text:
User:Egm6321.f10.team2.dube/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

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

$$  \displaystyle P_{2}(x)=\frac{1}{2}(3x^{2}-1) $$     (Eq 1.1) Legendre's equation where n=2
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (1-x^{2})y''-2xy'+6y=0 $$     (Eq 1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find

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

$$  \displaystyle Q_{2}(x)=\frac{1}{4}(3x^{2}-1)log(\frac{1+x}{1-x})-\frac{3}{2}x $$     (Eq 1.3) using the variation of parameter method.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Solution
To begin to solve Q_{2}(x) we must first solve the general form and then plug in.

We start with the equation:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=U(x)u_{1}(x) $$     (Eq 1.4) and take derivatives to get in the Legendre Equation form
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y=U*u_{1} $$     (Eq 1.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y'=U'*u_{1}+U*u'_{1} $$     (Eq 1.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y=U*u_{1}+U*u''_{1}+2U'u'_{1} $$     (Eq 1.7) Plugging Eqns. 1.5-1.7 into the general L2-ODE-VC we get:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle U*u_{1}+U*u_{1}+2U'u'_{1}+a_{1}(x)*(U'*u_{1}+U*u'_{1})+a_{0}(x)*(U*u_{1})=0 $$     (Eq 1.8) Combining terms:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle U(u_{1}+a_{1}(x)u'_{1}+a_{0}(x)u_{1})+U'(2u'_{1}(x)a_{1}(x)+a_{1}(x)u_{1}(x))+Uu_{1}(x)=0 $$     (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:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle U'(2u'_{1}(x)+a_{1}(x)u_{1}(x))+U''u_{1}(x)=0 $$     (Eq 1.10) To help solve, we set $$ \displaystyle Z=U' $$ so we can use the order reducing method.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Z(2u'_{1}(x)+a_{1}(x)u_{1}(x))+Z'u_{1}(x)=0 $$     (Eq 1.11) We divide by $$ \displaystyle u_{1}(x) $$ to get:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Z\frac{2u'_{1}(x)+a_{1}(x)u_{1}(x)}{u_{1}(x)}+Z'=0 $$     (Eq 1.12) We can now use the Integrating Factor Method
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h(x)=exp [\int (\frac{2u'_{1}(x)+a_{1}(x)u_{1}(x)}{u_{1}(x)})dx] $$     (Eq 1.13) Multiplying through by u_{1}(x):
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h(x)=exp [\int (a_1(x)+2\frac{u'_1(x)}{u_1(x)})dx] $$     (Eq 1.14) Separating:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h(x)=exp [\int a_1(x)dx+\int 2\frac{u'_1(x)}{u_1(x)}dx] $$     (Eq 1.15) We then can solve to get:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle h(x)=u_{1}^{2}(x)exp[\int a_1(x)dx] $$     (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 $$ \displaystyle (1-x^2) $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y''-\frac{2x}{1-x^2}y'+\frac{6}{1-x^2}y=0 $$     (Eq 1.17) From this we can see that:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle a_1(x)=\frac{-2x}{1-x^2} $$     (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:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle Q_2(x)=u_1(x)\int \frac{1}{h(x)}dx $$     (Eq 1.19)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{1}{u_{1}^{2}(x)exp[\int a_1(x)dx]}dx $$     (Eq 1.20)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{1}{(\frac{1}{2}(3x^2-1))^{2}exp[\int \frac{-2x}{1-x^2}dx]}dx $$     (Eq 1.21)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle exp\int \frac{-2x}{1-x^{2}}dx=exp[log(1-x^{2})]=1-x^{2} $$     (Eq 1.22) If we sustitute Eq(1.22) into Eq(1.21) we have
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle Q_2(x)=\frac{1}{2}(3x^2-1) \int \frac{4}{(3x^2-1)^{2}(1-x^{2})}dx $$     (Eq 1.23) Let's look at the fraction:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{1}{(3x^{2}-1)^{2}(1-x^{2})}=\frac{Ax+B}{(3x^{2}-1)}+\frac{Cx+D}{(3x^{2}-1)^{2}}+\frac{E}{(1+x)}+\frac{F}{(1-x)} $$     (Eq 1.24) Having compansated denominators we have;
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle (Ax+B)(3x^{2}-1)(1-x^{2})+(Cx+D)(1-x^{2})+E(1-x)(3x^{2}-1)^{2}+F(3x^{2}-1)^{2}(1+x)=1 $$     (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:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle A=0,B=\frac{3}{4},C=0,D=\frac{3}{2},E=\frac{1}{8},F=\frac{1}{8} $$     (Eq 1.26)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{1}{(3x^{2}-1)^{2}(1-x^{2})}=\frac{3}{4}\frac{1}{(3x^2-1)}+\frac{3}{2}\frac{1}{(3x^{2}-1)}+\frac{1}{8}\frac{1}{(x+1)}-\frac{1}{8}\frac{1}{(x-1)} $$     (Eq 1.27)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \int \frac{1}{(3x^{2}-1)(1-x^{2})}=\frac{1}{8}\left [ \frac{-6x}{(3x^{2}-1)} -log(1-x)+log(1+x)\right ] $$     (Eq 1.28) If we sustitute Eq(1.28) into Eq(1.23) we get the second Q
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle Q_{2}(x)=\frac{1}{4}(3x^{2}-1)log(\frac{1+x}{1-x})-\frac{3}{2}x $$     (Eq 1.29)
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= 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

