3.1E: Exercises

Exercise $\PageIndex{1}$

For each power series use Theorem $(3.1.3)$ to find the radius of convergence $R$. If $R>0$, find the open interval of convergence.

(a) $\displaystyle{\sum_{n=0}^\infty {(-1)^n\over2^nn}(x-1)^n}$

(b) $\displaystyle{\sum_{n=0}^\infty 2^nn(x-2)^n}$

(c) $\displaystyle{\sum_{n=0}^\infty {n!\over9^n}x^n}$

(d) $\displaystyle{\sum_{n=0}^\infty{n(n+1)\over16^n}(x-2)^n}$

(e) $\displaystyle{\sum_{n=0}^\infty (-1)^n{7^n\over n!}x^n}$

(f) $\displaystyle{\sum_{n=0}^\infty {3^n\over4^{n+1}(n+1)^2}(x+7)^n}$

Exercise $\PageIndex{2}$

Suppose there's an integer $M$ such that $b_m\ne0$ for $m\ge M$, and

$$\begin{eqnarray*} \lim_{m\to\infty}\left|b_{m+1}\over b_m\right|=L, \end{eqnarray*}$$

where $0\le L\le\infty$. Show that the radius of convergence of

$$\begin{eqnarray*} \sum_{m=0}^\infty b_m(x-x_0)^{2m} \end{eqnarray*}$$

is $R=1/\sqrt L$, which is interpreted to mean that $R=0$ if $L=\infty$ or $R=\infty$ if $L=0$.

Hint: Apply Theorem $(3.1.3)$ to the series $\sum_{m=0}^\infty b_mz^m$ and then let $z=(x-x_0)^2$.

Answer

From Theorem~3.1.3, $\sum_{m=0}^\infty b_mz^m$ converges
if $|z|<1/L$ and diverges if $|z|>1/L$ .
 Therefore,
$\sum_{m=0}^\infty b_m(x-x_0)^2$ converges if $|x-x_0|<1/\sqrt{L}$
and diverges if $|x-x_0|>1/\sqrt{L}$ .

Exercise $\PageIndex{3}$

For each power series, use the result of Exercise $(3.1E.2)$ to find the radius of convergence $R$. If $R>0$, find the open interval of convergence.

(a) $\displaystyle{\sum_{m=0}^\infty (-1)^m(3m+1)(x-1)^{2m+1}}$

(b) $\displaystyle{\sum_{m=0}^\infty (-1)^m{m(2m+1)\over2^m}(x+2)^{2m}}$

(c) $\displaystyle{\sum_{m=0}^\infty {m!\over(2m)!}(x-1)^{2m}}$

(d) $\displaystyle{\sum_{m=0}^\infty (-1)^m{m!\over9^m}(x+8)^{2m}}$

(e) $\displaystyle{\sum_{m=0}^\infty(-1)^m{(2m-1)\over3^m}x^{2m+1}}$

(f) $\displaystyle{\sum_{m=0}^\infty(x-1)^{2m}}$

Exercise $\PageIndex{4}$

Suppose there's an integer $M$ such that $b_m\ne0$ for $m\ge M$, and

$$\begin{eqnarray*} \lim_{m\to\infty}\left|b_{m+1}\over b_m\right|=L, \end{eqnarray*}$$

where $0\le L\le\infty$. Let $k$ be a positive integer. Show that the radius of convergence of

$$\begin{eqnarray*} \sum_{m=0}^\infty b_m(x-x_0)^{km} \end{eqnarray*}$$

is $R=1/\sqrt[k]L$, which is interpreted to mean that $R=0$ if $L=\infty$ or $R=\infty$ if $L=0$.

Hint: Apply Theorem $(3.1.3)$ to the series $\sum_{m=0}^\infty b_mz^m$ and then let $z=(x-x_0)^k$.

