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User:Egm6321.f10.team2.oztekin/HW6

=Problem 6.1 Using variation of parameters method to get $$\displaystyle {{Q}_{2}}(x)$$ from $$\displaystyle {{P}_{2}}(x)$$ =

From the lecture slide Mtg 32- 3

Given

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$$  \displaystyle P_{2}(x)=\frac{1}{2}(3x^{2}-1) $$     (Eq 1.1) Legendre's equation where n=2
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$$  \displaystyle (1-x^{2})y''-2xy'+6y=0 $$     (Eq 1.2)
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Find

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$$  \displaystyle Q_{2}(x)=\frac{1}{4}(3x^{2}-1)log(\frac{1+x}{1-x})-\frac{3}{2}x $$     (Eq 1.3) using the variation of parameter method.
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Solution
To begin to solve Q_{2}(x) we must first solve the general form and then plug in.

We start with the equation:
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$$  \displaystyle y(x)=U(x)u_{1}(x) $$     (Eq 1.4) and take derivatives to get in the Legendre Equation form
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$$  \displaystyle y=U*u_{1} $$     (Eq 1.5)
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$$  \displaystyle y'=U'*u_{1}+U*u'_{1} $$     (Eq 1.6)
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$$  \displaystyle y=U*u_{1}+U*u''_{1}+2U'u'_{1} $$     (Eq 1.7) Plugging Eqns. 1.5-1.7 into the general L2-ODE-VC we get:
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$$  \displaystyle U*u_{1}+U*u_{1}+2U'u'_{1}+a_{1}(x)*(U'*u_{1}+U*u'_{1})+a_{0}(x)*(U*u_{1})=0 $$     (Eq 1.8) Combining terms:
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$$  \displaystyle U(u_{1}+a_{1}(x)u'_{1}+a_{0}(x)u_{1})+U'(2u'_{1}(x)a_{1}(x)+a_{1}(x)u_{1}(x))+Uu_{1}(x)=0 $$     (Eq 1.9) Noting that $$ \displaystyle u''_{1}+a_{1}(x)u'_{1}+a_{0}(x)u_{1} $$ is a homogeneous solution to the Legendre solution and therefore =0 we get:
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$$  \displaystyle U'(2u'_{1}(x)+a_{1}(x)u_{1}(x))+U''u_{1}(x)=0 $$     (Eq 1.10) To help solve, we set $$ \displaystyle Z=U' $$ so we can use the order reducing method.
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$$  \displaystyle Z(2u'_{1}(x)+a_{1}(x)u_{1}(x))+Z'u_{1}(x)=0 $$     (Eq 1.11) We divide by $$ \displaystyle u_{1}(x) $$ to get:
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$$  \displaystyle Z\frac{2u'_{1}(x)+a_{1}(x)u_{1}(x)}{u_{1}(x)}+Z'=0 $$     (Eq 1.12) We can now use the Integrating Factor Method
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$$  \displaystyle h(x)=exp [\int (\frac{2u'_{1}(x)+a_{1}(x)u_{1}(x)}{u_{1}(x)})dx] $$     (Eq 1.13) Multiplying through by u_{1}(x):
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$$  \displaystyle h(x)=exp [\int (a_1(x)+2\frac{u'_1(x)}{u_1(x)})dx] $$     (Eq 1.14) Separating:
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$$  \displaystyle h(x)=exp [\int a_1(x)dx+\int 2\frac{u'_1(x)}{u_1(x)}dx] $$     (Eq 1.15) We then can solve to get:
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$$  \displaystyle h(x)=u_{1}^{2}(x)exp[\int a_1(x)dx] $$     (Eq 1.16) Now that we have the integrating factor, we need to set up Legendre's Equation (Eq 1.2) so that we can use the integrating factor. This is done below by dividing through by $$ \displaystyle (1-x^2) $$
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$$  \displaystyle y''-\frac{2x}{1-x^2}y'+\frac{6}{1-x^2}y=0 $$     (Eq 1.17) From this we can see that:
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$$  \displaystyle a_1(x)=\frac{-2x}{1-x^2} $$     (Eq 1.18) Combining this with the fact that $$ \displaystyle P_2(x)=\frac{1}{2}(3x^2-1) $$ and $$ \displaystyle Q_2(x)=u_1(x)\int \frac{1}{h(x)}dx $$ from above we can plug in out knowns and solve:
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$$  \displaystyle Q_2(x)=u_1(x)\int \frac{1}{h(x)}dx $$     (Eq 1.19)
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$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{1}{u_{1}^{2}(x)exp[\int a_1(x)dx]}dx $$     (Eq 1.20)
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$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{1}{(\frac{1}{2}(3x^2-1))^{2}exp[\int \frac{-2x}{1-x^2}dx]}dx $$     (Eq 1.21)
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$$  \displaystyle exp\int \frac{-2x}{1-x^{2}}dx=exp[log(1-x^{2})]=1-x^{2} $$     (Eq 1.22) If we sustitute Eq(1.22) into Eq(1.21) we have
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$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{4}{(3x^2-1)^{2}(1-x^{2})}dx $$     (Eq 1.23) Let's look at the fraction:
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$$  \displaystyle \frac{1}{(3x^{2}-1)^{2}(1-x^{2})}=\frac{Ax+B}{(3x^{2}-1)}+\frac{Cx+D}{(3x^{2}-1)^{2}}+\frac{E}{(1+x)}+\frac{F}{(1-x)} $$     (Eq 1.24) Having compansated denominators we have;
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$$  \displaystyle (Ax+B)(3x^{2}-1)(1-x^{2})+(Cx+D)(1-x^{2})+E(1-x)(3x^{2}-1)^{2}+F(3x^{2}-1)^{2}(1+x)=1 $$     (Eq 1.25) After unpacking the equation we will get five equations for x's orders. Then we can see the coeefficients of the terms as:
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$$  \displaystyle A=0,B=\frac{3}{4},C=0,D=\frac{3}{2},E=\frac{1}{8},F=\frac{1}{8} $$     (Eq 1.26)
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$$  \displaystyle \frac{1}{(3x^{2}-1)^{2}(1-x^{2})}=\frac{3}{4}\frac{1}{(3x^2-1)}+\frac{3}{2}\frac{1}{(3x^{2}-1)}+\frac{1}{8}\frac{1}{(x+1)}-\frac{1}{8}\frac{1}{(x-1)} $$     (Eq 1.27)
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$$  \displaystyle \int \frac{1}{(3x^{2}-1)(1-x^{2})}=\frac{1}{8}\left [ \frac{-6x}{(3x^{2}-1)} -log(1-x)+log(1+x)\right ] $$     (Eq 1.28) If we sustitute Eq(1.28) into Eq(1.23) we get the second Q
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$$  \displaystyle Q_{2}(x)=\frac{1}{4}(3x^{2}-1)log(\frac{1+x}{1-x})-\frac{3}{2}x $$     (Eq 1.29)
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= Problem 6.2 Solving Non-homogeneous L2-ODE-VC with different excitation functions =

