Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

7.1E: Review of Power Series (Exercises)

  • Page ID
    18309
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    [exer:7.1.1] For each power series use Theorem [thmtype:7.1.3} to find the radius of convergence \(R\). If \(R>0\), find the open interval of convergence.

    1. \({\displaystyle \sum_{n=0}^\infty {(-1)^n\over2^nn}(x-1)^n}\)
    2. \({\displaystyle \sum_{n=0}^\infty 2^nn(x-2)^n}\)
    3. \({\displaystyle \sum_{n=0}^\infty {n!\over9^n}x^n}\)
    4. \({\displaystyle \sum_{n=0}^\infty{n(n+1)\over16^n}(x-2)^n}\)
    5. \({\displaystyle \sum_{n=0}^\infty (-1)^n{7^n\over n!}x^n}\)
    6. \({\displaystyle \sum_{n=0}^\infty {3^n\over4^{n+1}(n+1)^2}(x+7)^n}\)

    [exer:7.1.2] Suppose there’s an integer \(M\) such that \(b_m\ne0\) for \(m\ge M\), and \[\lim_{m\to\infty}\left|b_{m+1}\over b_m\right|=L,\nonumber \] where \(0\le L\le\infty\). Show that the radius of convergence of \[\displaystyle \sum_{m=0}^\infty b_m(x-x_0)^{2m}\nonumber \] is \(R=1/\sqrt L\), which is interpreted to mean that \(R=0\) if \(L=\infty\) or \(R=\infty\) if \(L=0\).

    [exer:7.1.3] For each power series, use the result of Exercise [exer:7.1.2} to find the radius of convergence \(R\). If \(R>0\), find the open interval of convergence.

    1. \({\displaystyle \sum_{m=0}^\infty (-1)^m(3m+1)(x-1)^{2m+1}}\)
    2. \({\displaystyle \sum_{m=0}^\infty (-1)^m{m(2m+1)\over2^m}(x+2)^{2m}}\)
    3. \({\displaystyle \sum_{m=0}^\infty {m!\over(2m)!}(x-1)^{2m}}\)
    4. \({\displaystyle \sum_{m=0}^\infty (-1)^m{m!\over9^m}(x+8)^{2m}}\)
    5. \({\displaystyle \sum_{m=0}^\infty(-1)^m{(2m-1)\over3^m}x^{2m+1}}\)
    6. \({\displaystyle \sum_{m=0}^\infty(x-1)^{2m}}\)

    [exer:7.1.4] Suppose there’s an integer \(M\) such that \(b_m\ne0\) for \(m\ge M\), and \[\lim_{m\to\infty}\left|b_{m+1}\over b_m\right|=L,\nonumber \] where \(0\le L\le\infty\). Let \(k\) be a positive integer. Show that the radius of convergence of \[\displaystyle \sum_{m=0}^\infty b_m(x-x_0)^{km}\nonumber \] is \(R=1/\sqrt[k]L\), which is interpreted to mean that \(R=0\) if \(L=\infty\) or \(R=\infty\) if \(L=0\).

    [exer:7.1.5] For each power series use the result of Exercise [exer:7.1.4} to find the radius of convergence \(R\). If \(R>0\), find the open interval of convergence.

    1. \({\displaystyle \sum_{m=0}^\infty{(-1)^m\over(27)^m}(x-3)^{3m+2}}\)
    2. \({\displaystyle \sum_{m=0}^\infty{x^{7m+6}\over m}}\)
    3. \({\displaystyle \sum_{m=0}^\infty{9^m(m+1)\over(m+2)}(x-3)^{4m+2}}\)
    4. \({\displaystyle \sum_{m=0}^\infty(-1)^m{2^m\over m!}x^{4m+3}}\)
    5. \({\displaystyle \sum_{m=0}^\infty{m!\over(26)^m}(x+1)^{4m+3}}\)
    6. \({\displaystyle \sum_{m=0}^\infty{(-1)^m\over8^mm(m+1)}(x-1)^{3m+1}}\)

    [exer:7.1.6] Graph \(y=\sin x\) and the Taylor polynomial \[T_{2M+1}(x)=\displaystyle \sum_{n=0}^M{(-1)^nx^{2n+1}\over(2n+1)!}\nonumber \] on the interval \((-2\pi,2\pi)\) for \(M=1\), \(2\), \(3\), …, until you find a value of \(M\) for which there’s no perceptible difference between the two graphs.

    [exer:7.1.7] Graph \(y=\cos x\) and the Taylor polynomial \[T_{2M}(x)=\displaystyle \sum_{n=0}^M{(-1)^nx^{2n}\over(2n)!}\nonumber \] on the interval \((-2\pi,2\pi)\) for \(M=1\), \(2\), \(3\), …, until you find a value of \(M\) for which there’s no perceptible difference between the two graphs.

