User:Jiangpx/6321 hw7

=Problem1 - Plot Legendre Functions for n=0,1,2,3=

Given
Legendre functions

$$\begin{align} & \left\{ {{P}_{n}}(x),n=0,1,2,3 \right\} \\ & \left\{ {{Q}_{n}}(x),n=0,1,2,3 \right\} \\ \end{align}$$

Find
Plot

$$\left\{ {{P}_{n}}(x),n=0,1,2,3 \right\}$$

Plot

$$\left\{ {{Q}_{n}}(x),n=0,1,2,3 \right\}$$

Plot

$$\left\{ \begin{align} & {{P}_{0}}(x),{{Q}_{0}}(x) \\ & {{P}_{1}}(x),{{Q}_{1}}(x) \\ & {{P}_{2}}(x),{{Q}_{2}}(x) \\ & {{P}_{3}}(x),{{Q}_{3}}(x) \\ \end{align} \right\}$$

Observe

$${{P}_{n}}(x)\And {{Q}_{n}}(x)as-1<x<1$$

Observe even-ness/odd-ness of

$$\begin{align} & \left\{ {{P}_{n}}(x),n=0,1,2,3 \right\} \\ & \left\{ {{Q}_{n}}(x),n=0,1,2,3 \right\} \\ \end{align}$$

Guess

$$\int_{-1}^{1}{{{P}_{n}}(x){{Q}_{n}}(x)}dx$$

Solution
Legendre polynomials:

$$\begin{align} & {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ & {{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1) \\ & {{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

and the associated Legendre functions:

$$\begin{align} & {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ & {{Q}_{1}}(x)=\frac{1}{2}x\log \left( \frac{1+x}{1-x} \right)-1 \\ & {{Q}_{2}}(x)=\frac{1}{4}(3{{x}^{2}}-1)\log \left( \frac{1+x}{1-x} \right)-\frac{3}{2}x \\ & {{Q}_{3}}(x)=\frac{1}{4}(5{{x}^{3}}-3x)\log \left( \frac{1+x}{1-x} \right)-\frac{5}{2}{{x}^{2}}+\frac{2}{3} \\ \end{align}$$













For x approaches 1 or -1, Pn(x) and Qn(x) both have the steepest slopes in the domain of (-1 1).

For n being even, Pn(x) is even while Qn(x) is odd;

For n being odd, Pn(x) is odd while Qn(x) is even;

Guess $$\int_{-1}^{1}{{{P}_{n}}(x){{Q}_{n}}(x)}dx=0$$

=Problem2 - Prove even-ness/odd-ness of general functions=

Given
even $$\left\{ {{g}_{i}}(x) \right\}$$ and odd $$\left\{ {{h}_{i}}(x) \right\}$$

Find
$$\sum\limits_{i}{{{g}_{i}}(x)}$$ is even, and

$$\sum\limits_{i}{{{h}_{i}}(x)}$$ is odd.

Solution
In general, for even function g, we have

$$g(-x)=g(x)$$

Given $${{g}_{1}}(x)$$ is even, then

$${{g}_{1}}(-x)={{g}_{1}}(x)$$

Given $${{g}_{2}}(x)$$ is even, then

$${{g}_{2}}(-x)={{g}_{2}}(x)$$

so

$${{g}_{1}}(-x)+{{g}_{2}}(-x)={{g}_{1}}(x)+{{g}_{2}}(x)$$

that is

function $$G={{g}_{1}}+{{g}_{2}}$$ is even

Therefore, in general, given even $$\left\{ {{g}_{i}}(x) \right\}$$

we have function $$\sum\limits_{i}{{{g}_{i}}(x)}$$ also being even.

By the same token, in general, given odd $$\left\{ {{h}_{i}}(x) \right\}$$

we have function $$\sum\limits_{i}{{{h}_{i}}(x)}$$ also being odd.