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

$$ \displaystyle

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

(Eq 2.1)
 *

 * }

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

$$ \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:

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

$$ \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)
 *

 * }

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 $$\displaystyle f(x)=0$$:

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

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

$$ \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:

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

$$ \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 $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$.

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

$$ \displaystyle

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

(Eqs 2.6)
 *

 * }

Then,

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

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

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:

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

$$ \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:
 *

 * }

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

$$ \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:
 *

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

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

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

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

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

$$ \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:

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

$$ \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)
 *

 * }

Then,

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

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

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:

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

$$ \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:
 *

 * }

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

$$ \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:
 *

 * }

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

$$ \displaystyle

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

(Eq 2.17)
 *

 * }

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

$$ \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:
 *

 * }

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

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

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)
 *

 * }

= 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:

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

$$ \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
 *

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

$$ \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)
 *

 * }

Using the example from the King's book on page 34 we can use the variation of parameters formula that is

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

$$ \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)
 *

 * }

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

$$ \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)
 *

 * }

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

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

$$ \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:
 *

 * }

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

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

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)
 *

 * }

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

$$  \displaystyle x_{1}=rcos\theta cos\varphi =\xi _{1}cos\xi _{2}cos\xi _{3} $$     (Eq 4.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle x_{2}=rcos\theta sin\varphi $$     (Eq 4.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle x_{3}=rsin\theta $$     (Eq 4.3)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle ds^{2}=\vec{ds}.\vec{ds}=(dx_{i}dx_{j})(e_{i}e_{j}) $$     (Eq 4.4)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \delta _{ij}=\left\{\begin{matrix} 1 &for &i=j \\ 0 &for & i\neq j \end{matrix}\right. $$     (Eq 4.5) In astronomical convention angles are defined as
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle\begin{align} & -\frac{\pi }{2}\leq \theta \leq \frac{\pi }{2}\\ & 0\leq \varphi \leq 2\pi \\ \end{align} $$     (Eq 4.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Obtain the steps given in meeting 34

