# Problems for Chapter 3 -- Part 2-2
###### tags: `應用數學三`
$\require{cancel}$
## Problem 6
:::warning
Show that $$e^{\frac{x}{2}\left(t+\frac{1}{t}\right)}=\sum^{\infty}_{n=-\infty}I_n(x)t^n,\tag{1}$$ thus generating modified Bessel functions, $I_n(x)$.
:::
### Solution
The modified Bessel function, defined as $I_n(x)\equiv i^{-\nu}J_{\nu}(ix),$ has a series representation: $$\begin{align}I_n(x)&=i^{-\nu}\sum^\infty_{j=0}\dfrac{(-1)^j}{j!\,(\nu+j)!}\left(\dfrac{ix}{2}\right)^{2j+\nu}\\
&=\bcancel{i^{-\nu}}\bcancel{i^{\nu}}\sum^\infty_{j=0}\dfrac{\cancel{(-1)^j}\cancel{(i)^{2j}}}{j!\,(\nu+j)!}\left(\dfrac{x}{2}\right)^{2j+\nu}\\
&=\sum^\infty_{j=0}\dfrac{1}{j!\,(\nu+j)!}\left(\dfrac{x}{2}\right)^{2j+\nu}.\tag{2}\end{align}$$ Consider the Maclaurin seires of the function $e^{\frac{x}{2}\left(t+\frac{1}{t}\right)}$ given in Eqn. (1) $$\require{color}\begin{align}e^{\frac{x}{2}t}e^{\frac{x}{2}\frac{1}{t}}&=\left[\sum^\infty_{j=0}\dfrac{\left(\frac{x}{2}\right)^jt^j}{j!}\right]\left[\sum^\infty_{k=0}\dfrac{\left(\frac{x}{2}\right)^kt^{-k}}{k!}\right]\\&=\sum^\infty_{j=0}\sum^\infty_{k=0}\dfrac{\left(\frac{x}{2}\right)^{j\color{red}-k+2k}}{\left(j\color{red}-k+k \right)!\,k!}t^{j-k}\\&=\sum^\infty_{\color{blue}n=-k}\sum^\infty_{k=0}\dfrac{\left(\frac{x}{2}\right)^{\left.\color{blue}n\right.+2k}}{(\left.\color{blue}n\right.+k )!\,k!}t^{\color{blue}n}\\&=\sum^\infty_{n=-\infty} \sum^\infty_{k=0}\dfrac{\left(\frac{x}{2}\right)^{n+2k}}{(n+k)!\,k!}t^{n};\tag{3}\end{align}$$ Comparing Eqs. (2) & (3), we see that $$e^{\frac{x}{2}\left(t+\frac{1}{t}\right)}=\sum^{\infty}_{n=-\infty}I_n(x)t^n,$$ where $\nu\leftrightarrow n$, $j\leftrightarrow k$; but notice that in Eqn. (3), $n$ is integer, whilst $\nu$ is real. $\blacksquare$
## Problem 7
:::warning
Verify that $K_\nu(x)$ is given by $$\require{color}\require{cancel}K_\nu(x)=\frac{\pi}{2}\dfrac{I_{-\nu}(x)-I_{\nu}(x)}{\sin(\nu\pi)}\tag{1}$$ and from this show that $$K_\nu(x)=K_{-\nu}(x)\tag{2}$$
:::
### Solution
1. The definitions of $K_\nu$ and $I_\nu$ are (Eqn.(3.63) in the lecture note) $$I_\nu(x)\equiv i^{-\nu}J_\nu(ix),\tag{3}$$ $$K_\nu(x)\equiv\dfrac{\pi}{2}i^{\nu+1}H^{(1)}_\nu(ix),\tag{4}$$ where $H^{(1)}_\nu$ is one of the third kind of Bessel functions which relate to $J_\nu$ by (Eqn.(3.51) in the lecture note) $$H^{(1)}_\nu(x)=\dfrac{J_\nu(x)e^{-i\nu\pi}-J_{-\nu}(x)}{-i\sin(\nu\pi)}.\tag{5}$$
Then we use Eqn. (5) with $x$ replacd by $ix$ to rewirte Eqn. (4): $$\begin{align}K_\nu(x)&=\dfrac{\pi}{2}i^{\nu+\bcancel{1}}\dfrac{J_\nu(ix)e^{-i\nu\pi}-J_{-\nu}(ix)}{-\bcancel{i}\sin(\nu\pi)}\\
&=\dfrac{\pi}{2}\dfrac{-\left.\color{red}\bcancel{i^{\nu}}e^{i\nu\pi/2}\right.J_\nu(ix)e^{-i\nu\pi}+i^{\nu}J_{-\nu}(ix)}{\sin(\nu\pi)}\\
&=\dfrac{\pi}{2}\dfrac{i^{\nu}J_{-\nu}(ix)-i^{-\nu}J_\nu(ix)}{\sin(\nu\pi)}\\
&=\frac{\pi}{2}\dfrac{I_{-\nu}(x)-I_{\nu}(x)}{\sin(\nu\pi)},\end{align}$$ where we use Eqn. (3) to replace $J$ with $I$ and finally arrive at Eqn. (1). $\blacksquare$

