Examples E

Section 4.10 Examples E

Hermite’s Function.

Example 4.10.1.

Prove that

\begin{equation*} H_{n}(-x)=(-1)^{n} H_{n}(x) \end{equation*}

Solution.

We know that

\begin{equation*} \sum_{n=0}^{\infty}\frac{H_{n}(x)}{n!}=e^{2zx-z^{2}} =e^{2zx}\cdot e^{-z^{2}} \end{equation*}

\begin{equation*} =\sum_{n=0}^{\infty}\frac{(2x)^{n}z^{n}}{n!}\cdot \sum_{n=0}^{\infty}\frac{(-1)^{n}z^{2n}}{n!} \end{equation*}

\begin{equation*} =\sum_{n=0}^{\infty}\sum_{s=0}^{\frac{n}{2}}\frac{(-1)^{s}(2x)^{n-2s}}{s!(n-2s)!}z^{n} \end{equation*}

Equating the coefficient of \(\frac{z^{n}}{n!}\) on both sides, we get -

\begin{equation*} H_{n}(x)=\sum_{s=0}^{\frac{n}{2}}\frac{(-1)^{s}n!(2x)^{n-2s}}{s!(n-2s)!} \end{equation*}

Replacing \(x\) by \(-x\text{,}\) we get -

\begin{equation*} H_{n}(-x)=\sum_{s=0}^{\frac{n}{2}}\frac{(-1)^{s}n!(-2x)^{n-2s}}{s!(n-2s)!} = \sum_{s=0}^{\frac{n}{2}}\frac{(-1)^{s}(-1)^{n-2s}n!(2x)^{n-2s}}{s!(n-2s)!} \end{equation*}

\begin{equation*} =(-1)^{n}\sum_{s=0}^{\frac{n}{2}}\frac{(-1)^{s}n!(2x)^{n-2s}}{s!(n-2s)!}=(-1)^{n}H_{n}(x) \end{equation*}

\begin{equation*} \therefore H_{n}(-x)=(-1)^{n} H_{n}(x) \end{equation*}

Example 4.10.2.

Prove that

\begin{equation*} \int\limits_{-\infty}^{\infty}e^{-x^{2}}H_{m}(x)H_{n}(x)\,dx =\sqrt{\pi}\left[2^{n-1}n!\delta_{m,n-1}+2^{n}(n+1)!\delta_{m,n+1}\right] \end{equation*}

Solution.

We have

\begin{equation*} \int\limits_{-\infty}^{\infty}xe^{-x^{2}}H_{m}(x)H_{n}(x)\,dx = \int\limits_{-\infty}^{\infty}\frac{1}{x}\frac{d}{dx}\left(-\frac{1}{2}e^{-x^{2}}\right)xH_{m}(x)H_{n}(x)\,dx \end{equation*}

\begin{equation*} =\left.H_{m}(x)H_{n}(x)\left(-\frac{1}{2}e^{-x^{2}}\right)\right\vert_{-\infty}^{\infty}-\int\limits_{-\infty}^{\infty}\frac{d}{dx}[H_{m}(x)H_{n}(x)]\left(-\frac{1}{2}e^{-x^{2}}\right)\,dx \end{equation*}

\begin{equation*} =\left.-\frac{1}{2}e^{-x^{2}}H_{m}(x)H_{n}(x)\right\vert_{-\infty}^{\infty}+\frac{1}{2}\int\limits_{-\infty}^{\infty}e^{-x^{2}}\frac{d}{dx}[H_{m}(x)H_{n}(x)]\,dx \end{equation*}

[Integrating by parts]

\begin{equation*} =0+\frac{1}{2}\int\limits_{-\infty}^{\infty}e^{-x^{2}}\left[H'_{m}(x)H_{n}(x)+H_{m}(x)H'_{n}(x)\right]\,dx \end{equation*}

\begin{equation*} =\frac{1}{2}\int\limits_{-\infty}^{\infty}e^{-x^{2}}\left[2mH_{m-1}(x)H_{n}(x)+2nH_{m}(x)H_{n-1}(x)\right]\,dx \end{equation*}

( using recurrence relation 1.)

\begin{equation*} =m\int\limits_{-\infty}^{\infty}e^{-x^{2}}H_{m-1}(x)H_{n}(x)\,dx+n\int\limits_{-\infty}^{\infty}e^{-x^{2}}H_{m}(x)H_{n-1}(x)\,dx \end{equation*}

\begin{equation*} =m\sqrt{\pi}2^{n}n!\delta_{n,m-1}+n\sqrt{\pi}2^{n-1}(n-1)!\delta_{m,n-1} \end{equation*}

\begin{equation*} =\sqrt{\pi}\left[2^{n-1}n!\delta_{m,n-1}+2^{n}(n+1)!\delta_{m,n+1}\right] \end{equation*}

[\(\because \delta_{n,m-1}=\delta_{n+1,m}\)]