=Problem3 - Show the even-ness/odd-ness of Legendre polynomials=

Given
general expressions of Legendre polynomials

$${{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{{{(-1)}^{i}}\frac{(2n-2i)!{{x}^{n-2i}}}{{{2}^{n}}i!(n-i)!(n-2i)!}}$$

Find
(1)

$${{P}_{2k}}(x)$$ is even and

(2)

$${{P}_{2k+1}}(x)$$ is odd, for k=0,1,2,...

Solution
(1)

For general even function g, we have

$$g(-x)=g(x)$$

For n=2k, with k=0,1,2,...

$${{P}_{n}}(x)$$ becomes

$$ P_{2k}(x) = \sum_{0}^{k}(-1)^{i}\frac{(4k-2i)! \color{black}{(x)^{2k-2i}}}{2^{2k}i!(2k-i)!(2k-2i)!} $$

Consider $${{P}_{2k}}(-x)$$

$${{P}_{2k}}(-x)=\sum\limits_{0}^{k}\frac{(4k-2i)!{{(-x)}^{2k-2i}}}{{{2}^{2k}}i!(2k-i)!(2k-2i)!}={{P}_{2k}}(x)$$

Thus, it is clear that $${{P}_{2k}}(x)$$ is even for k=0,1,2,...

(2)

For general odd function h, we have

$$h(-x)=-h(x)$$

For n=2k+1, with k=0,1,2,...

$${{P}_{n}}(x)$$ becomes

$$ P_{2k+1}(x) = \sum_{0}^{[\frac{2k+1}{2}]}(-1)^{i}\frac{(4k+2-2i)! \color{black}{(x)^{2k+1-2i}}}{2^{2k+1}i!(2k+1-i)!(2k+1-2i)!} $$

Consider $${{P}_{2k+1}}(-x)$$

$${{P}_{2k+1}}(-x)=\sum\limits_{0}^{[\frac{2k+1}{2}]}\frac{(4k+2-2i)!{{(-x)}^{2k+1-2i}}}{{{2}^{2k+1}}i!(2k+1-i)!(2k+1-2i)!}=-{{P}_{2k+1}}(x)$$

Thus, it is clear that $${{P}_{2k+1}}(x)$$ is odd for k=0,1,2,...

=Problem4 - General polynomials vs Legendre polynomials=

Given
$$\begin{align} & \text{Polynomial} \\ & q(x)=\sum\limits_{i=0}^{5} \\ & \text{with} \\ & {{c}_{0}}=2,{{c}_{1}}=-5,{{c}_{2}}=-3,{{c}_{3}}=11,{{c}_{4}}=7,{{c}_{5}}=6 \\ & \\ \end{align}$$

Find
$$\begin{align} & \text{(1) }\left\{ {{a}_{i}} \right\}\text{ such that} \\ & q(x)=\sum\limits_{i=0}^{5}{{{a}_{i}}{{P}_{i}}(x)}\text{ } \\ & \text{where} \\ & {{P}_{i}}(x)-\text{Legendre Polynomials} \\ & \\  & \text{(2) Plot both expressions of }q(x)\text{ to verify} \\ \end{align}$$

Solution
Write c as

$$[c]=\left[ \begin{matrix} 2 & -5 & -3 & 11 & 7 & 6 \\ \end{matrix} \right]$$

Then

$$q=[c]{{[1\text{ }x\text{ }{{x}^{2}}\text{ }{{x}^{3}}\text{ }{{x}^{4}}\text{ }{{x}^{5}}]}^{T}}=[c]{{[{{x}^{i}}]}^{T}}=[c][X]$$

Also

$$q=[a]{{[{{P}_{0}}\text{ }{{P}_{1}}\text{ }{{P}_{2}}\text{ }{{P}_{3}}\text{ }{{P}_{4}}\text{ }{{P}_{5}}]}^{T}}=[a][P]$$