From the lecture slide From Mtg 32-3 and Mtg 33-1

Given

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$$ \displaystyle

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

(Eq 2.1)
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$$ \displaystyle

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

(Eq 2.2)
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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)
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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)
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To simplify the equation we will go through the following procedure:

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

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

(Eq 2.11)
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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:
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& 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)
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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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$$ \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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$$ \displaystyle
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y={{C}_{1}}\frac{x}+{{C}_{2}}\frac{1}{x{{e}^{ix}}}+\frac{x}\int{\frac{1}}\left[ \int{\frac{{{e}^{2ix}}\sin x}{x}dx} \right]dx$$

(Eq 2.19)
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= Problem 6.3 Solving Non-homogenous Legendre Equation using direct menthod =

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

(Eq 3.6)
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=Problem 6.4: Laplacian in Spherical Coordinate System=

From the lecture slide Mtg 34-1

Given
First we have to indentify our cartesian coorinates into spherical coordinates. Then we can aidentify line segment in spherical coordinate system.Our new coordinates ;
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$$  \displaystyle x_{1}=rcos\theta cos\varphi =\xi _{1}cos\xi _{2}cos\xi _{3} $$     (Eq 4.1)
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$$  \displaystyle x_{2}=rcos\theta sin\varphi $$     (Eq 4.2)
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$$  \displaystyle x_{3}=rsin\theta $$     (Eq 4.3)
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$$  \displaystyle ds^{2}=\vec{ds}.\vec{ds}=(dx_{i}dx_{j})(e_{i}e_{j}) $$     (Eq 4.4)
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$$  \displaystyle \delta _{ij}=\left\{\begin{matrix} 1 &for &i=j \\ 0 &for & i\neq j \end{matrix}\right. $$     (Eq 4.5) In astronomical convention angles are defined as
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$$  \displaystyle\begin{align} & -\frac{\pi }{2}\leq \theta \leq \frac{\pi }{2}\\ & 0\leq \varphi \leq 2\pi \\ \end{align} $$     (Eq 4.6)
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Find
Obtain the steps given in meeting 34

Solution

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$$  \displaystyle ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} $$     (Eq 4.7) Finding total derivative of first term
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$$  \displaystyle dx_{1}=\frac{\partial x_{1}}{\partial r}dr+\frac{\partial x_{1}}{\partial \theta }d\theta +\frac{\partial x_{1}}{\partial \varphi }d\varphi $$     (Eq 4.8)
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$$  \displaystyle \frac{\partial x_{1}}{\partial r}=cos\theta cos\varphi $$     (Eq 4.9)
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$$  \displaystyle \frac{\partial x_{1}}{\partial \theta }=-rsin\theta cos\theta $$     (Eq 4.10)
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$$  \displaystyle \frac{\partial x_{1}}{\partial \varphi }=-rcos\theta sin\varphi $$     (Eq 4.11)
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$$  \displaystyle dx_{1} =cos\theta cos\varphi dr-rsin\theta cos\varphi d\theta -rcos\theta sin\varphi d\varphi $$     (Eq 4.12)
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$$  \displaystyle dx_{1}^{2}=cos^{2}\theta cos^{2}\varphi dr^{2}+r^{2}sin^{2}\theta cos^{2}\varphi d\theta ^{2}+r^{2}cos^{2}\theta sin^{2}\varphi d\varphi ^{2}+2r^{2}sin\varphi cos\varphi sin\theta cos\theta d\theta d\varphi -2rcos\theta cos^{2}\varphi sin\theta drd\theta -2rcos^{2}\theta cos\varphi sin\varphi drd\varphi $$     (Eq 4.13) Finding total dervative of second term
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$$  \displaystyle dx_{2}=\frac{\partial x_{2}}{\partial r}dr+\frac{\partial x_{2}}{\partial \theta }d\theta +\frac{\partial x_{2}}{\partial \varphi }d\varphi $$     (Eq 4.14)
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$$  \displaystyle \frac{\partial x_{2}}{\partial r}=cos\theta sin\varphi $$     (Eq4.15)
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$$  \displaystyle \frac{\partial x_{2}}{\partial \theta }=-rsin\theta sin\varphi $$     (Eq 4.16)
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$$  \displaystyle \frac{\partial x_{2}}{\partial \varphi }=rcos\theta cos\varphi $$     (Eq 4.17)
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$$  \displaystyle dx_{2}=cos\theta sin\varphi dr-rsin\theta sin\varphi d\theta +rcos\theta cos\varphi d\varphi $$     (Eq 4.18)
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$$  \displaystyle dx_{2}^{2}=cos^{2}\theta sin^{2}\varphi dr^{2}+r^{2}sin^{2}\theta sin^{2}\varphi d\theta^{2} +r^{2}cos^{2}\theta cos^{2}\varphi d\varphi ^{2}+2rcos^{2}\theta sin\varphi cos\varphi drd\varphi -2r^{2}sin\theta sin\varphi cos\theta cos\varphi d\theta d\varphi -2rcos\theta sin\theta sin^{2}\varphi drd\theta $$     (Eq 4.19) Finding total derivative of third term
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$$  \displaystyle dx_{3}=\frac{\partial x_{3}}{\partial r}dr+\frac{\partial x_{3}}{\partial \theta }d\theta +\frac{\partial x_{3}}{\partial \varphi }d\varphi $$     (Eq 4.20)
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$$  \displaystyle \frac{\partial x_{3}}{\partial r}=sin\theta $$     (Eq 4.21)
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$$  \displaystyle \frac{\partial x_{3}}{\partial \theta }=rcos\theta $$     (Eq 4.22)
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$$  \displaystyle \frac{\partial x_{3}}{\partial \varphi \theta }=0 $$     (Eq 4.23)
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$$  \displaystyle dx_{3}=sin\theta dr+rcos\theta d\theta $$     (Eq 4.24)
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$$  \displaystyle dx_{3}^{2}=sin^{2}\theta dr^{2}+r^{2}cos^{2}\theta d\theta ^{2}+2rsin\theta cos\theta drd\theta $$     (Eq 4.25)
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Collecting all of the differentials Eq 4.13-Eq4.19-Eq4.25 and substitute into ds yields ;

