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# System prepended metadata

title: FINAL

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# FINAL
## Problem 1
By contour integration, 
y=(1-x^2)^{\frac{m}{2}} \frac{d^m P_\ell(x)}{dx^m}=\frac{1}{2^\ell}\frac{1}{2\pi i}\oint (1-x^2)^{\frac{m}{2}} \frac{d^m}{dx^m} \frac{(z^2-1)^{\ell}}{(z-x)^{\ell +1}} \, dz\\
=\frac{(\ell+m)!}{2^{\ell+1} \ell ! \pi i} \oint (1-x^2)^{\frac{m}{2}} \frac{(z^2-1)^{\ell}}{(z-x)^{\ell +m+1}} \, dz\\
Plugging the equation above into:
\left[\frac{d}{dx} (1-x^2)\frac{d}{dx} +\ell(\ell +1) -\frac{m^2}{1-x^2} \right] y(x) \\
=\frac{(\ell+m)!}{2^{\ell+1} \ell !  \pi i} \oint \left\lbrace
\frac{d}{dx} \left[ (1-x^2) \frac{(\ell+m+1)}{z-x} -mx \right ]+(1-x^2)^{-1} \left[ (1-x^2)\frac{(\ell+m+1)}{(z-x)}-mx \right]^2\right.\\

\ \left. +\left[\ell(\ell+1)- \frac{m^2}{1-x^2}]\right] \right\rbrace \frac{(1-x^2)^\frac{m}{2}(z^2-1)^\ell}{(z-x)^{\ell+m+1}}\, dz \\

=\frac{(\ell+m)! (1-x^2)^\frac{m}{2}}{2^{\ell+1} \ell !  \pi i} \oint \left\lbrace
\left[ (1-x^2) \frac{(\ell+m+1)}{(z-x)^2} -m -2x\frac{(\ell+m+1)}{(z-x)} \right ] \right. \\

\  +\left.(1-x^2)\frac{(\ell+m+1)^2}{(z-x)^2}-2mx\frac{(\ell+m+1)}{(z-x)} \right.\\

\ \left. +\frac{m^2 x^2}{1-x^2}+\ell(\ell+1)- \frac{m^2}{1-x^2}\right\rbrace \frac{(z^2-1)^\ell}{(z-x)^{\ell+m+1}}\, dz \\

=\frac{(\ell+m)! (1-x^2)^\frac{m}{2}}{2^{\ell+1} \ell !  \pi i} \oint \left\lbrace
\frac{2(\ell+m+1)(m+1)((z-x)-z)}{(z-x)} \right. \\

\  +\left.\frac{(\ell+m+1)(\ell+m+2)(1-(z-x)^2+2z(z-x)-z^2)}{(z-x)^2} \right.\\

\ \left. +\ell(\ell+1)-m^2-m\right\rbrace \frac{(z^2-1)^\ell}{(z-x)^{\ell+m+1}}\, dz \\

=\frac{(\ell+m+1)! (1-x^2)^\frac{m}{2}}{2^{\ell+1} \ell !  \pi i} \oint \left\lbrace
2(m+1)-(\ell+m+2)+(\ell-m) \right. \\

\  -\left.\frac{(m+1)2z}{(z-x)}+\frac{\ell+m+2}{(z-x)}+\frac{(\ell+m+2)(1-z^2)}{(z-x)^2} \right\rbrace \frac{(z^2-1)^\ell}{(z-x)^{\ell+m+1}}\, dz \\

=\frac{(\ell+m+1)! (1-x^2)^\frac{m}{2}}{2^{\ell+1} \ell !  \pi i} 
\oint \left[ \frac{(\ell+1)2z(z^2-1)^\ell}{(z-x)^{\ell+m+2}}-\frac{(\ell+m+2)(z^2-1)^{\ell+1}}{(z-x)^{\ell+m+3}} \right]\, dz\\

=\frac{(\ell+m+1)! (1-x^2)^\frac{m}{2}}{2^{\ell+1} \ell !  \pi i} 
\oint \left[ \frac{(z^2-1)^{\ell+1}}{(z-x)^{\ell+m+2}} \right]\, dz=0\\

if the branch cuts for (-\infty,-1) and the line segment from  z=1 to z=x are chosen.

## Problem 2
To solve this problem, we first define W(x)\equiv Y_{\nu}(x)Z'_{\nu}(x)-Y'_{\nu}(x)Z_{\nu}(x)
and we will get the relation as shown below:
W'(x)=Z''Y-Y''Z
=-Y[\frac{Z'}{x}+(1-\frac{\nu^{2}}{x^{2}})z]+Z{\frac{Y'}{x}+(1-\frac{\nu^{2}}{x^{2}})Y}
=\frac{1}{x}(-YZ'+ZY')=-\frac{1}{x}W
\therefore W(x)=\frac{A_{\nu}}{x}is the solution of the differential equation.

