W11 Stepwise

Due date: Thursday 3/26, 11:59pm

Power series as functions

01

01

Modifying geometric power series

Consider the geometric power series $\frac{1}{1-x}=1+x+x^2+x^3+\cdots ={\sum }_{n=0}^{\infty }{x}^n$ for $|x|<1$.

For this problem, you should modify the series for $\frac{1}{1-x}$.

(a) Write ${\displaystyle \frac{1}{5-x}}$ as a power series and determine its interval of convergence.

(b) Write ${\displaystyle \frac{1}{16+2x^3}}$ as a power series and determine its interval of convergence.

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Solution

Solutions - 0230-01

(a)

$$\begin{array}{c}\frac{1}{5-x}\gg \gg \frac{1}{5}\cdot \frac{1}{(1-\frac{x}{5})}\\ \\ \gg \gg \frac{1}{5}\sum _{n=0}^{\infty }{\left(\frac{x}{5}\right)}^n\gg \gg \sum _{n=0}^{\infty }\frac{1}{5^{n+1}}{x}^n\end{array}$$

The geometric series for $\frac{1}{1-x}$ converges when $|x|<1$ and diverges for $|x|\ge 1$. So ours will converge when $\left|{\displaystyle \frac{x}{5}}\right|<1$, which is when $|x|<5$, and diverge otherwise. The interval is therefore $(-5,+5)$.

One can check this in more detail by doing the ratio test:

$$|\frac{x^{n+1}}{5^{n+1}}\cdot \frac{5^n}{x^n}|\gg \gg \frac{1}{5}|x|$$

But we must be careful: the ratio test will not tell us what happens at the endpoints of the interval. If we apply the ratio test here, we would have to check the endpoint separately. But if we use the known result for geometric series, we know it diverges at both endpoints.

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(b)

$$\begin{array}{c}\frac{1}{16+2x^3}\gg \gg \frac{1}{16}\cdot \frac{1}{1-(-{\displaystyle \frac{x^3}{8}})}\\ \\ \gg \gg \frac{1}{16}\sum _{n=0}^{\infty }{(-\frac{x^3}{8})}^n\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{2^{3n+4}}{x}^{3n}\end{array}$$

The geometric series for $\frac{1}{1-x}$ converges when $|x|<1$. So our series will converge when $\left|{\displaystyle \frac{x^3}{8}}\right|<1$, which is when $|x|<2$, and diverges for $|x|\ge 2$. So the interval is $(-\mathrm{2,2})$.

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02

03

Finding a power series

Find a power series representation for these functions:

(a) $f(x)={\displaystyle \frac{x^2}{x^4+81}}$ $$ (b) $g(x)=x^2\ln (1+x)$

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Solution

Solutions - 0230-03

(a)

$$\begin{array}{c}\frac{x^2}{x^4+81}\gg \gg \frac{x^2}{81}\cdot \frac{1}{1-(-\frac{x^4}{81})}\\ \\ \gg \gg \frac{x^2}{81}\sum _{n=0}^{\infty }{(-\frac{x^4}{81})}^n\gg \gg f(x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{3^{4n+4}}{x}^{4n+2}\end{array}$$

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Another approach:

$$\begin{array}{ccc}& \frac{x^2}{x^4+81}\gg \gg \frac{x^2}{a}\frac{a}{{\left(x^2\right)}^2+a^2},a=9& \text{(A)}\end{array}$$

We know that:

$$\begin{array}{c}\frac{a}{u^2+a^2}=\frac{d}{du}{\tan }^{-1}\left(\frac{u}{a}\right)\gg \gg \frac{d}{du}\sum _{n=0}^{\infty }{(-1)}^n\frac{{(u/a)}^{2n+1}}{2n+1}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n{\left(\frac{u}{a}\right)}^{2n}\frac{1}{a}\end{array}$$

Plug in $u=x^2$:

$$\gg \gg \sum _{n=0}^{\infty }{(-1)}^n{\left(\frac{x^2}{a}\right)}^{2n}\frac{1}{a}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{a^{2n+1}}{x}^{4n}$$

