User:Egm6321.f10.team2.reserve/HW7

= Problem 7.4 =

From the lecture slide Mtg 38-3

Given

 * {| style="width:100%" border="0"

$$ \displaystyle

f(\theta )={{T}_{0}}{{\cos }^{6}}\theta $$

(Eq 4.1)
 * 
 * }


 * {| style="width:100%" border="0"

$$ \displaystyle

f(\theta )={{T}_{0}}\exp \left[ -{{(\frac{2\theta }{\pi })}^{2}} \right]$$

(Eq 4.2)
 * 
 * }

Find
i) Without calculation find property of $$\displaystyle {{A}_{n}}$$


 * $$\displaystyle \left. \begin{align}

& {{A}_{2k}}=0 \\ & {{A}_{2k+1}}=0 \\ \end{align} \right\}k=0,1,2,3...$$

ii) Evaluate non-zero coefficients.

Part a

 * $$\displaystyle f(\mu )={{T}_{0}}{{(1-{{\mu }^{2}})}^{3}}$$

where $$\displaystyle \mu =\sin \theta $$.

When $$\displaystyle n=2k+1$$ and when $$n\displaystyle \ge 7$$, we have,


 * $$\displaystyle \left. \begin{align}

& k=0\Rightarrow n=1 \\ & k=1\Rightarrow n=3 \\ & k=2\Rightarrow n=5 \\ & k=3\Rightarrow n=7 \\ \end{align} \right\}{{A}_{n}}=0$$

Then for evaluating non-zero coefficients.

For simplicity setting $$\displaystyle \mu =x$$ we get $$\displaystyle \mu =x$$.

Therefore we derive coefficients using WolframAlpha ,


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{A}_{0}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{0}}(x)dx}=\frac{1}{2}\int\limits_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}*1*dx=}0.457{{T}_{0}} \\ & {{A}_{2}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{2}}(x)dx}=\frac{5}{2}\int\limits_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}*(\frac{3{{x}^{2}}}{2}-\frac{1}{2})*dx}=-0.762{{T}_{0}} \\ & {{A}_{4}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{4}}(x)dx=}\frac{9}{2}\int\limits_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}*(\frac{35{{x}^{4}}}{8}-\frac{15{{x}^{2}}}{4}+\frac{3}{8})*dx}=0.374{{T}_{0}} \\ & {{A}_{6}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{0}}(x)dx}=\frac{13}{2}\int\limits_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}*(\frac{231{{x}^{6}}}{16}-\frac{315{{x}^{4}}}{16}+\frac{105{{x}^{2}}}{16}-\frac{5}{16})*dx}=0.069{{T}_{0}} \\ \end{align}$$

(Eqs 4.3)
 * 
 * }

Part b

 * $$\displaystyle f(\theta )={{T}_{0}}\exp \left[ -{{(\frac{2\theta }{\pi })}^{2}} \right]$$

Then,


 * $$\displaystyle f(\mu )={{T}_{0}}\exp \left[ -{{(\frac{2\mu }{\pi })}^{2}} \right]={{T}_{0}}\exp \left[ -(\frac{4{{\mu }^{2}}}) \right]$$

where $$\displaystyle \mu =\sin \theta $$.

Since f(x) is an even function for k=0,1,2,3… we get $$\displaystyle {{A}_{2k}}$$.

As to evaluating non-zero coefficients we will repeat the same procedure by setting $$\displaystyle \mu =x$$ to get $$\displaystyle f(x)={{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]$$.


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{A}_{0}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{0}}(x)dx}=\frac{1}{2}\int\limits_{-1}^{1}{{{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]*1*dx=}1.173{{T}_{0}} \\ & {{A}_{2}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{2}}(x)dx}=\frac{5}{2}\int\limits_{-1}^{1}{{{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]*(\frac{3{{x}^{2}}}{2}-\frac{1}{2})*dx}=-0.639{{T}_{0}} \\ & {{A}_{4}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{4}}(x)dx=}\frac{9}{2}\int\limits_{-1}^{1}{{{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]*(\frac{35{{x}^{4}}}{8}-\frac{15{{x}^{2}}}{4}+\frac{3}{8})*dx}=0.714{{T}_{0}} \\ & {{A}_{6}}=\frac{2n+1}{2}\int\limits_{-1}^{1}{f(x){{P}_{0}}(x)dx}=\frac{13}{2}\int\limits_{-1}^{1}{{{T}_{0}}\exp \left[ -(\frac{4{{x}^{2}}}) \right]*(\frac{231{{x}^{6}}}{16}-\frac{315{{x}^{4}}}{16}+\frac{105{{x}^{2}}}{16}-\frac{5}{16})*dx}=-0.726{{T}_{0}} \\ \end{align}$$