2. Replacing $\nu$ by $-\nu$ in Eqn. (1), we obtain Eqn. (2): $$K_{-\nu}(x)=\dfrac{\pi}{2}\dfrac{I_{\nu}(x)-I_{-\nu}(x)}{\sin(-\nu\pi)}=\dfrac{\pi}{2}\dfrac{I_{\nu}(x)-I_{-\nu}(x)}{-\sin(\nu\pi)}=K_\nu(x).\ \blacksquare$$
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## Problem 8
:::warning
The Rodrigues’s form for the Laguerre polynomial is $$L_n(x)=\dfrac{e^x}{n!}\dfrac{\text{d}^n}{\text{d}x^n}\left(x^ne^{-x}\right),\quad n=0,1,2,3,\dots\tag{1}$$
- **(a)** Show that $$\sum^{\infty}_{n=0}L_n(x)t^n=\dfrac{e^{-\frac{xt}{1-t}}}{1-t}.\tag{2}$$
- **(b)** Show that $L_n(x)$ satisfies the Laguerre equation $$x\dfrac{\text{d}^2L_n(x)}{\text{d}x^2}+(1-x)\dfrac{\text{d}L_n(x)}{\text{d}x}+nL_n(x)=0\tag{3}$$
- **(c\)** Show that $$\begin{align}L'_n(0)&=-n\tag{4a}\\L''_n(0)&=\dfrac{1}{2}n(n-1)\tag{4b}\end{align}$$
:::
### Solution of (a)
Let $C$ be a positively oriented simple closed contour which encloses the origin ($z=0$) and a fixed point on the real axis ($z=x\in\mathbb{R})$.

Since the complex function $z^ne^{-z}$ is analytic everywhere on $\mathbb{C}$, thus the ***[Cauchy's differentiation formula](https://hackmd.io/@ulynx/complex0)*** gives $$\begin{align}L_n(x)&=\dfrac{e^x}{n!}\dfrac{\text{d}^n}{\text{d}x^n}\left(x^ne^{-x}\right)=\frac{e^x}{\cancel{n!}}\dfrac{\cancel{n!}}{2\pi i}\oint_{C}\dfrac{s^ne^{-s}}{(s-x)^{n+1}}\text{d}s\\&=\frac{e^x}{2\pi i}\oint_{C}\dfrac{s^ne^{-s}}{(s-x)^{n+1}}\text{d}s,\tag{5}\end{align}$$ where $s$ denotes points on $C$. Changing the integration variable to $\color{blue}t=\frac{s-x}{s}$, so that $\color{blue}s=\frac{x}{1-t}$, $\color{blue}\text{d}s=\frac{x}{(1-t)^2}\text{d}t$, we can recast Eqn. (5) as $$\begin{align}L_n(x)&=\frac{e^x}{2\pi i}\oint_{C}\left[\left(\color{blue}\frac{s}{s-x}\right)^{n\color{red}+1}\dfrac{e^{-\color{blue}s}}{\color{red}s}\right]\color{blue}\text{d}s\\
&=\frac{e^x}{2\pi i}\oint_{C}\left[\left(\color{blue}\dfrac{1}{t}\right)^{n+1}\dfrac{e^{-\color{blue}\frac{x}{1-t}}}{\color{blue}\frac{\bcancel{x}}{\bcancel{1-t}}}\right]\left[\color{blue}\frac{\bcancel{x}}{(1-t)^\bcancel{2}}\text{d}t\right]\\&=\frac{1}{2\pi i}\oint_{C}\dfrac{e^{-\frac{xt}{1-t}}}{(1-t)t^{n+1}}\text{d}t.\tag{6}\quad \end{align}$$ Since the function $\frac{1}{(1-z)}e^{-\frac{xz}{1-z}}$ is analytic elsewhere than $z=1$, if we ==restrict $C$ from enclosing $z=1$==, then using Cauchy's differentiation formula again, Eqn. (6) becomes $$L_n(x)=\left.\dfrac{1}{n!}\dfrac{\text{d}^n}{\text{d}x^n}\dfrac{e^{-\frac{xz}{1-z}}}{1-z}\right|_{z=0}.\tag{7}$$ But Eqn. (7) happens to be the coefficient of the Maclaurin series of $\frac{1}{(1-z)}e^{-\frac{xz}{1-z}}$, i.e. $$\sum^{\infty}_{n=0}L_n(x)z^n=\dfrac{e^{-\frac{xz}{1-z}}}{1-z}.$$ (Eqn. (2)) Therefore, we recognize $\frac{1}{(1-z)}e^{-\frac{xz}{1-z}}$ as the generating function of the Laguerre polynomial. $\blacksquare$