Express $${{[{{P}_{0}}\text{ }{{P}_{1}}\text{ }{{P}_{2}}\text{ }{{P}_{3}}\text{ }{{P}_{4}}\text{ }{{P}_{5}}]}^{T}}$$ as $$[A][X]$$

$$[P]=[A][X]=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 \\   0 & 1 & 0 & 0 & 0 & 0  \\   -1/2 & 0 & 3/2 & 0 & 0 & 0  \\   0 & -3/2 & 0 & 5/2 & 0 & 0  \\   3/8 & 0 & -15/4 & 0 & 35/8 & 0  \\   0 & 15/8 & 0 & -35/4 & 0 & 63/8  \\ \end{matrix} \right]\left[ \begin{matrix} 1 \\   x  \\ {{x}^{2}} \\ {{x}^{3}} \\ {{x}^{4}} \\ {{x}^{5}} \\ \end{matrix} \right]$$

From above derivations, we have

$$\begin{align} & q=[a][A][X] \\ & q=[c][X] \\ \end{align}$$

So for the above to be valid for all x, we must have

$$[a][A]=[c]$$

Therefore

$$[a]=[c]{{[A]}^{-1}}$$

With [c] and [A] known, [a] can be obtained as

$${{[a]}^{T}}=\left[ \begin{align} & 2.4000000000 \\ & 4.1714285714 \\  & 2.0000000000 \\  & 7.0666666666 \\  & 1.6000000000 \\  & 0.7619047619 \\ \end{align} \right]$$

The polynomial q(x) has been plot in terms of [c] and [a]



=Problem5 - Find The Coefficients An in Legendre-Fourier Series=

Given
Boundary condition

$$\begin{align} & (a)\text{ }\psi (r=1,\theta )=f(\theta )={{T}_{0}}{{\cos }^{6}}\theta  \\ & (b)\text{ }\psi (r=1,\theta )=f(\theta )={{T}_{0}}\exp \left( -\frac{4{{\theta }^{2}}} \right) \\ \end{align}$$

and expression of An

$${{A}_{n}}=\frac{2n+1}{2}$$

Find
(1) Properties of An, i.e.

$$\begin{align} & {{A}_{2k}}=0\text{ }? \\ & {{A}_{2k+1}}=0\text{  }? \\ & for\text{  }k=0,1,2,... \\ \end{align}$$

(2) Calculate non-zero

$$\left\{ {{A}_{n}} \right\}$$

Solution
(a).(1)

Define $$x=\sin \theta $$

Then $$f(x)={{T}_{0}}{{(1-{{x}^{2}})}^{3}}$$ and it is a even function

We know Legendre polynomial $${{P}_{n}}$$ is even with even n and odd with odd n

Hence, for n=2k+1, k=0,1,2,...

$$=0$$, i.e.,

$${{A}_{2k+1}}=0$$

Since

$$f(x)\in {{\Pi }_{6}},\text{ }a\text{  }set\text{  }of\text{  }all\text{  }polynomials\text{  }with\text{  }order\text{ }\le 6$$

and the linear independence of Legendre polynomials, we have

$$\begin{align} & {{P}_{n}}\bot f\text{ }for\text{  }n\ge 7,\text{  }i.e., \\ & =0 \\ & so\text{ } \\ & {{A}_{n}}=0,\text{ }for\text{  }n\ge 7 \\ \end{align}$$

$$Based\text{ }on\text{  }that,\text{  }only\text{  }{{A}_{0}},\text{  }{{A}_{2}},\text{  }{{A}_{4}},\text{  }and\text{  }{{A}_{6}}\text{  }are\text{  }non-zero$$

(a).(2)

We know that

$$\begin{align} & {{P}_{0}}(x)=1 \\ & {{P}_{2}}(x)=\frac{3}{2}{{x}^{2}}-\frac{1}{2} \\ & {{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ & {{P}_{6}}(x)=\frac{231}{16}{{x}^{6}}-\frac{315}{16}{{x}^{4}}+\frac{105}{16}{{x}^{2}}-\frac{5}{16} \\ & \\ \end{align}$$