Solution

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

$$  \displaystyle ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} $$     (Eq 4.7) Finding total derivative of first term
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle dx_{1}=\frac{\partial x_{1}}{\partial r}dr+\frac{\partial x_{1}}{\partial \theta }d\theta +\frac{\partial x_{1}}{\partial \varphi }d\varphi $$     (Eq 4.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial x_{1}}{\partial r}=cos\theta cos\varphi $$     (Eq 4.9)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{\partial x_{1}}{\partial \theta }=-rsin\theta cos\theta $$     (Eq 4.10)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{\partial x_{1}}{\partial \varphi }=-rcos\theta sin\varphi $$     (Eq 4.11)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle dx_{1} =cos\theta cos\varphi dr-rsin\theta cos\varphi d\theta -rcos\theta sin\varphi d\varphi $$     (Eq 4.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dx_{1}^{2}=cos^{2}\theta cos^{2}\varphi dr^{2}+r^{2}sin^{2}\theta cos^{2}\varphi d\theta ^{2}+r^{2}cos^{2}\theta sin^{2}\varphi d\varphi ^{2}+2r^{2}sin\varphi cos\varphi sin\theta cos\theta d\theta d\varphi -2rcos\theta cos^{2}\varphi sin\theta drd\theta -2rcos^{2}\theta cos\varphi sin\varphi drd\varphi $$     (Eq 4.13) Finding total dervative of second term
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle dx_{2}=\frac{\partial x_{2}}{\partial r}dr+\frac{\partial x_{2}}{\partial \theta }d\theta +\frac{\partial x_{2}}{\partial \varphi }d\varphi $$     (Eq 4.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial x_{2}}{\partial r}=cos\theta sin\varphi $$     (Eq4.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial x_{2}}{\partial \theta }=-rsin\theta sin\varphi $$     (Eq 4.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial x_{2}}{\partial \varphi }=rcos\theta cos\varphi $$     (Eq 4.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dx_{2}=cos\theta sin\varphi dr-rsin\theta sin\varphi d\theta +rcos\theta cos\varphi d\varphi $$     (Eq 4.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dx_{2}^{2}=cos^{2}\theta sin^{2}\varphi dr^{2}+r^{2}sin^{2}\theta sin^{2}\varphi d\theta^{2} +r^{2}cos^{2}\theta cos^{2}\varphi d\varphi ^{2}+2rcos^{2}\theta sin\varphi cos\varphi drd\varphi -2r^{2}sin\theta sin\varphi cos\theta cos\varphi d\theta d\varphi -2rcos\theta sin\theta sin^{2}\varphi drd\theta $$     (Eq 4.19) Finding total derivative of third term
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle dx_{3}=\frac{\partial x_{3}}{\partial r}dr+\frac{\partial x_{3}}{\partial \theta }d\theta +\frac{\partial x_{3}}{\partial \varphi }d\varphi $$     (Eq 4.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial x_{3}}{\partial r}=sin\theta $$     (Eq 4.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial x_{3}}{\partial \theta }=rcos\theta $$     (Eq 4.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial x_{3}}{\partial \varphi \theta }=0 $$     (Eq 4.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dx_{3}=sin\theta dr+rcos\theta d\theta $$     (Eq 4.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dx_{3}^{2}=sin^{2}\theta dr^{2}+r^{2}cos^{2}\theta d\theta ^{2}+2rsin\theta cos\theta drd\theta $$     (Eq 4.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Collecting all of the differentials Eq 4.13-Eq4.19-Eq4.25 and substitute into ds yields ;

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

$$  \displaystyle ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} $$     (Eq 4.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \begin{align} &ds^{2}=cos^{2}\theta cos^{2}\varphi dr^{2}+r^{2}sin^{2}\theta cos^{2}\varphi d\theta ^{2}+r^{2}cos^{2}\theta sin^{2}\varphi d\varphi ^{2}+cos^{2}\theta sin^{2}\varphi dr^{2}+r^{2}sin^{2}\theta sin^{2}\varphi d\theta^{2} +r^{2}cos^{2}\theta cos^{2}\varphi d\varphi ^{2}\\ &+sin^{2}\theta dr^{2}+r^{2}cos^{2}\theta d\theta ^{2}+2r^{2}sin\varphi cos\varphi sin\theta cos\theta d\theta d\varphi -2rcos\theta cos^{2}\varphi sin\theta drd\theta -2rcos^{2}\theta cos\varphi sin\varphi drd\varphi\\ &+2rcos^{2}\theta sin\varphi cos\varphi drd\varphi -2r^{2}sin\theta sin\varphi cos\theta cos\varphi d\theta d\varphi -2rcos\theta sin\theta sin^{2}\varphi drd\theta +2rsin\theta cos\theta drd\theta\\ \end{align} $$     (Eq 4.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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.