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$$  \displaystyle ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} $$     (Eq 4.26)
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$$  \displaystyle \begin{align} &ds^{2}=cos^{2}\theta cos^{2}\varphi dr^{2}+r^{2}sin^{2}\theta cos^{2}\varphi d\theta ^{2}+r^{2}cos^{2}\theta sin^{2}\varphi d\varphi ^{2}+cos^{2}\theta sin^{2}\varphi dr^{2}+r^{2}sin^{2}\theta sin^{2}\varphi d\theta^{2} +r^{2}cos^{2}\theta cos^{2}\varphi d\varphi ^{2}\\ &+sin^{2}\theta dr^{2}+r^{2}cos^{2}\theta d\theta ^{2}+2r^{2}sin\varphi cos\varphi sin\theta cos\theta d\theta d\varphi -2rcos\theta cos^{2}\varphi sin\theta drd\theta -2rcos^{2}\theta cos\varphi sin\varphi drd\varphi\\ &+2rcos^{2}\theta sin\varphi cos\varphi drd\varphi -2r^{2}sin\theta sin\varphi cos\theta cos\varphi d\theta d\varphi -2rcos\theta sin\theta sin^{2}\varphi drd\theta +2rsin\theta cos\theta drd\theta\\ \end{align} $$     (Eq 4.27)
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Now we can collect$$ \displaystyle dr,d\theta,d\varphi$$ together. And we can see that some terms cancel each other and for other terms we hate to manupilate the equation little more.

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$$  \displaystyle \begin{align} &ds^{2}=cos^{2}\theta (cos^{2}\varphi +sin^{2}\varphi )dr^{2}+sin^{2}\theta dr^{2}+r^{2}sin^{2}\theta (cos^{2}\varphi +sin^{2}\varphi )d\theta ^{2}+r^{2}cos^{2}\theta d\theta ^{2}+r^{2}cos^{2}\theta (sin^{2}\varphi +cos^{2}\varphi )d\varphi ^{2}\\ &+2rsin\theta cos\theta drd\theta -2rcos\theta sin\theta (cos^{2}\varphi +sin^{2}\varphi )drd\theta \\ \end{align} $$     (Eq 4.28) We can obtain the equation given in the lecture 34-1
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$$  \displaystyle ds^{2}=dr^{2}+r^{2}d\theta ^{2}+r^{2}cos^{2}\theta d\phi ^{2} $$     (Eq 4.29)
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$$  \displaystyle \Delta \psi =\frac{1}{h_{1}h_{2}h_{3}}\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\partial \psi }{\partial \xi _{i}}\right ] $$     (Eq 4.30)
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$$  \displaystyle h_{1}.h_{2}.h_{3}=r^{2}cos\theta $$     (Eq 4.31)
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$$  \displaystyle i=1,\xi _{1}=r\Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial r}\left [ \frac{r^{2}cos\theta }{1}\frac{\partial \psi }{\partial r} \right ] $$     (Eq 4.32)
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Since $$\displaystyle cos\theta $$is constant for r we can take it out from the derivative.
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$$  \displaystyle i=1,\xi _{1}=r\Rightarrow \frac{1}{r^{2} }\frac{\partial }{\partial r}\left [ \frac{r^{2} }{1}\frac{\partial \psi }{\partial r} \right ] $$     (Eq 4.33)
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$$  \displaystyle i=2,\xi _{2}=\theta \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }\left [ \frac{r^{2}cos\theta }{r^{2}}\frac{\partial \psi }{\partial \theta } \right ] $$     (Eq 4.34)
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$$  \displaystyle i=2,\xi _{2}=\theta \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }\left [ \frac{cos\theta }{1}\frac{\partial \psi }{\partial \theta } \right ] $$     (Eq 4.35)
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$$  \displaystyle i=3,\xi _{3}=\varphi \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \varphi }\left [ \frac{r^{2}cos\theta }{r^{2}cos^{2}\theta }\frac{\partial \varphi }{\partial\varphi } \right ] $$     (Eq 4.36)
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Since $$\displaystyle cos\theta $$ is constant for \varphi we can take it out from the derivative.

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$$  \displaystyle i=3,\xi _{3}=\varphi \Rightarrow \frac{1}{r^{2}cos^{2}\theta }\frac{\partial }{\partial \varphi }\left [ \frac{\partial^{2}\psi }{\partial\varphi ^{2} } \right ] $$     (Eq 4.37)
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Then we can collect all terms into the $$\displaystyle \Delta \psi $$ equation:

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$$  \displaystyle \Delta \psi =\frac{1}{r^{2}}(r^{2}\frac{\partial \psi }{\partial r})+\frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }(cos\theta \frac{\partial \psi }{\partial \theta })+\frac{1}{r^{2}cos^{2}\theta }\frac{\partial ^2\psi }{\partial \varphi ^2} $$     (Eq 4.38)
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=Problem 6.5: Laplacian in Cylindrical Coordinate System=

From the lecture slide Mtg 35-3 and Mtg 36-3

Given
Three dimensional cylindrical coordinates were given in the meeting 35-3 as follows:
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$$  \displaystyle x_{1}=rcos\theta =\xi _{1}cos\xi _{2} $$     (Eq 5.1)
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$$  \displaystyle x_{2}=rsin\theta =\xi _{1}sin\xi _{2} $$     (Eq 5.2)
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$$  \displaystyle x_{3}=z=\xi _{3} $$     (Eq 5.3) Now,after defining our new curvilinear coordinates,we can define our cartesian coordinates in terms of our new independent variables.
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$$  \displaystyle\begin{align} & x_{1}=x_{1}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ &x_{2}=x_{2}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ &x_{3}=x_{3}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ \end{align}$$ (Eq 5.4)
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Find
Repeat steps for cylindrical coordinates and conclude with Bessel's Eq.