(Eq 4.4)
 * 
 * }

= Problem 7.5 Finding second homogenous solution of Legendre equations of order 0 and 1 =

From the lecture slide Mtg39-1

Given
The Legendre equation is,


 * $$\displaystyle (1-{{x}^{2}}){y}''-2x{y}'+n(n+1)y=0$$

with its first homogeneous solution with respect to order 0 and 1,


 * $$\displaystyle \begin{align}

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

Find
Prove the second homogenous solutions of Legendre equation with respect to order 0 and 1 are,


 * $$\displaystyle \begin{align}

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

Solution
During the derivation of Legendre equation we assumed that $$\displaystyle x=\sin \theta $$ meaning $$\displaystyle x$$ has the domain of $$\displaystyle [-1,+1]$$ in which +1 and -1 are its singularity point.

Verification of $$\displaystyle {{Q}_{0}}(x)$$
When the order of Legendre equation equals 0,


 * {| style="width:100%" border="0"

$$ \displaystyle

(1-{{x}^{2}}){y}''-2x{y}'=0$$

(Eq 5.1)
 * 
 * }

Eq 5.1 can be immediately reduced to a linear first ODE with varing coefficients by taking $$\displaystyle {y}'=z$$,


 * {| style="width:100%" border="0"

$$ \displaystyle

{z}'+\frac{2x}{{{x}^{2}}-1}z=0$$

(Eq 5.2)
 * 
 * }

The integrating factor is,


 * {| style="width:100%" border="0"

$$ \displaystyle

\begin{align} h(x) & =\exp \left[ \int_ – ^{x}{\frac{2x}{{{x}^{2}}-1}dx} \right] \\ & =\exp \left[ \int_ – ^{x}{\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] \\ & =\exp [\log (1-{{x}^{2}})] \\ & =1-{{x}^{2}} \\ \end{align}$$

(Eq 5.3)
 * 
 * }

Hence,


 * {| style="width:100%" border="0"

$$ \displaystyle

\begin{align} z & ={{h}^{-1}}(x)\int_ – ^{x}{h(x)*0dx} \\ & =\frac{1-{{x}^{2}}} \\ \end{align}$$

(Eq 5.4)
 * 
 * }

Then we can derive $$\displaystyle y$$,


 * {| style="width:100%" border="0"

$$ \displaystyle

\begin{align} y & =\int{zdx}+{{k}_{1}} \\ & ={{k}_{1}}+\int_ – ^{}{\frac{1-{{x}^{2}}}dx} \\ & ={{k}_{1}}+\frac{2}\int_ – ^{}{\left( \frac{1}{1+x}+\frac{1}{1-x} \right)dx} \\ & ={{k}_{1}}+{{k}_{2}}\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ \end{align}$$

(Eq 5.5)
 * 
 * }

Since $$\displaystyle {{P}_{0}}(x)$$ and $$\displaystyle {{Q}_{0}}(x)$$ are two homogeneous solutions of Eq 5.1, we can say,


 * {| style="width:100%" border="0"

$$ \displaystyle

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

(Eq 5.6)
 * 
 * }

On the other hand,


 * {| style="width:100%" border="0"

$$ \displaystyle

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

(Eq 5.7)
 * 
 * }

Thus $$\displaystyle {{Q}_{0}}(x)$$ can also be written as


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

{{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x$$

(Eq 5.8)
 * 
 * }

Verification of $$\displaystyle {{Q}_{1}}(x)$$
When the order of Legendre is set to 1,