### Solution of (b)
#### Method 1
Substitute Eqn. (6) into Eqn. (3): $$\begin{align}&\quad~ x\dfrac{\text{d}^2L_n(x)}{\text{d}x^2}+(1-x)\dfrac{\text{d}L_n(x)}{\text{d}x}+nL_n(x)\\
&=\frac{1}{2\pi i}\oint_{C}\dfrac{\small x\frac{\partial^2}{\partial x^2}\left(e^{-\frac{xt}{1-t}}\right)+(1-x)\frac{\partial}{\partial x}\left(e^{-\frac{xt}{1-t}}\right)+ne^{-\frac{xt}{1-t}}}{(1-t)t^{n+1}}\text{d}t\\
&=\frac{1}{2\pi i}\oint_{C}\dfrac{\Big[ x\left(\frac{t}{1-t}\right)^2-(1-x)\left(\frac{t}{1-t}\right)+n\Big]e^{-\frac{xt}{1-t}}}{(1-t)t^{n+1}}\text{d}t\\
&=\frac{1}{2\pi i}\oint_{C}\left\{\dfrac{x\left[\frac{t^2}{(1-t)^2}+\frac{t}{1-t}\right]e^{-\frac{xt}{1-t}}}{(1-t)t^{n+1}}+\dfrac{\left(-\frac{t}{1-t}+n\right)e^{-\frac{xt}{1-t}}}{(1-t)t^{n+1}}\right\}\text{d}t\\
&=\frac{-1}{2\pi i}\oint_{C}\left\{\dfrac{-x\left[\frac{t}{(1-t)^2}+\frac{1}{t}\right]e^{-\frac{xt}{1-t}}}{(1-t)t^{n}}+\left[\small\dfrac{1}{(1-t)^2t^{n}}+\dfrac{-n}{(1-t)t^{n+1}}\right]e^{-\frac{xt}{1-t}}\right\}\text{d}t\\
&=\frac{-1}{2\pi i}\oint_{C}\left\{\frac{\partial}{\partial t}\left[e^{-\frac{xt}{1-t}}\right]\dfrac{1}{(1-t)t^{n}}+e^{-\frac{xt}{1-t}}\frac{\partial}{\partial t}\left[\dfrac{1}{(1-t)t^{n}}\right]\right\}\text{d}t\\&=\frac{-1}{2\pi i}\oint_{C}\left[\frac{\partial}{\partial t}\dfrac{e^{-\frac{xt}{1-t}}}{(1-t)t^{n}}\right]\text{d}t.\end{align}$$

As we require that the contour $C$ not enclose $z=1$, the integral above vanishes, and thus Eqn.(3) is proved. $\blacksquare$

#### Method 2

### Solution of (c\)
1. To evaluate $L'_n(0)$, we differentiate Eqn. (2) with respect to $x$ and set $x=0$: $$\sum^{\infty}_{n=0}L'_n(0)t^n=\left.\dfrac{\partial}{\partial x}\left(\dfrac{e^{-\frac{xt}{1-t}}}{1-t}\right)\right|_{x=0}=\left.\dfrac{-te^{-\frac{xt}{1-t}}}{(1-t)^2}\right|_{x=0}=\dfrac{-t}{(1-t)^2}.\tag{8}$$ Comparing $$\dfrac{t}{(1-t)^2}=t\dfrac{\text{d}}{\text{d}t}\dfrac{1}{1-t}=t\dfrac{\text{d}}{\text{d}t}\sum^{\infty}_{n=0}t^n=\sum^{\infty}_{n=0}nt^{n}$$ with Eqn. (8), we can identify $L'_n(0)=-n$. (Eqn.(4a))

