Use the recursion formula (Equation $\mathbf{2}\mathbf{.}\mathbf{85}\mathbf{)}$ to work out ${\mathit{H}}_{\mathbf{5}}\mathbf{\left(}\mathit{\xi }\mathbf{\right)}$ and ${\mathit{H}}_{\mathbf{6}}\mathbf{\left(}\mathit{\xi }\mathbf{\right)}$ Invoke the convention that the coefficient of the highest power of role="math" localid="1657778520591" $\mathit{\xi }$ is ${\mathbf{2}}^{\mathbf{\text{t}}}$ to fix the overall constant.

The required formula are given by:

${\mathit{H}}_{\mathbf{5}}\mathbf{\left(}\mathit{\xi }\mathbf{\right)}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{15}}\left(15\xi -20{\xi }^3+4{\xi }^5\right)$

${\mathit{H}}_{\mathbf{6}}\mathbf{\left(}\mathit{\xi }\mathbf{\right)}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{-}\mathbf{6}{\mathit{a}}_{\mathbf{0}}{\mathit{\xi }}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathit{a}}_{\mathbf{0}}{\mathit{\xi }}^{\mathbf{4}}\mathbf{-}\frac{\mathbf{8}}{\mathbf{15}}{\mathit{\xi }}^{\mathbf{6}}{\mathit{a}}_{\mathbf{0}}$

${\mathit{a}}_{\mathbf{j}\mathbf{+}\mathbf{2}}\mathbf{=}\frac{\mathbf{-}\mathbf{2}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{j}\mathbf{)}}{\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{2}\mathbf{)}}{\mathit{a}}_{\mathbf{j}}$

Let,

$n=5$And $j=1$

Then $H_{\tilde{s}}\left(\xi \right)$is expressed as,

$H_5\left(\xi \right)=\frac{a_1}{15}\left(15\xi -20{\xi }^3+4{\xi }^5\right)$...(1)

And

$a_{j12}=\frac{-2(n-j)}{(j+1)(j+2)}{a}_j$

(2)

Substitute the $n=5$and $j=1$in the equation (2).

$a_{1-2}=\frac{-2(5-1)}{(1+1)(1+2)}{a}_1$

$a_3=-\frac{4}{3}{a}_1$

For $j=3$

$a_{3+2}=\frac{-2(5-3)}{(3+1)(3+2)}{a}_3$

$a_{\mathrm{f}}=-\frac{1}{5}{a}_3$

$a_5=-\frac{4}{15}{a}_1$

For $j=3$

$a_{5+2}=\frac{-2(5-5)}{(5+1)(5+2)}{a}_5$

$a_7=0$

So, from the equation (1).

$H_s\left(\xi \right)=\frac{a_1}{15}\left(15\xi -20{\xi }^3+4{\xi }^5\right)$

So,

$H_5\left(\xi \right)=a_1\xi -\frac{4}{3}{a}_1{\xi }^3+\frac{4}{15}{a}_1{\xi }^5$

$-\frac{a_1}{15}\left(15\xi -20{\xi }^3+4{\xi }^5\right)$

Thus, the value of $H_s\left(\xi \right)$is $\frac{a_1}{15}\left(15\dot{\xi }-20{\xi }^3+4{\xi }^5\right)$

By convention of the coefficient of${\xi }^5$is $2^5$, so $a_1=15.8$

$H_5\left(\xi \right)=\left(120\xi -160{\xi }^3+32{\xi }^5\right)$

For,$n=6$And $j=0$

$a_{60}=\frac{-2(6-0)}{(0+1)(0+2)}{a}_0$

$a_6=-6a_0$

For $n=6$And$j=2$

$a_{62}=\frac{-2(6-2)}{(2+1)(2+2)}{a}_2$

$a_6=\frac{2}{3}{a}_2$

For $n=6$And $j=4$

$a_{64}=\frac{-2(6-4)}{(4+1)(4+2)} a_{4}$

$a_{10}=\frac{2}{15}{a}_4$

For $n=6Andj=6$

$a_{66}=\frac{-2(6-6)}{(6+1)(6+2)}{a}_6$

$a_{12}=0$

So, the value of $H_6\left(\xi \right)$ is expressed as,

The coefficient of ${\xi }^6$ is $2^8$, so $2^8=2^6=-\frac{8}{15}{a}_0\to a_0=-120$ Therefore the value of $H_6\left(\xi \right)=120+720{\xi }^2-480{\xi }^4+64{\xi }^8$.