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

$\displaystyle 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.:

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

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

$\displaystyle 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

$\displaystyle 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

$\displaystyle 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.,

    $\displaystyle \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

    $\displaystyle \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:

    $\displaystyle J_0(x)$ $\displaystyle = \sum_{n=1}^{\oo } \frac{(-1)^n x^{2n}}{2^{2n}(n!)^2}$    
      $\displaystyle =1 - \frac{1}{4}x^{2} + \frac{1}{64}x^{4} - \frac{1}{2304}x^{6} + \frac{1}{147456}x^{8} - \frac{1}{14745600}x^{10} + \cdots.$    

    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)$.

William Stein 2006-03-15