91影视

The required formula are given by:

$\begin{array}{c}{\mathbf{H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{尉}\mathbf{\right)}\mathbf{=}{\left(-1\right)}^{\mathbf{n}}{\mathbf{e}}^{{\mathbf{尉}}^{\mathbf{2}}}{\mathbf{\left(}\frac{\mathbf{d}}{\mathbf{诲尉}}\mathbf{\right)}}^{\mathbf{n}}{\mathbf{e}}^{\mathbf{鈭�}{\mathbf{尉}}^{\mathbf{2}}}\\ {\mathbf{H}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{尉}\mathbf{\right)}\mathbf{=}\mathbf{2}{\mathbf{尉贬}}_{\mathbf{n}}\mathbf{\left(}\mathbf{尉}\mathbf{\right)}\mathbf{鈭�}\mathbf{2}{\mathbf{nH}}_{\mathbf{n}\mathbf{鈭�}\mathbf{1}}\mathbf{\left(}\mathbf{尉}\mathbf{\right)}\\ \frac{{\mathbf{dH}}_{\mathbf{n}}}{\mathbf{诲尉}}\mathbf{=}\mathbf{2}{\mathbf{nH}}_{\mathbf{n}\mathbf{鈭�}\mathbf{1}}\mathbf{\left(}\mathbf{尉}\mathbf{\right)}\\ {\mathbf{e}}^{\mathbf{鈭�}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{尉锄}}\mathbf{=}\underset{n=0}{\overset{\mathrm{鈭�}}{鈭�}}\frac{z^n}{n!}{\mathbf{H}}_{\mathbf{n}}\mathbf{\left(}\mathbf{尉}\mathbf{\right)}\end{array}$

(a)

In the question given that, the Rodrigues formula

$H_n\left(尉\right)={(鈭�1)}^n{e}^{{尉}^2}{\left(\frac{d}{d尉}\right)}^n{e}^{鈭�{尉}^2}$ ...(1)

Now, $e^{鈭�{尉}^2}$differentiate the with respect to the $尉$.

$\frac{d}{d尉}\left(e^{鈭�{尉}^2}\right)=鈭�2尉e^{鈭�{尉}^2}$

Again, differentiate the with respect to the $尉$.

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

Again, differentiate the with respect to the $尉$.

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

Again, differentiate the with respect to the$尉$.

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

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

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

$H_3\left(尉\right)=鈭�e^{{尉}^2}{\left(\frac{d}{d尉}\right)}^3{e}^{鈭�{尉}^2}$

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

$H_3\left(尉\right)=鈭�12尉+8{尉}^3$

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

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

$H_4\left(尉\right)=e^{{尉}^2}{\left(\frac{d}{d尉}\right)}^4{e}^{鈭�{尉}^2}$

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

$H_4\left(尉\right)=12鈭�48{尉}^2+16{尉}^4$

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

b)

According to the question, the given Hermie polynomial:

${\mathrm{H}}_{\mathrm{n}+1}\left(\mathrm{尉}\right)=2{\mathrm{尉贬}}_{\mathrm{n}}\left(\mathrm{尉}\right)鈭�2{\mathrm{nH}}_{\mathrm{n}鈭�1}\left(\mathrm{尉}\right)$ ...(2)

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

${\mathrm{H}}_5\left(\mathrm{尉}\right)=2{\mathrm{尉贬}}_4\left(\mathrm{尉}\right)鈭�8{\mathrm{H}}_3\left(\mathrm{尉}\right)$

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

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

Hence the value of $H_5$ is, $120尉鈭�160{尉}^3+32{尉}^5$.

And

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

$H_6\left(尉\right)=2尉H_5\left(尉\right)鈭�10H_4\left(尉\right)$

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

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

Hence the value of $H_6$ is, $鈭�120+720{尉}^2鈭�480{尉}^4+64{尉}^6$.

(c)

The value of $H_5$ is differentiate with respect to $尉$.

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

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

And

The value of$H_6$ is differentiate with respect to$尉$ .

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

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

(d)

Differentiate the value of $e^{鈭�z^2+2z尉}$with respect to z.

$\frac{d}{dz}\left(e^{鈭�z^2+2z尉}\right)=(鈭�2z+2尉)\left(e^{鈭�z^2+2z尉}\right)$ ...(3)

Setting $z=0$,

$H_0\left(尉\right)=2尉$

Equation (3), differentiate with respect to z.

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

Setting$z=0$,

$H_1\left(尉\right)=鈭�2+4{尉}^2$

Equation (4), differentiate with respect to z.

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

Setting$z=0$

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

Hence

The value of $H_0$,$H_1$ and$H_2$ are respectively $2尉$, $(鈭�2+4{尉}^2)$ and $(鈭�12尉+8{尉}^3)$.