91Ó°ÊÓ

The relationship of recurrence is. The polynomials fulfil the second-order differential equation $\mathbf{H}\mathbf{n}\mathbf{+}\mathbf{1}\left(x\right)\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{H}\mathbf{n}\left(x\right)\mathbf{2}\mathbf{n}\mathbf{H}\mathbf{n}\mathbf{1}\left(x\right)\mathbf{,}$and the polynomials satisfy the second-order differential equation$\mathbf{H}\mathbf{n}+\mathbf{1}\left(x\right).$

(a)

The function,$\mathrm{Φ}(x,h)=\frac{e^{−x/l−h}}{1−h}$

Proof:Differentiate the function $\mathrm{Φ}(x,h)=\frac{e^{−xtll−h}}{1−h}$with respect to $x$

$\begin{array}{c}\frac{∂\mathrm{Φ}}{∂x}=\frac{∂}{∂x}\left(\frac{e^{−xh(1−h)}}{1−h}\right)\\ \frac{∂\mathrm{Φ}}{∂x}=\frac{1}{1−h}\frac{∂}{∂x}\left(e^{−xh/(1−h)}\right)\\ \frac{∂\mathrm{Φ}}{∂x}=\frac{1}{1−h}{e}^{−xh/(1−h)}\frac{∂}{∂x}\left(−\frac{xh}{1−h}\right)\frac{∂\mathrm{Φ}}{∂x}=\frac{1}{1−h}{e}^{−xh/(1−h)}\left(−\frac{h}{1−h}\right)\end{array}$

Simplify further as follows:

$\begin{array}{c}\frac{∂\mathrm{Φ}}{∂x}=\frac{1}{1−h}{e}^{−xh/(1−h)}\left(−\frac{h}{1−h}\right)\\ \frac{∂\mathrm{Φ}}{∂x}=−\frac{h}{(1−h)}\frac{e^{−xh(1−h)}}{1−h}\\ \frac{∂\mathrm{Φ}}{∂x}=−\frac{h}{(1−h)}\mathrm{Φ}\\ \frac{∂\mathrm{Φ}}{∂x}=\frac{h}{(h−1)}\mathrm{Φ}\end{array}$

Therefore,$h\mathrm{Φ}=(h−1)(∂\mathrm{Φ}/∂x)$

(b)

Given information

The function,$\mathrm{Φ}(x,h)=\frac{e^{−xd/1−h}}{1−h}$

Proof:Differentiate the function $\mathrm{Φ}(x,h)=\frac{e^{−xd/1−h}}{1−h}$with respect to $h$.

$\begin{array}{c}\frac{∂\mathrm{Φ}}{∂h}=\frac{∂}{∂h}\left(\frac{e^{−xh(1−h)}}{1−h}\right)\\ \frac{∂\mathrm{ϕ}}{∂h}=\frac{(1−h)\frac{∂}{dh}{e}^{−xh(1−h)}−e^{−xh(1−h)}\frac{∂}{dh}(1−h)}{{(1−h)}^2}\\ \frac{∂\mathrm{ϕ}}{∂h}=\frac{(1−h)e^{−xH(1−h)}\frac{∂}{dh}\left[\frac{−xt}{1−h}\right]−e^{−xh(1−h)}(0−1)}{{(1−h)}^2}\\ \frac{∂\mathrm{ϕ}}{∂h}=\frac{(1−h)e^{−xH(1−h)}\frac{(1−h)(−x)−\left(x\right)\left)\right(0−1)}{{(1−h)}^2}−e^{−xH(1−h)}(0−1)}{{(1−h)}^2}\end{array}$

Simplify further as follows:

$\begin{array}{c}\frac{∂\mathrm{ϕ}}{∂h}=\frac{(−x+xh−xh)\frac{e^{−xh(1−h)}}{(1−h)}−e^{−xH(1−h)}(0−1)}{{(1−h)}^2}\\ \frac{∂\mathrm{Φ}}{∂h}=\frac{−x\mathrm{Φ}+e^{−xh(1−h)}}{{(1−h)}^2}\\ \frac{∂\mathrm{Φ}}{∂h}=\frac{−x\mathrm{ϕ}+(1−h)\mathrm{Φ}}{{(1−h)}^2}\\ \frac{∂\mathrm{Φ}}{∂h}=\frac{(1−h−x)\mathrm{Φ}}{{(1−h)}^2}\end{array}$

Therefore,${(1−h)}^2(∂\mathrm{Φ}/∂h)=(1−h−x)\mathrm{Φ}$

Step 3 Combine results from part (a) and (b) then substitute

(c

Given information: $h\mathrm{Φ}=(h−1)(∂\mathrm{Φ}/∂x)$and${(1−h)}^2(∂\mathrm{Φ}/∂h)=(1−h−x)\mathrm{Φ}$

Proof:

$\begin{array}{c}h\mathrm{Φ}=(h−1)(∂\mathrm{Φ}/∂x)\\ ∂\mathrm{Φ}/∂x=\frac{h}{h−1}\\ −x(∂\mathrm{Φ}/∂x)=\frac{xh}{1−h}\end{array}$...... (1)

Now, simplify

$\begin{array}{c}{\left(1-h\right)}^2(∂\mathrm{Φ}/∂h)=(1−h−x)\mathrm{Φ}\\ (1−h)(∂\mathrm{Φ}/∂h)=\frac{(1−h−x)}{1−h}\mathrm{Φ}\\ h(1−h)(∂\mathrm{Φ}/∂h)=\frac{(1−h−x)}{1−h}h\mathrm{Φ}\end{array}$

Simplify further as follows:

$\begin{array}{c}h(1−h)(∂\mathrm{Φ}/∂h)=\left(1−\frac{x}{1−h}\right)h\mathrm{Φ}\\ h(1−h)(∂\mathrm{Φ}/∂h)=h\mathrm{Φ}−\frac{xh}{1−h}\mathrm{Φ}\\ h(1−h)(∂\mathrm{Φ}/∂h)=h\mathrm{Φ}+x(∂\mathrm{Φ}/∂x)\end{array}$

Since, $\frac{xh}{1−h}=−x(∂\mathrm{Φ}/∂x)$, from (1),

$x(∂\mathrm{Φ}/∂x)+h\mathrm{Φ}−h(1−h)(∂\mathrm{Φ}/∂h)=0$