Some calculus with power series.

Theorem 10.4.1   Suppose that the power series $ \underset{n=0}{\overset{+\infty }{\sum}} c_n (x-x_0)^n$ is convergent on the interval $ (x_0-R, x_0+R)$ . Its sum defines on that interval a function $ f$ of the variable $ x$ .

The function $ f$ has derivatives of any order $ n$ , and these derivatives are obtained by term-by-term differentiation, namely:

$\displaystyle f'(x)$ $\displaystyle =\underset{n=0}{\overset{+\infty }{\sum}} n c_n (x-x_0)^{n-1}$    
$\displaystyle f''(x)$ $\displaystyle =\underset{n=0}{\overset{+\infty }{\sum}} n(n-1)c_n (x-x_0)^{n-2} \dots$ $\displaystyle \quad$    

These power series have the same interval of convergence as thge original power series.

Theorem 10.4.2   Suppose that the power series $ \underset{n=0}{\overset{+\infty }{\sum}} c_n (x-x_0)^n$ is convergent on the interval $ (x_0-R, x_0+R)$ and denote its sum by $ f(x)$ . Then the power series $ \underset{n=0}{\overset{+\infty }{\sum}} \frac {c_n}{n+1} (x-x_0)^{n+1}$ is convergent on the same interval and we have:

$\displaystyle \forall x \in (x_0-R,x_0+R), \; \int f(x) \; dx = \underset{n=0}{\overset{+\infty }{\sum}} \frac {c_n}{n+1} (x-x_0)^{n+1} +C$    

Example 10.4.3 (A series for $ \arctan x$ )   Let $ f(x)= \underset{n=0}{\overset{+\infty }{\sum}} \frac {(-1)^n}{x^{2n}} = 1 - \frac {1}{x^2} + \frac {1}{x^4} - \dots$ . This series is abolutely convergent for $ \vert x\vert<1$ (v.s. Thm prop geometric series cv and Def def absolute cv). thus it is convergent; its sum is $ f(x)=\frac {1}{1-(-x^2)}=\frac {1}{1+x^2}$ .

We integrate the function:

$\displaystyle \int f(x) \; dx = \int \frac {dx}{1+x^2} = \arctan x +C.$    

Integrating the series term-by-term and equating both sides to 0 for $ x=0$ , we have:

$\displaystyle \forall x \in (-1,1), \; \arctan x = x - \frac 13 x^3 + \frac 15 ...
...- \dots = \underset{n=1}{\overset{+\infty }{\sum}} \frac {(-1)^{n+1}}{x^{2n+1}}$    

Theorem 10.4.4 (The product of two power series)   Suppose that the two power series $ F(x)=\underset{n=0}{\overset{+\infty }{\sum}} a_nx^n$ and $ G(x)=\underset{n=0}{\overset{+\infty }{\sum}} b_nx^n$ are absolutely convergent for $ \vert x\vert<R$ . Denote $ c_n=\underset{k=0}{\overset{n}{\sum}} a_kb_{n-k} = a_0b_n+a_1b_{n-1}+ \dots + a_nb_0$ .

Then the series $ \underset{n=0}{\overset{+\infty }{\sum}} c_n x^n$ is absolutely convergent for $ \vert x\vert<R$ and its sum is equal to $ F(x)G(x)$ ..

Example 10.4.5   Consider the two geometric series $ F(x)= \frac {1}{1+x} = \underset{n=0}{\overset{+\infty }{\sum}} (-1)^{n+1}x^n$ and $ G(x)= \frac {1}{1-x} = \underset{n=0}{\overset{+\infty }{\sum}} x^n$ , i.e.

$\displaystyle \forall n \in \mathbb{N}, \; a_n=(-1)^{n+1}$    and $\displaystyle b_n=1.$    

The product of these series is given by:

$\displaystyle c_n=a_0b_n+a_1b_{n-1}+ \dots + a_{n-1}b_1+a_nb_0 = -1+1-1+ \dots + (-1)^{n+1}$    

i.e. $ c_n=0$ for even $ n$ and $ c_n=1$ for odd $ n$ . We have:

$\displaystyle \frac {1}{1-x^2}=F(x)G(x)=1-x^2+x^4-x^6 + \dots$    

This could have been obtained either by a direct computation or by noting that

$\displaystyle \forall x \in \mathbb{R}-{-1,1}, \; \frac {1}{1-x^2} = \frac 12 \left( \frac {1}{1+x} +\frac {1}{1-x} \right).$    

Noah Dana-Picard 2007-12-28