Radius of Convergence

Section: 
9.5
Date: 
Friday, October 8, 2010 - 14:00 - 15:00
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fall2010math2300_power-series.pdf59.93 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Power Series and Radius of Converges – October 8, 2010

Power Series

Definition: A power series is an infinite series of the form

f(x)=\sum _{{n=0}}^{\infty}c_{n}(x-a)^{n}=c_{0}+c_{1}(x-a)^{1}+c_{2}(x-a)^{2}+\cdots

where

  • c_{n} represents the coefficient of the nth term.

  • a is some (fixed) real number.

  • x varies around a, and so we sometimes say that the power series is “centered” at a.

Radius of Convergence

Theorem: For a power series \sum _{{n=0}}^{\infty}c_{n}(x-a)^{n}, exactly one of the following happens:

  1. The series converges for a single value, when x=a. (Zero radius of convergence)

  2. The series converges for all x. (Infinite radius of convergence)

  3. There is a positive number R such that the series converges whenever |x-a|<R and diverges whenever |x-a|>R. (Finite non-zero radius of convergence. Check the endpoints of your interval of convergence to determine if the endpoints converge.)

Finding the Radius

  • We use the ratio test to determine the radius of convergence. Specifically, the ratio test says that the series will converge if

    \lim _{{n\rightarrow\infty}}\frac{\displaystyle|c_{{n+1}}(x-a)^{{n+1}}|}{\displaystyle|c_{n}(x-a)^{n}|}<1.
  • Notice that this is the same as

    \lim _{{n\rightarrow\infty}}\frac{\displaystyle|c_{{n+1}}|\cdot|(x-a)^{n}|\cdot|(x-a)|}{\displaystyle|c_{n}|\cdot|(x-a)^{n}|}=|(x-a)|\lim _{{n\rightarrow\infty}}\frac{|c_{{n+1}}|}{|c_{n}|}<1
  • The key point to understand here is that we know the series will always converge when x=a, since this makes all but the first terms in the series zero. As the series is “centered at a,” the radius of convergence is “centered at a.” That is, the quantity |x-a| in the calculation above denotes distance between x and a. If the limit is some fixed finite number (it often is), the multiplication by |x-a| will determine if the inequality is satisfied…depending on how big |x-a| is. Thus, the radius r satisfying

    \lim _{{n\rightarrow\infty}}\frac{|c_{{n+1}}|}{|c_{n}|}=\frac{1}{r}

    will be the maximum value that |x-a| may take on and still satisfy the inequality. This value of r is the “radius of convergence.”

  • Notice that you only need to consider the coefficients c_{n} in order to determine the radius of convergence. That is you don't always need to plug in the (x-a)^{n} portion of the series terms into the ratio test in order to calculate the radius of convergence.

Examples:

  1. Recall that 0!=1. Find the radius of convergence of

    \sum _{{n=0}}^{\infty}\frac{x^{n}}{n!}

    Solution: This is a power series with center zero. That is

    \sum _{{n=0}}^{\infty}\frac{x^{n}}{n!}=\sum _{{n=0}}^{\infty}\frac{(x-0)^{n}}{n!}=\sum _{{n=0}}^{\infty}c_{n}(x-0)^{n}\qquad c_{n}=\frac{1}{n!}.

    By the ratio test (as described above), we have

    \lim _{{n\rightarrow\infty}}\frac{|c_{{n+1}}|}{|c_{n}|}=\lim _{{n\rightarrow\infty}}\frac{\left|\frac{1}{(n+1)!}\right|}{\left|\frac{1}{n!}\right|}=\lim _{{n\rightarrow\infty}}\frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}=\lim _{{n\rightarrow\infty}}\frac{n!}{(n+1)!}=\lim _{{n\rightarrow\infty}}\frac{1}{n+1}=0.

    This implies

    0=\frac{1}{r}

    where the radius r is some positive number, implying that r=\infty. In this situation, the radius of convergence is infinite. This implies that the series converges everywhere on the real number line.

  2. Find the radius of convergence of

    \sum _{{n=0}}^{\infty}\frac{(-3)^{n}x^{n}}{\sqrt{n+1}}
  3. Find the radius of convergence of

    1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^{2}+\frac{1}{16}(x-1)^{3}-\frac{1}{32}(x-1)^{4}+\cdots

© 2011 Jason B. Hill. All Rights Reserved.