## Problem 3

(a)

\because \rho J_v(k\rho)\frac{dJ_v(k'\rho)}{d\rho}-\rho\frac{dJ_v(k\rho)}{d\rho}J_v(k'\rho)
=\sum_{n,m}{\frac{(-1)^{n+m}}{n!m!(n+v)!(m+v)!}}[(2m+v)(\frac{k\rho}{2})^{2n+v}(\frac{k'\rho}{2})^{2m+v}-(2n+v)(\frac{k\rho}{2})^{2n+v}(\frac{k'\rho}{2})^{2m+v}];
Therefore, by assuming \rho<<0, we get \frac{-2(\frac{k}{2})^{v}(\frac{k'}{2})^{v}}{v!(v+1)!}((\frac{k'}{2})^{2}-(\frac{k}{2})^{2})\rho^{2v+2};
If we let  \rho \rightarrow0 and Re(v)>-1, we get 
lim_{\rho\rightarrow0}(\frac{-2(\frac{k}{2})^{v}(\frac{k'}{2})^{v}}{v!(v+1)!}((\frac{k'}{2})^{2}-(\frac{k}{2})^{2})\rho^{2v+2} )\rightarrow0;

(b)

By letting \psi_{1}(\rho) = J_{\nu}(k'\rho) and \psi_{2}(\rho) = J_{\nu}(k\rho)
We obtain \psi_{2}\frac{1}{\rho}\frac{d}{d\rho}(\rho\frac{d\psi_{1}}{d\rho})+(k'^{2}-\frac{\nu^{2}}{\rho^{2}})\psi_{1}\psi_{2}=0
and \psi_{1}\frac{1}{\rho}\frac{d}{d\rho}(\rho\frac{d\psi_{2}}{d\rho})+(k'^{2}-\frac{\nu^{2}}{\rho^{2}})\psi_{1}\psi_{2}=0
\Rightarrow (k'^{2}-k^{2})\int_{0}^{a}\rho\psi_{1}\psi{2}d\rho= \int_{0}^{a}\psi_{1}\frac{d}{d\rho}(\rho\frac{d\psi_{2}}{d\rho})-\psi_{2}\frac{d}{d\rho}(\rho\frac{d\psi_{1}}{d\rho})d\rho;
=\int_{0}^{a}\frac{d}{d\rho}(\rho\psi_{1}\psi'_{2}-\rho\psi_{2}\psi'_{1})d\rho=[\rho\psi_{1}\psi'_{2}-\rho\psi_{2}\psi'_{1}]|_{0}^{a};
\Rightarrow J'_{\nu}(ka)=-\frac{\lambda}{a}J_{\nu}(ka)
Similarly,J'_{\nu}(k'a)=-\frac{\lambda}{a}J_{\nu}(k'a)becomes 
a(-\frac{\lambda}{a}J_{\nu}(k'a)J_{\nu}(ka)+\frac{\lambda}{a}J_{\nu}(k'a)J_{\nu}(ka))=0;
Therefore, \int_{0}^{a}\rho J_{\nu}(k'\rho)J_{\nu}(k\rho)d\rho=0;

