7.7: Problems

7.1. Consider the set of vectors \((-1,1,1),(1,-1,1),(1,1,-1)\).

a. Use the Gram-Schmidt process to find an orthonormal basis for \(R^{3}\) using this set in the given order.
b. What do you get if you do reverse the order of these vectors?

7.2. Use the Gram-Schmidt process to find the first four orthogonal polynomials satisfying the following:

a. Interval: \((-\infty, \infty)\) Weight Function: \(e^{-x^{2}}\).
b. Interval: \((0, \infty)\) Weight Function: \(e^{-x}\).

7.3. Find \(P_{4}(x)\) using

a. The Rodrigues Formula in Equation (7.12).
b. The three term recursion formula in Equation (7.14).

7.4. Use the generating function for Legendre polynomials to derive the recursion formula \(P_{n+1}^{\prime}(x)-P_{n-1}^{\prime}(x)=(2 n+1) P_{n}(x)\). Namely, consider \(\dfrac{\partial g(x, t)}{\partial x}\) using Equation (7.18) to derive a three term derivative formula. Then use three term recursion formula (7.14) to obtain the above result.

7.5. Use the recursion relation (7.14) to evaluate \(\int_{-1}^{1} x P_{n}(x) P_{m}(x) d x, n \leq m\).

7.6. Expand the following in a Fourier-Legendre series for \(x \in(-1,1)\).

a. \(f(x)=x^{2}\).

b. \(f(x)=5 x^{4}+2 x^{3}-x+3 .\)

c. \(f(x)=\left\{\begin{array}{l}-1,-1<x<0 \\ 1, \quad 0<x<1\end{array}\right.\)

d. \(f(x)=\left\{\begin{array}{l}x,-1<x<0 \\ 0,0<x<1\end{array}\right.\)

7.7. Use integration by parts to show \(\Gamma(x+1)=x \Gamma(x)\).

7.8. 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.

a. \(\int_{0}^{\infty} x^{2 / 3} e^{-x} d x\)
b. \(\int_{0}^{\infty} x^{5} e^{-x^{2}} d x\)
c. \(\int_{0}^{1}\left[\ln \left(\dfrac{1}{x}\right)\right]^{n} d x\)

7.9. The Hermite polynomials, \(H_{n}(x)\), satisfy the following:

i. \(<H_{n}, H_{m}>=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}\).
ii. \(H_{n}^{\prime}(x)=2 n H_{n-1}(x)\).
iii. \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\).
iv. \(H_{n}(x)=(-1)^{n} e^{x^{2}} \dfrac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right)\).

Using these, show that

a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\). [Use properties ii. and iii.]
b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right]\). [Use properties i. and iii.]
c. \(H_{n}(0)=\left\{\begin{array}{cc}0, & n \text { odd, } \\ (-1)^{m} \dfrac{(2 m) !}{m !}, & n=2 m\end{array}\right.\). [Let \(x=0\) in iii. and iterate. Note from iv. that \(H_{0}(x)=1\) and \(\left.H_{1}(x)=1 .\right]\)

7.10. In Maple one can type simplify(LegendreP \(\left.\left(2^{*} \mathbf{n}-\mathbf{2}, \mathbf{0}\right)-\operatorname{Legendre} \mathbf{P}\left(\mathbf{2}^{*} \mathbf{n}, \mathbf{0}\right)\right)\); 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 7.6 for Fourier-Legendre series, the value is given in terms of double factorials! So, we have

\[P_{2 n-2}(0)-P_{2 n}(0)=\dfrac{\sqrt{\pi}(4 n-1)}{2 \Gamma(n+1) \Gamma\left(\dfrac{3}{2}-n\right)}=(-1)^{n} \dfrac{(2 n-3) ! !}{(2 n-2) ! !} \dfrac{4 n-1}{2 n} . \nonumber \]

You will verify that both results are the same by doing the following

a. Prove that \(P_{2 n}(0)=(-1)^{n} \dfrac{(2 n-1) ! !}{(2 n) ! !}\) using the generating function and a binomial expansion.
b. Prove that \(\Gamma\left(n+\dfrac{1}{2}\right)=\dfrac{(2 n-1) ! !}{2^{n}} \sqrt{\pi}\) using \(\Gamma(x)=(x-1) \Gamma(x-1)\) and iteration.
c. Verify the result from Maple that \(P_{2 n-2}(0)-P_{2 n}(0)=\dfrac{\sqrt{\pi}(4 n-1)}{2 \Gamma(n+1) \Gamma\left(\dfrac{3}{2}-n\right)}\).
d. Can either expression for \(P_{2 n-2}(0)-P_{2 n}(0)\) be simplified further?

