In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials.

a. The Rodrigues formula says that${\mathbf{H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\mathbf{=}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathbf{e}}^{{\mathbf{\xi }}^{\mathbf{2}}}{\mathbf{\left(}\frac{\mathbf{d}}{\mathbf{d\xi }}\mathbf{\right)}}^{\mathbf{n}}{\mathbf{e}}^{\mathbf{-}{\mathbf{\xi }}^{\mathbf{2}}}$

Use it to derive ${\mathbf{H}}_{\mathbf{3}}$ and ${\mathbf{H}}_{\mathbf{4}}$ .

b. The following recursion relation gives you ${\mathbf{H}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$ in terms of the two preceding Hermite polynomials: ${\mathbf{H}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\mathbf{=}\mathbf{2}{\mathbf{\xi H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\mathbf{-}\mathbf{2}{\mathbf{nH}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}$

Use it, together with your answer in (a), to obtain ${\mathbf{H}}_{\mathbf{5}}$ and ${\mathbf{H}}_{\mathbf{6}}$ .

(c) If you differentiate an nth-order polynomial, you get a polynomial of

Order (n-1). For the Hermite polynomials, in fact,

$\frac{{\mathbf{dH}}_{\mathbf{n}}}{\mathbf{d\xi }}\mathbf{=}\mathbf{2}{\mathbf{nH}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}$

Check this, by differentiating${\mathit{H}}_{\mathbf{5}}$and ${\mathit{H}}_{\mathbf{6}}$.

d. ${\mathit{H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}$is the nth z-derivative, at z = 0, of the generating function $\mathbf{exp}\mathbf{(}\mathbf{-}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{\xi }\mathbf{z}\mathbf{)}$or, to put it another way, it is the coefficient of$\frac{{\mathbf{z}}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}$ in the Taylor series expansion for this function: ${\mathbf{e}}^{\mathbf{-}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{\xi z}}\mathbf{=}\sum _{n=0}^{\infty }\frac{z^n}{n!}{\mathbf{H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}$

Use this to obtain ${\mathit{H}}_{\mathbf{0}\mathbf{,}\mathbf{}}{\mathit{H}}_{\mathbf{1}}$and ${\mathit{H}}_{\mathbf{2}}$.

The required formula are given by:

$\begin{array}{c}{\mathbf{H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\mathbf{=}{\left(-1\right)}^{\mathbf{n}}{\mathbf{e}}^{{\mathbf{\xi }}^{\mathbf{2}}}{\mathbf{\left(}\frac{\mathbf{d}}{\mathbf{d\xi }}\mathbf{\right)}}^{\mathbf{n}}{\mathbf{e}}^{\mathbf{-}{\mathbf{\xi }}^{\mathbf{2}}}\\ {\mathbf{H}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\mathbf{=}\mathbf{2}{\mathbf{\xi H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\mathbf{-}\mathbf{2}{\mathbf{nH}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\\ \frac{{\mathbf{dH}}_{\mathbf{n}}}{\mathbf{d\xi }}\mathbf{=}\mathbf{2}{\mathbf{nH}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\\ {\mathbf{e}}^{\mathbf{-}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{\xi z}}\mathbf{=}\sum _{n=0}^{\infty }\frac{z^n}{n!}{\mathbf{H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{\xi }\mathbf{\right)}\end{array}$

(a)

In the question given that, the Rodrigues formula

$H_n\left(\xi \right)={(-1)}^n{e}^{{\xi }^2}{\left(\frac{d}{d\xi }\right)}^n{e}^{-{\xi }^2}$ ...(1)

Now, $e^{-{\xi }^2}$differentiate the with respect to the $\xi$.

$\frac{d}{d\xi }\left(e^{-{\xi }^2}\right)=-2\xi e^{-{\xi }^2}$

Again, differentiate the with respect to the $\xi$.