Then

$${{A}_{n}}=\frac{2n+1}{2}=\frac{2n+1}{2}\int_{-1}^{1}{f(x){{P}_{n}}(x)dx}$$

So $$\begin{align} & {{A}_{0}}=\frac{1}{2}=\frac{1}{2}\int_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}dx} \\ & {{A}_{2}}=\frac{5}{2}=\frac{5}{2}\int_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}\left( \frac{3}{2}{{x}^{2}}-\frac{1}{2} \right)dx} \\ & {{A}_{4}}=\frac{9}{2}=\frac{9}{2}\int_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}\left( \frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \right)dx} \\ & {{A}_{6}}=\frac{13}{2}=\frac{13}{2}\int_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}\left( \frac{231}{16}{{x}^{6}}-\frac{315}{16}{{x}^{4}}+\frac{105}{16}{{x}^{2}}-\frac{5}{16} \right)dx} \\ \end{align}$$

Evaluating the above definite integrals via Matlab, we may get

$$\begin{align} & {{A}_{0}}=\frac{16}{35}{{T}_{0}} \\ & {{A}_{2}}=-\frac{16}{21}{{T}_{0}} \\ & {{A}_{4}}=\frac{144}{385}{{T}_{0}} \\ & {{A}_{6}}=-\frac{16}{231}{{T}_{0}} \\ & \\ \end{align}$$

(b).(1) Define $$x=\theta $$

then $$f(x)={{T}_{0}}\exp \left( -\frac{4{{x}^{2}}} \right)$$

and

$$\begin{align} & \mu =\sin \theta =\sin x \\ & {{P}_{n}}(\mu )={{P}_{n}}(\sin x) \\ \end{align}$$

Note that $$f(x)$$ is an even function, along with the properties of $${{P}_{n}}(\sin x)$$, we have that

$$\begin{align} & =0 \\ & i.e.,\text{ }{{A}_{2k+1}}=0 \\ \end{align}$$

for k=0,1,2,...

Since $$f(x)$$ is not any kind of polynomials,

$$\begin{align} & \ne 0 \\ & so \\ & {{A}_{2k}}\text{ }exist\text{  }for\text{  }k=0,1,2,... \\ \end{align}$$

(b).(2) We have

$${{A}_{n}}=\frac{2n+1}{2}=\frac{2n+1}{2}\int_{-\pi /2}^{\pi /2}{f(x){{P}_{n}}(\sin x)dx}$$

Given

$$f(x)={{T}_{0}}\exp \left( -\frac{4{{x}^{2}}} \right)$$

and

$$\begin{align} & {{P}_{0}}(\sin x)=1 \\ & {{P}_{2}}(\sin x)=3/2{{\left( \sin \left( x \right) \right)}^{2}}-1/2 \\ & {{P}_{4}}(\sin x)=\frac{35}{8}{{\left( \sin \left( x \right) \right)}^{4}}-\frac{15}{4}{{\left( \sin \left( x \right) \right)}^{2}}+3/8 \\ & {{P}_{6}}(\sin x)=\frac{231}{16}{{\left( \sin \left( x \right) \right)}^{6}}-\frac{315}{16}{{\left( \sin \left( x \right) \right)}^{4}}+\frac{105}{16}{{\left( \sin \left( x \right) \right)}^{2}}-\frac{5}{16} \\ \end{align}$$

Evaluating the above definite integrals via Matlab, we may get

$$\begin{align} & {{A}_{0}}=1.1731086046{{T}_{0}} \\ & {{A}_{2}}=0.6393928815{{T}_{0}} \\ & {{A}_{4}}=0.7149018027{{T}_{0}} \\ & {{A}_{6}}=0.7261792127{{T}_{0}} \\ \end{align}$$

Note that we only calculated $${{A}_{0}}\to {{A}_{6}}$$ for demonstration.