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

$$  \displaystyle \begin{align} &ds^{2}=cos^{2}\theta (cos^{2}\varphi +sin^{2}\varphi )dr^{2}+sin^{2}\theta dr^{2}+r^{2}sin^{2}\theta (cos^{2}\varphi +sin^{2}\varphi )d\theta ^{2}+r^{2}cos^{2}\theta d\theta ^{2}+r^{2}cos^{2}\theta (sin^{2}\varphi +cos^{2}\varphi )d\varphi ^{2}\\ &+2rsin\theta cos\theta drd\theta -2rcos\theta sin\theta (cos^{2}\varphi +sin^{2}\varphi )drd\theta \\ \end{align} $$     (Eq 4.28) We can obtain the equation given in the lecture 34-1
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle ds^{2}=dr^{2}+r^{2}d\theta ^{2}+r^{2}cos^{2}\theta d\phi ^{2} $$     (Eq 4.29)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Delta \psi =\frac{1}{h_{1}h_{2}h_{3}}\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\partial \psi }{\partial \xi _{i}}\right ] $$     (Eq 4.30)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle h_{1}.h_{2}.h_{3}=r^{2}cos\theta $$     (Eq 4.31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle i=1,\xi _{1}=r\Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial r}\left [ \frac{r^{2}cos\theta }{1}\frac{\partial \psi }{\partial r} \right ] $$     (Eq 4.32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since $$\displaystyle cos\theta $$is constant for r we can take it out from the derivative.
 * {| style="width:100%" border="0"

$$  \displaystyle i=1,\xi _{1}=r\Rightarrow \frac{1}{r^{2} }\frac{\partial }{\partial r}\left [ \frac{r^{2} }{1}\frac{\partial \psi }{\partial r} \right ] $$     (Eq 4.33)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle i=2,\xi _{2}=\theta \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }\left [ \frac{r^{2}cos\theta }{r^{2}}\frac{\partial \psi }{\partial \theta } \right ] $$     (Eq 4.34)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle i=2,\xi _{2}=\theta \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }\left [ \frac{cos\theta }{1}\frac{\partial \psi }{\partial \theta } \right ] $$     (Eq 4.35)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle i=3,\xi _{3}=\varphi \Rightarrow \frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \varphi }\left [ \frac{r^{2}cos\theta }{r^{2}cos^{2}\theta }\frac{\partial \varphi }{\partial\varphi } \right ] $$     (Eq 4.36)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since $$\displaystyle cos\theta $$ is constant for \varphi we can take it out from the derivative.

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

$$  \displaystyle i=3,\xi _{3}=\varphi \Rightarrow \frac{1}{r^{2}cos^{2}\theta }\frac{\partial }{\partial \varphi }\left [ \frac{\partial^{2}\psi }{\partial\varphi ^{2} } \right ] $$     (Eq 4.37)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then we can collect all terms into the $$\displaystyle \Delta \psi $$ equation:

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

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

$$  \displaystyle \Delta \psi =\frac{1}{r^{2}}(r^{2}\frac{\partial \psi }{\partial r})+\frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }(cos\theta \frac{\partial \psi }{\partial \theta })+\frac{1}{r^{2}cos^{2}\theta }\frac{\partial ^2\psi }{\partial \varphi ^2} $$     (Eq 4.38)
 *

 * }

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

$$  \displaystyle x_{1}=rcos\theta =\xi _{1}cos\xi _{2} $$     (Eq 5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle x_{2}=rsin\theta =\xi _{1}sin\xi _{2} $$     (Eq 5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle x_{3}=z=\xi _{3} $$     (Eq 5.3) Now,after defining our new curvilinear coordinates,we can define our cartesian coordinates in terms of our new independent variables.
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle\begin{align} & x_{1}=x_{1}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ &x_{2}=x_{2}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ &x_{3}=x_{3}\left \{ \xi _{1},\xi _{2},\xi _{3} \right \}\\ \end{align}$$ (Eq 5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Repeat steps for cylindrical coordinates and conclude with Bessel's Eq.