Finding dxi's

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$$  \displaystyle dx_{1}=\frac{\partial x_{1}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{1}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{1}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.5)
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$$  \displaystyle\begin{align} &\frac{\partial x_{1}}{\partial \xi _{1}}=cos\xi _{2}\\ &\frac{\partial x_{1}}{\partial \xi _{2}}=-\xi _{1}sin\xi _{2}\\ &\frac{\partial x_{1}}{\partial \xi _{3}}=0\\ \end{align}$$ (Eq 5.6)
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$$  \displaystyle dx_{1}=cos\xi _{2}d\xi _{1}-\xi _{1}sin\xi _{2}d\xi _{2} $$     (Eq 5.7)
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$$  \displaystyle dx_{2}=\frac{\partial x_{2}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{2}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{2}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.8)
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$$  \displaystyle\begin{align} &\frac{\partial x_{2}}{\partial \xi _{1}}=sin\xi _{2}\\ &\frac{\partial x_{2}}{\partial \xi _{2}}=\xi _{1}cos\xi _{2}\\ &\frac{\partial x_{2}}{\partial \xi _{3}}=0\\ \end{align}$$ (Eq 5.9)
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$$  \displaystyle dx_{2}=sin\xi _{2}d\xi _{1}+\xi _{1}cos\xi _{2}d\xi _{2} $$     (Eq 5.10)
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$$  \displaystyle dx_{3}=\frac{\partial x_{3}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{3}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{3}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.11)
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$$  \displaystyle\begin{align} &\frac{\partial x_{3}}{\partial \xi _{1}}=0\\ &\frac{\partial x_{3}}{\partial \xi _{2}}=0\\ &\frac{\partial x_{3}}{\partial \xi _{3}}=1\\ \end{align}$$ (Eq 5.12)
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Finding ds^2

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$$  \displaystyle dx_{1}^{2}=cos^{2}\xi _{2}d\xi _{1}^{2}+\xi _{1}^{2}sin^{2}\xi _{2}d\xi _{2}^{2}-2\xi _{1}cos\xi _{2}sin\xi _{2}d\xi _{1}d\xi _{2} $$     (Eq 5.13)
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$$  \displaystyle dx_{2}^{2}=sin^{2}\xi _{2}d\xi _{1}^{2}+\xi _{1}^{2}cos^{2}\xi _{2}d\xi _{2}^{2}+2\xi _{1}cos\xi _{2}sin\xi _{2}d\xi _{1}d\xi _{2} $$     (Eq 5.14)
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$$  \displaystyle dx_{3}^{2}=d\xi _{3}^{2} $$     (Eq 5.15)
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$$  \displaystyle ds^{2}=d\xi _{1}^{2}+\xi _{1}^{2}(sin^{2}\xi _{2}+cos^{2}\xi _{2})d\xi _{2}^{2}+d\xi _{3}^{2}=d\xi _{1}^{2}+\xi _{1}^{2}d\xi _{2}^{2}+d\xi _{3}^{2} $$     (Eq 5.16)
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$$  \displaystyle\begin{align} & h_{1}^{2}=1^{2}\Rightarrow h_{1}=1\\ & h_{2}^{2}=\xi _{1}^{2}\Rightarrow h_{2}=\xi _{1}\\ & h_{3}^{2}=1^{2}\Rightarrow h_{3}=1\\ \end{align}$$ (Eq 5.17)
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$$  \displaystyle h_{1}h_{2}h_{3}=\xi _{1} $$     (Eq 5.18)
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$$  \displaystyle \Delta \psi =\frac{1}{h_{1}h_{2}h_{3}}\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\partial \psi }{\partial \xi _{i}}\right ] $$     (Eq 5.19)
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$$  \displaystyle i=1\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{1}} \right ) $$     (Eq 5.20)
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$$  \displaystyle i=2\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{2}}\left ( \frac{\xi _{1}}{\xi _{1}^{2}}\frac{\partial \psi }{\partial \xi _{2}} \right )\Rightarrow \frac{1}{\xi _{1}^{2}}\frac{\partial ^2\psi }{\partial \xi _{2}} $$     (Eq 5.21)
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$$  \displaystyle i=3\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{3}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{3}} \right )\Rightarrow \frac{\partial^2 \psi }{\partial \xi _{3}^2} $$     (Eq 5.22)
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$$  \displaystyle \Delta \psi =\frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{1}} \right )+\frac{1}{\xi _{1}^{2}}\frac{\partial^2 \psi }{\partial \xi _{2}^2}+\frac{\partial^2 \psi }{\partial \xi _{3}^2}=0 $$     (Eq 5.23)
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Obtaining Bessel's Eq
To obtain bessel equation we have to use separation of variables.We can identify $$\displaystyle \psi $$ as multiplacation of three functions depending their own variables.
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$$  \displaystyle \psi (\xi _{1},\xi _{2},\xi _{3})=X(\xi _{1})Y(\xi _{2})Z(\xi _{3}) $$     (Eq 5.24)
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$$  \displaystyle \frac{Y(\xi _{2})Z(\xi _{3})}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X(\xi _{1})}{\partial \xi _{1}} \right )+\frac{X(\xi _{1})Z(\xi _{3})}{\xi _{1}^{2}}\frac{\partial^2Y(\xi _{2})}{\partial^2 \xi _{2}}+X(\xi _{1})Y(\xi _{2})\frac{\partial^2 Z(\xi _{3})}{\partial \xi _{3}^2}=0 $$     (Eq 5.25) Now, dividing through $$ \displaystyle XYZ$$ we can obtain more useful form.
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$$  \displaystyle \frac{1}{X\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+\frac{1}{Y\xi _{1}^{2}}\frac{\partial^2 Y}{\partial \xi _{2}^2}+\frac{1}{Z}\frac{\partial^2 Z}{\partial \xi _{3}^2}=0 $$     (Eq 5.26) Assuming that
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$$  \displaystyle \frac{\partial^2 Z}{\partial \xi _{3}^2}=\alpha ^{2}Z $$     (Eq 5.27) After that our equation forms as
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$$  \displaystyle \frac{1}{X\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+\frac{1}{Y\xi _{1}^{2}}\frac{\partial^2 Y}{\partial \xi _{2}^2}+\alpha ^{2}=0 $$     (Eq 5.28) We can see that after our asssumption $$\displaystyle\triangledown ^{2}\psi $$ is no longer three dimensional instead two dimensional.The Eq(5.28) is more closely to Helmhotz's equation.Multiply whole eq by $$\displaystyle \xi _{1}^{2}$$.
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In addition let's assume
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$$  \displaystyle \frac{\partial^2 Y}{\partial \xi _{2}^2}=-n^{2}Y $$     (Eq 5.29) Which has solutions $$\displaystyle Y=exp(+,-in\xi _{2})$$ where n is an integer. It is now closely related to Bessel's equation.
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$$  \displaystyle \xi _{1}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.30)
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$$  \displaystyle \xi _{1}^{2}\frac{\partial^2 X}{\partial \xi_{1} ^2}+\xi _{1}\frac{\partial X}{\partial \xi _{1}}+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.31) Say that
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$$  \displaystyle x=\alpha \xi _{1} $$     (Eq 5.32)
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$$  \displaystyle \frac{\partial X}{\partial \xi _{1}}=\frac{\partial X}{\partial x}\frac{\partial x}{\partial \xi _{1}}=\alpha \frac{\partial X}{\partial x} $$ (Eq 5.33)
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$$  \displaystyle \frac{\partial }{\partial \xi _{1}}\left ( \alpha \frac{\partial X}{\partial x} \right )=\alpha ^{2}\frac{\partial X}{\partial x} $$ (Eq 5.34) Substituting Eq(5.33) and Eq (5.34) in to Eq (Eq 5.32) We observe
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$$  \displaystyle \xi _{1}^{2}\alpha ^{2}\frac{\partial^2 X}{\partial x^2}+\xi _{1}\alpha \frac{\partial X}{\partial x}+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.35)
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$$  \displaystyle x^{2}\frac{\partial^2 X}{\partial x^2}+x \frac{\partial X}{\partial x}+(x^{2}-n^{2})X=0 $$     (Eq 5.36)
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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)
 *