Answer

From Theorem~3.1.3, $\sum_{m=0}^\infty b_mz^m$ converges
if $|z|<1/L$ and diverges if $|z|>1/L$ .
 Therefore,
$\sum_{m=0}^\infty b_m(x-x_0)^{km}$ converges if
$|x-x_0|<1/\sqrt[k]{L}$ and diverges if $|x-x_0|>1/\sqrt[k]{L}$ .

Exercise $\PageIndex{5}$

For each power series use the result of Exercise $(3.1E.4)$ to find the radius of convergence $R$. If $R>0$, find the open interval of convergence.

(a) $\displaystyle{\sum_{m=0}^\infty{(-1)^m\over(27)^m}(x-3)^{3m+2}}$

(b) $\displaystyle{\sum_{m=0}^\infty{x^{7m+6}\over m}}$

(c) $\displaystyle{\sum_{m=0}^\infty{9^m(m+1)\over(m+2)}(x-3)^{4m+2}}$

(d) $\displaystyle{\sum_{m=0}^\infty(-1)^m{2^m\over m!}x^{4m+3}}$

(e) $\displaystyle{\sum_{m=0}^\infty{m!\over(26)^m}(x+1)^{4m+3}}$

(f) $\displaystyle{\sum_{m=0}^\infty{(-1)^m\over8^mm(m+1)}(x-1)^{3m+1}}$

Exercise $\PageIndex{6}$

Graph $y=\sin x$ and the Taylor polynomial

$$\begin{eqnarray*} T_{2M+1}(x)=\sum_{n=0}^M{(-1)^nx^{2n+1}\over(2n+1)!} \end{eqnarray*}$$

on the interval $(-2\pi,2\pi)$ for $M=1$, $2$, $3$, $\dots$, until you find a value of $M$ for which there's no perceptible difference between the two graphs.

Exercise $\PageIndex{7}$

Graph $y=\cos x$ and the Taylor polynomial

$$\begin{eqnarray*} T_{2M}(x)=\sum_{n=0}^M{(-1)^nx^{2n}\over(2n)!} \end{eqnarray*}$$

on the interval $(-2\pi,2\pi)$ for $M=1$, $2$, $3$, $\dots$, until you find a value of $M$ for which there's no perceptible difference between the two graphs.

Exercise $\PageIndex{8}$

Graph $y=1/(1-x)$ and the Taylor polynomial

$$\begin{eqnarray*} T_N(x)=\sum_{n=0}^Nx^n \end{eqnarray*}$$

on the interval $[0,.95]$ for $N=1$, $2$, $3$, $\dots$, until you find a value of $N$ for which there's no perceptible difference between the two graphs. Choose the scale on the $y$-axis so that $0\le y\le20$.

Exercise $\PageIndex{9}$

Graph $y=\cosh x$ and the Taylor polynomial

$$\begin{eqnarray*} T_{2M}(x)=\sum_{n=0}^M{x^{2n}\over(2n)!} \end{eqnarray*}$$

on the interval $(-5,5)$ for $M=1$, $2$, $3$, $\dots$, until you find a value of $M$ for which there's no perceptible difference between the two graphs. Choose the scale on the $y$-axis so that $0\le y\le75$.

Exercise $\PageIndex{10}$

Graph $y=\sinh x$ and the Taylor polynomial

$$\begin{eqnarray*} T_{2M+1}(x)=\sum_{n=0}^M{x^{2n+1}\over(2n+1)!} \end{eqnarray*}$$

on the interval $(-5,5)$ for $M=0$, $1$, $2$, $\dots$, until you find a value of $M$ for which there's no perceptible difference between the two graphs. Choose the scale on the $y$-axis so that $-75~\le~y\le~75$.

In Exercises $(3.1E.11)$ to $(3.1E.15)$, find a power series solution $y(x)=\sum_{n=0}^\infty a_nx^n$.