7.11. 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 \(a_{j}=\dfrac{-1}{j(2 n+j)} a_{j-2}\). 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 \(j=2 k\):

\[J_{n}(x)=\sum_{k=0}^{\infty} \dfrac{(-1)^{k}}{k !(n+k) !}\left(\dfrac{x}{2}\right)^{n+2 k} \nonumber \]

7.12. Use the infinite series in the last problem to derive the derivative identities (7.41) and (7.42):

a. \(\dfrac{d}{d x}\left[x^{n} J_{n}(x)\right]=x^{n} J_{n-1}(x)\).
b. \(\dfrac{d}{d x}\left[x^{-n} J_{n}(x)\right]=-x^{-n} J_{n+1}(x)\).

7.13. 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.

a. Put this differential equation into Sturm-Liouville form.
b. Prove that solutions corresponding to different eigenvalues are orthogonal by first writing the corresponding Green's identity using these Bessel functions.
c. Prove that

\(\int_{0}^{1} x J_{p}(\lambda x) J_{p}(\mu x) d x=\dfrac{1}{2} J_{p+1}^{2}(\lambda)=\dfrac{1}{2} J_{p}^{\prime 2}(\lambda)\)

Note that \(\lambda\) is a zero of \(J_{p}(x)\).

7.14. We can rewrite our Bessel function in a form which will allow the order to be non-integer by using the gamma function. You will need the results from Problem 7.10b for \(\Gamma\left(k+\dfrac{1}{2}\right)\).

a. Extend the series definition of the Bessel function of the first kind of order \(\nu, J_{\nu}(x)\), for \(\nu \geq 0\) by writing the series solution for \(y(x)\) in Problem 7.11 using the gamma function.
b. Extend the series to \(J_{-\nu(x)}\), for \(\nu \geq 0\). Discuss the resulting series and what happens when \(\nu\) is a positive integer.
c. Use these results to obtain closed form expressions for \(J_{1 / 2}(x)\) and \(J_{-1 / 2}(x)\). Use the recursion formula for Bessel functions to obtain a closed form for \(J_{3 / 2}(x)\).

7.15. In this problem you will derive the expansion

\[x^{2}=\dfrac{c^{2}}{2}+4 \sum_{j=2}^{\infty} \dfrac{J_{0}\left(\alpha_{j} x\right)}{\alpha_{j}^{2} J_{0}\left(\alpha_{j} c\right)}, \quad 0<x<c, \nonumber \]

where the \(\alpha_{j}^{\prime} s\) are the positive roots of \(J_{1}(\alpha c)=0\), by following the below steps.

a. List the first five values of \(\alpha\) for \(J_{1}(\alpha c)=0\) using the Table 7.4 and Figure 7.7. [Note: Be careful determining \(\alpha_{1}\).]
b. Show that \(\left\|J_{0}\left(\alpha_{1} x\right)\right\|^{2}=\dfrac{c^{2}}{2}\). Recall,

\(\left\|J_{0}\left(\alpha_{j} x\right)\right\|^{2}=\int_{0}^{c} x J_{0}^{2}\left(\alpha_{j} x\right) d x\)

c. Show that \(\left\|J_{0}\left(\alpha_{j} x\right)\right\|^{2}=\dfrac{c^{2}}{2}\left[J_{0}\left(\alpha_{j} c\right)\right]^{2}, j=2,3, \ldots\) (This is the most involved step.) First note from Problem 7.13 that \(y(x)=J_{0}\left(\alpha_{j} x\right)\) is a solution of

\(x^{2} y^{\prime \prime}+x y^{\prime}+\alpha_{j}^{2} x^{2} y=0 .\)

i. Show that the Sturm-Liouville form of this differential equation is \(\left(x y^{\prime}\right)^{\prime}=-\alpha_{j}^{2} x y\)
ii. Multiply the equation in part i. by \(y(x)\) and integrate from \(x=0\) to \(x=c\) to obtain

\[\begin{aligned}
\int_{0}^{c}\left(x y^{\prime}\right)^{\prime} y d x &=-\alpha_{j}^{2} \int_{0}^{c} x y^{2} d x \\
&=-\alpha_{j}^{2} \int_{0}^{c} x J_{0}^{2}\left(\alpha_{j} x\right) d x
\end{aligned} \label{7.67} \]

iii. 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.

1. \(J_{0}^{\prime}(x)=-J_{1}(x)\) from Equation (7.42).
2. Equation (7.45).
3. \(J_{2}\left(\alpha_{j} c\right)+J_{0}\left(\alpha_{j} c\right)=0\) from Equation (7.43).

iv. Now you should have enough information to complete this part.

d. Use the results from parts \(b\) and \(c\) to derive the expansion coefficients for

\(x^{2}=\sum_{j=1}^{\infty} c_{j} J_{0}\left(\alpha_{j} x\right)\)

in order to obtain the desired expansion.

7.16. Use the derivative identities of Bessel functions, (7.41)-(7.42), and integration by parts to show that

\[\int x^{3} J_{0}(x) d x=x^{3} J_{1}(x)-2 x^{2} J_{2}(x) \nonumber \]