 * }

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

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

$$ \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)
 *

 * }

Hence,

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

$$ \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)
 *

 * }

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

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

$$ \displaystyle

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

(Eq 6.6)
 *

 * }

Then we have scale factors,

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

$$ \displaystyle

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

(Eqs 6.7)
 *

 * }

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

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

$$ \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)
 *

 * }

Substitute Eqs 6.7 into Eq 6.8 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" |

\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)
 *

 * }

= 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:

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

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

(Eqs 7.1)
 *

 * }

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.

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

$$ \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)
 *

 * }

We can derive scale factors from Eq 7.2,

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

$$ \displaystyle

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

(Eq 7.3)
 *

 * }

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

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

$$ \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)
 *

 * }

and in two-dimensional case:

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

$$ \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)
 *

 * }

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,

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

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

 * }

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 $$\displaystyle \nu \in [0, 2\pi)$$.

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

$$ \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)
 *

 * }

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

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

$$ \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)
 *

 * }

Then we have scale factors,

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

$$ \displaystyle

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

(Eq 7.9)
 *

 * }

From Eq 7.4 we have

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

$$ \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)
 *

 * }

Substitute Eq 7.9 into Eq 7.10 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" |

\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)
 *

 * }

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,

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

$$ \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)
 *

 * }

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

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

$$ \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)
 *

 * }

Then we have scale factors,

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

$$ \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)
 *

 * }

Substitute Eq 7.14 into Eq 7.10 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" |

\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)
 *

 * }

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 Verifying different expressions of Legendre Polynominals =

From the lecture slide Mtg 36-2

Given

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

$$ \displaystyle

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

(Eq 8.1)
 *

 * }

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

$$ \displaystyle

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

(Eq 8.2)
 *

 * }

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

$$ \displaystyle

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

(Eq 8.3)
 *

 * }

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

$$ \displaystyle

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

(Eq 8.4)
 *

 * }

 * {| 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)
 *

 * }

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

$$ \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)
 *

 * }

 * {| 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)
 *

 * }

 * $$\displaystyle [n/2]\equiv $$ 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,

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

$$ \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)
 *

 * }

into Eq8.7,

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

$$ \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)
 *

 * }

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,

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

$$ \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)
 *

 * }

into Eq8.7,

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

$$ \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)
 *

 * }

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,

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

$$ \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)
 *

 * }

into Eq8.7,

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

$$ \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)
 *

 * }

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,

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

$$ \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)
 *

 * }

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)
 *

 * }

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,

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

$$ \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)
 *

 * }

into Eq8.7,

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

$$ \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)
 *

 * }

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.