 * $$\displaystyle (1-{{x}^{2}}){y}''-2x{y}'+2y=0$$

or


 * {| style="width:100%" border="0"

$$ \displaystyle

{y}''+\frac{2x}{({{x}^{2}}-1)}{y}'+\frac{2}{(1-{{x}^{2}})}y=0$$

(Eq 5.9)
 * 
 * }

By direct method in Mtg29-3 and Mtg30-1, we derive,


 * {| style="width:100%" border="0"

$$ \displaystyle

\begin{align} & h(x)={{u}_{1}}^{2}(x)\exp \left[ \int_ – ^ – {{{a}_{1}}(x)dx} \right] \\ & ={{[{{P}_{1}}(x)]}^{2}}\exp \left[ \int_ – ^ – {\frac{2x}{{{x}^{2}}-1}dx} \right] \\ & ={{x}^{2}}\exp \left[ \int_ – ^ – {\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] \\ & ={{x}^{2}}(1-{{x}^{2}}) \\ \end{align}$$

(Eq 5.10)
 * 
 * }

and then,


 * {| style="width:100%" border="0"

$$ \displaystyle

\begin{align} {{Q}_{1}}(x) & ={{u}_{2}}(x) \\ & ={{u}_{1}}(x)\int_ – ^ – {\frac{1}{h(x)}dx} \\ & =x\int{\frac{1}{{{x}^{2}}(1-{{x}^{2}})}dx} \\ & =x\int{\frac{1}+\frac{1}{(1-{{x}^{2}})}dx} \\ & =x\int{\frac{1}{(1-{{x}^{2}})}dx+}x\int{\frac{1}dx} \\ & =\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \\ \end{align}$$

(Eq 5.11)
 * 
 * }

By Eq 5.7,


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

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

(Eq 5.12)
 * 
 * }

= Problem 7.6 The oddness of $$\displaystyle {{Q}_{n}}(x)$$ with respect to $$\displaystyle n$$ =

From the lecture slide Mtg 39-1

Given
In general the second catogary of Legendre functions have such expression,


 * {| style="width:100%" border="0"

$$ \displaystyle

{{Q}_{n}}(x)={{P}_{n}}(x){{\tanh }^{-1}}(x)-2\sum\nolimits_{j=1,3,5\cdots }^{J}{\frac{2n-2j+1}{(2n-j+1)j}{{P}_{n-j}}(x)}$$

(Eq 6.1)
 * <p style="text-align:right">
 * }

where $$\displaystyle J=1+2\left[ \frac{n-1}{2} \right]$$.

Find
Prove that $$\displaystyle {{Q}_{n}}(x)$$ has oddness depending on the value $$\displaystyle n$$.

When $$\displaystyle n$$ is even
From problem 7.4 we have known that if $$\displaystyle n$$ is even then $$\displaystyle {{P}_{n}}(x)$$ is even and vice versa.

On the other hand $$\displaystyle {{\tanh }^{-1}}(x)$$ is always odd, then the first tern of Eq 6.1 is odd.

Since $$\displaystyle J=1+2\left[ \frac{n-1}{2} \right]$$, it's clear that $$\displaystyle J$$ is always odd.

Also $$\displaystyle j$$ is always odd making every $$\displaystyle n-j$$ in the summation odd.

Therefore every $$\displaystyle {{P}_{n-j}}(x)$$ in the summation is odd.

Drawn together, when $$\displaystyle n$$ is even, $$\displaystyle {{Q}_{n}}(x)$$ consists of several odd terms. By problem 7.2 we know that $$\displaystyle {{Q}_{n}}(x)$$ is eventually odd.

When $$\displaystyle n$$ is odd
Going the same process like above, we know that:

$$\displaystyle {{P}_{n}}(x)$$ is odd;

$$\displaystyle {{\tanh }^{-1}}(x)$$ is always odd;

the first tern of Eq 6.1 is even;

$$\displaystyle J$$ is always odd;

every $$\displaystyle n-j$$ in the summation even;

every $$\displaystyle {{P}_{n-j}}(x)$$ in the summation is even;

$$\displaystyle {{Q}_{n}}(x)$$ consists of several even terms thus it's even.