As stated in (b).(1) that $${{A}_{2k}}\text{ }exist\text{  }for\text{  }k=0,1,2,...$$

=Problem6 - Derivation of Inverse hyperbolic tangent function=

Given
hyperbolic tangent function

$$\tanh (x)=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}$$

Find
Inverse hyperbolic tangent function as

$${{\tanh }^{-1}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$

Solution
Give $$\tanh (x)=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}$$

Consider

$$\tanh (\log (x))=\frac{{{e}^{\log x}}-{{e}^{-\log x}}}{{{e}^{\log x}}+{{e}^{-\log x}}}=\frac{x-\frac{1}{x}}{x+\frac{1}{x}}=\frac{{{x}^{2}}-1}{{{x}^{2}}+1}$$

and it follows that

$$\begin{align} & \tanh \left[ \frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right]=\tanh \left[ \log \sqrt{\frac{1+x}{1-x}} \right]=\frac{{{\left( \sqrt{\frac{1+x}{1-x}} \right)}^{2}}-1}{{{\left( \sqrt{\frac{1+x}{1-x}} \right)}^{2}}+1} \\ & =\frac{\frac{1+x}{1-x}-1}{\frac{1+x}{1-x}+1}=\frac{(1+x)-(1-x)}{(1+x)+(1-x)}=\frac{2x}{2}=x \\ \end{align}$$

Therefore

$${{\tanh }^{-1}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$

=Problem7 - Even-ness and odd-ness of Qn(x)=

Given
$$\begin{align} & {{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{{{(-1)}^{i}}\frac{(2n-2i)!{{x}^{n-2i}}}{{{2}^{n}}i!(n-i)!(n-2i)!}}\text{                                           (1)} \\ & {{Q}_{n}}(x)={{P}_{n}}(x){{\tanh }^{-1}}(x)-2\sum\limits_{j=1,3,5}^{J}{\frac{2n-2j+1}{(2n-j+1)j}}{{P}_{n-j}}(x)\text{                   (2)} \\ & J:=1+2\left[ \frac{n-1}{2} \right]\text{                                                                          (3)} \\ \end{align}$$

Find
Even-ness and odd-ness of Qn(x) depending on n

Solution
We know that $${{P}_{n}}(x)$$ is even when n is even (n=2k, k=0,1,2,...), and $${{P}_{n}}(x)$$ is odd when n is odd (n=2k+1, k=0,1,2,....)

Also note that $${{\tanh }^{-1}}(x)$$ is an odd function, because

$$\begin{align} & {{\tanh }^{-1}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ & {{\tanh }^{-1}}(-x)=\frac{1}{2}\log \left( \frac{1-x}{1+x} \right) \\ & =-\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)=-{{\tanh }^{-1}}(x) \\ \end{align}$$

(1) Consider when n is odd, then

$${{P}_{n}}(x)\text{ }is\text{  }odd,\text{  }so\text{  }{{P}_{n}}(x){{\tanh }^{-1}}(x)\text{  }is\text{  }even$$,

n is odd and J is odd ---> n-j is even and $${{P}_{n-j}}(x)\text{ }is\text{  }even$$

Therefore, equation (2) is sum of even functions ---> $${{Q}_{n}}(x)$$ is even.

(2) Consider when n is even, then

$${{P}_{n}}(x)\text{ }is\text{  even},\text{  }so\text{  }{{P}_{n}}(x){{\tanh }^{-1}}(x)\text{  }is\text{  odd}$$,

n is even and J is odd ---> n-j is odd and $${{P}_{n-j}}(x)\text{ }is\text{  }odd$$

Therefore, equation (2) is sum of odd functions ---> $${{Q}_{n}}(x)$$ is odd.

Conclusion

(1) for n=2k, k=0,1,2,... $${{P}_{n}}(x)$$ is even and $${{Q}_{n}}(x)$$ is odd;

(2) for n=2k+1, k=0,1,2,... $${{P}_{n}}(x)$$ is odd and $${{Q}_{n}}(x)$$ is even.