2. Likewise, to evaluate $L''_n(0)$, we differentiate Eqn. (2) with respect to $x$ twice and set $x=0$: $$\sum^{\infty}_{n=0}L''_n(0)t^n=\left.\dfrac{\partial^2}{\partial x^2}\left(\dfrac{e^{-\frac{xt}{1-t}}}{1-t}\right)\right|_{x=0}=\left.\dfrac{t^2e^{-\frac{xt}{1-t}}}{(1-t)^3}\right|_{x=0}=\dfrac{t^2}{(1-t)^3}.\tag{9}$$ Comparing $$\dfrac{t^2}{(1-t)^3}=\dfrac{t^2}{2}\dfrac{\text{d}^2}{\text{d}t^2}\dfrac{1}{1-t}=\dfrac{t^2}{2}\dfrac{\text{d}^2}{\text{d}t^2}\sum^{\infty}_{n=0}t^n=\sum^{\infty}_{n=0}\dfrac{1}{2}n(n-1)t^{n}$$ with Eqn. (9), we can identify $L''_n(0)=\frac{1}{2}n(n-1)$. (Eqn.(4b)) $\blacksquare$

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## Problem 9
:::warning
The associated Laguerre polynomial is defined as $$L^k_n(x)\equiv(-1)^k\dfrac{\text{d}^kL_{n+k}(x)}{\text{d}x^k}.\tag{1}$$
- **(a)** Show that $$x\dfrac{\text{d}^2L^k_n(x)}{\text{d}x^2}+(k+1-x)\dfrac{\text{d}L^k_n(x)}{\text{d}x}+nL^k_n(x)=0\tag{2}$$ (that is to say Eqn.(1) satisifies ***associated Laguerre equation***.)
- **(b)** Show that (the ***orthonormality*** of associated Laguerre polynomial) $$\int^\infty_0e^{-x}x^kL^k_n(x)\,L^k_m(x)\,\text{d}x=\dfrac{(n+k)!}{n!}\delta_{nm}\tag{3}$$
:::
### Solution of (a)
#### Method 1
Define $D\equiv\frac{\text{d}}{\text{d}x}$ as the differentiatial operator.
The (ordinary) Laguerre polynomial of $L_n(x)$ satisfies $$\left[xD^2+(1-x)D+n\right]L_n(x)=0.\tag{4}$$
We introduce three procedures:
1. Apply $D$ on Eqn. (4): $$\begin{align} [xD^3+D^2+(1-x)D^2-D+nD]L_n(x)&=0\\\implies\left[xD^3+(2-x)D^2+(n-1)D\right]L_n(x)&=0.\end{align}$$
2. Replace $n$ by $n+1$: $$\left[xD^3+(2-x)D^2+nD\right]L_{n+1}(x)=0.$$
3. Factor out one $-D$ from the operator to its operand (this brings nothing to the right-hand side): $$\left[xD^2+(\color{orangered}1\color{black}+1-x)D+n\right]\left[(-D)^{\color{orangered}1}L_{n+\color{orangered}1}(x)\right]=0.\tag{4}$$

Repeating these steps another $k-1$ times, we may arrive at $$\left[xD^2+(\color{orangered}k\color{black}+1-x)D+n\right]\underbrace{\left[(-D)^{\color{orangered}k}L_{n+\color{orangered}k}(x)\right]}_{\large L^k_n(x)}=0.\tag{5}$$

Suppose Eqn. (5) is true for $k=p$, as long as we follow the three procedures, we can obtain Eqn. (5) for $k=p+1$: $$\begin{align}D\left[xD^2+(\color{orangered}p\color{black}+1-x)D+n\right]\left[(-D)^{\color{orangered}p}L_{n+\color{orangered}p}(x)\right]&=0\\
\implies\left[xD^3+D^2+(\color{orangered}p\color{black}+1-x)D^2-D+nD\right]\left[(-D)^{\color{orangered}p}L_{n+\color{orangered}p}(x)\right]&=0\\
\implies\left[xD^3+(\color{orangered}p+1\color{black}+1-x)D^2+(n-1)D\right]\left[(-D)^{\color{orangered}p}L_{n+\color{orangered}p}(x)\right]&=0\\
\implies\left[xD^3+(\color{orangered}p+1\color{black}+1-x)D^2+nD\right]\left[(-D)^{\color{orangered}p}L_{n+\color{orangered}p+1}(x)\right]&=0\\
\implies\left[xD^2+(\color{orangered}p+1\color{black}+1-x)D+n\right]\left[(-D)^{\color{orangered}p+1}L_{n+\color{orangered}p+1}(x)\right]&=0\end{align}\tag{6}$$ Eqn. (5) is justified by mathematical induction and Eqs. (4) & (6).