Exercise $\PageIndex{11}$

$(2+x)y''+xy'+3y$

Exercise $\PageIndex{12}$

$(1+3x^2)y''+3x^2y'-2y$

Answer

$(1+3x^2)y''+3x^2y'-2y= \sum_{n=2}^\infty n(n-1)a_nx^{n-2} +3\sum_{n=2}^\infty n(n-1)a_nx^n +3\sum_{n=1}^\infty na_nx^{n+1}-2\sum_{n=0}^\infty a_nx^n =\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n +3\sum_{n=1}^\infty n(n-1)na_nx^n +3\sum_{n=1}^\infty (n-1)a_{n-1}x^n-2\sum_{n=0}^\infty a_nx^n =2a_2-2a_0+\sum_{n=1}^\infty [(n+2)(n+1)a_{n+2}+(3n(n-1)-2)a_n+3(n-1)a_{n-1}]x^n$.

 

Exercise $\PageIndex{13}$

$(1+2x^2)y''+(2-3x)y'+4y$

Answer

$(1+2x^2)y''+(2-3x)y'+4y= \sum_{n=2}^\infty n(n-1)a_nx^{n-2} +2\sum_{n=2}^\infty n(n-1)a_nx^n +2\sum_{n=1}^\infty na_nx^{n-1}-2\sum_{n=0}^\infty a_nx^n +4\sum_{n=0}^\infty a_nx^n =\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n +2\sum_{n=0}^\infty n(n-1)a_nx^n +2\sum_{n=0}^\infty (n+1)a_{n+1}x^n-3\sum_{n=0}^\infty a_nx^n +4\sum_{n=0}^\infty a_nx^n= \sum_{n=0}^\infty \left[(n+2)(n+1)a_{n+2}+2(n+1)a_{n+1}+(2n^2-5n+4)a_n\right]x^n$

Exercise $\PageIndex{14}$

$(1+x^2)y''+(2-x)y'+3y$

Answer

$(1+x^2)y''+(2-x)y'+3y= \sum_{n=2}^\infty n(n-1)a_nx^{n-2} +\sum_{n=2}^\infty n(n-1)a_nx^n +2\sum_{n=1}^\infty na_nx^{n-1}-\sum_{n=1}^\infty na_nx^n +3\sum_{n=0}^\infty a_nx^n =\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n \\+\sum_{n=0}^\infty n(n-1)a_nx^n +2\sum_{n=0}^\infty (n+1)a_{n+1}x^n-\sum_{n=0}^\infty na_nx^n +3\sum_{n=0}^\infty a_nx^n \\=\sum_{n=0}^\infty \left[(n+2)(n+1)a_{n+2}+2(n+1)a_{n+1}+(n^2-2n+3)a_n\right]x^n$.

Exercise $\PageIndex{15}$

$(1+3x^2)y''-2xy'+4y$

Exercise $\PageIndex{16}$

Suppose $y(x)=\sum_{n=0}^\infty a_n(x+1)^n$ on an open interval that contains $x_0=-1$. Find a power series in $x+1$ for

$$\begin{eqnarray*} xy''+(4+2x)y'+(2+x)y. \end{eqnarray*}$$

Answer

Let $t=x+1$; then $xy''+(4+2x)y'+(2+x)y=(-1+t)y''+(2+2t)y'+(1+t)y =-\sum_{n=2}^\infty n(n-1)a_nt^{n-2} +\sum_{n=2}^\infty n(n-1)a_nt^{n-1} +2\sum_{n=1}^\infty n a_nt^{n-1} +2\sum_{n=1}^\infty n a_nt^n +\sum_{n=0}^\infty  a_nt^n +\sum_{n=0}^\infty  a_nt^{n+1} =-\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}t^n +\sum_{n=0}^\infty (n+1)na_{n+1}t^n +2\sum_{n=0}^\infty (n+1) a_{n+1}t^n +2\sum_{n=0}^\infty n a_nt^n +\sum_{n=0}^\infty  a_nt^n +\sum_{n=1}^\infty  a_{n-1}t^n =(-2a_2+2a_1+a_0) +\sum_{n=1}^\infty  \left[-(n+2)(n+1)a_{n+2}+(n+1)(n+2)a_{n+1}+(2n+1)a_n+a_{n-1}\right] (x+1)^n$.

