User:Egm6321.f10.team2.reserve/HW6

= Problem 6.2 =

From the lecture slide Mtg 33-1

Given

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$$ \displaystyle

(x-1)y''-xy'+y=f(x)$$

(Eq 2.1)
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 * }


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$$ \displaystyle

xy''+2y'+xy=f(x)$$

(Eq 2.2)
 * 
 * }

Find
Solve the equations above for $$\displaystyle f(x)=0$$ and $$\displaystyle f(x)=\sin x$$; so


 * a)	Eq 2.1 with $$\displaystyle f(x)=0$$


 * b)	Eq 2.1 with $$\displaystyle f(x)=\sin x$$


 * c)	Eq 2.2 with $$\displaystyle f(x)=0$$


 * d)	Eq 2.2 with $$\displaystyle f(x)=\sin x$$

Solution
First of all applying the generic form of $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$ and then taking the first and the second derivatives will result in the following:


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$$ \displaystyle

\begin{align} & y'=c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}} \\ & y''=c(c-1){{x}^{c-2}}{{e}^{rx}}+c{{x}^{c-1}}r{{e}^{rx}}+c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}rr{{e}^{rx}} \\ & y''=c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}} \\ \end{align}$$

(Eqs 2.3)
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 * }

The procedure above will be the same for both part(a) and part(c) parts of the problem.

Part a
Now that we found the derivatives, the solutions can be plugged into the Eq 2.1 that is set $$\displaystyle f(x)=0$$:


 * $$\displaystyle (x-1)y''-xy'+y=0$$


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$$ \displaystyle

(x-1)*\underbrace{[c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}}]}_{y''}-x*\underbrace{[c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}}]}_{y'}+\underbrace{[{{x}^{c}}{{e}^{rx}}]}_{y}=0$$

(Eq 2.4)
 * 
 * }

Since every term in the equation has $$\displaystyle {{e}^{rx}}$$, it can be taken out of the parenthesis to be eliminated:


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$$ \displaystyle

(x-1)*{{e}^{rx}}[c(c-1){{x}^{c-2}}+2c{{x}^{c-1}}r+{{x}^{c}}{{r}^{2}}]-x*{{e}^{rx}}[c{{x}^{c-1}}+{{x}^{c}}r]+{{e}^{rx}}[{{x}^{c}}]=0$$

(Eq 2.5)
 * 
 * }

To simplify the equation we will go through the following procedure:


 * $$\displaystyle \begin{align}

& (x-1)[c(c-1){{x}^{c-2}}+2c{{x}^{c-1}}r+{{x}^{c}}{{r}^{2}}]-x[c{{x}^{c-1}}+{{x}^{c}}r]+[{{x}^{c}}]=0 \\ & ({{c}^{2}}-c){{x}^{c-1}}+2c{{x}^{c}}r+{{x}^{c+1}}{{r}^{2}}-({{c}^{2}}-c){{x}^{c-2}}-2c{{x}^{c-1}}r-{{x}^{c}}{{r}^{2}}-c{{x}^{c}}-{{x}^{c+1}}r+{{x}^{c}}=0 \\ & {{x}^{c+1}}({{r}^{2}}-r)+{{x}^{c}}(2cr-{{r}^{2}}-c+1)+{{x}^{c-1}}({{c}^{2}}-c-2cr)+{{x}^{c-2}}(c-{{c}^{2}})=0 \\ \end{align}$$

When $$\displaystyle c=1\Rightarrow r=0$$ or $$\displaystyle c=0\Rightarrow r=1$$, which will yield us two homogenous solutions plugging into $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$.


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$$ \displaystyle

\begin{align} & {{y}_{H1}}={{x}^{0}}{{e}^{1x}}={{e}^{x}} \\ & {{y}_{H2}}={{x}^{1}}{{e}^{0x}}=x \\ \end{align}$$

(Eqs 2.6)
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 * }

Then,


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$$ \displaystyle
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y={{C}_{1}}{{e}^{x}}+{{C}_{2}}x$$

(Eq 2.7)
 * 
 * }

Part b
When we set $$\displaystyle f(x)=\sin x$$ for Eq 2.1, we get:


 * $$\displaystyle (x-1)y''-xy'+y=\sin x$$

We will rewrite the equation to put in the proper form and find the integrating factor:


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$$ \displaystyle

y''-\frac{x}{(x-1)}y'+\frac{1}{(x-1)}y=\frac{\sin x}{(x-1)}$$

(Eq 2.8) Plugging in all the values into the integrating factor formula, we get:
 * 
 * }


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$$ \displaystyle

h(x)=\underbrace_{y_{H1}^{2}}{{e}^{-\int{\frac{x}{(x-1)}dx}}}={{e}^{2x}}{{e}^{-\ln (x-1)-x}}={{e}^{x-\ln (x-1)}}={{e}^{x}}{{e}^{\ln \frac{1}{(x-1)}}}=\frac{(x-1)}$$

(Eq 2.9) Now to find the particular solution we will apply the following formula:
 * 
 * }


 * $$\displaystyle {{y}_{P}}(x)={{u}_{1}}(x)\int{\frac{1}{h(x)}}\left[ \int{h(x)F(x)dx} \right]dx$$

$$ \displaystyle

{{y}_{P}}(x)={{e}^{x}}\int{\frac{(x-1)}}\left[ \int{\frac{{{e}^{x}}\sin x}{(x-1)(x-1)}dx} \right]dx={{e}^{x}}\int{\frac{(x-1)}}\left[ \int{\frac{{{e}^{x}}\sin x}dx} \right]dx$$

(Eq 2.10) Finally combining homogenous solution with the particular solution we get:
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 * }
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$$ \displaystyle
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y={{C}_{1}}{{e}^{x}}+{{C}_{2}}x+{{e}^{x}}\int{\frac{(x-1)}}\left[ \int{\frac{{{e}^{x}}\sin x}dx} \right]dx$$

(Eq 2.11)
 * 
 * }

Part c

 * $$\displaystyle xy''+2y'+xy=0$$


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$$ \displaystyle

x*\underbrace{[c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}}]}_{y''}+2*\underbrace{[c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}}]}_{y'}+x*\underbrace{[{{x}^{c}}{{e}^{rx}}]}_{y}=0$$

(Eq 2.12) Just like the way it was done in part a, the $$\displaystyle {{e}^{rx}}$$ term will be taken out and the equation will be simplified:
 * 
 * }
 * $$\begin{align}

& x*[c(c-1){{x}^{c-2}}+2c{{x}^{c-1}}r+{{x}^{c}}{{r}^{2}}]+2*[c{{x}^{c-1}}+{{x}^{c}}r]+x*[{{x}^{c}}]=0 \\ & ({{c}^{2}}-c){{x}^{c-1}}+2cr{{x}^{c}}+{{r}^{2}}{{x}^{c+1}}+2c{{x}^{c-1}}+2{{x}^{c}}r+{{x}^{c+1}}=0 \\ & {{x}^{c+1}}({{r}^{2}}+1)+{{x}^{c}}(2cr+2r)+{{x}^{c-1}}({{c}^{2}}+c)=0 \\ \end{align}$$

By solving the above equation we get $$\displaystyle c=-1\Rightarrow r=\pm i$$

Plugging these values into $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$, will give us the following equations:


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$$ \displaystyle

\begin{align} & {{y}_{H1}}={{x}^{-1}}{{e}^{ix}}=\frac{x} \\ & {{y}_{H2}}={{x}^{-1}}{{e}^{-ix}}=\frac{1}{x{{e}^{ix}}} \\ \end{align}$$

(Eqs 2.13)
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 * }

Then,


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$$ \displaystyle
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y={{C}_{1}}\frac{x}+{{C}_{2}}\frac{1}{x{{e}^{ix}}}$$

(Eq 2.14)
 * 
 * }

Part d
At last but not least Eq 2.2 will be solved with $$\displaystyle f(x)=\sin x$$.


 * $$\displaystyle xy''+2y'+xy=\sin x$$

Rearranging the equation to put in the proper form and find the integrating factor:


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$$ \displaystyle

y''+\frac{2}{x}y'+y=\frac{\sin x}{x}$$

(Eq 2.15) Plugging in all the values into the integrating factor formula, we get:
 * 
 * }


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$$ \displaystyle

h(x)=\frac{{e}^{\int{\frac{2}{x}dx}}}=\frac{{e}^{\ln {{x}^{2}}}}=\frac{{x}^{2}}={{e}^{2ix}}$$

(Eq 2.16) Now to find the particular solution we will apply the following formula:
 * 
 * }