If we define $L^k_n(x)$ as in Eqn. (1), then Eqn. (5) is equivalent ot Eqn. (2). $\blacksquare$
#### Method 2
### Solution of (b)
#### Method 1
We will keep using the operator notation $D\equiv\frac{\text{d}}{\text{d}x}$.
1. In order to evaluate the integral in Eqn. (3), we need the Rodrigues’ formula for the associated Laguerre polynomial $$L^k_n(x)=\dfrac{e^xx^{-k}}{n!}D^n\left[x^{n+k}e^{-x}\right]\tag{7}$$ The derivation of Eqn. (7) is given as follows.

1. L

2. Orthogonality
Without loss of generality, let $n\leq m$ $\begin{align}&\quad\int^\infty_0e^{-x}x^kL^k_n(x)\,L^k_m(x)\,\text{d}x\\
&=\int^\infty_0\left[\cancel{e^{-x}}\cancel{x^k}\left(\dfrac{\cancel{e^{x}}\cancel{x^{-k}}}{n!}D^n\left[x^{n+k}e^{-x}\right]\right)\left(\dfrac{e^xx^{-k}}{m!}D^m\left[x^{m+k}e^{-x}\right]\right)\right]\,\text{d}x\\
&=\frac{1}{n!\,m!}\int^\infty_0\left(D^n\left[x^{n+k}e^{-x}\right]\right)\left(e^xx^{-k}D^m\left[x^{m+k}e^{-x}\right]\right)\,\text{d}x\\
&=\frac{(-1)^n}{n!\,m!}\int^\infty_0\left(x^{n+k}e^{-x}\right)D^n\left[e^xx^{-k}D^m\left[x^{m+k}e^{-x}\right]\right]\,\text{d}x\\
&=\frac{(-1)^n}{n!\cancel{m!}}\int^\infty_0\left(x^{n+k}e^{-x}\right)D_n\left[e^xx^{-k}\sum^{m}_{j=0}\dfrac{\cancel{m!}D^j\left[x^{m+k}\right]D^{n-j}\left[e^{-x}\right]}{j!\,(m-j)!}\right]\,\text{d}x\\
&=\frac{\bcancel{(-1)^n}}{n!}\int^\infty_0\left(x^{n+k}e^{-x}\right)D_n\left[\cancel{e^x}\bcancel{x^{-k}}\sum^{m}_{j=0}\dfrac{\frac{(m+k)!}{(m+k-j)!}x^{m+\bcancel{k}-j}(-1)^{\bcancel{n}-j}\cancel{e^{-x}}}{j!\,(m-j)!}\right]\,\text{d}x\\
&=\frac{1}{n!}\int^\infty_0\left(x^{n+k}e^{-x}\right)D_n\left[\sum^{\color{red}m}_{j=0}\dfrac{(m+k)!(-1)^{-j}x^{m-j}}{j!\,(m-j)!\,(m+k-j)!}\right]\,\text{d}x\\
&=\frac{1}{n!}\int^\infty_0\left(x^{n+k}e^{-x}\right)\left[\sum^{\color{red}m-n+1}_{j=0}\dfrac{(m+k)!(-1)^{-j}\frac{\bcancel{(m-j)!}}{(m-j-n)!}x^{m-j-n}}{j!\bcancel{(m-j)!}(m+k-j)!}\right]\,\text{d}x\\
&=\frac{(m+k)!}{n!}\left[\sum^{m-n+1}_{j=0}\dfrac{(-1)^{j}\int^\infty_0\left(x^{m+k-j}e^{-x}\right)\text{d}x}{j!\,(m-j-n)!\,(m+k-j)!}\right]\\
&=\frac{(m+k)!}{n!}\left[\sum^{m-n+1}_{j=0}\dfrac{(-1)^{j}\,\cancel{\Gamma(m+k-j+1)}}{j!\,(m-j-n)!\cancel{(m+k-j)!}}\right]\\
&=\frac{(m+k)!}{n!}\sum^{m-n+1}_{j=0}\dfrac{(-1)^{j}}{j!\,(m-j-n)!}=\frac{(m+k)!}{n!}\sum^{m-n+1}_{j=0}\dfrac{(-1)^{j}}{j!\,(m-j-n)!}\\&=\left\{\begin{array}{ll}\frac{(m+k)!}{n!}&\text{if} ~m= n,\\0&\text{if} ~m\neq n,\end{array}\right.\\&=\frac{(m+k)!}{n!}\delta_{mn}=\frac{(n+k)!}{n!}\delta_{mn}.~\blacksquare\end{align}$

#### Method 2
Instead of using Eqn. (2), we rewrite it as $$x^{-k}e^x\dfrac{\text{d}}{\text{d}x}\left[x^{k+1}e^{-x}\frac{\text{d}L^k_n(x)}{\text{d}x}\right]+nL^k_n(x)$$