Finding dxi's

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

$$  \displaystyle dx_{1}=\frac{\partial x_{1}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{1}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{1}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle\begin{align} &\frac{\partial x_{1}}{\partial \xi _{1}}=cos\xi _{2}\\ &\frac{\partial x_{1}}{\partial \xi _{2}}=-\xi _{1}sin\xi _{2}\\ &\frac{\partial x_{1}}{\partial \xi _{3}}=0\\ \end{align}$$ (Eq 5.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dx_{1}=cos\xi _{2}d\xi _{1}-\xi _{1}sin\xi _{2}d\xi _{2} $$     (Eq 5.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dx_{2}=\frac{\partial x_{2}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{2}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{2}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle\begin{align} &\frac{\partial x_{2}}{\partial \xi _{1}}=sin\xi _{2}\\ &\frac{\partial x_{2}}{\partial \xi _{2}}=\xi _{1}cos\xi _{2}\\ &\frac{\partial x_{2}}{\partial \xi _{3}}=0\\ \end{align}$$ (Eq 5.9)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle dx_{2}=sin\xi _{2}d\xi _{1}+\xi _{1}cos\xi _{2}d\xi _{2} $$     (Eq 5.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dx_{3}=\frac{\partial x_{3}}{\partial \xi _{1}}d\xi _{1}+\frac{\partial x_{3}}{\partial \xi _{2}}d\xi _{2}+\frac{\partial x_{3}}{\partial \xi _{3}}d\xi _{3} $$     (Eq 5.11)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle\begin{align} &\frac{\partial x_{3}}{\partial \xi _{1}}=0\\ &\frac{\partial x_{3}}{\partial \xi _{2}}=0\\ &\frac{\partial x_{3}}{\partial \xi _{3}}=1\\ \end{align}$$ (Eq 5.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finding ds^2

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

$$  \displaystyle dx_{1}^{2}=cos^{2}\xi _{2}d\xi _{1}^{2}+\xi _{1}^{2}sin^{2}\xi _{2}d\xi _{2}^{2}-2\xi _{1}cos\xi _{2}sin\xi _{2}d\xi _{1}d\xi _{2} $$     (Eq 5.13)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle dx_{2}^{2}=sin^{2}\xi _{2}d\xi _{1}^{2}+\xi _{1}^{2}cos^{2}\xi _{2}d\xi _{2}^{2}+2\xi _{1}cos\xi _{2}sin\xi _{2}d\xi _{1}d\xi _{2} $$     (Eq 5.14)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle dx_{3}^{2}=d\xi _{3}^{2} $$     (Eq 5.15)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle ds^{2}=d\xi _{1}^{2}+\xi _{1}^{2}(sin^{2}\xi _{2}+cos^{2}\xi _{2})d\xi _{2}^{2}+d\xi _{3}^{2}=d\xi _{1}^{2}+\xi _{1}^{2}d\xi _{2}^{2}+d\xi _{3}^{2} $$     (Eq 5.16)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle\begin{align} & h_{1}^{2}=1^{2}\Rightarrow h_{1}=1\\ & h_{2}^{2}=\xi _{1}^{2}\Rightarrow h_{2}=\xi _{1}\\ & h_{3}^{2}=1^{2}\Rightarrow h_{3}=1\\ \end{align}$$ (Eq 5.17)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle h_{1}h_{2}h_{3}=\xi _{1} $$     (Eq 5.18)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Delta \psi =\frac{1}{h_{1}h_{2}h_{3}}\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\partial \psi }{\partial \xi _{i}}\right ] $$     (Eq 5.19)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle i=1\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{1}} \right ) $$     (Eq 5.20)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle i=2\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{2}}\left ( \frac{\xi _{1}}{\xi _{1}^{2}}\frac{\partial \psi }{\partial \xi _{2}} \right )\Rightarrow \frac{1}{\xi _{1}^{2}}\frac{\partial ^2\psi }{\partial \xi _{2}} $$     (Eq 5.21)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle i=3\Rightarrow \frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{3}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{3}} \right )\Rightarrow \frac{\partial^2 \psi }{\partial \xi _{3}^2} $$     (Eq 5.22)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Delta \psi =\frac{1}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial \psi }{\partial \xi _{1}} \right )+\frac{1}{\xi _{1}^{2}}\frac{\partial^2 \psi }{\partial \xi _{2}^2}+\frac{\partial^2 \psi }{\partial \xi _{3}^2}=0 $$     (Eq 5.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \psi (\xi _{1},\xi _{2},\xi _{3})=X(\xi _{1})Y(\xi _{2})Z(\xi _{3}) $$     (Eq 5.24)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{Y(\xi _{2})Z(\xi _{3})}{\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X(\xi _{1})}{\partial \xi _{1}} \right )+\frac{X(\xi _{1})Z(\xi _{3})}{\xi _{1}^{2}}\frac{\partial^2Y(\xi _{2})}{\partial^2 \xi _{2}}+X(\xi _{1})Y(\xi _{2})\frac{\partial^2 Z(\xi _{3})}{\partial \xi _{3}^2}=0 $$     (Eq 5.25) Now, dividing through $$ \displaystyle XYZ$$ we can obtain more useful form.
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{1}{X\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+\frac{1}{Y\xi _{1}^{2}}\frac{\partial^2 Y}{\partial \xi _{2}^2}+\frac{1}{Z}\frac{\partial^2 Z}{\partial \xi _{3}^2}=0 $$     (Eq 5.26) Assuming that
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{\partial^2 Z}{\partial \xi _{3}^2}=\alpha ^{2}Z $$     (Eq 5.27) After that our equation forms as
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{1}{X\xi _{1}}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+\frac{1}{Y\xi _{1}^{2}}\frac{\partial^2 Y}{\partial \xi _{2}^2}+\alpha ^{2}=0 $$     (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 $$\displaystyle \xi _{1}^{2}$$.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