### (c.)

As \int_{0}^{a}\rho J^{2}_{\nu}(k\rho)d\rho=\frac{\rho^{2}-\frac{\nu^{2}}{k^{2}}}{2}J^{2}_{\nu}(k\rho)|_{0}^{a}-k\int_{0}^{a}(\rho^{2}-\frac{\nu^{2}}{k^{2}})J_{\nu}(k\rho)J'_{\nu}(k\rho)d\rho;
=\frac{\rho^{2}-\frac{\nu^{2}}{k^{2}}}{2}J^{2}_{\nu}(k\rho)|_{0}^{a}-k\int_{0}^{a}(\rho^{2}-\frac{\nu^{2}}{k^{2}})J'_{\nu}(k\rho)\frac{\frac{1}{\rho}\frac{d}{d\rho}(\rho\frac{d}{d\rho}J_{\nu}(k\rho))}{\frac{\nu^{2}}{\rho^{2}}-k^{2}}d\rho
=\frac{\rho^{2}-\frac{\nu^{2}}{k^{2}}}{2}J^{2}_{\nu}(k\rho)|_{0}^{a}-k\int_{0}^{a}(\frac{\rho^{2}}{k^{2}})J'_{\nu}(k\rho)\frac{1}{\rho}\frac{d}{d\rho}(\rho J'_{\nu}(k\rho))d\rho;
=\frac{\rho^{2}-\frac{\nu^{2}}{k^{2}}}{2}J^{2}_{\nu}(k\rho)|_{0}^{a}-k\int_{0}^{a}\frac{d}{d\rho}(\frac{1}{2}\rho J'_{\nu}(k\rho)^{2})d\rho;
=\frac{a^{2}-\frac{\nu^{2}}{k^{2}}}{2}J^{2}_{\nu}(ka)+\frac{1}{2}a^{2}(J'_{\nu}(ka))^{2};
=\frac{a^{2}}{2}[(1-\frac{v^2}{a^{2}k^{2}})J_{\nu}^{2}(ka)+(J'_{\nu}(ka))^{2}];
=\frac{a^{2}}{2}[(1-\frac{v^2}{y_{m}})J_{\nu}^{2}(y_{m})+(J'_{\nu}(y_{m}))^{2}];
Thus, \int_{0}^{a}\rho f(\rho)J_{\nu}(\frac{y_{m}}{a}\rho)d\rho=\sum_{n=1}^{\infty}A_{m}\int_{0}^{a}\rho J^{2}_{\nu}(\frac{y_{m}}{a}\rho)d\rho;
=\frac{a^{2}}{2}[(1-\frac{v^2}{y_{m}})J_{\nu}^{2}(y_{m})+(J'_{\nu}(y_{m}))^{2}]A_{m};
Thus, A_{m}=\int_{0}^{a}\rho f(\rho)J_{\nu}(\frac{y_{m}}{a}\rho)d\rho / \frac{a^{2}}{2}[(1-\frac{v^2}{y_{m}})J_{\nu}^{2}(y_{m})+(J'_{\nu}(y_{m}))^{2}]

## Problem 4
Using the Rodriguez's form of Legendre polynomial:
P_n(x)=\frac{1}{2^n n!} \frac{d^n}{dx^n}(x^2-1)^n
Define, D\equiv\frac{d}{dx}
(x^2-1)^\ell has roots x=1 and x=-1 with multplicity \ell.
Next,
D(x^2-1)^\ell\\=D[(x-1)^\ell(x+1)^\ell]\\=\ell[(x-1)^{\ell-1}(x+1)^\ell+(x+1)^{\ell-1}(x-1)^\ell]\\=\ell(x-1)^{\ell-1}(x+1)^{\ell-1}[(x+1)^\ell+(x-1)^\ell]
Therefore by Rolle's theorem,D(x^2-1)^\ell must have one root in (-1,1)and (\ell-1) multiple roots at x=\pm1

Similiarly, D^2(x^2-1)^\ell must have two different roots in (-1,1) and (\ell-2)  multiple roots at x=\pm1.

Therefore,  D^\ell(x^2-1)^\ell must have \ell different roots in (-1,1). 
\therefore P_\ell(x) has\ell different zeroes at the interval (-1,1).