Exercise $\PageIndex{17}$

Suppose $y(x)=\sum_{n=0}^\infty a_n(x-2)^n$ on an open interval that contains $x_0=2$. Find a power series in $x-2$ for

$$\begin{eqnarray*} x^2y''+2xy'-3xy. \end{eqnarray*}$$

Exercise $\PageIndex{18}$

Do the following experiment for various choices of real numbers $a_0$ and $a_1$.

(a) Use differential equations software to solve the initial value problem

$$\begin{eqnarray*} (2-x)y''+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \end{eqnarray*}$$

numerically on $(-1.95,1.95)$. Choose the most accurate method your software package provides. (See Section 3.1 for a brief discussion of one such method.)

(b) For $N=2$, $3$, $4$, $\dots$, compute $a_2$, $\dots$, $a_N$ from Equation $(3.1.18)$ and graph

$$\begin{eqnarray*} T_N(x)=\sum_{n=0}^N a_nx^n \end{eqnarray*}$$

and the solution obtained in part (a) on the same axes. Continue increasing $N$ until it's obvious that there's no point in continuing. (This sounds vague, but you'll know when to stop.)

Exercise $\PageIndex{19}$

Follow the directions of Exercise $(3.1.18)$ for the initial value problem

$$\begin{eqnarray*} (1+x)y''+2(x-1)^2y'+3y=0,\quad y(1)=a_0,\quad y'(1)=a_1, \end{eqnarray*}$$

on the interval $(0,2)$. Use Equations $(3.1.24) \mbox{ and } (3.1.25)$ to compute $\{a_n\}$.

Exercise $\PageIndex{20}$

Suppose the series $\sum_{n=0}^\infty a_nx^n$ converges on an open interval $(-R,R)$, let $r$ be an arbitrary real number, and define

$$\begin{eqnarray*} y(x)=x^r\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^{n+r} \end{eqnarray*}$$

on $(0,R)$. Use Theorem $(3.1.4)$ and the rule for differentiating the product of two functions to show that

$$\begin{eqnarray*} y'(x)&=&\displaystyle{\sum_{n=0}^\infty (n+r)a_nx^{n+r-1}},\\ y''(x)&=&\displaystyle{\sum_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}},\\ &\vdots&\\ y^{(k)}(x)&=&\displaystyle{\sum_{n=0}^\infty(n+r)(n+r-1)\cdots(n+r-k)a_nx^{n+r-k}} \end{eqnarray*}$$

on $(0,R)$

Answer

$y'(x)=\displaystyle x^r\sum_{n=0}^\infty na_nx^{n-1}+rx^{r-1}\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty (n+r)x^{n+r-1}$

$y''=\displaystyle{d\over dx}y'(x)={d\over dx}\left[x^{r-1}\sum_{n=0}^\infty (n+r)a_nx^n\right]=x^{r-1}\sum_{n=0}^\infty (n+r)na_nx^{n-1}+ (r-1)x^{r-2}\sum_{n=0}^\infty (n+r)a_nx^n=\sum_{n=0}^\infty (n+r)(n+r-1)a_nx^{n+r-2}$.

In Exercises $(3.1E.21)$ to $(3.1E.26)$, let $y$ be as defined in Exercise $(3.1E.20)$, and write the given expression in the form $x^r\sum_{n=0}^\infty b_nx^n$.