$\begin{array}{c}{\left(\frac{d}{d\xi }\right)}^2{e}^{-{\xi }^2}=\frac{d}{d\xi }(-2\xi e^{-{\xi }^2})\\ =(-2+4{\xi }^2)e^{-{\xi }^2}\end{array}$

Again, differentiate the with respect to the $\xi$.

$\begin{array}{c}{\left(\frac{d}{d\xi }\right)}^3{e}^{-{\xi }^2}=\frac{d}{d\xi }\left[(-2+4{\xi }^2)e^{-{\xi }^2}\right]\\ =[8\xi +(-2+4{\xi }^2)(-2\xi )]e^{-{\xi }^2}\\ =(12\xi -8{\xi }^3)e^{-{\xi }^2}\end{array}$

Again, differentiate the with respect to the$\xi$.

$\begin{array}{c}{\left(\frac{d}{d\xi }\right)}^4{e}^{-{\xi }^2}=\frac{d}{d\xi }\left[(12-8{\xi }^3)e^{-{\xi }^2}\right]\\ =[12-24{\xi }^2+(12-8{\xi }^3)(-2\xi )]e^{-{\xi }^2}\\ =(12-48{\xi }^2+16{\xi }^4)e^{-{\xi }^2}\end{array}$

Using the Rodrigues formula,$H_3\left(\xi \right)$and$H_4\left(\xi \right)$are expressed as,

In equation (1), substitute n equal to 3.

$H_3\left(\xi \right)=-e^{{\xi }^2}{\left(\frac{d}{d\xi }\right)}^3{e}^{-{\xi }^2}$

Substitute the value of ${\left(\frac{d}{d\xi }\right)}^3{e}^{-{\xi }^2}$in the above equation.

$H_3\left(\xi \right)=-12\xi +8{\xi }^3$

Hence the value of localid="1659003042297" $H_3\left(\xi \right)$is, $H_3\left(\xi \right)=-12\xi +8{\xi }^3$.

In equation (1), substitute n equal to 4.

$H_4\left(\xi \right)=e^{{\xi }^2}{\left(\frac{d}{d\xi }\right)}^4{e}^{-{\xi }^2}$

Substitute the value of ${\left(\frac{d}{d\xi }\right)}^4{e}^{-{\xi }^2}$ in the above equation.

$H_4\left(\xi \right)=12-48{\xi }^2+16{\xi }^4$

Hence the value of $H_4\left(\xi \right)$is, $H_4\left(\xi \right)=12-48{\xi }^2+16{\xi }^4$.

b)

According to the question, the given Hermie polynomial:

${\mathrm{H}}_{\mathrm{n}+1}\left(\mathrm{\xi }\right)=2{\mathrm{\xi H}}_{\mathrm{n}}\left(\mathrm{\xi }\right)-2{\mathrm{nH}}_{\mathrm{n}-1}\left(\mathrm{\xi }\right)$ ...(2)

In equation (2), substitute n equal to 4.

${\mathrm{H}}_5\left(\mathrm{\xi }\right)=2{\mathrm{\xi H}}_4\left(\mathrm{\xi }\right)-8{\mathrm{H}}_3\left(\mathrm{\xi }\right)$

Substitute the value of $H_3\left(\xi \right)$and $H_4\left(\xi \right)$ in the above equation.

$\begin{array}{c}{\mathrm{H}}_5=2\mathrm{\xi }(12-48{\mathrm{\xi }}^2+16{\mathrm{\xi }}^4)-8(-12\mathrm{\xi }+8{\mathrm{\xi }}^3)\\ {\mathrm{H}}_5=120\mathrm{\xi }-160{\mathrm{\xi }}^3+32{\mathrm{\xi }}^5\end{array}$

Hence the value of $H_5$ is, $120\xi -160{\xi }^3+32{\xi }^5$.

And

In equation (2), substitute n equal to 5.

$H_6\left(\xi \right)=2\xi H_5\left(\xi \right)-10H_4\left(\xi \right)$

Substitute the value of $H_5$ and $H_4\left(\xi \right)$in the above equation.

