7.E: Power series methods (Exercises)
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)7.1: Power Series
Is the power series \( \sum_{k=0}^\infty e^k x^k\) convergent? If so, what is the radius of convergence?
Is the power series \( \sum_{k=0}^\infty k x^k\) convergent? If so, what is the radius of convergence?
Is the power series \( \sum_{k=0}^\infty k! x^k\) convergent? If so, what is the radius of convergence?
Is the power series \( \sum_{k=0}^\infty \frac{1}{(2k)!} {(x-10)}^k\) convergent? If so, what is the radius of convergence?
Determine the Taylor series for \(\sin x\) around the point \(x_0 = \pi\).
Determine the Taylor series for \(\ln x\) around the point \(x_0 = 1\), and find the radius of convergence.
Determine the Taylor series and its radius of convergence of \(\dfrac{1}{1+x}\) around \(x_0 = 0\).
Determine the Taylor series and its radius of convergence of \(\dfrac{x}{4-x^2}\) around \(x_0 = 0\). Hint: You will not be able to use the ratio test.
Expand \(x^5+5x+1\) as a power series around \(x_0 = 5\).
Suppose that the ratio test applies to a series \( \sum_{k=0}^\infty a_k x^k\). Show, using the ratio test, that the radius of convergence of the differentiated series is the same as that of the original series.
Suppose that \(f\) is an analytic function such that \(f^{(n)}(0) = n\). Find \(f(1)\).
Is the power series \( \sum_{n=1}^\infty {(0.1)}^n x^n\) convergent? If so, what is the radius of convergence?
- Answer
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Yes. Radius of convergence is \(10\).
Is the power series \( \sum_{n=1}^\infty \frac{n!}{n^n} x^n\) convergent? If so, what is the radius of convergence?
- Answer
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Yes. Radius of convergence is \(e\).
Using the geometric series, expand \(\frac{1}{1-x}\) around \(x_0=2\). For what \(x\) does the series converge?
- Answer
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\(\frac{1}{1-x}=-\frac{1}{1-(2-x)}\) so \(\frac{1}{1-x}=\sum\limits_{n=0}^\infty (-1)^{n+1}(x-2)^{n}\), which converges for \(1<x<3\).
Find the Taylor series for \(x^7 e^x\) around \(x_0 = 0\).
- Answer
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\(\sum\limits_{n=7}^\infty \frac{1}{(n-7)!} x^{n}\)
Imagine \(f\) and \(g\) are analytic functions such that \(f^{(k)}(0) = g^{(k)}(0)\) for all large enough \(k\). What can you say about \(f(x)-g(x)\)?
- Answer
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\(f(x)-g(x)\) is a polynomial. Hint: Use Taylor series.
7.2: Series solutions of linear second order ODEs
In the following exercises, when asked to solve an equation using power series methods, you should find the first few terms of the series, and if possible find a general formula for the \(k^{\text{th}}\) coefficient.
Use power series methods to solve \(y''+y = 0\) at the point \(x_0 = 1\).
Use power series methods to solve \(y''+4xy = 0\) at the point \(x_0 = 0\).
Use power series methods to solve \(y''-xy = 0\) at the point \(x_0 = 1\).
Use power series methods to solve \(y''+x^2y = 0\) at the point \(x_0 = 0\).
The methods work for other orders than second order. Try the methods of this section to solve the first order system \(y'-xy = 0\) at the point \(x_0 = 0\).
Chebyshev’s equation of order \(p\):
- Solve \((1-x^2)y''-xy' + p^2y = 0\) using power series methods at \(x_0=0\).
- For what \(p\) is there a polynomial solution?
Find a polynomial solution to \((x^2+1) y''-2xy'+2y = 0\) using power series methods.
- Use power series methods to solve \((1-x)y''+y = 0\) at the point \(x_0 = 0\).
- Use the solution to part a) to find a solution for \(xy''+y=0\) around the point \(x_0=1\).
Use power series methods to solve \(y'' + 2 x^3 y = 0\) at the point \(x_0 =0\).
