration

- 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