Conclusion
The oddness of $$\displaystyle {{Q}_{n}}(x)$$ depend on the value of $$\displaystyle n$$, and $$\displaystyle {{P}_{n}}(x)$$ and $$\displaystyle {{Q}_{n}}(x)$$ have opposote oddness.

= Problem 7.7 Distance Between Two Spheres =

From the lecture slide Mtg 40-3

Given
$$\displaystyle P\left( {{r}_{P}},{{\varphi }_{P}},{{\theta }_{P}} \right)$$ and $$\displaystyle Q\left( {{r}_{Q}},{{\varphi }_{Q}},{{\theta }_{Q}} \right)$$ for spherical coordinate system.

In Astronomical Convention we have,


 * $$\displaystyle \begin{align}

& x_{P}^{1}=r_{P}^ – \cos {{\theta }_{P}}\cos {{\varphi }_{P}} \\ & x_{P}^{2}=r_{P}^ – \cos {{\theta }_{P}}\sin {{\varphi }_{P}} \\ & x_{P}^{3}=r_{P}^ – \sin {{\theta }_{P}} \\ \end{align}$$


 * $$\displaystyle \begin{align}

& x_{Q}^{1}=r_{Q}^ – \cos {{\theta }_{Q}}\cos {{\varphi }_{Q}} \\ & x_{Q}^{2}=r_{Q}^ – \cos {{\theta }_{Q}}\sin {{\varphi }_{Q}} \\ & x_{Q}^{3}=r_{Q}^ – \sin {{\theta }_{Q}} \\ \end{align}$$


 * $$\displaystyle \cos \gamma =\cos {{\theta }_{Q}}\cos {{\theta }_{P}}\cos ({{\varphi }_{Q}}-{{\varphi }_{P}})+\sin {{\theta }_{Q}}\sin {{\theta }_{P}}$$


 * $$\displaystyle {{\left( r_{PQ}^ – \right)}^{2}}={{\left\| \overrightarrow{PQ} \right\|}^{2}}=\sum\limits_{i=1}^{3}={{\left( x_{Q}^{1}-x_{P}^{1} \right)}^{2}}+{{\left( x_{Q}^{2}-x_{P}^{2} \right)}^{2}}+{{\left( x_{Q}^{3}-x_{P}^{3} \right)}^{2}}$$

Find
Show that,


 * $$\displaystyle {{\left( r_{PQ}^ – \right)}^{2}}={{\left( r_{Q}^ –  \right)}^{2}}\left[ {{\left( \frac \right)}^{2}}+1-2\left( \frac \right)\cos \gamma  \right]$$

Solution
Since,


 * {| style="width:100%" border="0"

$$ \displaystyle

{{\left( r_{PQ}^ – \right)}^{2}}={{\left( x_{Q}^{1}-x_{P}^{1} \right)}^{2}}+{{\left( x_{Q}^{2}-x_{P}^{2} \right)}^{2}}+{{\left( x_{Q}^{3}-x_{P}^{3} \right)}^{2}}$$

(Eq 7.1)
 * <p style="text-align:right">
 * }

so we need to calculate each terms.

First Term:


 * $$\displaystyle {{\left( x_{Q}^{1}-x_{P}^{1} \right)}^{2}}={{\left( r_{Q}^ – \cos {{\theta }_{Q}}\cos {{\varphi }_{Q}}-r_{P}^ – \cos {{\theta }_{P}}\cos {{\varphi }_{P}} \right)}^{2}}$$


 * $$\displaystyle {{\left( x_{Q}^{1}-x_{P}^{1} \right)}^{2}}=r_{Q}^{2}{{\cos }^{2}}{{\theta }_{Q}}{{\cos }^{2}}{{\varphi }_{Q}}+r_{P}^{2}{{\cos }^{2}}{{\theta }_{P}}{{\cos }^{2}}{{\varphi }_{P}}-2r_{Q}^ – r_{P}^ – \cos {{\theta }_{Q}}\cos {{\varphi }_{Q}}\cos {{\theta }_{P}}\cos {{\varphi }_{P}}$$

Second Term:


 * $$\displaystyle {{\left( x_{Q}^{2}-x_{P}^{2} \right)}^{2}}={{\left( r_{Q}^ – \cos {{\theta }_{Q}}\sin {{\varphi }_{Q}}-r_{P}^ – \cos {{\theta }_{P}}\sin {{\varphi }_{P}} \right)}^{2}}$$


 * $$\displaystyle {{\left( x_{Q}^{2}-x_{P}^{2} \right)}^{2}}=r_{Q}^{2}{{\cos }^{2}}{{\theta }_{Q}}{{\sin }^{2}}{{\varphi }_{Q}}+r_{P}^{2}{{\cos }^{2}}{{\theta }_{P}}{{\sin }^{2}}{{\varphi }_{P}}-2r_{Q}^ – r_{P}^ – \cos {{\theta }_{Q}}\sin {{\varphi }_{Q}}\cos {{\theta }_{P}}\sin {{\varphi }_{P}}$$

Third Term:


 * $$\displaystyle {{\left( x_{Q}^{3}-x_{P}^{3} \right)}^{2}}={{\left( r_{Q}^ – \sin {{\theta }_{Q}}-r_{P}^ – \sin {{\theta }_{P}} \right)}^{2}}$$


 * $$\displaystyle {{\left( x_{Q}^{3}-x_{P}^{3} \right)}^{2}}=r_{Q}^{2}{{\sin }^{2}}{{\theta }_{Q}}+r_{P}^{2}{{\sin }^{2}}{{\theta }_{P}}-2\sin {{\theta }_{Q}}\sin {{\theta }_{P}}$$

Add all three terms together, then we get


 * {| style="width:100%" border="0"

$$ \displaystyle

\begin{align} {{\left( r_{PQ}^ – \right)}^{2}}= & r_{Q}^{2}\left[ {{\cos }^{2}}{{\theta }_{Q}}\left( {{\cos }^{2}}{{\varphi }_{Q}}+{{\sin }^{2}}{{\varphi }_{Q}} \right)+{{\sin }^{2}}{{\theta }_{Q}} \right]+r_{P}^{2}\left[ {{\cos }^{2}}{{\theta }_{P}}\left( {{\cos }^{2}}{{\varphi }_{P}}+{{\sin }^{2}}{{\varphi }_{P}} \right)+{{\sin }^{2}}{{\theta }_{P}} \right] \\ & -2r_{Q}^ – r_{P}^ – \left[ \cos {{\theta }_{Q}}\cos {{\theta }_{P}}\left( \cos {{\varphi }_{Q}}\cos {{\varphi }_{P}}+\sin {{\varphi }_{Q}}\sin {{\varphi }_{P}} \right)+\sin {{\theta }_{Q}}\sin {{\theta }_{P}} \right] \\ \end{align}$$

(Eq 7.2)
 * <p style="text-align:right">
 * }

From trigonometric equations, we know that


 * $$\displaystyle \left( {{\cos }^{2}}{{\varphi }_{Q}}+{{\sin }^{2}}{{\varphi }_{Q}} \right)=1$$


 * $$\displaystyle \left( {{\cos }^{2}}{{\varphi }_{P}}+{{\sin }^{2}}{{\varphi }_{P}} \right)=1$$


 * $$\displaystyle \left[ {{\cos }^{2}}{{\theta }_{Q}}+{{\sin }^{2}}{{\theta }_{Q}} \right]=1$$


 * $$\displaystyle \left[ {{\cos }^{2}}{{\theta }_{P}}+{{\sin }^{2}}{{\theta }_{P}} \right]=1$$


 * $$\displaystyle \left( \cos {{\varphi }_{Q}}\cos {{\varphi }_{P}}+\sin {{\varphi }_{Q}}\sin {{\varphi }_{P}} \right)=\cos \left( {{\varphi }_{Q}}-{{\varphi }_{P}} \right)$$

So putting equations above in the $$\displaystyle {{\left( r_{PQ}^ – \right)}^{2}}$$, we obtain