In addition let's assume
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial^2 Y}{\partial \xi _{2}^2}=-n^{2}Y $$     (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.
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \xi _{1}\frac{\partial }{\partial \xi _{1}}\left ( \xi _{1}\frac{\partial X}{\partial \xi _{1}} \right )+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.30)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \xi _{1}^{2}\frac{\partial^2 X}{\partial \xi_{1} ^2}+\xi _{1}\frac{\partial X}{\partial \xi _{1}}+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.31) Say that
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle x=\alpha \xi _{1} $$     (Eq 5.32)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{\partial X}{\partial \xi _{1}}=\frac{\partial X}{\partial x}\frac{\partial x}{\partial \xi _{1}}=\alpha \frac{\partial X}{\partial x} $$ (Eq 5.33)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{\partial }{\partial \xi _{1}}\left ( \alpha \frac{\partial X}{\partial x} \right )=\alpha ^{2}\frac{\partial X}{\partial x} $$ (Eq 5.34) Substituting Eq(5.33) and Eq (5.34) in to Eq (Eq 5.32) We observe
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \xi _{1}^{2}\alpha ^{2}\frac{\partial^2 X}{\partial x^2}+\xi _{1}\alpha \frac{\partial X}{\partial x}+(\alpha ^{2}\xi _{1}^{2}-n^{2})X=0 $$     (Eq 5.35)
 * style="width:95%" |
 * style="width:95%" |
 *

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

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

$$  \displaystyle x^{2}\frac{\partial^2 X}{\partial x^2}+x \frac{\partial X}{\partial x}+(x^{2}-n^{2})X=0 $$     (Eq 5.36)
 *

 * }

= 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 ,

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

$$ \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)
 *

 * }

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

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

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

$$ \displaystyle

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

(Eq 6.2)
 *

 * }

Solution
Sustitute Eq 6.2 into Eq 6.1 we get,

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

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

\begin{align} \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)
 * style="width:95%" |
 * 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.dube/HW6


text:
Program evaluation

Program evaluation is a systematic method for collecting, analysing, and using information to answer questions about projects, policies and programs

Program evaluation guides

 * McNamara, C. (nd). Basic guide to program evaluation. pdf
 * McNamara, C. (nd). Checklist for program evaluation planning

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

title: Program evaluation


text: File:3.B.Magnetic Sensor.new.pdf

title: File:3.B.Magnetic Sensor.new.pdf


text:
User talk:EdEveridge

 Hello EdEveridge, 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
 * 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) 04:38, 13 November 2010 (UTC)

title: User talk:EdEveridge


text:
User:Alchemy Heels I

Anagram
Alchemy Heels I is an anagram.

title: User:Alchemy Heels I


text:
User talk:Alchemy Heels I

 Hello Alchemy Heels I, 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
 * 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) 05:26, 13 November 2010 (UTC)

title: User talk:Alchemy Heels I


text:
User:Sergejs.akulics

Sergejs Akulics (latv. - Sergejs Akuličs, born in Riga, Latvia, 26.12.1985). Now - Senior Expert at the State Agency of Medicine of Latvia.

In 2008 graduated Riga Stradins University the Faculcy of Pharmacy with grade - Pharmacist.