= Problem 6.9: Verification that the Legendre Polynomials are Solutions to the Legendre Equation =

From the lecture slide Mtg 36-3

Given
The following equations are Legendre Polynomials, Pn(x), where n=0,1,2,3,4:

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

$$ \displaystyle {{P}_{0}}\left( x \right)=1 $$ (Eq 9.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle {{P}_{1}}\left( x \right)=x $$ (Eq 9.2)
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 * style="width:95%" |
 *

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

$$ \displaystyle {{P}_{2}}\left( x \right)=\frac{1}{2}\left(3x^{2}-1\right) $$ (Eq 9.3)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle {{P}_{3}}\left( x \right)=\frac{1}{2}\left(5x^{3}-3x\right) $$ (Eq 9.4)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle {{P}_{4}}\left( x \right)=\frac{1}{8}\left(35x^{4}-30x^{2}+3\right) $$    (Eq 9.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Verify that Equations 9.1, 9.2, 9.3, 9.4, and 9.5 are all solutions of the following Legendre Equation:
 * {| style="width:100%" border="0"

$$ \displaystyle \left(1-x^2 \right)y'' -2x y' + n\left(n+1\right)y=0 $$ (Eq 9.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
For each equation the following calculations are made:
 * {| style="width:100%" border="0"

$$ \displaystyle y={{P}_{n}}\left( x \right) $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle y'=\frac{d}{dx}[{{P}_{n}}\left( x \right)] $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle y''=\frac{d}{dx}[\frac{d}{dx}[{{P}_{n}}\left( x \right)]] $$ Beginning with P0(x) {where n=0}, y=1 and y' and y" are both 0. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \cancel{\left(1-x^2 \right)[0]}^0 \cancel{-2x [0]}^0 + \cancel{(0)\left((0)+1\right)[1]}^0=0 $$ (Eq 9.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For P1(x) {where n=1}, y=x, y'=1 and y"=0. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * {| style="width:100%" border="0"

$$ \displaystyle \cancel{\left(1-x^2 \right)[0]}^0 -2x [1]+(1)\left((1)+1\right)[x]=0 $$ (Eq 9.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle \cancel{(-2+2)x}^0=0 $$ (Eq 9.9) For P2(x) {where n=2}, y=1/2(3x2-1), y'=3x and y"=3. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \left(1-x^2 \right)[3] -2x [3x]+(2)\left((2)+1\right)[\frac{1}{2}\left(3x^2-1\right)]=0 $$ (Eq 9.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle \left(3-3x^2\right)-(6x^2)+(9x^2-3)=0 $$ (Eq 9.11)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \cancel{(-3-6+9)x^2}^0+\cancel{(3-3)}^0=0 $$ (Eq 9.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For P3(x) {where n=3}, y=1/2(5x3-3x), y'=1/2(15x2-3) and y"=15x. Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * {| style="width:100%" border="0"

$$ \displaystyle \left(1-x^2 \right)[15x] -2x \frac{1}{2}\left(15x^2-3\right)+(3)\left((3)+1\right)[\frac{1}{2}\left(5x^3-3x\right)]=0 $$ (Eq 9.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle \left(15x-15x^3\right)-(15x^3-3x)+(30x^3-18x)=0 $$ (Eq 9.14)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \cancel{(-15-15+30)x^3}^0+\cancel{(15+3-18)x}^0=0 $$ (Eq 9.15) For P4(x) {where n=4}, y=1/8(35x4-30x2+3), y'=1/8(140x3-60x) and y"=1/8(420x2-60). Plugging these values into Equation 9.6 verifies it as a solution to the Legendre Equation as shown:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \left(1-x^2 \right)[\frac{1}{8}\left(420x^2-60\right)] -2x \frac{1}{8}\left(140x^3-60x\right)+(4)\left((4)+1\right)[\frac{1}{8}\left(35x^4-30x^2+3\right)]=0 $$ (Eq 9.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle \left(-\frac{420}{8}x^4+60x^2-\frac{60}{8})\right)+(-35x^4+15x^2)+(\frac{175}{2}x^4-75x+\frac{15}{2})=0 $$ (Eq 9.17)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \cancel{(-\frac{420}{8}-35+\frac{175}{2})x^4}^0+\cancel{(60+15-75)x^2}^0+\cancel{(-\frac{60}{8}+\frac{15}{2})}^0=0 $$ (Eq 9.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

=Contributions Of Team Members=

 * Problem 1:    Solved by: Dube

Author: Dube

Proofread by: Dube,Oztekin

 * Problem 2:    Solved by: Abudaram

Author: Abudaram, Zou

Proofread by: Ismail,Zou,Abudaram

 * Problem 3:    Solved by: Abudaram

Author: Abudaram, Zou

Proofread by: Ismail,Zou,Abudaram

 * Problem 4:    Solved by: Oztekin

Author: Oztekin

Proofread by: Zou,Oztekin

 * Problem 5:    Solved by: Oztekin

Author: Oztekin

Proofread by: Zou,Oztekin

 * Problem 6:    Solved by: Zou

Author: Zou

Proofread by: Ismail,Zou,Abudaram

 * Problem 7:    Solved by: Zou

Author: Zou

Proofread by: Ismail,Zou,Abudaram

 * Problem 8:    Solved by: Sahin

Author: Sahin, Zou

Proofread by: Ismail,Zou,Abudaram

 * Problem 9:    Solved by: Patterson

Author: Patterson

Proofread by: Oztekin,Patterson

= References =

title: User:Egm6321.f10.team2.oztekin/HW6


text:
Fibonacci sequences, binary numbers and compositions

Fibonacci numbers
The Fibonacci numbers can be found in different ways in the sequence of binary strings. Due to a bijection between binary strings and compositions, every definition in terms of strings can also be given in terms of compositions, and vice versa.
 * The number of binary words of length n (length from first digit to last value 1 digit) without consecutive 0s is the Fibonacci number Fn+1. E.g., out of the 8 binary words of length 4, there are F5 = 5 without consecutive 0s - they are 0101, 1101, 1011, 0111 and 1111.
 * The number of strings of length n without consecutive 1s is the Fibonacci number Fn+2. E.g., out of the 16 binary strings of length 4, there are F6 = 8 without consecutive 1s - they are 0000, 1000, 0100, 0010, 1010, 0001, 1001 and 0101. (By symmetry, the number of strings of length n without consecutive 0s is also Fn+2.) This corresponds to the following statement about compositions, quoted from OEIS A000045:    F(n)=number of compositions of n-1 with no part greater than 2.    Example: F(4)=3 because we have 3 = 1+1+1=1+2=2+1.  00 corresponds to 1+1+1, 01 corresponds to 1+2, 10 corresponds to 2+1.