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$$ \displaystyle

{{y}_{P}}(x)={{u}_{1}}(x)\int{\frac{1}{h(x)}}\left[ \int{h(x)F(x)dx} \right]dx$$

(Eq 2.17)
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 * }


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$$ \displaystyle

{{y}_{P}}(x)=\frac{x}\int{\frac{1}}\left[ \int{\frac{{{e}^{2ix}}\sin x}{x}dx} \right]dx$$

(Eq 2.18) Finally combining homogenous solution with the particular solution we get:
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 * }


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$$ \displaystyle
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y={{C}_{1}}\frac{x}+{{C}_{2}}\frac{1}{x{{e}^{ix}}}+\frac{x}\int{\frac{1}}\left[ \int{\frac{{{e}^{2ix}}\sin x}{x}dx} \right]dx$$

(Eq 2.19)
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 * }

= Problem 6.3 =

From the lecture slide Mtg 33-1

Given

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

Find
Solve equation using direct method.

Solution
The two homogeneous were given in class lecture 30-2 as the following:


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$$ \displaystyle

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

(Eqs 3.1) $$\displaystyle {{y}_{H}}$$ can be expressed as
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 * }
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$$ \displaystyle

{{y}_{H}}={{C}_{1}}x+{{C}_{2}}\left[ \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right]$$

(Eq 3.2)
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 * }

Using the example from the textbook on page 34 we can use the variation of parameters formula that is


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$$ \displaystyle

y(x)=\int_ – ^{x}{f(s)\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{W(s)} \right\}}ds$$

(Eq 3.3)
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 * }


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$$ \displaystyle

{{y}_{P}}=\int_ – ^{x}{\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{(1-{{s}^{2}})[{{u}_{1}}(s)u_{2}^{'}(s)-u_{1}^{'}(s)u_{2}^ – (s)} \right\}}ds$$

(Eq 3.4)
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 * }


 * $$\displaystyle W=\frac{1-{{s}^{2}}}$$


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$$ \displaystyle

\begin{align} {{y}_{P}}(x) & ={{u}_{2}}(x)\int{{{u}_{1}}(x)dx-}{{u}_{1}}(x)\int{{{u}_{2}}(x)dx} \\ & =\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)\int{xdx-}x\int{\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)dx} \\ & =\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)\frac{2}-\left( \frac{{{x}^{3}}-x}{4}\log \left( \frac{1+x}{1-x} \right)-\frac{2} \right) \\ & =\frac{x}{4}\log \left( \frac{1+x}{1-x} \right) \\ \end{align}$$

(Eq 3.5) So we can express the general solution as the following:
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 * }


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$$ \displaystyle
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y={{C}_{1}}x+{{C}_{2}}\left[ \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right]+\frac{x}{4}\log \left( \frac{1+x}{1-x} \right)$$

(Eq 3.6)
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 * }

= Problem 6.6 Finding Laplacian in spherical coordinates using Math/Physics convention =

From the lecture slide Mtg 35-4

Given
The Laplacian in spherical coordinates using astronomy convention is ,


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$$ \displaystyle

\Delta \Psi =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\cos }^{2}}\theta }\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\cos \theta }\frac{\partial }{\partial \theta }\left( \cos \theta \frac{\partial \Psi }{\partial \theta } \right)$$

(Eq 6.1)
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 * }

where $$\displaystyle \theta \in [-\pi /2,+\pi /2]$$

Find
Laplacian in spherical coordinates using Math/Physics convention: $$\displaystyle r,\bar{\theta },\varphi $$ where,


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$$ \displaystyle

\bar{\theta }=\frac{\pi }{2}-\theta $$

(Eq 6.2)
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Solution
Sustitute Eq 6.2 into Eq 6.1 we get,


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$$ \displaystyle
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\begin{align} \Delta \Psi & =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\cos }^{2}}(\pi /2-\bar{\theta })}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\cos (\pi /2-\bar{\theta })}\frac{\partial }{\partial (\pi /2-\bar{\theta })}\left( \cos (\pi /2-\bar{\theta })\frac{\partial \Psi }{\partial (\pi /2-\bar{\theta })} \right) \\ & =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\sin }^{2}}\bar{\theta }}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\sin \bar{\theta }}\frac{\partial }{\partial \bar{\theta }}\left( \sin \bar{\theta }\frac{\partial \Psi }{\partial \bar{\theta }} \right) \\ \end{align}$$

