W11 Regular

Due date: Sunday 3/29, 11:59pm

Power series as functions

01

02

Power series of a derivative

Suppose that a function $f(x)$ has power series given by:

$$f(x)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots =\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n+2}}{n+1}$$

The radius of convergence of this series is $R=1$.

What is the power series of $f^{\prime }(x)$ and what is its interval of convergence?

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Solution

Solutions - 0230-02

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

If $R=1$ for $f(x)$, then we know $R=1$ for $f^{\prime }(x)$.

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02

04

Modifying and integrating a power series

(a) Modify the power series $\frac{1}{1-x}=1+x+x^2+x^3+\cdots ={\sum }_{n=0}^{\infty }{x}^n$ for $|x|<1$ to obtain the power series for $f(x)=\frac{1}{1+x^4}$.

(b) Now integrate this series to find the power series for $\int f(x)dx$.

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Solution

Solutions - 0230-04

(a)

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

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(b)

$$\int \frac{1}{1+x^4}dx=\int \sum _{n=0}^{\infty }{(-1)}^n{x}^{4n}dx\gg \gg C+\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{4n+1}{x}^{4n+1}$$Link to original

Taylor and Maclaurin series

03 $★$

02

Maclaurin series

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

(a) $\sin (3x^2)$ $$ (b) $x^2{e}^{5x}$

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Solution

Solutions - 0240-02

(a)

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

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(b)

$$\begin{array}{c}{e}^u=\sum _{n=0}^{\infty }\frac{u^n}{n!}\\ \\ \gg \gg e^{5x}=\sum _{n=0}^{\infty }\frac{{(5x)}^n}{n!}\\ \\ \gg \gg x^2{e}^{5x}=\sum _{n=0}^{\infty }\frac{5^n}{n!}{x}^{n+2}\end{array}$$Link to original

04

03

Taylor series of $1/x$

Find the Taylor series for the function $f(x)=\frac{1}{x}$, centered at $c=1$, by differentiating repeatedly to determine the coefficients.

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Solution

Solutions - 0240-03

Calculate derivatives. Use $a_n={\displaystyle \frac{f^{(n)}(1)}{n!}}$.

$n$	$f^{(n)}(x)$	$f^{(n)}(1)$	$a_n$
0	$+x^{-1}$	$+1$	$+1$
1	$-x^{-2}$	$-1$	$-1$
2	$+2\cdot x^{-3}$	$+2$	$+1$
3	$-3\cdot 2\cdot x^{-4}$	$-3!$	$-1$
4	$+4\cdots 2\cdot x^{-5}$	$+4!$	$+1$
5	$-5\cdots 2\cdot x^{-6}$	$-5!$	$-1$
6	$+6\cdots 2\cdot x^{-7}$	$+6!$	$+1$

Each new derivative takes down the next power as a factor, and switches the sign. The accumulation of powers follows a factorial pattern, and these factorials cancel those added to the denominator to make $a_n$.

So we have:

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

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Alternate method:

We can derive this Taylor series using some algebraic tricks with the standard geometric series:

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

05

05

Discovering the function from its Maclaurin series

For each of these series, identify the function of which it is the Maclaurin series, and evaluate the function at an appropriate choice of $x$ to find the total sum for the series.

(a) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{\pi }^{2n+1}}{4^{2n+1}(2n+1)!}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{2^{2n}}{n!}}$ $$ (c) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{\pi }^{2n+2}}{3^{2n+1}(2n)!}}$

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Solution

Solutions - 0240-05

(a)

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^n\frac{{\pi }^{2n+1}}{4^{2n+1}(2n+1)!}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{{(\pi /4)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg \sin x{|}_{x=\pi /4}\gg \gg \frac{\sqrt{2}}{2}\end{array}$$

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(b)

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{2^{2n}}{n!}\gg \gg \sum _{n=0}^{\infty }\frac{{(4)}^n}{n!}\\ \\ \gg \gg e^x{|}_{x=4}\gg \gg e^4\end{array}$$

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(c)

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

06 $★$

06

Summing a Maclaurin series by guessing its function

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

(a) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{5x^{4n+2}}{(2n+1)!}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-5x)}^{n+1}}{n+1}}$

Now find the total sums for these series:

(c) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-5)}^n}{n!}}$ $$ (d) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{9^n(2n)!}}$

(Hint: for (c)-(d), do the process in (a)-(b), then evaluate the resulting function somewhere.)

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Solution

Solutions - 0240-06

(a)

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

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(b)

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

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(c)

$$\sum _{n=0}^{\infty }\frac{{(-5)}^n}{n!}\gg \gg e^x{|}_{x=-5}\gg \gg e^{-5}$$

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(d)

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n}}{9^n(2n)!}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{{(\pi /3)}^{2n}}{(2n)!}\\ \\ \gg \gg \cos x{|}_{x=\pi /3}\gg \gg \cos \frac{\pi }{3}\gg \gg \frac{1}{2}\end{array}$$Link to original

07

07

Data of a Taylor series

Assume that $f(3)=1$, $f^{\prime }(3)=2$, $f^{\prime \prime }(3)=12$, and $f^{\prime \prime \prime }(3)=3$.

Find the first four terms of the Taylor series of $f(x)$ centered at $c=3$.