- Answer
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\(a_{2}=0\), \(a_{3}=0\), \(a_{4}=0\), recurrence relation (for \(k\geq 5\)): \(a_{k}=\frac{-2a_{k-5}}{k(k-1)}\), so \(y(x)=a_{0}+a_{1}x-\frac{a_{0}}{10}x^{5}-\frac{a_{1}}{15}x^{6}+\frac{a_{0}}{450}x^{10}+\frac{a_{1}}{825}x^{11}-\frac{a_{0}}{47250}x^{15}-\frac{a_{1}}{99000}x^{16}+\cdots\)
We can also use power series methods in nonhomogeneous equations.
- Use power series methods to solve \(y'' - x y = \frac{1}{1-x}\) at the point \(x_0 = 0\). Hint: Recall the geometric series.
- Now solve for the initial condition \(y(0)=0\), \(y'(0) = 0\).
- Answer
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- \(a_{2}=\frac{1}{2}\), and for \(k\geq 1\) we have \(a_{k}=\frac{a_{k-3}+1}{k(k-1)}\), so \(y(x)=a_{0}+a_{1}x+\frac{1}{2}x^{2}+\frac{a_{0}+1}{6}x^{3}+\frac{a_{1}+1}{12}x^{4}+\frac{3}{40}x^{5}+\frac{a_{0}+2}{30}x^{6}+\frac{a_{1}+2}{42}x^{7}+\frac{5}{112}x^{8}+\frac{a_{0}+3}{72}x^{9}+\frac{a_{1}+3}{90}x^{10}+\cdots\)
- \(y(x)=\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{12}x^{4}+\frac{3}{40}x^{5}+\frac{1}{15}x^{6}+\frac{1}{21}x^{7}+\frac{5}{112}x^{8}+\frac{1}{24}x^{9}+\frac{1}{30}x^{10}+\cdots\)
Attempt to solve \(x^2 y'' - y = 0\) at \(x_0 = 0\) using the power series method of this section (\(x_0\) is a singular point). Can you find at least one solution? Can you find more than one solution?
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Applying the method of this section directly we obtain \(a_{k}=0\) for all \(k\) and so \(y(x)=0\) is the only solution we find.
7.3: Singular points and the method of Frobenius
Find a particular (Frobenius-type) solution of \(x^2 y'' + x y' + (1+x) y = 0\).
Find a particular (Frobenius-type) solution of \(x y'' - y = 0\).
Find a particular (Frobenius-type) solution of \(y'' +\frac{1}{x}y' - xy = 0\).
Find the general solution of \(2 x y'' + y' - x^2 y = 0\).
Find the general solution of \(x^2 y'' - x y' -y = 0\).
In the following equations classify the point \(x=0\) as ordinary, regular singular, or singular but not regular singular.
- \(x^2(1+x^2)y''+xy=0\)
- \(x^2y''+y'+y=0\)
- \(xy''+x^3y'+y=0\)
- \(xy''+xy'-e^xy=0\)
- \(x^2y''+x^2y'+x^2y=0\)
In the following equations classify the point \(x=0\) as ordinary, regular singular, or singular but not regular singular.
- \(y''+y=0\)
- \(x^3y''+(1+x)y=0\)
- \(xy''+x^5y'+y=0\)
- \(\sin(x)y''-y=0\)
- \(\cos(x)y''-\sin(x)y=0\)
- Answer
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- ordinary,
- singular but not regular singular,
- regular singular,
- regular singular,
- ordinary.
Find the general solution of \(x^2 y'' -y = 0\).
- Answer
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\(y=Ax^{\frac{1+\sqrt{5}}{2}}+Bx^{\frac{1-\sqrt{5}}{2}}\)
Find a particular solution of \(x^2 y'' +(x-\frac{3}{4})y = 0\).
- Answer
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\(y=x^{3/2}\sum\limits_{k=0}^\infty \frac{(-1)^{-1}}{k!(k+2)!}x^{k}\) (Note that for convenience we did not pick \(a_{0}=1\).)
Find the general solution of \(x^2 y'' - x y' +y = 0\).
- Answer
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\(y=Ax+Bx\ln (x)\)