(Eq6.3)
 * <p style="text-align:right">
 * }

If we don't know Laplacian in astronomy convention at first
Then the transmation between Cartesian coodinates and spherical coordinates(Math/Physics convention) is,


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$$ \displaystyle

\begin{align} & x=r\sin \bar{\theta }\cos \varphi \\ & y=r\cos \bar{\theta }\sin \varphi \\ & z=r\cos \bar{\theta } \\ \end{align}$$

(Eqs 6.4)
 * <p style="text-align:right">
 * }

Hence,


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$$ \displaystyle

\begin{align} & dx=\sin \bar{\theta }\cos \varphi dr+r\cos \bar{\theta }\cos \varphi d\bar{\theta }-r\sin \bar{\theta }\sin \varphi d\varphi \\ & dy=\sin \bar{\theta }\sin \varphi dr+r\cos \bar{\theta }\sin \varphi d\bar{\theta }+r\sin \bar{\theta }\cos \varphi d\varphi \\ & dz=\cos \bar{\theta }dr-r\sin \bar{\theta }d\bar{\theta } \\ \end{align}$$

(Eqs 6.5)
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 * }

Consider the magnitude of an infinitesimal segment $$\displaystyle ds$$,


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$$ \displaystyle

d{{s}^{2}}=d{{r}^{2}}+{{r}^{2}}d{^{2}}+{{r}^{2}}{{\sin }^{2}}\bar{\theta }d{{\varphi }^{2}}$$

(Eq 6.6)
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 * }

Then we have scale factors,


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$$ \displaystyle

\begin{align} & {{h}_{r}}=1 \\ & {{h}_}=r \\ & {{h}_{\varphi }}=r\sin \bar{\theta } \\ \end{align}$$

(Eqs 6.7)
 * <p style="text-align:right">
 * }

From lecture slide 34-2 we know the Laplacian for curvilinear coordinates is:


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$$ \displaystyle

\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\xi }_{i}}}\left[ \frac{h_{i}^{2}}\frac{\partial \Phi }{\partial {{\xi }_{i}}} \right]}$$

(Eq 6.8)
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 * }

Substitute Eqs 6.7 into Eq 6.8 we have,


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$$ \displaystyle
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\Delta \Psi =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\sin }^{2}}\bar{\theta }}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\sin \bar{\theta }}\frac{\partial }{\partial \bar{\theta }}\left( \sin \bar{\theta }\frac{\partial \Psi }{\partial \bar{\theta }} \right)$$

(Eq 6.9)
 * <p style="text-align:right">
 * }

= Problem 6.7 Verify expression for $$\displaystyle \Delta \Psi $$ in elliptic coordinates from wikipedia article =

From the lecture slide Mtg 36-1

Given
The most common two-dimension definition of elliptic coordinates $$\displaystyle (\mu, \nu)$$ is



x = a \ \cosh \mu \ \cos \nu $$



y = a \ \sinh \mu \ \sin \nu $$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, 2\pi].$$

Find
The Laplacian $$\displaystyle \Delta \Psi $$ expressed in elliptic coordinates.

Solution
From the information given we have:


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$$ \displaystyle \begin{align} & dx=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ & dy=a\sinh \mu \cos \nu d\mu +a\cosh \mu \sin \nu dv \\ \end{align}$$

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

Consider magnitude of an infinitesimal segment $$\displaystyle ds$$. Before that we should aware that since ellipic coordinates system is among ORTHOGONAL curvilinear coordinates which never have off-diagonal terms in their metric tensor, meaning terms like $$\displaystyle dxdy$$ will never show up in the magnitude of $$\displaystyle ds$$ thus no need to be calculated.


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$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}} \\ & =({{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu +a{{\cosh }^{2}}\mu {{\sin }^{2}}\nu )(d{{\mu }^{2}}+d{{v}^{2}}) \\ & =\left[ ({{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu )+(a{{\cosh }^{2}}\mu {{\sin }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu ) \right](d{{\mu }^{2}}+d{{v}^{2}}) \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}} \\ \end{align}$$

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

We can derive scale factors from Eq 7.2,


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$$ \displaystyle

{{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}$$

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

From lecture slide 34-2 we know the Laplacian for curvilinear coordinates is:


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$$ \displaystyle

\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\xi }_{i}}}\left[ \frac{h_{i}^{2}}\frac{\partial \Psi }{\partial {{\xi }_{i}}} \right]}$$

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

and in two-dimensional case:


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$$ \displaystyle

\Delta \Psi =\frac{1}\left[ \frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial u}+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial \nu }+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}} \right]$$

(Eq 7.5)
 * <p style="text-align:right">
 * }

from which we can see that the sign before scale factors won't matter since the they will be cancelled.