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Solution

Solutions - 0240-07

Use the formula $a_n={\displaystyle \frac{f^{(n)}(3)}{n!}}$:

$$\begin{array}{cc}{a}_0& =\frac{f^{(0)}(3)}{0!}=1\\ & \\ a_1& =\frac{f^{(1)}(3)}{1!}=2\\ & \\ a_2& =\frac{f^{(2)}(3)}{2!}=6\\ & \\ a_3& =\frac{f^{(3)}(3)}{3!}=\frac{1}{2}\end{array}$$

Therefore:

$$f(x)=1+2(x-3)+6{(x-3)}^2+\frac{1}{2}{(x-3)}^3+\cdots$$Link to original

08

08

Evaluating series

Find the total sums for these series:

(a) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{\sqrt{3}}^{2n+1}}{3^{2n+1}(2n+1)}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^{n+1}\frac{1}{5^{n+1}(n+1)}}$

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Solution

Solutions - 0240-08

(a)

Recall the series for ${\tan }^{-1}$:

$${\tan }^{-1}(x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n+1}}{2n+1}$$

This matches our series if we set $x={\displaystyle \frac{\sqrt{3}}{3}}$. So the total sum is:

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^n\frac{{\sqrt{3}}^{2n+1}}{3^{2n+1}(2n+1)}\gg \gg {\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\\ \\ \gg \gg {\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\gg \gg \frac{\pi }{6}\end{array}$$

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(b)

$$\begin{array}{c}\sum _{n=0}^{\infty }{(-1)}^{n+1}\frac{1}{5^{n+1}(n+1)}\gg \gg \sum _{n=0}^{\infty }\frac{{(-1/5)}^{n+1}}{n+1}\\ \\ \gg \gg -\sum _{n=0}^{\infty }-\frac{{(-1/5)}^{n+1}}{n+1}\gg \gg {-\sum _{n=0}^{\infty }-\frac{x^{n+1}}{n+1}|}_{x=-1/5}\\ \\ \gg \gg -\ln (1-x){|}_{x=-1/5}\gg \gg -\ln (1+1/5)\\ \\ \gg \gg -\ln (6/5)\gg \gg \ln (5/6)\end{array}$$Link to original

09 $★$

09

Large derivative at $x=0$ using pattern of Maclaurin series

Consider the function $f(x)=x^2\sin \left(5x^3\right)$. Find the value of $f^{(35)}(0)$.

(Hint: find the rule for coefficients of the Maclaurin series of $f(x)$ and then plug in $0$.)

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Solution

Solutions - 0240-09

$$\begin{array}{c}{x}^2\sin \left(5x^3\right)=x^2\sum _{n=0}^{\infty }{(-1)}^n\frac{{\left(5x^3\right)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{5^{2n+1}}{(2n+1)!}{x}^{6n+5}\end{array}$$

The Coefficient-Derivative Identity says that $f^{(k)}(0)={\displaystyle \frac{a_k}{k!}}$ where $a_k$ is the coefficient of the $k^{\text{th}}$ power $x^k$.

Solve:

$$x^{35}=x^{6n+5}\gg \gg 35=6n+5\gg \gg n=5$$

So, for the coefficient $a_{35}$:

$$\begin{array}{c}{(-1)}^5\frac{5^{2\cdot 5+1}}{(2\cdot 5+1)!}\gg \gg -\frac{5^{11}}{11!}\\ \\ \gg \gg f^{(35)}(0)=35!a_{35}\gg \gg -5^{11}\frac{35!}{11!}\end{array}$$Link to original

Applications of Taylor series

10

02

Some estimates using series

Without a calculator, estimate $\cos (0.02)$ (angle in radians) with an error below ${10}^{-6}$.

(Use the error bound formula for alternating series.)

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Solution

Solutions - 0250-02

Write the alternating series for $\cos (0.02)$:

$$\begin{array}{c}\cos (x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n}}{(2n)!}\\ \\ \gg \gg \cos (0.02)=\sum _{n=0}^{\infty }(-1)\frac{{(0.02)}^{2n}}{(2n)!}\\ \\ \gg \gg 1-\frac{{(0.02)}^2}{2!}+\frac{{(0.02)}^4}{4!}-\frac{{(0.02)}^6}{6!}\end{array}$$

This is an alternating series, so we can apply the “Next Term Bound” rule. Calculate some terms:

$$\frac{{(0.02)}^2}{2!}=2.0\times {10}^{-4},\frac{{(0.02)}^4}{4!}=6.7\times {10}^{-9}$$

(Without a calculator, we can see that ${(0.02)}^4=2^4\times {10}^{-8}=1.6\times {10}^{-7}$. Dividing by $4!$ will only decrease this value.)

So we add up the prior terms:

$$\gg \gg 1-\frac{{(0.02)}^2}{2!}=0.9998$$Link to original

11 $★$

03

Some estimates using series

Find an infinite series representation of ${\displaystyle {\int }_0^1\sin (x^2)dx}$ and then use your series to estimate this integral to within an error of ${10}^{-3}$.

(Use the error bound formula for alternating series.)

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Solution

Solutions - 0250-03

Write the series of the integrand:

$$\begin{array}{c}\sin \left(x^2\right)=\sum _{n=0}^{\infty }{(-1)}^n\frac{{\left(x^2\right)}^{2n+1}}{(2n+1)!}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+2}}{(2n+1)!}\end{array}$$

Integrate:

$$\begin{array}{c}{\int }_0^1\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+2}}{(2n+1)!}dx\\ \\ \gg \gg {\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{4n+3}}{(2n+1)!(4n+3)}|}_0^1\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{(2n+1)!(4n+3)}\\ \\ \gg \gg \frac{1}{3}-\frac{1}{3!(7)}+\frac{1}{5!(11)}-\cdots \end{array}$$

Now apply the “Next Term Bound” and look for the first term below ${10}^{-3}$:

$$\frac{1}{3!(7)}=2.4\times {10}^{-2},\frac{1}{5!(11)}=7.6\times {10}^{-4}$$

So we simply add the first two terms:

$$\frac{1}{3}-\frac{1}{3!(7)}\approx 0.3095$$Link to original