5.7: Problems
- Page ID
- 90948
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider the set of vectors \((-1,1,1),(1,-1,1),(1,1,-1)\).
- Use the Gram-Schmidt process to find an orthonormal basis for \(R^{3}\) using this set in the given order.
- What do you get if you do reverse the order of these vectors?
Use the Gram-Schmidt process to find the first four orthogonal polynomials satisfying the following:
- Interval: \((-\infty, \infty)\) Weight Function: \(e^{-x^{2}}\).
- Interval: \((0, \infty)\) Weight Function: \(e^{-x}\).
In Equations (5.3.18)-(5.3.25) we provide several identities for Legendre polynomials. Derive the results in Equations (5.3.19)-(5.3.25) as described in the text. Namely,
- Differentiating Equation (5.3.18) with respect to \(x\), derive Equation (5.3.19).
- Derive Equation (5.3.20) by differentiating \(g(x, t)\) with respect to \(x\) and rearranging the resulting infinite series.
- Combining the last result with Equation (5.3.18), derive Equations (5.3.21)-(5.3.22).
- Adding and subtracting Equations (5.3.21)-(5.3.22), obtain Equations (5.3.23)-(5.3.24).
- Derive Equation (5.3.25) using some of the other identities.
Use the recursion relation (5.3.5) to evaluate \(\int_{-1}^{1} x P_{n}(x) P_{m}(x) d x, n \leq m\).
Expand the following in a Fourier-Legendre series for \(x \in(-1,1)\).
- \(f(x)=x^{2}\).
- \(f(x)=5 x^{4}+2 x^{3}-x+3\).
- \(f(x)=\left\{\begin{array}{cc}-1, & -1<x<0, \\ 1, & 0<x<1 .\end{array}\right.\)
- \(f(x)=\left\{\begin{array}{cc}x, & -1<x<0, \\ 0, & 0<x<1 .\end{array}\right.\)
Use integration by parts to show \(\Gamma(x+1)=x \Gamma(x)\).
Prove the double factorial identities:
\[(2 n) ! !=2^{n} n !\nonumber \]
and
\[(2 n-1) ! !=\frac{(2 n) !}{2^{n} n !} .\nonumber \]
Express the following as Gamma functions. Namely, noting the form \(\Gamma(x+1)=\int_{0}^{\infty} t^{x} e^{-t} d t\) and using an appropriate substitution, each expression can be written in terms of a Gamma function.
- \(\int_{0}^{\infty} x^{2 / 3} e^{-x} d x\).
- \(\int_{0}^{\infty} x^{5} e^{-x^{2}} d x\)
- \(\int_{0}^{1}\left[\ln \left(\frac{1}{x}\right)\right]^{n} d x\)
The coefficients \(C_{k}^{p}\) in the binomial expansion for \((1+x)^{p}\) are given by
\[C_{k}^{p}=\frac{p(p-1) \cdots(p-k+1)}{k !} .\nonumber \]
- Write \(C_{k}^{p}\) in terms of Gamma functions.
- For \(p=1 / 2\) use the properties of Gamma functions to write \(C_{k}^{1 / 2}\) in terms of factorials.
- Confirm you answer in part \(b\) by deriving the Maclaurin series expansion of \((1+x)^{1 / 2}\).
The Hermite polynomials, \(H_{n}(x)\), satisfy the following:
- \(\left\langle H_{n}, H_{m}\right\rangle=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}\).
- \(H_{n}^{\prime}(x)=2 n H_{n-1}(x)\).
- \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\).
- \(H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right)\).
Using these, show that
- \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\). [Use properties ii. and iii.]
- \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right]\). [Use properties i. and iii.]
- \(H_{n}(0)=\left\{\begin{array}{cc}0, & n \text { odd, } \\ (-1)^{m} \frac{(2 m) !}{m !}, & n=2 m .\end{array} \quad\right.\) [Let \(x=0\) in iii. and iterate. Note from iv. that \(H_{0}(x)=1\) and \(H_{1}(x)=2 x\). ]
In Maple one can type simplify(LegendreP \(\left(2^{*} \mathrm{n}-2,0\right)\)-LegendreP \(\left(2^{*} \mathrm{n}, 0\right)\) ); to find a value for \(P_{2 n-2}(0)-P_{2 n}(0)\). It gives the result in terms of Gamma functions. However, in Example 5.3.8 for Fourier-Legendre series, the value is given in terms of double factorials! So, we have
\[P_{2 n-2}(0)-P_{2 n}(0)=\frac{\sqrt{\pi}(4 n-1)}{2 \Gamma(n+1) \Gamma\left(\frac{3}{2}-n\right)}=(-1)^{n} \frac{(2 n-3) ! !}{(2 n-2) ! !} \frac{4 n-1}{2 n} .\nonumber \]
You will verify that both results are the same by doing the following:
- Prove that \(P_{2 n}(0)=(-1)^{n} \frac{(2 n-1) ! !}{(2 n) ! !}\) using the generating function and a binomial expansion.