Exercise $\PageIndex{21}$

$x^2(1-x)y''+x(4+x)y'+(2-x)y$

Exercise $\PageIndex{22}$

$x^2(1+x)y''+x(1+2x)y'-(4+6x)y$

Answer

$x^2(1+x)y''+x(1+2x)y'-(4+6x)y= (x^2y''+xy'-4y)+x(x^2y''+2xy'-6y)=\displaystyle \sum_{n=0}^\infty [(n+r)(n+r-1)+(n+r)-4]a_nx^{n+r} +\sum_{n=0}^\infty [(n+r)(n+r-1)+2(n+r)-6]a_nx^{n+r+1} =\sum_{n=0}^\infty (n+r-2)(n+r+2)a_nx^{n+r} +\sum_{n=0}^\infty (n+r+3)(n+r-2)a_nx^{n+r+1} =\sum_{n=0}^\infty (n+r-2)(n+r+2)a_nx^{n+r} +\sum_{n=1}^\infty (n+r+2)(n+r-3)a_{n-1}x^{n+r} =x^r\sum_{n=0}^\infty  b_nx^n$ with
$b_0=(r-2)(r+2)a_0$ and
$b_n=(n+r-2)(n+r+2)a_n+(n+r+2)(n+r-3)a_{n-1}, n\ge1$.

Exercise $\PageIndex{23}$

$x^2(1+x)y''-x(1-6x-x^2)y'+(1+6x+x^2)y$

Exercise $\PageIndex{24}$

$x^2(1+3x)y''+x(2+12x+x^2)y'+2x(3+x)y$

Answer

$x^2(1+3x)y''+x(2+12x+x^2)y'+2x(3+x)y= (x^2y''+2xy')+x(3x^2y''+12xy'+6y)+x^2(xy'+2y)=\displaystyle \sum_{n=0}^\infty [(n+r)(n+r-1)+2(n+r)]a_nx^{n+r} +\sum_{n=0}^\infty [3(n+r)(n+r-1)+12(n+r)+6]a_nx^{n+r+1} +\sum_{n=0}^\infty[(n+r)+2]a_nx^{n+r+2 } =\sum_{n=0}^\infty (n+r)(n+r+1)a_nx^{n+r} +3\sum_{n=0}^\infty (n+r+1)(n+r+2)a_nx^{n+r+1} +\sum_{n=0}^\infty (n+r+2)a_nx^{n+r+2} =\sum_{n=0}^\infty (n+r)(n+r+1)a_nx^{n+r} +3\sum_{n=1}^\infty (n+r)(n+r+1)a_{n-1}x^{n+r} +\sum_{n=2}^\infty (n+r)a_{n-2}x^{n+r} =x^r\sum_{n=0}^\infty  b_nx^n$ with
$b_0=r(r+1)a_0, b_1=(r+1)(r+2)a_1+3(r+1)(r+2)a_0, b_n=(n+r)(n+r+1)a_n+3(n+r)(n+r+1)a_{n-1}+(n+r)a_{n-2}, n\ge2$.

Exercise $\PageIndex{25}$

$x^2(1+2x^2)y''+x(4+2x^2)y'+2(1-x^2)y$

Exercise $\PageIndex{26}$

$x^2(2+x^2)y''+2x(5+x^2)y'+2(3-x^2)y$

Answer

$x^2(2+x^2)y''+2x(5+x^2)y'+2(3-x^2)y= (2x^2y''+10xy'+6y)+x^2(x^2y''+2xy'-2y)=\displaystyle \sum_{n=0}^\infty [2(n+r)(n+r-1)+10(n+r)+6]a_nx^{n+r} +\sum_{n=0}^\infty [(n+r)(n+r-1)+2(n+r)-2]a_nx^{n+r+2} =2\sum_{n=0}^\infty (n+r+1)(n+r+3)a_nx^{n+r} +\sum_{n=0}^\infty (n+r-1)(n+r+2)a_nx^{n+r+2} =2\sum_{n=0}^\infty (n+r+1)(n+r+3)a_nx^{n+r} +\sum_{n=2}^\infty (n+r-3)(n+r)a_{n-2}x^{n+r} =x^r\sum_{n=0}^\infty  b_nx^n$ with
$b_0=2(r+1)(r+3)a_0$,
$b_1=2(r+2)(r+4)a_1$
$b_n=2(n+r+1)(n+r+3)a_n+(n+r-3)(n+r)a_{n-2}, n\ $.