 * The number of binary words of length n (length from first digit to last value 1 digit) without even numbers of consecutive 0s or 1s is the Fibonacci number Fn. E.g. out of 32 binary words of length 6 there are F6 = 8 with all runlengths odd - they are 000001, 010001, 000101, 010101, 011101, 000111, 010111 and 011111.

 * The number of strings of length n without an odd number of consecutive 1s is the Fibonacci number Fn+1. This corresponds to the following statement about compositions, quoted from OEIS:    F(n) = number of compositions of n into odd parts;    e.g. F(6) counts 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 1+5, 3+1+1+1, 3+3, 5+1.

 * Further, the number of compositions of n without addends equal to 1 is the Fibonacci number Fn$-$1. E.g. out of the 8 compositions of 4 there are F3 = 2 without 1s - they are 2+2 and the 4 itself. Out of the 16 compositions of 5 there are F4 = 3 without 1s - they are 2+3, 3+2 and the 5 itself.

Fibonacci numbers of higher order
About the files showing two matrices:
 * In the top matrices blocks of n or more consecutive 0s are indicated by green dots. The number of columns without green dots between two green trigons is an element of the Fibonacci sequence of order n.
 * The number of compositions of natural numbers with no addend greater than n is a Fibonacci sequence of order n. Due to the bijection between compositions and binary numbers, this is what the bottom matrices show: The blocks of n or more consecutive 1s are indicated by strong red squares. The number of columns without strong red squares, from the left side to a green trigon, is an element of the Fibonacci sequence of order n.

Tribonacci numbers

Tetranacci numbers

Pentanacci numbers

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
User:Vdo-ed

 * Vdo-ed has completed the following assignments in the course on Basic filmmaking (pre-production) for narrative motion pictures.

 * [[Image:Blancovinkjeup.png|16px]] Formatting the Script [[Image:Nuvola apps package wordprocessing.png|28px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Pop Quiz - What is the first shot? [[Image:Crystal Clear app clock.svg|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Creating the thumbnail storyboards [[Image:Crystal Clear app gimp.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Quick Exercise - Over-the-shoulder shot [[Image:Applications-accessories.svg|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Creating the 3D storyboards [[Image:Crystal Clear app Login Manager.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] A simple DVD-Video Editing Workshop [[Image:Crystal Clear device cdwriter unmount.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Learning about camera lenses [[Image:Crystal Clear app camera.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Record the dialog for the animatic [[Image:Crystal Clear app krec.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Intro to narrative filmmaking [[Image:Clapboard-Small.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Draw the movie poster for "Star Wars" [[Image:Crystal Clear app kcoloredit.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Understanding dailies [[Image:Clapboard-Small.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Create a matte painting [[Image:Crystal Clear app kcoloredit.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Understanding editing [[Image:Clapboard-Small.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Pop Quiz - Creating the moods (film scoring) [[Image:Crystal Clear app clock.svg|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Understanding "L-Cuts" [[Image:Clapboard-Small.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Create the animatic [[Image:Clapboard-Small.png|28px]]

title: User:Vdo-ed


text:
User talk:Vdo-ed

 Hello Vdo-ed, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Vdo-ed


text:
User talk:Anne-Maree Shepherd

 Hello Anne-Maree Shepherd, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Anne-Maree Shepherd


text:
User:Soowjo

http://pilotlab.net/aircraft-manufacture/boeing/wingtip-devices-boeing.pdf XXXXXXXXXXXXXXX

http://www.icas-proceedings.net/ICAS2002/PAPERS/122.PDF XXXXXXXXXXXXXXXXXXXXX

http://www.boeing.com/commercial/aeromagazine/aero_17/winglets.pdf

http://events.theatlantic.com/static/files/green_pdfs/boeing_environment_brochures.pdf xxxxxxxxxxxxxxxx

http://www.waset.org/journals/ijame/v6/v6-1-2.pdf xxxxxxxxxxxxxxxxxxxx

http://www.aer.ita.br/~bmattos/download/winglet_aiaa.pdf xxxxxxxxxxxxxxxxxxxxx

http://www.enviro.aero/Content/Upload/File/ATAG_BeginnersGuidetoAviationEfficiency_MIDRESO(1).pdf xxxxxxxxxxxxxxxxxxx

http://freeportproceedings.info/2009/Hussaini_2009_FHS.pdf xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

https://engineering.purdue.edu/AAE/Academics/Courses/aae490t/teams2003/rakedtip/Review%20of%20Literature.doc

http://www.aoe.vt.edu/~mason/Mason_f/WingtipDevicesS04.pdf

http://www.ijens.org/103503-0909%20IJMME-IJENS.pdf xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

http://www.boeing.com/commercial/aeromagazine/articles/qtr_03_09/pdfs/AERO_Q309_article03.pdf

http://pdf.aiaa.org/preview/CDReadyMAAC09_2124/PV2009_4107.pdf xxxxxxxxxxxxxxxxxxxxxxx

http://www.springerlink.com/index/n0h682181865v621.pdf xxxxxxxxxxxxxxxxxx

http://www.onera.fr/daap/reduction-trainee-civil/drag-reduction-technologies-for-civil-transport-aircraft-reneaux.pdf xxxxxxxxxxxxx

http://www.buet.ac.bd/me/icme/icme2007/Proceedings/PDF/ICME07-FL-05.pdf xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

http://www.aer.bris.ac.uk/contact/academic/friswell/PDF_Files/C240.pdf xxxxxxxxxxxxxxxxxxxxxxxx

http://www.star-tech-inc.com/papers/aiaa/aiaa.pdf xxxxxxxxxxxxxxxxxxxxxxxx

http://www.mae.ufl.edu/~sforza/EAS4700/Wing%20and%20Tailplane%20Design%2008-05.doc xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

http://www.aoe.vt.edu/~mason/Mason_f/WingletsS04.ppt xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

http://www.nap.edu/openbook.php?record_id=11839&page=18 xxxxxxxxxxxxxxxxxxxxxxx

http://www.smartcockpit.com/data/pdfs/flightops/aerodynamics/Understanding_Winglets_Technology.pdf