## Problem 5
### (a)
\frac{1}{x}\frac{d}{dx} x \frac{d\Omega_{\nu}^{(0)}}{dx}+(1-\frac{\nu^2}{x^2})\Omega_{\nu}^{(0)}(x)\\
=\frac{1}{2\pi i}\int_{C_0} (\frac{\partial^2}{\partial x^2} +\frac{\partial}{x\partial x}+1-\frac{\nu^2}{x^2})e^{\frac{x}{2}(t-\frac{1}{t})}t^{-\nu-1}dt\\
=\frac{1}{2\pi i}\int_{C_0} ((\frac{1}{2})^2(t-\frac{1}{t})^2+\frac{1}{2x}(t-\frac{1}{t})+1-\frac{\nu^2}{x^2})e^{\frac{x}{2}(t-\frac{1}{t})}t^{-\nu-1}dt\\
=\frac{1}{2\pi ix^2}\int_{C_0} t^{-\nu-1}[(\frac{x}{2})^2(t+\frac{1}{t})^2+t\frac{\partial}{\partial t}[\frac{x}{2}(t+\frac{1}{t})]-\nu^2]e^{\frac{x}{2}(t-\frac{1}{t})}dt\\
=\frac{1}{2\pi ix^2}\int_{C_0} t^{-\nu-1}[(t\frac{\partial}{\partial t})^2-\nu^2]e^{\frac{x}{2}(t-\frac{1}{t})}dt\\
=\frac{1}{2\pi ix^2}\int_{C_0} [t^{-\nu}\frac{\partial}{\partial t} t\frac{\partial e^{\frac{x}{2}(t-\frac{1}{t})}}{\partial t}-e^{\frac{x}{2}(t-\frac{1}{t})}\frac{\partial}{\partial t}t\frac{\partial t^{-\nu}}{\partial t}]dt\\
=\frac{1}{2\pi ix^2}\int_{C_0} [\frac{\partial}{\partial t}[ t\frac{\partial e^{\frac{x}{2}(t-\frac{1}{t})}}{\partial t}t^{-\nu}]-\frac{\partial}{\partial t}[e^{\frac{x}{2}(t-\frac{1}{t})}t\frac{\partial t^{-\nu}}{\partial t}]]dt\\
=\frac{1}{2\pi ix^2}\int_{C_0} \frac{\partial}{\partial t}[ e^{\frac{x}{2}(t-\frac{1}{t})}t^{-\nu}[\frac{x}{2}(t+\frac{1}{t})+\nu]    ] dt\\
=\frac{1}{2\pi ix^2}\int_{C_0} d[ e^{\frac{x}{2}(t-\frac{1}{t})}t^{-\nu}[\frac{x}{2}(t+\frac{1}{t})+\nu]    ] \\
The branch cut is chosen to be negative real axis and let s=xt/2\\
\int_{C_0} \frac{\partial}{\partial t}[ e^{\frac{x}{2}(t-\frac{1}{t})}t^{-\nu}[\frac{x}{2}(t+\frac{1}{t})-\nu]] dt\\
=(\frac{x}{2})^{\nu}\int_{C_0} [\frac{\partial}{\partial s} e^{(s-\frac{x^2}{4s})}s^{-\nu}(s+\frac{x^2}{4s}+\nu)]ds \\
=(\frac{x}{2})^{\nu}\lim_{\rho \to \infty}[\frac{\partial}{\partial s} e^{(s-\frac{x^2}{4s})}s^{-\nu}(s+\frac{x^2}{4s}+\nu)]_{s=\rho e^{-i\pi}}^{s=\rho e^{i\pi}}\\
=0 

### (b)
\lim_{x\rightarrow 0}\Omega ^{(0)}_{\nu}(x)=\lim_{x\rightarrow 0}\frac{1}{2\pi i}\int_{C_0}\frac{e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}}{t^{\nu+1}}dt
With the change of variable s=\frac{\:xt}{2}
=\lim_{x\rightarrow0}\frac{1}{2\pi i}(\frac{x}{2})^{\nu}\int_{C_{0}}\frac{e^{s-\frac{x^2}{4s}}}{s^{\nu+1}}ds
\rightarrow\frac{1}{2\pi i}\left(\frac{x}{2}\right)^{\nu}\int_{C_0}e^ss^{-\left(\nu+1\right)}ds
And switching to polar coordinates,
=\frac{1}{2\pi i}\left(\frac{x}{2}\right)^{\nu}\left[-\int^0_{\infty}e^{-\rho}\rho^{-\left(\nu+1\right)}e^{+i\pi\left(\nu+1\right)}d\rho-\int_0^{\infty}e^{-\rho}\rho^{-\left(\nu+1\right)}e^{-i\pi\left(\nu+1\right)}d\rho\right]
=\frac{1}{2\pi i}\left(\frac{x}{2}\right)^{\nu}\int^{\infty}_0e^{-\rho}\rho^{-\left(\nu+1\right)}\left(e^{i\pi\left(\nu+1\right)}-e^{-i\pi\left(\nu+1\right)}\right)d\rho
=-\frac{\sin\nu\pi}{\:\pi}\left(\frac{x}{2}\right)^{\nu}\Gamma\left(-\nu\right)=\frac{-\Gamma\left(-\nu\right)}{\Gamma\left(\nu\right)\Gamma\left(1-\nu\right)}\left(\frac{x}{2}\right)^{\nu}
=\frac{-\Gamma\left(-\nu\right)}{\Gamma\left(\nu\right)\left(-\nu\right)\Gamma\left(-\nu\right)}\left(\frac{x}{2}\right)^{\nu}=\frac{1}{\Gamma\left(\nu+1\right)}\left(\frac{x}{2}\right)^{\nu}=\frac{1}{\nu!}\left(\frac{x}{2}\right)^{\nu}



### (c)
The difference between \Omega and H is only that the contour for H is the contour of \Omega separated at \epsilon. So the exact differential \int d\left\{e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}t^{-\nu}\left[\frac{x}{2}\left(t+\frac{1}{t}\right)+\nu\right]\right\} that appeared in (a) will still vanish as t\rightarrow0 because of the e^{-\frac{x}{2t}}\rightarrow0 factor. Hence the separate portions (\epsilon to -\infty+i\epsilon) and (-\infty-i\epsilon to \epsilon) is still a solution of Bessel's equation.