Personal information of Sergejs Akulics you can find at facebook.com Professional curriculum of Sergejs Akulics you can find at linkedin.com

Personal blog of Sergejs Akulics on livejournal.com

title: User:Sergejs.akulics


text:
User talk:Sergejs.akulics

I am glad I have foun this division. It is fantastic. Sergejs.Akulics 10:00, 13 November 2010 (UTC)

Welcome
 Hello Sergejs.akulics, 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
 * 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) 14:11, 13 November 2010 (UTC)

title: User talk:Sergejs.akulics


text: File:DFT.1.A.Complex Phase.pdf

title: File:DFT.1.A.Complex Phase.pdf


text:
First World War History

General Overview

 * World War I: A course divided into 13 lessons, incomplete

Specific topics

 * Africa and the First World War: Research area

title: First World War History


text:
Africa and the First World War

This is a research project.

Introduction and Goals
This project is aimed at collecting factual information to support further research regarding Africa and the First World War. It will include the involvement of Africans of the Diaspora, Africa's role in events leading up to the war, and the impact of the consequences of the war.

Prelude to war

 * Agadir Crisis
 * Scramble for Africa

During the war

 * Volta-Bani War

Campaigns in Africa

 * East African Campaign
 * North African Campaign
 * South-West Africa Campaign
 * West Africa Campaign

Aftermath

 * Paris Peace Conference

title: Africa and the First World War


text: User:Ha98574/Railroad/Korea/Korea Metropolian Subway Gyeonggiseobu Line


text: User:Ha98574/Railroad/Korea/Korea Metropolian Subway Gyeonggiseobu Line/Template


text:
User:Robertparten

My name is Robert Maris Parten II and I reside in Durham, NC. I am a Network Engineer with a focus on Cisco and Juniper devices. I am also a Linux and Windows systems admin (more the Jr Level). I love to teach networking to others and participate in teaching other the importance of being a dual environment capable administrator and avoid the cliche Linux Vs Windows arguments.

title: User:Robertparten


text:
User talk:Robertparten

 Hello Robertparten, 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
 * 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) 17:33, 13 November 2010 (UTC)

title: User talk:Robertparten


text: User:Ha98574/Railroad/Korea/Gyeongbu Line


text:
User talk:Quad~enwikiversity

the existing energy of electricity and others comes mainly from the energy of the sun,how? if we see closy the hydropower station are working only because of the hydrocircle that work because of the sun .The energy which is created by the fossil fules is nothing but the chamicle energy created with the use of the suns energy .But what is the biggest energy sources in the world and the universe around us .Man will reach to it's highest hights when he will be able to use these energies .Because of these abandunt energy sources the world will change for ever .For the answer we shall look arroung us aske the question what makes a coin from our hand fall down oh that is gravity ...a force against us really but the same force is making the earth spin arround the sun and made everything possible arround us .Humans shall use this gravitational energy should conduct experiments on this energy this is a huge force of energy if realy used can change a lot of things .This doesn't mean direct use of the gravitational force no but could be derived wit the phenomenon that exists becuase of the gravitational energy ...

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 Quad. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Quad~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 23:48, 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) 06:18, 19 April 2015 (UTC)

title: User talk:Quad~enwikiversity


text:
Talk:Africa and the First World War

For more about wikiversity research see Research

title: Talk:Africa and the First World War


text: User:Ha98574/Railroad/Korea/Gyeongbu Line/Template


text:
File:AfhFigure 2-3.JPG

Inside the cockpit.

title: File:AfhFigure 2-3.JPG


text:
Ideas in Geometry/Constructible Numbers

This lesson was based around Group Project 8 that we did in class on November 8th.

The packet began by introducing the idea of constructible numbers, it said

“Given a unit of measure (like inches or meters or miles), what distances can we precisely construct using a compass and straightedge? These distances are what we call constructible numbers.”

Then the packet had us create a number line when we were given the distance between 0 and 1.

To do this you should set your compass so that one leg of the compass was touching point 0 and the other leg was touching point 1. This was the distance. Then you should place the non-pencil leg on point 1 and swing the pencil side around to mark the point that was the same distance from 0 to 1 on the right of point 1. This created point 2. This is what the number line looks like now:

You could use this to create the next consecutive point for infinity.