= Problem 9 - Generate Legendre Polynomials Using Recurrence Relation=

Given
Reccurence relation of Legendre polynomials:

$$\left( n+1 \right){{P}_{n+1}}-\left( 2n+1 \right)x{{P}_{n}}+n{{P}_{n-1}}=0 (1)$$

Legendre polynomials for n=0,1:

$$\begin{align} & {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ \end{align}$$

Find
Generate $$\displaystyle \left\{ {{P}_{2}},{{P}_{3}},{{P}_{4}},{{P}_{5}},{{P}_{6}} \right\}$$ using recurrence relation starting from $$\displaystyle {{P}_{0}}$$ and $$\displaystyle {{P}_{1}}$$ and compare with [[media:2010_11_09_15_00_14.djvu|Eq.4-6 in page-2 Mtg-36]]

Solution
(1) can be written as:

$${{P}_{n+1}}=\frac{\left( 2n+1 \right)x{{P}_{n}}-n{{P}_{n-1}}}{\left( n+1 \right)}$$

Generate $$\displaystyle {{P}_{2}}$$

Let n=1 in  (1), we get

$${{P}_{2}}=\frac{\left( 2\times 1+1 \right)x{{P}_{1}}-1\times {{P}_{0}}}{\left( 1+1 \right)}$$

$${{P}_{2}}=\frac{3x\cdot x-1\times 1}{2}$$

$${{P}_{2}}=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$

Generate $$\displaystyle {{P}_{3}}$$

Let n=2 in  (1), we get

$${{P}_{3}}=\frac{\left( 2\times 2+1 \right)x{{P}_{2}}-2{{P}_{1}}}{\left( 2+1 \right)}$$

$${{P}_{3}}=\frac{5x\cdot \frac{1}{2}\left( 3{{x}^{2}}-1 \right)-2x}{3}$$

$${{P}_{3}}=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$

Generate $$\displaystyle {{P}_{4}}$$

Let n=3 in  (1), we get

$${{P}_{4}}=\frac{\left( 2\times 3+1 \right)x{{P}_{3}}-3{{P}_{2}}}{\left( 3+1 \right)}$$

$${{P}_{4}}=\frac{7x\cdot \frac{1}{2}\left( 5{{x}^{3}}-3x \right)-3\times \frac{1}{2}\left( 3{{x}^{2}}-1 \right)}{4}$$

$${{P}_{4}}=\frac{35{{x}^{4}}-21{{x}^{2}}-9{{x}^{2}}+3}{8}$$

$${{P}_{4}}=\frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)$$

Generate $$\displaystyle {{P}_{5}}$$

Let n=4 in  (1), we get

$${{P}_{5}}=\frac{\left( 2\times 4+1 \right)x{{P}_{4}}-4{{P}_{3}}}{\left( 4+1 \right)}$$

$${{P}_{5}}=\frac{9x\cdot \frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)-4\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)}{5}$$

$${{P}_{5}}=\frac{63{{x}^{5}}}{8}-\frac{54{{x}^{3}}}{8}+\frac{27x}{8\times 5}-2{{x}^{3}}+\frac{6x}{5}$$

$${{P}_{5}}=\frac{1}{8}\left( 63{{x}^{5}}-70{{x}^{3}}+15x \right)$$

Generate $$\displaystyle {{P}_{6}}$$

Let n=5 in  (1), we get

$${{P}_{6}}=\frac{\left( 2\times 5+1 \right)x{{P}_{5}}-5{{P}_{4}}}{\left( 5+1 \right)}$$

$${{P}_{6}}=\frac{11x\cdot \frac{1}{8}\left( 63{{x}^{5}}-70{{x}^{3}}+15x \right)-5\times \frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)}{6}$$

$${{P}_{6}}=\frac{231{{x}^{6}}}{16}-\frac{770{{x}^{4}}}{48}+\frac{55{{x}^{2}}}{16}-\frac{175{{x}^{4}}}{48}+\frac{50{{x}^{2}}}{16}-\frac{5}{16}$$

$${{P}_{6}}=\frac{1}{16}\left( 231{{x}^{6}}-315{{x}^{4}}+105{{x}^{2}}-5 \right)$$

It is seen that $$\left\{ {{P}_{2}}\text{ }{{P}_{3}}\text{ }{{P}_{4}}\text{ }{{P}_{5}}\text{ }{{P}_{6}} \right\}$$ are the same as Eqs.4-6 in 36-2