 * {| style="width:100%" border="0"

$$ \displaystyle

\begin{align} {{\left( r_{PQ}^ – \right)}^{2}}= & r_{Q}^{2}+r_{P}^{2}-2r_{Q}^ – r_{P}^ – \left[ \cos {{\theta }_{Q}}\cos {{\theta }_{P}}\left( \cos ({{\varphi }_{Q}}-{{\varphi }_{P}}) \right)+\sin {{\theta }_{Q}}\sin {{\theta }_{P}} \right] \\ & =r_{Q}^{2}+r_{P}^{2}-2r_{Q}^ – r_{P}^ – \cos \gamma \\ \end{align}$$

(Eq 7.3)
 * <p style="text-align:right">
 * }

If we take common parenthesis of $$\displaystyle {{\left( r_{Q}^ – \right)}^{2}}$$, it will be shown


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

{{\left( r_{PQ}^ – \right)}^{2}}={{\left( r_{Q}^ –  \right)}^{2}}\left[ {{\left( \frac \right)}^{2}}+1-2\left( \frac \right)\cos \gamma  \right]$$

(Eq 7.4)
 * <p style="text-align:right">
 * }

= Problem 7.9 Deriving Legendre polynominals using recurrence relation =

From the lecture slide Mtg 41-2

Given
The second recurrence relation of Legendre polynominals is,


 * {| style="width:100%" border="0"

$$ \displaystyle

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

(Eq 9.1)
 * <p style="text-align:right">
 * }

And Legendre polynominal of order 0 and 1 are,


 * $$\displaystyle \begin{align}

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

Find
Legendre polynominals of order 2,3,4,5,6.

Solution
From Eq 9.1 we have,


 * {| style="width:100%" border="0"

$$ \displaystyle

{{P}_{n+1}}=\frac{2n+1}{n+1}x{{P}_{n}}-\frac{n}{n+1}{{P}_{n-1}}$$

(Eq 9.2)
 * <p style="text-align:right">
 * }

Take $$\displaystyle n$$ as $$\displaystyle 2,3,4,5,6$$ we have,


 * {| style="width:100%" border="0"

$$ \displaystyle

\begin{align} & {{P}_{2}}=\frac{3}{2}x{{P}_{1}}-\frac{1}{2}{{P}_{0}} \\ & {{P}_{3}}=\frac{5}{3}x{{P}_{2}}-\frac{2}{3}{{P}_{1}} \\ & {{P}_{4}}=\frac{7}{4}x{{P}_{3}}-\frac{3}{4}{{P}_{2}} \\ & {{P}_{5}}=\frac{9}{5}x{{P}_{4}}-\frac{4}{5}{{P}_{3}} \\ & {{P}_{6}}=\frac{11}{6}x{{P}_{5}}-\frac{5}{6}{{P}_{4}} \\ \end{align}$$

(Eqs 9.3)
 * <p style="text-align:right">
 * }

Solve Eqs 9.3 by sequence we have,


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{P}_{2}}=\frac{3}{2}x*x-\frac{1}{2}*1=\frac{3}{2}{{x}^{2}}-\frac{1}{2} \\ & {{P}_{3}}=\frac{5}{3}x\left( \frac{3}{2}{{x}^{2}}-\frac{1}{2} \right)-\frac{2}{3}x=\frac{5}{2}{{x}^{3}}-\frac{3}{2}x \\ & {{P}_{4}}=\frac{7}{4}x\left( \frac{5}{2}{{x}^{3}}-\frac{3}{2}x \right)-\frac{3}{4}\left( \frac{3}{2}{{x}^{2}}-\frac{1}{2} \right)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ & {{P}_{5}}=\frac{9}{5}x\left( \frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \right)-\frac{4}{5}\left( \frac{5}{2}{{x}^{3}}-\frac{3}{2}x \right)=\frac{63}{8}{{x}^{5}}-\frac{35}{4}{{x}^{3}}+\frac{15}{8}x \\ & {{P}_{6}}=\frac{11}{6}x\left( \frac{63}{8}{{x}^{5}}-\frac{35}{4}{{x}^{3}}+\frac{15}{8}x \right)-\frac{5}{6}\left( \frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \right)=\frac{231}{16}{{x}^{6}}-\frac{315}{16}{{x}^{4}}+\frac{105}{16}{{x}^{2}}-5 \\ \end{align}$$

(Eq 9.4)
 * <p style="text-align:right">
 * }

= References =