Alternative. From recurrence relation 2, we have

\begin{equation*} xH_{n}=nH_{n-1}+\frac{1}{2}H_{n+1} \end{equation*}

\begin{equation*} \therefore \int\limits_{-\infty}^{\infty}x e^{-x^{2}}H_{m}(x)H_{n}(x)\,dx=\int\limits_{-\infty}^{\infty} e^{-x^{2}}[nH_{n-1}+\frac{1}{2}H_{n+1}]H_{m}(x)\,dx \end{equation*}

\begin{equation*} =n\int\limits_{-\infty}^{\infty} e^{-x^{2}}H_{n-1}(x)H_{m}(x)\,dx+\frac{1}{2}\int\limits_{-\infty}^{\infty}e^{-x^{2}}H_{n+1}(x)H_{m}(x)\,dx \end{equation*}

\begin{equation*} =n\sqrt{\pi}2^{n-1}(n-1)!\delta_{m,n-1}+\frac{1}{2}\sqrt{\pi}2^{n+1}(n+1)!\delta_{m,n+1} \end{equation*}

\begin{equation*} =\sqrt{\pi}2^{n-1}n!\delta_{m,n-1}+\sqrt{\pi}2^{n}(n+1)!\delta_{m,n+1} \end{equation*}

Example 4.10.3.

Prove that

\begin{equation*} H_{2n}(0)=(-1)^{n}\frac{(2n)!}{n!} \end{equation*}
\begin{equation*} H_{2n+1}(0)=0 \end{equation*}

Solution.

We have

\begin{equation*} H_{n}(x)=\sum\limits_{s=0}^{\frac{n}{2}}(-1)^{s}\frac{n!(2x)^{n-2s}}{s!(n-2s)!} \end{equation*}

\begin{equation*} H_{2n}(x)=\sum\limits_{s=0}^{n}(-1)^{s}\frac{(2n)!(2x)^{n-2s}}{s!(n-2s)!} \end{equation*}

putting \(x=0\text{,}\) we get -

\begin{equation*} H_{2n}(0)=(-1)^{n}\frac{(2n)!}{n!} \end{equation*}
\begin{equation*} H_{2n+1}(x)=\sum\limits_{s=0}^{(2n+1)/2}(-1)^{s}\frac{(2n+1)!(2x)^{2n+1-2s}}{s!(2n+1-2s)!} \end{equation*}

\begin{equation*} =\sum\limits_{s=0}^{(n+1)/2}(-1)^{s}\frac{(2n+1)!(2x)^{2n-2s+1}}{s!(2n-2s+1)!} \end{equation*}

putting \(x=0\text{,}\) we get -

\begin{equation*} H_{2n+1}(0)=(-1)^{n}\frac{(2n+1)!}{n!}\cdot 0 = 0 \end{equation*}

Example 4.10.4.

Prove that

\begin{equation*} P_{n}(x)=\frac{2}{\sqrt{\pi}n!}\int\limits_{0}^{\infty}t^{n}e^{-t^{2}}H_{n}(xt)\,dt \end{equation*}

Solution.

We have

\begin{equation*} H_{n}(x)=\sum\limits_{r=0}^{n/2}(-1)^{r}\frac{n!}{r!(n-2r)!}(2x)^{n-2r} \end{equation*}

or,

\begin{equation*} H_{n}(xt)=\sum\limits_{r=0}^{n/2}(-1)^{r}\frac{n!}{r!(n-2r)!}(2xt)^{n-2r} \end{equation*}

Therefore,

\begin{equation*} \frac{2}{\sqrt{\pi}n!}\int\limits_{0}^{\infty}t^{n}e^{-t^{2}}H_{n}(xt)\,dt \end{equation*}

\begin{equation*} =\frac{2}{\sqrt{\pi}n!}\int\limits_{0}^{\infty}t^{n}e^{-t^{2}}\sum\limits_{r=0}^{n/2}(-1)^{r}\frac{n!}{r!(n-2r)!}(2xt)^{n-2r}\,dt \end{equation*}

\begin{equation*} =\sum\limits_{r=0}^{n/2}\frac{2^{n-2r+1}(-1)^{r}x^{n-2r}}{\sqrt{\pi}r!(n-2r)!}\int\limits_{0}^{\infty}e^{-t^{2}}t^{2n-2r}\,dt \end{equation*}

\begin{equation*} = \sum\limits_{r=0}^{n/2}\frac{2^{n-2r+1}(-1)^{r}x^{n-2r}}{\sqrt{\pi}r!(n-2r)!}\frac{1}{2}\Gamma\left(n-r+\frac{1}{2}\right) \end{equation*}

\begin{equation*} =\sum\limits_{r=0}^{n/2}\frac{2^{n-2r}(-1)^{r}x^{n-2r}\{2(n-r)\}}{\sqrt{\pi}r!(n-2r)!}\sqrt{\pi} \end{equation*}

\begin{equation*} =\sum\limits_{r=0}^{n/2}(-1)^{r}\frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)!}x^{n-2r}=P_{n}(x). \end{equation*}

Since,

\begin{equation*} t^{2n-2r}=t^{2(n-r+\frac{1}{2})-1}; \end{equation*}

\begin{equation*} 2 \int\limits_{0}^{\infty}e^{-t^{2}}t^{2n-1}\,dt =\Gamma {(n)} \end{equation*}

and

\begin{equation*} \Gamma(x+\frac{1}{2})=\frac{(2x)!}{2^{2x}x!}\sqrt{\pi}. \end{equation*}

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