Complete:

$$\begin{array}{c}\text{A:}\frac{x^2}{a}\frac{a}{{\left(x^2\right)}^2+a^2}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{a^{2n+2}}{x}^{4n+2}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{9^{2n+2}}{x}^{4n+2}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{3^{4n+4}}{x}^{4n+2}\end{array}$$

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(b)

Notice:

$$\frac{d}{dx}\ln (1+x)=\frac{1}{1+x}\gg \gg \sum _{n=0}^{\infty }{(-x)}^n$$

Integrate:

$$\ln (1+x)=\int \sum _{n=0}^{\infty }{(-x)}^{n}dx\gg \gg C+\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+1}$$

Plug in $x=0$ to solve and find $C=0$.

Now then:

$$\begin{array}{c}{x}^2\ln (1+x)\gg \gg x^2\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+1}\\ \\ \gg \gg g(x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+3}\end{array}$$Link to original

Taylor and Maclaurin series

03

01

Maclaurin series

For each of these functions, find the Maclaurin series, and the interval on which the expansion is valid.

(a) $x\ln (1-5x)$ $$ (b) $x^2\cos (x^3)$

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Solution

Solutions - 0240-01

(a)

$$\begin{array}{c}\ln (1-u)=\sum _{n=0}^{\infty }-\frac{u^{n+1}}{n+1}\\ \\ \gg \gg \ln (1-5x)=\sum _{n=0}^{\infty }-\frac{{(5x)}^{n+1}}{n+1}\\ \\ \gg \gg x\ln (1-5x)=\sum _{n=0}^{\infty }-\frac{5^{n+1}}{n+1}{x}^{n+2}\end{array}$$

$I=[-1/5,+1/5)$

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(b)

$$\begin{array}{c}\cos (u)=\sum _{n=0}^{\infty }{(-1)}^n\frac{u^{2n}}{(2n)!}\\ \\ \gg \gg \cos (x^3)=\sum _{n=0}^{\infty }{(-1)}^n\frac{{(x^3)}^{2n}}{(2n)!}\\ \\ \gg \gg x^2\cos (x^3)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{(2n)!}{x}^{6n+2}\end{array}$$

$I=(-\infty ,\infty )$

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04

04

Discovering the function from its Maclaurin series

For each of these series, identify the function of which it is the Maclaurin series.

(a) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n{2}^n{x}^n}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{x^{3n}}{n!}}$

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Solution

Solutions - 0240-04

(a) Notice the matching powers. Collect powers and then observe the geometric series pattern:

$$\begin{array}{c}\gg \gg \sum _{n=0}^{\infty }{(-2x)}^n\gg \gg \sum _{n=0}^{\infty }{u}^n\text{for }u=-2x\\ \\ \gg \gg \frac{1}{1-u}\gg \gg \frac{1}{1+2x}\end{array}$$

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(b)

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{3n}}{n!}\gg \gg \sum _{n=0}^{\infty }\frac{{(-x^3)}^n}{n!}\\ \\ \gg \gg e^{-x^3}\end{array}$$Link to original

Applications of Taylor series

05

01

Approximating $1/e$

Using the series representation of $e^x$, show that:

$$\frac{1}{e}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\cdots$$

Now use the alternating series error bound to approximate $\frac{1}{e}$ to an error within ${10}^{-3}$.

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Solution

Solutions - 0250-01

Notice that ${\displaystyle \frac{1}{e}}=e^{-1}$.

We have this series for $e^x$:

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

Therefore:

$$\begin{array}{c}{e}^{-1}=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!}=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}\\ \\ \gg \gg \frac{1}{e}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\cdots \end{array}$$

Now we use the “Next Term Bound” rule. Calculate terms until we find a term less than ${10}^{-3}$:

$$\begin{array}{c}\frac{1}{4!}=4.2\times {10}^{-2},\frac{1}{5!}=8.3\times {10}^{-3},\\ \\ \frac{1}{6!}=1.4\times {10}^{-3},\frac{1}{7!}=2.4\times {10}^{-4}\end{array}$$

So we take the following partial sum approximation:

$$\frac{1}{e}\approx \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}=0.3681$$Link to original