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3.1E: Exercises

This page is a draft and is under active development. 

( \newcommand{\kernel}{\mathrm{null}\,}\)

Exercise 3.1E.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) n=0(1)n2nn(x1)n

(b) n=02nn(x2)n

(c) n=0n!9nxn

(d) n=0n(n+1)16n(x2)n

(e) n=0(1)n7nn!xn

(f) n=03n4n+1(n+1)2(x+7)n

Exercise 3.1E.2

Suppose there's an integer M such that bm0 for mM, and

limm|bm+1bm|=L,

where 0L. Show that the radius of convergence of

m=0bm(xx0)2m

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

Hint: Apply Theorem (3.1.3) to the series m=0bmzm and then let z=(xx0)2.

Answer

From Theorem~3.1.3, m=0bmzm converges
if |z|<1/L and diverges if |z|>1/L .
 Therefore,
m=0bm(xx0)2 converges if |xx0|<1/L
and diverges if |xx0|>1/L .

Exercise 3.1E.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) m=0(1)m(3m+1)(x1)2m+1

(b) m=0(1)mm(2m+1)2m(x+2)2m

(c) m=0m!(2m)!(x1)2m

(d) m=0(1)mm!9m(x+8)2m

(e) m=0(1)m(2m1)3mx2m+1

(f) m=0(x1)2m

Exercise 3.1E.4

Suppose there's an integer M such that bm0 for mM, and

limm|bm+1bm|=L,

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

m=0bm(xx0)km

is R=1/kL, which is interpreted to mean that R=0 if L= or R= if L=0.

Hint: Apply Theorem (3.1.3) to the series m=0bmzm and then let z=(xx0)k.

Answer

From Theorem~3.1.3, m=0bmzm converges
if |z|<1/L and diverges if |z|>1/L .
 Therefore,
m=0bm(xx0)km converges if
|xx0|<1/kL and diverges if |xx0|>1/kL .

Exercise 3.1E.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) m=0(1)m(27)m(x3)3m+2

(b) m=0x7m+6m

(c) m=09m(m+1)(m+2)(x3)4m+2

(d) m=0(1)m2mm!x4m+3

(e) m=0m!(26)m(x+1)4m+3

(f) m=0(1)m8mm(m+1)(x1)3m+1

Exercise 3.1E.6

Graph y=sinx and the Taylor polynomial

T2M+1(x)=Mn=0(1)nx2n+1(2n+1)!

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

Exercise 3.1E.7

Graph y=cosx and the Taylor polynomial

T2M(x)=Mn=0(1)nx2n(2n)!

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

Exercise 3.1E.8

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

TN(x)=Nn=0xn

on the interval [0,.95] for N=1, 2, 3, , 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 0y20.

Exercise 3.1E.9

Graph y=coshx and the Taylor polynomial

T2M(x)=Mn=0x2n(2n)!

on the interval (5,5) for M=1, 2, 3, , 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 0y75.

Exercise 3.1E.10

Graph y=sinhx and the Taylor polynomial

T2M+1(x)=Mn=0x2n+1(2n+1)!

on the interval (5,5) for M=0, 1, 2, , 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  y 75.

In Exercises (3.1E.11) to (3.1E.15), find a power series solution y(x)=n=0anxn.

Exercise 3.1E.11

(2+x)y

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\ .


This page titled 3.1E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by William F. Trench.

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