$\begin{array}{c}{H}_6=2\xi (120\xi -160{\xi }^3+32{\xi }^5)-10(12-48{\xi }^2+16{\xi }^4)\\ H_6=-120+720{\xi }^2-480{\xi }^4+64{\xi }^6\end{array}$

Hence the value of $H_6$ is, $-120+720{\xi }^2-480{\xi }^4+64{\xi }^6$.

(c)

The value of $H_5$ is differentiate with respect to $\xi$.

$\begin{array}{c}\frac{{\mathrm{dH}}_5}{\mathrm{d\xi }}=\frac{\mathrm{d}}{\mathrm{d\xi }}(120\mathrm{\xi }-160{\mathrm{\xi }}^3+32{\mathrm{\xi }}^5)\\ \frac{{\mathrm{dH}}_5}{\mathrm{d\xi }}=120-480{\mathrm{\xi }}^2+160{\mathrm{\xi }}^4\\ \frac{{\mathrm{dH}}_5}{\mathrm{d\xi }}=10(12-48{\mathrm{\xi }}^2+16{\mathrm{\xi }}^4)\\ \frac{{\mathrm{dH}}_5}{\mathrm{d\xi }}=\left(2\right)\left(5\right){\mathrm{H}}_4\end{array}$

Hence the value of $\frac{d{H}_5}{d\xi }$ is, $\left(2\right)\left(5\right)H_4$ .

And

The value of$H_6$ is differentiate with respect to$\xi$ .

$\begin{array}{c}\frac{d{H}_6}{d\xi }=\frac{d}{d\xi }(-120+720{\xi }^2-480{\xi }^4+64{\xi }^6)\\ \frac{d{H}_6}{d\xi }=1440\xi -1920{\xi }^3+384{\xi }^5\\ \frac{d{H}_6}{d\xi }=12(120\xi -160{\xi }^3+32{\xi }^5)\\ \frac{d{H}_6}{d\xi }=\left(2\right)\left(6\right)H_5\end{array}$

Hence the value of $\frac{d{H}_6}{d\xi }$ is, $\left(2\right)\left(6\right)H_5$.

(d)

Differentiate the value of $e^{-z^2+2z\xi }$with respect to z.

$\frac{d}{dz}\left(e^{-z^2+2z\xi }\right)=(-2z+2\xi )\left(e^{-z^2+2z\xi }\right)$ ...(3)

Setting $z=0$,

$H_0\left(\xi \right)=2\xi$

Equation (3), differentiate with respect to z.

$\begin{array}{c}{\left(\frac{d}{dz}\right)}^2\left(e^{-z^2+2z\xi }\right)=\frac{d}{dz}\left[(-2z+2\xi )\left(e^{-z^2+2z\xi }\right)\right]\\ {\left(\frac{d}{dz}\right)}^2\left(e^{-z^2+2z\xi }\right)=[-2+{(-2z+2\xi )}^2]\left(e^{-z^2+2z\xi }\right)\end{array}$ ...(4)

Setting$z=0$,

$H_1\left(\xi \right)=-2+4{\xi }^2$

Equation (4), differentiate with respect to z.

$\begin{array}{c}{\left(\frac{d}{dz}\right)}^3\left(e^{-z^2+2z\xi }\right)=\frac{d}{dz}\left\{\left[-2+{\left(-2z+2\xi \right)}^2\right]\left(e^{-z^2+2z\xi }\right)\right\}\\ =\left\{2(-2z+2\xi )(-2)+[-2+{(-2z+2\xi )}^2](-2z+2\xi )\right\}\left(e^{-z^2+2z\xi }\right)\end{array}$

Setting$z=0$

$\begin{array}{c}{H}_2\left(\xi \right)=-8\xi +(-2+4{\xi }^2)\left(2\xi \right)\\ =-12\xi +8{\xi }^3\end{array}$

Hence

The value of $H_0$,$H_1$ and$H_2$ are respectively $2\xi$, $(-2+4{\xi }^2)$ and $(-12\xi +8{\xi }^3)$.