- Prove that \(\Gamma\left(n+\frac{1}{2}\right)=\frac{(2 n-1) ! !}{2^{n}} \sqrt{\pi}\) using \(\Gamma(x)=(x-1) \Gamma(x-1)\) and iteration.
- Verify the result from Maple that \(P_{2 n-2}(0)-P_{2 n}(0)=\frac{\sqrt{\pi}(4 n-1)}{2 \Gamma(n+1) \Gamma\left(\frac{3}{2}-n\right)}\).
- Can either expression for \(P_{2 n-2}(0)-P_{2 n}(0)\) be simplified further?
A solution Bessel’s equation, \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-n^{2}\right) y=0,\), can be found using the guess \(y(x)=\sum_{j=0}^{\infty} a_{j} x^{j+n}\). One obtains the recurrence relation \(a_{j}=\frac{-1}{(2 n+i)} a_{j-2}\). Show that for \(a_{0}=\left(n ! 2^{n}\right)^{-1}\) we get the Bessel function of the first kind of order \(n\) from the even values \(j=2 k\) :
\[J_{n}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !(n+k) !}\left(\frac{x}{2}\right)^{n+2 k} .\nonumber \]
Prove the following identities based on those in the last problem.
- \(J_{p-1}(x)+J_{p+1}(x)=\frac{2 p}{x} J_{p}(x)\).
- \(J_{p-1}(x)-J_{p+1}(x)=2 J_{p}^{\prime}(x)\).
Use the generating function to find \(J_{n}(0)\) and \(J_{n}^{\prime}(0)\).
Bessel functions \(J_{p}(\lambda x)\) are solutions of \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(\lambda^{2} x^{2}-p^{2}\right) y=0\). Assume that \(x \in(0,1)\) and that \(J_{p}(\lambda)=0\) and \(J_{p}(0)\) is finite.
- This is the standard Sturm-Liouville form for Bessel’s equation.
- by considering
\[\int_{0}^{1}\left[J_{p}(\mu x) \frac{d}{d x}\left(x \frac{d}{d x} J_{p}(\lambda x)\right)-J_{p}(\lambda x) \frac{d}{d x}\left(x \frac{d}{d x} J_{p}(\mu x)\right)\right] d x .\nonumber \]
We can rewrite Bessel functions, \(J_{v}(x)\), in a form which will allow the order to be non-integer by using the gamma function. You will need the results from Problem \(\PageIndex{12}\)b for \(\Gamma\left(k+\frac{1}{2}\right)\).
- Extend the series definition of the Bessel function of the first kind of order \(v, J_{v}(x)\), for \(v \geq 0\) by writing the series solution for \(y(x)\) in Problem \(\PageIndex{13}\) using the gamma function.
- Extend the series to \(J_{-v}(x)\), for \(v \geq 0\). Discuss the resulting series and what happens when \(v\) is a positive integer.
- Use the results in part \(c\) with the recursion formula for Bessel functions to obtain a closed form for \(J_{3 / 2}(x)\).
In this problem you will derive the expansion
\[x^{2}=\frac{c^{2}}{2}+4 \sum_{j=2}^{\infty} \frac{J_{0}\left(\alpha_{j} x\right)}{\alpha_{j}^{2} J_{0}\left(\alpha_{j} c\right)}, \quad 0<x<c,\nonumber \]
where the \(\alpha_{j}^{\prime} s\) are the positive roots of \(J_{1}(\alpha c)=0\), by following the below steps.
- List the first five values of \(\alpha\) for \(J_{1}(\alpha c)=0\) using the Table 5.5.1 and Figure 5.5.1. [Note: Be careful determining \(\alpha_{1}\).]
- Show that \(\left\|J_{0}\left(\alpha_{j} x\right)\right\|^{2}=\frac{c^{2}}{2}\left[J_{0}\left(\alpha_{j} c\right)\right]^{2}, j=2,3, \ldots\). (This is the most involved step.) First note from Problem \(\PageIndex{18}\) that \(y(x)=J_{0}\left(\alpha_{j} x\right)\) is a solution of
\[x^{2} y^{\prime \prime}+x y^{\prime}+\alpha_{j}^{2} x^{2} y=0 .\nonumber \]
- Verify the Sturm-Liouville form of this differential equation: \(\left(x y^{\prime}\right)^{\prime}=-\alpha_{j}^{2} x y .\)
- Noting that \(y(x)=J_{0}\left(\alpha_{j} x\right)\), integrate the left hand side by parts and use the following to simplify the resulting equation.
- Now you should have enough information to complete this part.
- in order to obtain the desired expansion.