Power Series

Final exam: Wednesday, March 22, 7-10pm in PCYNH 109. Bring ID!
Quiz 4: This Friday
Today: 11.8 Power Series, 11.9 Functions defined by power series
Next: 11.10 Taylor and Maclaurin series

Recall that a polynomial is a function of the form

$$f(x) = c_0 + c_1 x + c_2 x^2 + \cdots + c_k x^k.$$

Polynomials are easy!!!

They are easy to integrate, differentiate, etc.:

$$\frac{d}{dx} \left(\sum_{n=0}^k c_n x^n\right) = \sum_{n=1}^k n c_n x^{n-1}$$

$$\int \sum_{n=0}^k c_n x^n dx = C + \sum_{n=0}^k c_n \frac{x^{n+1}}{n+1}.$$

Definition 6.5.1 (Power Series)   A power series is a series of the form

$$f(x) = \sum_{n=0}^{\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + \cdots,$$

where $x$ is a variable and the $c_n$ are coefficients.

A power series is a function of $x$ for those $x$ for which it converges.

Example 6.5.2   Consider

$$f(x) = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots.$$

When $\vert x\vert < 1$, i.e., $-1 < x < 1$, we have

$$f(x) = \frac{1}{1-x}.$$

But what good could this possibly be? Why is writing the simple function $\frac{1}{1-x}$ as the complicated series $\sum_{n=0}^{\infty} x^n$ of any value?

1. Power series are relatively easy to work with. They are ``almost'' polynomials. E.g.,
   $$\frac{d}{dx} \sum_{n=0}^{\oo } x^n = \sum_{n=1}^{\oo } nx^{n-1} = 1 + 2x + 3x^2 + \cdots = \sum_{m=0}^{\oo } (m+1)x^m,$$
   where in the last step we ``re-indexed'' the series. Power series are only ``almost'' polynomials, since they don't stop; they can go on forever. More precisely, a power series is a limit of polynomials. But in many cases we can treat them like a polynomial. On the other hand, notice that
   $$\frac{d}{dx}\left( \frac{1}{1-x} \right)= \frac{1}{(1-x)^2} = \sum_{m=0}^{\infty} (m+1)x^m.$$

2. For many functions, a power series is the best explicit representation available.
   Example 6.5.3   Consider $J_0(x)$, the Bessel function of order 0. It arises as a solution to the differential equation $x^2 y'' + x y' + x^2 y = 0$, and has the following power series expansion:

   $$J_0(x) = \sum_{n=1}^{\oo } \frac{(-1)^n x^{2n}}{2^{2n}(n!)^2}$$

   $$=1 - \frac{1}{4}x^{2} + \frac{1}{64}x^{4} - \frac{1}{2304}x^{6} + \frac{1}{147456}x^{8} - \frac{1}{14745600}x^{10} + \cdots.$$

   
   

   This series is nice since it converges for all $x$ (one can prove this using the ratio test). It is also one of the most explicit forms of $J_0(x)$.