Then we derive $$\displaystyle \Delta \Psi $$ in two dimensional elliptic coordinates by substituting Eq7.3 into Eq7.5,


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

\Delta \Psi =\frac{1}{{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}\left( \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}} \right)$$

(Eq 7.6)
 * <p style="text-align:right">
 * }

Compare Eq 7.6 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

The Laplacian in three dimensinal elliptic coordinate systems
The wikipedia article required for verification also provided some outer links to some 3-D elliptic coordinate systems, hence one may want to verify Laplacians in different 3-D elliptic coordinate systems.

Verification of Laplacian in Elliptic cylindrical coordinates
The most common definition of elliptic cylindrical coordinates $$\displaystyle (\mu, \nu, z)$$ is ,



x = a \ \cosh \mu \ \cos \nu $$



y = a \ \sinh \mu \ \sin \nu $$



z = z \!$$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, 2\pi)$$.


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$$ \displaystyle

\begin{align} & dx=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ & dy=a\sinh \mu \cos \nu d\mu +a\cosh \mu \sin \nu dv \\ & dz=dz \\ \end{align}$$

(Eq 7.7)
 * <p style="text-align:right">
 * }

Then the corresponding magnitude of $$\displaystyle ds$$ is


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$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}}+d{{z}^{2}} \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}}+d{{z}^{2}} \\ \end{align}$$

(Eq 7.8)
 * <p style="text-align:right">
 * }

Then we have scale factors,


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$$ \displaystyle

\begin{align} & {{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )} \\ & {{h}_{z}}=1 \\ \end{align}$$

(Eq 7.9)
 * <p style="text-align:right">
 * }

From Eq 7.4 we have


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$$ \displaystyle

\Delta \Psi =\frac{1}\left[ \frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial u}+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial \nu }+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial z}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}} \right]$$

(Eq 7.10)
 * <p style="text-align:right">
 * }

Substitute Eq 7.9 into Eq 7.10 we have,


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

\Delta \Psi =\frac{1}{{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}\left( \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}} \right)$$

(Eq 7.11)
 * <p style="text-align:right">
 * }

Compare Eq 7.11 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

Verification of Laplacian in Prolate spheroidal coordinates
The most common definition of prolate spheroidal coordinates $$\displaystyle (\mu, \nu, \phi)$$ is



x = a \ \sinh \mu \ \sin \nu \ \cos \phi $$



y = a \ \sinh \mu \ \sin \nu \ \sin \phi $$



z = a \ \cosh \mu \ \cos \nu $$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, \pi]$$. The azimuthal angle $$\displaystyle \phi$$ belongs to the interval $$\displaystyle [0, 2\pi)$$.

Then we have,


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$$ \displaystyle

\begin{align} & dx=-a\sin \phi \sinh \mu \cos \nu d\phi +a\cos \phi \cosh \mu \sin \nu d\mu +a\cos \phi \sinh \mu \cos \nu dv \\ & dy=a\cos \phi \sinh \mu \sin \nu d\phi +a\sin \phi \cosh \mu \sin \nu d\mu +a\sin \phi \sinh \mu \cos \nu dv \\ & dz=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ \end{align}$$

(Eq 7.12)
 * <p style="text-align:right">
 * }

Then the corresponding magnitude of $$\displaystyle ds$$ is


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$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}}+d{{z}^{2}} \\ & =({{a}^{2}}{{\cosh }^{2}}\mu {{\sin }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu )(d{{\mu }^{2}}+d{{v}^{2}})+{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu d{{\phi }^{2}} \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}}+{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu d{{\phi }^{2}} \\ \end{align}$$

(Eq 7.13)
 * <p style="text-align:right">
 * }

Then we have scale factors,


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$$ \displaystyle

\begin{align} & {{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )} \\ & {{h}_{\phi }}=a\sinh \mu \sin \nu \\ \end{align}$$