    [exer:7.1.8] Graph \(y=1/(1-x)\) and the Taylor polynomial \[T_N(x)=\displaystyle \sum_{n=0}^Nx^n\nonumber \] 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 \(0\le y\le20\).

    [exer:7.1.9] Graph \(y=\cosh x\) and the Taylor polynomial \[T_{2M}(x)=\displaystyle \sum_{n=0}^M{x^{2n}\over(2n)!}\nonumber \] 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 \(0\le y\le75\).

    [exer:7.1.10] Graph \(y=\sinh x\) and the Taylor polynomial \[T_{2M+1}(x)=\displaystyle \sum_{n=0}^M{x^{2n+1}\over(2n+1)!}\nonumber \] 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 \(-75~\le~y\le~75\).

    [exer:7.1.11] \(\vspace*{-5pt}(2+x)y''+xy'+3y\)

    [exer:7.1.12] \((1+3x^2)y''+3x^2y'-2y\)

    [exer:7.1.13] \((1+2x^2)y''+(2-3x)y'+4y\)

    [exer:7.1.14] \((1+x^2)y''+(2-x)y'+3y\)

    [exer:7.1.15] \((1+3x^2)y''-2xy'+4y\)

    [exer:7.1.16] Suppose \(y(x)=\displaystyle \sum_{n=0}^\infty a_n(x+1)^n\) on an open interval that contains \(x_0~=~-1\). Find a power series in \(x+1\) for \[xy''+(4+2x)y'+(2+x)y.\nonumber \]

    [exer:7.1.17] Suppose \(y(x)=\displaystyle \sum_{n=0}^\infty a_n(x-2)^n\) on an open interval that contains \(x_0~=~2\). Find a power series in \(x-2\) for \[x^2y''+2xy'-3xy.\nonumber \]

    [exer:7.1.18] Do the following experiment for various choices of real numbers \(a_0\) and \(a_1\).

    Use differential equations software to solve the initial value problem

    \[(2-x)y''+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1,\nonumber \]

    numerically on \((-1.95,1.95)\). Choose the most accurate method your software package provides. (See Section 10.1 for a brief discussion of one such method.)

    For \(N=2\), \(3\), \(4\), …, compute \(a_2\), …, \(a_N\) from Eqn.Equation \ref{eq:7.1.18} and graph \[T_N(x)=\displaystyle \sum_{n=0}^N a_nx^n\nonumber \] and the solution obtained in (a) on the same axes. Continue increasing \(N\) until it is obvious that there’s no point in continuing. (This sounds vague, but you’ll know when to stop.)

    [exer:7.1.19] Follow the directions of Exercise [exer:7.1.18} for the initial value problem \[(1+x)y''+2(x-1)^2y'+3y=0,\quad y(1)=a_0,\quad y'(1)=a_1,\nonumber \] on the interval \((0,2)\). Use Eqns. Equation \ref{eq:7.1.24} and Equation \ref{eq:7.1.25} to compute \(\{a_n\}\).

    [exer:7.1.20] Suppose the series \(\displaystyle \sum_{n=0}^\infty a_nx^n\) converges on an open interval \((-R,R)\), let \(r\) be an arbitrary real number, and define \[y(x)=x^r\displaystyle \sum_{n=0}^\infty a_nx^n=\displaystyle \sum_{n=0}^\infty a_nx^{n+r}\nonumber \] on \((0,R)\). Use Theorem [thmtype:7.1.4} and the rule for differentiating the product of two functions to show that \[\begin{aligned} y'(x)&=&{\displaystyle \sum_{n=0}^\infty (n+r)a_nx^{n+r-1}},\\[10pt] y''(x)&=&{\displaystyle \sum_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}},\\ &\vdots&\\ y^{(k)}(x)&=&{\displaystyle \sum_{n=0}^\infty(n+r)(n+r-1)\cdots(n+r-k)a_nx^{n+r-k}}\end{aligned}\nonumber \] on \((0,R)\)

    [exer:7.1.21] \(x^2(1-x)y''+x(4+x)y'+(2-x)y\)

    [exer:7.1.22] \(x^2(1+x)y''+x(1+2x)y'-(4+6x)y\)

    [exer:7.1.23] \(x^2(1+x)y''-x(1-6x-x^2)y'+(1+6x+x^2)y\)

    [exer:7.1.24] \(x^2(1+3x)y''+x(2+12x+x^2)y'+2x(3+x)y\)

    [exer:7.1.25] \(x^2(1+2x^2)y''+x(4+2x^2)y'+2(1-x^2)y\)

    [exer:7.1.26] \(x^2(2+x^2)y''+2x(5+x^2)y'+2(3-x^2)y\)