\frac{1}{2}\left(H^{\left(1\right)}_{\nu}\left(x\right)+H^{\left(2\right)}_{\nu}\left(x\right)\right)=\frac{1}{2\pi i}\int_{C_1}\frac{e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}}{t^{\nu+1}}dt+\frac{1}{2\pi i}\int_{C_2}\frac{e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}}{t^{\nu+1}}dt

=\frac{1}{2\pi i}\int^{-\infty+i\epsilon}_{\epsilon}\frac{e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}}{t^{\nu+1}}dt+\frac{1}{2\pi i}\int^{\epsilon}_{-\infty-i\epsilon}\frac{e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}}{t^{\nu+1}}dt

=\frac{1}{2\pi i}\int^{-\infty+i\epsilon}_{-\infty-i\epsilon}\frac{e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}}{t^{\nu+1}}dt=\Omega_{\nu}^{\left(0\right)}




## Problem 6
For the Laguerre equation
xy''+(1-x)y'+\lambda y=0\\
y''+\frac{(1-x)}{x}y'+\frac{\lambda}{x}y=0
since x=0 is a regular singular point and y can be express as a solution with the series expansion
y=\sum_{k=0}^{\infty}a_kx^{k+\alpha}
where a_0\neq0
\Rightarrow xy''+(1-x)y'+\lambda y\\=\sum_{k=0}^{\infty}[(k+\alpha)(k+\alpha-1)+(1-x)(k+\alpha)]a_kx^{k+\alpha-1}+\sum_{k=0}^{\infty}\lambda a_kx^{k+\alpha}\\=\sum_{k=0}^{\infty}(k+\alpha)^2a_kx^{k+\alpha-1}+\sum_{k=0}^{\infty}(\lambda -k-\alpha)a_kx^{k+\alpha}\\=\alpha^{2}a_0x^{\alpha-1}+\sum_{k=0}^{\infty}[(k+1+\alpha)^2a_{k+1}+(\lambda -k-\alpha)a_k]x^{k+\alpha}----(1)
Where \alpha^{2}a_0x^{\alpha-1} is the k=0 term of
\sum_{k=0}^{\infty}(k+\alpha)^2a_kx^{k+\alpha-1}
(1) leads that \alpha=0 and a_{k+1}=\frac{(k-\lambda)}{(k+1)^2}a_k when the equation (1)=0
For large k>m\gg 1 and m\gg \lambda
a_{k+1}\sim \frac{1}{k+1}a_k\sim \frac{1}{(k+1)k}a_{k-1}\sim \frac{(k-2)!}{(k+1)!}a_{k-2}\sim\cdots \sim \frac{m!a_m}{(k+1)!}
since k>m\gg 1 and m\gg \lambda, such that the terms of [0,m] vanish
\Rightarrow y\sim\sum_{k=m}^{\infty}a_kx^k\sim m!a_m\sum_{k=m}^{\infty}\frac{x^k}{k!}\sim m!a_me^x
When x\longrightarrow \infty
xy''+(1-x)y'+\lambda y=0\sim x(y''-y')=0
where x=0 and y''-y'=0
such that y’’-y’=0 has the solution of y\sim e^x where disverges as x\longrightarrow \infty unless \lambda=n is a non-negative integer which leads a_m=0
such that a_{n+1}=a_{n+2}=a_{n+3}=\cdots =0 and y is a polynomial of order n.
 
a_k=\frac{(k-1-n)}{k^2}a_{k-1}=\frac{(k-1-n)(k-2-n)}{(k^2)(k-1)^2}a_{k-2}=\cdots \\
=\frac{(k-1-n)(k-2-n)\cdots(-n)}{(k!)^2}a_0=\frac{(-1)^kn!}{(k!)^2(n-k)!}a_0
\Rightarrow y=\sum_{k=0}^{\infty}a_kx^k=\sum_{k=0}^{\infty}\frac{(-1)^kn!}{(k!)^2(n-k)!}a_0x^k
Where if a_0 define as a_0=1 will obtain the Laguerre polynomial
\Rightarrow L_n(x)=\sum_{k=0}^{\infty}\frac{(-1)^kn!}{(k!)^2(n-k)!}x^k
which is the solution of
xL_n^{''}+(1-x)L_n^{'}+\lambda L_n=0
