Power Series

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 11.8 – Power Series

• Section 11.9 – Representations of Functions as Power Series

Expected Educational Results

• Objective 24–1: I can use the Geometric Series to rewrite functions as power series.

• Objective 24–2: I can differentiate a power series written in summation notation.

• Objective 24–3: I can explain why the index changes when differentiating a power series written in summation notation.

• Objective 24–4: I can integrate a power series written in summation notation.

• Objective 24–5: I can use the geometric series to find the radius of convergence, R, of a power series.

• Objective 24–6: I can use the geometric series to find the interval of convergence, I, of a power series.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Power Series

Derivatives and Antiderivatives of Power Series

Definition: Derivative of a Power Series

${\displaystyle {\displaystyle \frac{d}{dx}}\left[\sum _{n=0}^{\mathrm{\infty }}{c}_n(x-a{)}^n\right]=\sum _{n=1}^{\mathrm{\infty }}n{c}_n(x-a{)}^{n-1}}$$\displaystyle\dfrac{d}{dx}\left[\sum_{n=0}^{\infty}{c_n(x-a)^n}\right]=\sum_{n=1}^{\infty}{nc_n(x-a)^{n-1}}$

NOTE: Differentiating a power series shifts the index of the series.

Definition: Integral of a Power Series

${\displaystyle \int \left[\sum _{n=0}^{\mathrm{\infty }}{c}_n(x-a{)}^n\right]dx=C+\sum _{n=0}^{\mathrm{\infty }}{c}_n\frac{(x-a{)}^{n+1}}{n+1}}$$\displaystyle\int{\left[\sum_{n=0}^{\infty}{c_n(x-a)^n}\right]\,dx}=C+\sum_{n=0}^{\infty}{c_n\frac{(x-a)^{n+1}}{n+1}}$

NOTE: Remember: Antiderivatives require the $+C$$+C$.

Investigation 07

Find the derivatives of the following power series:

1. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{\sqrt{n}}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{x^n}{\sqrt{n}}}$

2. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n^3}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{x^n}{n^3}}$

3. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}(-1{)}^n{4}^n{x}^n}$$\displaystyle \sum_{n=0}^{\infty}{(-1)^n4^nx^n}$

4. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(x-3{)}^n}{2n+1}}$$\displaystyle \sum_{n=0}^{\infty}{\frac{(-1)^n(x-3)^n}{2n+1}}$

5. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(2x-1{)}^n}{5^n\sqrt{n}}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{(2x-1)^n}{5^n\sqrt{n}}}$

Investigation 08

Find the antiderivatives of the following power series:

1. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{\sqrt{n}}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{x^n}{\sqrt{n}}}$

2. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n^3}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{x^n}{n^3}}$

3. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}(-1{)}^n{4}^n{x}^n}$$\displaystyle \sum_{n=0}^{\infty}{(-1)^n4^nx^n}$

4. ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(x-3{)}^n}{2n+1}}$$\displaystyle \sum_{n=0}^{\infty}{\frac{(-1)^n(x-3)^n}{2n+1}}$

5. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(2x-1{)}^n}{5^n\sqrt{n}}}$$\displaystyle \sum_{n=1}^{\infty}{\frac{(2x-1)^n}{5^n\sqrt{n}}}$

Investigation 09

Find the power series for $f^{\prime }(x)$$f^{\,\prime}(x)$:

1. ${\displaystyle f(x)={\tan }^{-1}(x)}$$\displaystyle f(x)=\tan^{-1}(x)$

2. ${\displaystyle f(x)=\ln (1+x)}$$\displaystyle f(x)=\ln{(1+x)}$

Investigation 10

Find the power series for $f(x)$$f(x)$:

1. ${\displaystyle f(x)={\tan }^{-1}(x)}$$\displaystyle f(x)=\tan^{-1}(x)$

2. ${\displaystyle f(x)=\ln (1+x)}$$\displaystyle f(x)=\ln{(1+x)}$

Investigation 11

Find the power series for $\int f(x)dx$$\int{f(x)\,dx}$:

1. ${\displaystyle f(x)=\frac{1}{(1+x{)}^2}}$$\displaystyle f(x)=\frac{1}{(1+x)^2}$

Investigation 12

Find the power series for $f(x)$$f(x)$:

1. ${\displaystyle f(x)=\frac{1}{(1+x{)}^2}}$$\displaystyle f(x)=\frac{1}{(1+x)^2}$

Homework

At this time, you should be able to complete the following assignments:

• Section 11.9: # 13, 17, 25.

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Created: Tuesday, 1 December 2020 7:38 EDT Last Modified: Tuesday, 30 May 2023 – 23:12 (EDT)