Power Series
- Page ID
- 219492
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definition of a Power Series
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Definition of a Power Series |
More generally, if f(x) is represented by the series
\( \displaystyle\sum_{n=0}^{\infty} a_n (x - c)^n \)
Then we call f(x) a power series centered at x = c. The domain of f(x) is called the Interval of Convergence and half the length of the domain is called the Radius of Convergence.
The Radius of Convergence
To compute the radius of convergence, we use the ratio test.
Example: Find the radius of convergence of
\( f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(x-3)^n}{2^n} \)
Solution: We use the Ratio Test:
\( \lim\limits_{n\to \infty} |\frac{(x-3)^{n+1}}{2^{n+1}}\frac{2^n}{(x-3)^n)}| = \lim\limits_{n\to \infty} |\frac{x-3}{2}| \)
We solve
or
|x - 3| < 2
so that
1 < x < 5
Since
1
(5 - 1) = 2
2
the radius of convergence is 2. Notice that we could have use the geometric series test and obtained the same result. The ratio test is the most likely test to work, but occasionally another test such as the geometric series test or the root test is easier to use.
Exercise:
Find the radius of convergence of
\( f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(x-2)^n}{n+1} \)
Interval of Convergence
To find the interval of convergence we follow the three steps:
-
Use the ratio test to find the interval where the series is absolutely convergent.
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Plug in the left endpoint to see if it converges at the left endpoint. (AST may be useful).
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Plug in the right endpoint to see if it converges at the right endpoint. (AST may be useful).
Example:
Find the interval of convergence for the previous example:
\( f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(x-3)^n}{2^n} \)
Solution:
-
We have already done this step and found that the series converges absolutely
for 1 < x < 5. -
We plug in x = 1 to get
\( f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(1-3)^n}{2^n} = \displaystyle\sum_{n=0}^{\infty}(-1)^n \)
This series diverges by the limit test. -
We plug in x = 5 to get
\( f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(5-3)^n}{2^n} = \displaystyle\sum_{n=0}^{\infty}1 \)
This series also diverges by the limit test.
Hence the endpoints are not included in the interval of convergence. We can conclude that the interval of convergence is
1 < x < 5
Exercise
Find the interval of convergence of the previous exercise:
\( f(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(x-2)^n}{n+1} \)
Differentiation and Integration of Power Series
Since a power series is a function, it is natural to ask if the function is continuous, differentiable or integrable. The following theorem answers this question.
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Theorem \[ \frac{d}{dx}f(x) = \frac{d}{dx}\displaystyle\sum_{n=0}^{\infty} a_n (x - c)^n \] \[ = \displaystyle\sum_{n=0}^{\infty} \frac{d}{dx}a_n (x-c)^n = \displaystyle\sum_{n=1}^{\infty} n a_n (x-c)^{n-1} \]
and \[ \int f(x)dx = \int\displaystyle\sum_{n=0}^{\infty} a_n (x - c)^n dx \] \[ = \displaystyle\sum_{n=0}^{\infty} \int a_n (x-c)^n dx = \displaystyle\sum_{n=1}^{\infty} a_n \frac{(x-c)^{n+1}}{n+1} + C \]
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Example:
Consider the series
\( f(x) = \displaystyle\sum_{n=0}^{\infty} x^n \)
by the GST this series converges for |x| < 1, hence the center of convergence is 0 and the radius is 1. By the above theorem
\( f'(x) = \displaystyle\sum_{n=1}^{\infty} nx^{n-1} \)
has center of convergence 0 and radius of convergence 1 also. We can also say that
\( \int f(x) dx = \displaystyle\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} + C \)
also has center of convergence 0 and radius of convergence 1.