http://www.ultraligero.net/Cursos/varios/todo_sobre_los_winglets.pdf xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

http://www.aviationpartnersboeing.com/news/pdf/news/07.6.29_whatswithwinglets-1586-1724.pdf xxxxxxxxxxxxxxxxxxxxxxxxxxx

http://www.engbrasil.eng.br/index_arquivos/art60.pdf

http://www.mandhsoaring.com/Why%20Winglets/WL-Soaring.pdf xxxxxxxxxxxxxxxxxxxxxxxxx

http://quietwing.com/wp-content/uploads/Winglet-article.pdf xxxxxxxxxxxxxxxxxxxx

http://www.nasa.gov/centers/dryden/pdf/88033main_H-1193.pdf

http://www.icas-proceedings.net/ICAS2008/PAPERS/536.PDF xxxxxxxxxxxxxxxxxxxxxxx

http://www.postfrontal.com/PDF/Winglets_02.pdf xxxxxxxxxxxxxxxxxxxxx

http://www.nasa.gov/centers/dryden/pdf/89234main_TF-2004-15-DFRC.pdf xxxxxxxxxxxxxxxxx

http://books.google.com/books?id=N4iZCl7dtO0C&pg=PA22&lpg=PA22&dq=NLF+concept&source=bl&ots=TGly3q95ED&sig=b7z7CaavYJZNKoav32ZS2zOVZMY&hl=en&ei=cLn_TPKJA4WglAfnk4WpCA&sa=X&oi=book_result&ct=result&resnum=5&ved=0CDAQ6AEwBA#v=onepage&q=NLF%20concept&f=false

http://books.google.com/books?id=wsSl_-Qo5JAC&pg=PA205&lpg=PA205&dq=%22flow+control%22+drag&source=bl&ots=9DNHZj4bE5&sig=hloxYzQw2mUjCxYDwVG0uvVwuyU&hl=en&ei=7UAATdebMsqclgeqouTDCw&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCAQ6AEwATge#v=onepage&q=%22flow%20control%22%20drag&f=false

title: User:Soowjo


text:
User talk:Lysandros~enwikiversity

 Hello Lysandros, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:17, 18 November 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Lysandros. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Lysandros~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:37, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:14, 19 April 2015 (UTC)

title: User talk:Lysandros~enwikiversity


text:
User:Nisar bhat

Nisar Ahmed Bhat, Kashmir,India,doing MSc. Biotech, from Lovely Professional University.

title: User:Nisar bhat


text:
User talk:Nisar bhat

 Hello Nisar bhat, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
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title: User talk:Nisar bhat


text:
User talk:Lotje

 Hello Lotje, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 15:47, 18 November 2010 (UTC)

Thnks, but I do not twitter and do not glitter :) Lotje ツ 15:00, 12 May 2011 (UTC)

abbreviated interwiki prefix
. As this is the English Wikiversity, you can use w: as a prefix, you only need to specify the language code if it is not en. But it doesn't hurt anything to have the code. --Abd (discuss • contribs) 22:33, 20 May 2014 (UTC)
 * Thanks Abd, I was not sure about that one. In the future I'll make the interwikilink. Lotje (discuss • contribs) 03:12, 21 May 2014 (UTC)

title: User talk:Lotje


text:
User talk:Riotr

 Hello Riotr, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 15:47, 18 November 2010 (UTC)

Human Legacy Course
Hello Riotr! I am very glad that you have decided to join my Human Legacy course and I'm very sorry that it took so long for me to respond. If you would like to have an orientation session so that you can get the basics of what will be going on with this course and any other questions that you may have, please tell me when you are free and I will create the class. In order to come to these classes, you must be a member of WiZiQ. I hope to hear from you very soon and if you have any questions at all, please do not hesitate to ask! :)

--MrABlair23 16:27, 13 June 2011 (UTC)

Hello Riotr! I know that it has been a while, however, I would like to invite you to join my Human Legacy Course. If you are interested, please contact me on my talk page and we can set up an orientation session. I hope to see you soon! :) --MrABlair23 15:51, 3 September 2011 (UTC)

title: User talk:Riotr


text:
User talk:FormerEditor2718

 Hello Khfan93, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
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title: User talk:FormerEditor2718


text:
C++/Function overloading

In object oriented programming language the flexibility of using the polymorphism features are achieved by function overloading.Basically it is the process of using the same function name for two or more functions,but the functions need to have different types of parameter or different numbers of parameters to be used in the program. Examples :: #include using namespace std; int sqr(int); float sqr(float); main {  int a=15; float b=2.5; cout<<"square= "sqr(a)<<"\n"; cout<<"square= "<<sqr(b)<<"\n"; return 0; } int sqr(int s) { return (s*s); } int sqr(float j) { return (j*j); }

OUTPUT: square=225 square=6.25

title: C++/Function overloading


text:
User talk:Debdash

 Hello Debdash, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 21:20, 18 November 2010 (UTC)

title: User talk:Debdash


text:
User talk:Soowjo

 Hello Soowjo, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 21:46, 18 November 2010 (UTC)

title: User talk:Soowjo


text:
User talk:Alborzagros

 Hello Alborzagros, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --La comadreja formerly AFriedman RESEARCH (talk) 22:29, 18 November 2010 (UTC)

 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

title: User talk:Alborzagros


text: User:Picopa

title: User:Picopa


text: User talk:Picopa

title: User talk:Picopa


text:
Template:Motivation and emotion/E-portfolio/Feedback/2010

E-portfolio Feedback for
Hi ,

Thanks for participating in the Motivation and emotion unit. I have read |your e-portfolio and have provided some feedback below. You marks for the e-portfolio are available via Moodle. Feel free to let me know if you have any comments, concerns, questions. I also welcome your feedback.

Contributions/Statistics

 * Contributions
 * Statistics
 * Page size: User: kb | Talk page: kb | User+Talk: kb

Sincerely, James Neill - talk - email

P.S. If you are dissatisfied with the marking and feedback, please see the marking dispute process.