(Eq 7.14)
 * <p style="text-align:right">
 * }

Substitute Eq 7.14 into Eq 7.10 we have,


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

\Delta \Psi =\frac{1}{{{a}^{3}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )\sinh \mu \sin \nu }\left( \begin{align} & a\cosh \mu \sin \nu \frac{\partial \Psi }{\partial \mu }+a\sinh \mu \cos \nu \frac{\partial \Psi }{\partial \nu }+a\sinh \mu \sin \nu \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}} \\  & +a\sinh \mu \sin \nu \frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+\frac{a({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}{\sinh \mu \sin \nu }\frac{{{\partial }^{2}}\Psi }{\partial {{\phi }^{2}}} \\ \end{align} \right)$$

(Eq 7.15)
 * <p style="text-align:right">
 * }

Compare Eq 7.15 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

= Problem 6.8 =

From the lecture slide Mtg 36-2

Given

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$$ \displaystyle

{{P}_{0}}(x)=1$$

(Eq 8.1)
 * <p style="text-align:right">
 * }


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$$ \displaystyle

{{P}_{1}}(x)=x$$

(Eq 8.2)
 * <p style="text-align:right">
 * }


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$$ \displaystyle

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

(Eq 8.3)
 * <p style="text-align:right">
 * }


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$$ \displaystyle

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

(Eq 8.4)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

{{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$

(Eq 8.5)
 * <p style="text-align:right">
 * }


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$$ \displaystyle

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

(Eq 8.6)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle

{{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{\frac{1.3.....(2n-2i-1)}{{{2}^{i}}i!(n-2i)!}{{(-1)}^{i}}}{{x}^{n-2i}}$$

(Eq 8.7)
 * <p style="text-align:right">
 * }


 * $$\displaystyle [n/2]\equiv $$ integer part of n/2

Find
Verify that Eq8.1 through Eq8.5 can be written as Eq8.6 or Eq8.7

Solution
If we expand Eq8.1 through Eq8.5 by using Eq8.6 and Eq8.7 for each “n” according to 0,1,2,3,4, we obtain:

For n=0, and $$\displaystyle [n/2]=0\equiv 0$$
into Eq8.6,


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$$ \displaystyle

\begin{align} {{P}_{0}}(x) & =\sum\limits_{i=0}^{[0/2]}{{{(-1)}^{i}}\frac{(2\times 0-2i)!{{x}^{0-(2i)}}}{{{2}^{0}}i!(0-i)!(0-2i)!}}={{(-1)}^{0}}\frac{(0-2\times 0)!{{x}^{0-(2\times 0)}}}{1\times 0!(0)!(0-2\times 0)!}=1\times \frac{1\times 1}{1\times 1\times 1\times 1} \\ & =1 \\ \end{align}$$

(Eq 8.8)
 * <p style="text-align:right">
 * }

into Eq8.7,


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$$ \displaystyle

\begin{align} {{P}_{0}}(x) & =\sum\limits_{i=0}^{[0/2]}{\frac{1.3....(2\times 0-2i-1)}{{{2}^{i}}i!(0-2i)!}{{(-1)}^{i}}}{{x}^{0-(2i)}}=\frac{1....(0-2\times 0-1)}{{{2}^{0}}0!(0-2\times 0)!}{{(-1)}^{0}}{{x}^{0-(2\times 0)}}=\frac{1}{1\times 1\times 1}\times 1\times 1 \\ & =1 \\ \end{align}$$

(Eq 8.9)
 * <p style="text-align:right">
 * }

As seen above, Eq8.8, and Eq8.9 are equal to Eq8.1 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=0.

For n=1, and $$\displaystyle [n/2]=0.5\equiv 0$$
into Eq8.6,


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$$ \displaystyle

\begin{align} {{P}_{1}}(x) & =\sum\limits_{i=0}^{[1/2]}{{{(-1)}^{i}}\frac{(2\times 1-2i)!{{x}^{1-(2i)}}}{{{2}^{1}}i!(1-i)!(1-2i)!}}={{(-1)}^{0}}\frac{(2-2\times 0)!{{x}^{1-(2\times 0)}}}{2\times 0!(1-0)!(1-2\times 0)!}=1\times \frac{2x}{2\times 1\times 1\times 1} \\ & =x \\ \end{align}$$

(Eq 8.10)
 * <p style="text-align:right">
 * }

into Eq8.7,


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$$ \displaystyle

\begin{align} {{P}_{1}}(x) & =\sum\limits_{i=0}^{[1/2]}{\frac{1.3...(2\times 1-2i-1)}{{{2}^{i}}i!(1-2i)!}{{(-1)}^{i}}}{{x}^{1-(2i)}}=\frac{1}{{{2}^{0}}0!(1-2\times 0)!}{{(-1)}^{0}}{{x}^{1-(2\times 0)}}=\frac{1}{1\times 1\times 1}x \\ & =x \\ \end{align}$$

(Eq 8.11)
 * <p style="text-align:right">
 * }

As seen above, Eq8.10 and Eq8.11 are equal to Eq8.2 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=1.