This technique can also be used to create negative points on the number line. If the same distance measure was kept on the compass, the non-pencil leg could be placed on 0 and the pencil leg could be swung to the left to find the point -1. This is what the number line looks like now:

This could also be continued to negative infinity.

Next you are given two segments (a and b) and are asked to have them add together and subtract.

To add segments a and b together you should first draw a straight line using your straightedge. Mark an arbitrary point on the lawn you have drawn. This will be the left end point of the new segment you are creating. Then set the compass to length a. Place the non-pencil leg on the point on the line and swing the pencil leg to find the other endpoint of segment a.

Then set your compass to the length of b. Place the non-pencil leg on the right endpoint of segment a on the line you have drawn and swing the pencil side the mark the other endpoint of segment b. You have now created segment a+b.

To create a-b you must once again draw a line with your straightedge and mark a point to be the left end of segment a. Then set your compass to distance a. Place the non-pencil side on the point you have marked and swing it so that the pencil leg marks the other end of a.

Then set your compass to segment b. Place the non-pencil leg of the compass on the right endpoint of a and swing it to mark the left endpoint of b. The part of a that is not also encompassed in b is a-b.

Next you learn to create fractions by using the concept of similar triangles. Similar triangles are triangles that have the same angle measures and whose sides are proportional. This can be done using the following example:

You are given segments a and b, how can you construct a line segment of length a/b. This can be done be creating a right triangle where a is the vertical leg and b is the horizontal leg.

Now add a point that is 1 away from the right angle and create a triangle so that the vertical side is of length a/b. It is of side length a/b because the side lengths are proportional.

The main goal of this lesson was to introduce the idea of constructible numbers and how they can be created.

title: Ideas in Geometry/Constructible Numbers


text:
User talk:Brylie/Student Profile

I like what you have started Brylie, and look forward to seeing it evolve. Do you have any thoughts on how you'll measure your progress? --Charles Jeffrey Danoff 02:38, 14 November 2010 (UTC)

title: User talk:Brylie/Student Profile


text:
User:Yourstruly~enwikiversity

Well this is me..Yours Truly..get it? Good! Well, I'm me, if we're just sticking to simplistic matters. Now, if we care to go in depth, then read on. I'm relatively young in the harsh world we exist in, yet I'm able to maintain my own self--with the help of my parents, of course. I'm obsessed with airplanes, psychology, and technical theatre. I know a ton about aviation (study of airplanes, their workings, and so on so forth), and I've only met one person who's been able to outsmart me in a general aspect. Psychology is merely a hobby of mine, I try to read people, and mess with their heads. Some people call me completely nuts and psychopathical...I prefer the term "talented". I don't always aim to hurt, I aim to learn knowledge about the working of people. Harm happens by accident. Then technical theatre. I'm not a genius in this category, but I'm pretty close. I run 24 channel musicals, all wireless mics, and my current project I'm working on is "The Sound of Music". Show opens on the 16th..so it's like insanely close to show. I think we're ready. I know I am.

title: User:Yourstruly~enwikiversity


text:
Web Design/Getting Your Site On the Web/XAMPP

XAMPP is an easy to install Apache distribution containing MySQL, PHP and Perl. XAMPP have different packages for MAC,Windows,Linux And Solaris. XAMPP is easy to install and use.

Install
XAMPP supports Portable and Normal installation.

Links
http://www.apachefriends.org/en/xampp.html


text:
User:Storm~enwikiversity/Multimedia Presentation

Emotion and Sleep


text:
The periodic table/Gold

Discovery
Gold was discovered in ancient times. One reason for why Gold is so precious is the fact that Gold can stay perfect/shiny for a long time. Pop a penny in a dusty place in your car, that penny is disgusting! While a Roman gold coin is still shiny after years and years.

Quick Facts
Name: gold atom

Symbol: Au

Atomic Mass: 196.9665

Classification: transition metal

Protons: 79

Neutrons: 118

Electrons: 79

Colour: gold

Discovery in: ancient

Density: 19.3 g/cc

Crystal Structure: cubic

Melting Point: 1064.18 °C (1337.33 K, 1947.52 °F)

Boiling Point: 2856 °C (3129 K, 5173 °F)

Common Uses: jewelery, dentistry, electronics, medicine, glass, awards, decorations

title: The periodic table/Gold