For n=2, and $$\displaystyle [n/2]=1\equiv 1$$
into Eq8.6,


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$$ \displaystyle

\begin{align} {{P}_{2}}(x) & =\sum\limits_{i=0}^{[2/2]}{{{(-1)}^{i}}\frac{(2\times 2-2i)!{{x}^{2-(2i)}}}{{{2}^{2}}i!(2-i)!(2-2i)!}} \\ & =\left( {{(-1)}^{0}}\frac{(4-2\times 0)!{{x}^{2-(2\times 0)}}}{4\times 0!(2-0)!(2-2\times 0)!} \right)+\left( {{(-1)}^{1}}\frac{(4-2\times 1)!{{x}^{2-(2\times 1)}}}{4\times 1!(2-1)!(2-2\times 1)!} \right) \\ & =\left( 1\times \frac{4!{{x}^{2}}}{4\times 0!2!2!} \right)+\left( (-1)\frac{2!{{x}^{0}}}{4\times 1!1!0!} \right) \\ & =\left( \frac{24{{x}^{2}}}{4\times 1\times 2\times 2} \right)-\left( \frac{2\times 1}{4\times 1\times 1\times 1} \right) \\ & =\frac{24{{x}^{2}}}{16}-\frac{2}{4} \\ & =\frac{1}{2}(3{{x}^{2}}-1) \\ \end{align}$$

(Eq 8.12)
 * <p style="text-align:right">
 * }

into Eq8.7,


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$$ \displaystyle

\begin{align} {{P}_{2}}(x) & =\sum\limits_{i=0}^{[2/2]}{\frac{1.3.....(2\times 2-2i-1)}{{{2}^{i}}i!(2-2i)!}{{(-1)}^{i}}}{{x}^{2-(2i)}} \\ & =\frac{1....(4-2\times 0-1)}{{{2}^{0}}.0!(2-2\times 0)!}{{(-1)}^{0}}{{x}^{2-(2\times 0)}}+\frac{1....(4-2\times 1-1)}{{{2}^{1}}.1!(2-2\times 1)!}{{(-1)}^{1}}{{x}^{2-(2\times 1)}} \\ & =\frac{1\times 3}{1\times 1\times 2}{{x}^{2}}-\frac{1}{2\times 1\times 1} \\ & =\frac{1}{2}(3{{x}^{2}}-1) \\ \end{align}$$

(Eq 8.13)
 * <p style="text-align:right">
 * }

As seen above, Eq8.12 and Eq8.13 are equal to Eq8.3 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=2.

For n=3, and $$\displaystyle [n/2]=1.5\equiv 1$$
into Eq8.6,


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$$ \displaystyle

\begin{align} {{P}_{3}}(x) & =\sum\limits_{i=0}^{[3/2]}{{{(-1)}^{i}}\frac{(2\times 3-2i)!{{x}^{3-(2i)}}}{{{2}^{3}}i!(3-i)!(3-2i)!}} \\ & =\left( {{(-1)}^{0}}\frac{(6-2\times 0)!{{x}^{3-(2\times 0)}}}{8\times 0!(3-0)!(3-2\times 0)!} \right)+\left( {{(-1)}^{1}}\frac{(6-2\times 1)!{{x}^{3-(2\times 1)}}}{8\times 1!(3-1)!(3-2\times 1)!} \right) \\ & =\left( 1\times \frac{6!{{x}^{3}}}{8\times 0!3!3!} \right)+\left( (-1)\frac{4!{{x}^{1}}}{8\times 1!2!1!} \right)=\left( \frac{720{{x}^{3}}}{8\times 1\times 6\times 6} \right)-\left( \frac{24x}{8\times 1\times 2\times 1} \right) \\ & =\frac{720{{x}^{3}}}{288}-\frac{24x}{16} \\ & =\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

(Eq 8.14)
 * <p style="text-align:right">
 * }

into Eq8.7,


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

$$ \displaystyle

\begin{align} {{P}_{3}}(x) & =\sum\limits_{i=0}^{[3/2]}{\frac{1.3.....(2\times 3-2i-1)}{{{2}^{i}}i!(3-2i)!}{{(-1)}^{i}}}{{x}^{3-(2i)}} \\ & =\frac{1.3...(6-2\times 0-1)}{{{2}^{0}}.0!(3-2\times 0)!}{{(-1)}^{0}}{{x}^{3-(2\times 0)}}+\frac{1....(6-2\times 1-1)}{{{2}^{1}}.1!(3-2\times 1)!}{{(-1)}^{1}}{{x}^{3-(2\times 1)}} \\ & =\frac{1\times 3\times 5}{1\times 1\times 6}{{x}^{3}}-\frac{1\times 3}{2\times 1\times 1}x \\ & =\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

(Eq 8.15)
 * <p style="text-align:right">
 * }

As seen above, Eq8.14 and Eq8.15 are equal to Eq8.4 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=3.

For n=4, and $$\displaystyle [n/2]=2\equiv 2$$
into Eq8.6,


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$$ \displaystyle

\begin{align} {{P}_{4}}(x) & =\sum\limits_{i=0}^{[4/2]}{{{(-1)}^{i}}\frac{(2\times 4-2i)!{{x}^{4-(2i)}}}{{{2}^{4}}i!(4-i)!(4-2i)!}} \\ & ={{(-1)}^{0}}\frac{(8-2\times 0)!{{x}^{4-(2\times 0)}}}{16\times 0!(4-0)!(4-2\times 0)!}+{{(-1)}^{1}}\frac{(8-2\times 1)!{{x}^{4-(2\times 1)}}}{16\times 1!(4-1)!(4-2\times 1)!}+{{(-1)}^{2}}\frac{(8-2\times 2)!{{x}^{4-(2\times 2)}}}{16\times 2!(4-2)!(4-2\times 2)!} \\ & =\left( 1\times \frac{8!{{x}^{4}}}{16\times 0!\times 4!\times 4!} \right)-\left( 1\times \frac{6!{{x}^{2}}}{16\times 1!\times 3!\times 2!} \right)+\left( 1\times \frac{4!{{x}^{0}}}{16\times 2!\times 2!\times 0!} \right) \\ & =\left( \frac{40320{{x}^{4}}}{16\times 1\times 24\times 24} \right)-\left( \frac{720{{x}^{2}}}{16\times 1\times 6\times 2} \right)+\left( \frac{24}{16\times 2\times 2\times 1} \right) \\ & =\frac{40320{{x}^{4}}}{9216}-\frac{720{{x}^{2}}}{192}+\frac{24}{64} \\ & =\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ \end{align}$$

(Eq 8.16)
 * <p style="text-align:right">
 * }

into Eq8.7,


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$$ \displaystyle

\begin{align} {{P}_{4}}(x) & =\sum\limits_{i=0}^{[4/2]}{\frac{1.3.....(2\times 4-2i-1)}{{{2}^{i}}i!(4-2i)!}{{(-1)}^{i}}}{{x}^{4-(2i)}} \\ & =\frac{1.3....(8-2\times 0-1)}{{{2}^{0}}.0!(4-2\times 0)!}{{(-1)}^{0}}{{x}^{4-(2\times 0)}}+\frac{1.3....(8-2\times 1-1)}{{{2}^{1}}.1!(4-2\times 1)!}{{(-1)}^{1}}{{x}^{4-(2\times 1)}}+\frac{1....(8-2\times 2-1)}{{{2}^{2}}.2!(4-2\times 2)!}{{(-1)}^{2}}{{x}^{4-(2\times 2)}} \\ & =\frac{1\times 3\times 5\times 7}{1\times 1\times 24}{{x}^{4}}-\frac{1\times 3\times 5}{2\times 1\times 2}{{x}^{2}}+\frac{1\times 3}{4\times 4\times 1}{{x}^{0}} \\ & =\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ \end{align}$$

(Eq 8.17)
 * <p style="text-align:right">
 * }

As seen above, Eq8.16 and Eq8.17 are equal to Eq8.5 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=4.

As a result, Eq8.8 through Eq8.17 for n=0,1,2,3,4 are equal to Eq8.1 through Eq8.5. Therefore, it is verified that Eq8.1 through Eq8.5 can be written as Eq8.6 or Eq8.7.

= References =