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User:EGM6321.f10.team6.cook/hw7

= Problem 1 - Plot Legendre Functions for $$n=0,1,2 \text{ and } 3$$ =

Plot the Legendre polynomials and functions together, namely

Also plot the Legendre polynomials with the function for each $$n$$.

Observe $$P_n(\mu)$$ and $$Q_n(\mu)$$ as $$\scriptstyle \mu \rightarrow \pm 1$$ and observe the even and oddness.

Guess the value of

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

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

Legendre Function Plots
The solutions are plotted below. Note that if $$P_n(x)$$ is even $$Q_n(x)$$ is odd, and vice versa.

$$n=16$$
A higher order set is shown where Runge's phenomenon can be seen at the edges.

Observations
For $$\scriptstyle x \rightarrow \pm 1$$, $$P_n(x)\text{ and }Q_n(x)$$ both have the steepest slopes in the domain (-1 1).

For $$n$$ being even, $$P_n(x)$$ is even while $$Q_n(x)$$ is odd;

For $$n$$ being odd, $$P_n(x)$$ is odd while $$Q_n(x)$$ is even;

The integral is thus guessed to have a zero value,

as the product of an even and odd function will be odd, with a symmetric integral having zero value.

MATLAB Script
The following MATLAB script was used to generate and plot $$P_n(x),Q_n(x)$$. The plots were exported as an EPS file format and then converted to PNG using GIMP.

The expression for $$P_n(x)$$ was taken from Lecture 36 page 2 and $$Q_n(x)$$ was taken from  Lecture 39, page 1.

= Problem 2 - Show the Character of Even and Odd Functions =

From class note lecture 38_1

Problem Statement
Let

Required
Show that: 1. if the $$ \begin{align} \{g_i\} \end{align}$$ is odd, then $$ \begin{align} f \end{align}$$ is odd.

2. if the $$ \begin{align} \{g_i\} \end{align}$$ is even, then $$ \begin{align} f \end{align}$$ is even.

==Solution ==

1. The definition of an odd function is that $$\begin{align} f(x) = -f(-x) \end{align}$$.

If $$\begin{align}\sum_{i=0}^{n} g_i(x)\end{align}$$ is odd function,

 Odd functions, $$\begin{align} y = x^{2n+1}\end{align}$$

$$\begin{align} y= sin (nx) \end{align}$$

2. The definition of an even function is that $$\displaystyle f(x) = f(-x)$$.

If $$\begin{align}\sum_{i=0}^{n} g_i(x)\end{align}$$ is even function,  Even functions, $$\begin{align} y = x^{2n}\end{align}$$

$$\begin{align} y= cos (nx) \end{align}$$

Problem Statement
A solution to the Legendre equation may be written as :

Required
1. Determine if equation 3.1 is even or odd.

2. Find {a} such that the following two relations are equal for i=0 to 5.

Part 1
A function is considered to be even if it satisfies equation 3.2 and odd if it satisfies equation 3.3.

These conditions are essentially restrictions on the symmetry of a function. If a function is symmetric about the y-axis it will be even and if it is symmetric about the origin. Consider the sine and cosine.

It can be seen in equation 3.4 and 3.5 that the cosine and sine are even and odd functions, respectively.

In order to determine if equation 3.1 is even or odd we will apply the same test. Let us re-examine equation 3.1 to see if it can be simplified before we apply the test.

Most of the quantities in the equation are always positive, so they will have no impact on the evenness or oddness of the function. As a result we will neglect these terms and reduce the equation to the following:

Now, if we examine the argument we see that the quantity 2i is always an even number:

Recall that an even number subtracted from an even number always results in an even number, and an even number subtracted from an odd number always results in an odd number. Therefore, two cases that must be considered, namely when n is even and when n is odd. First let us examine when n is even:

We can see in equation 3.9 that if n is even that the sign of the argument will always be positive. Therefore, for n=even the Legendre polynomials of the first kind are even functions.

Now consider when n is odd:

It is clear in equation 3.10 that for n=odd the Legendre polynomials of the first kind are odd.

So, for n=even the Legendre polynomials are even and for n=odd the Legendre polynomials are odd.

Part 2
First let us write the terms for both sides of the equation.

Substituting in the functions for the Legendre polynomials we obtain:

We can now simplify and equate terms of like power in x to obtain the following equations:

This is a linear system of equations so we can solve for the values of $${{a}_{i}}$$ in many different ways, but for simplicity let’s rewrite equation 3.13 in matrix format:

Rewriting equation 3.14 as an augmented matrix we have:

We obtain the solution by putting the matrix in reduced row echelon form:

Substituting the following values:

This yields:

Given
Boundary condition

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

and expression of An

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

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

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

(2) Calculate non-zero

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

(a).(1)
Define $$\begin{align}x=\sin \theta \end{align}$$

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

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

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

$$\begin{align}=0\end{align}$$, i.e.,

$$\begin{align}{{A}_{2k+1}}=0\end{align}$$

Since

$$f(x)\in {{\Pi }_{6}}$$,a set of polynomials with order less than 6

and the linear independence of Legendre polynomials, we have

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

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

(a).(2)
We know that

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

Then

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

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

Evaluating the above definite integrals via Matlab, we may get

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

(b).(1)
Define $$\begin{align}x=\theta \end{align}$$

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

and

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

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

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

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

Since $$\begin{align}f(x)\end{align}$$ is not any kind of polynomials,

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

(b).(2)
We have

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

Given

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

and

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

Evaluating the above definite integrals via Matlab, we may get

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

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

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

=Problem 5 - Derivation of Inverse hyperbolic tangent function=

Given
hyperbolic tangent function

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

Find
Inverse hyperbolic tangent function as

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

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

Consider

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

and it follows that

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

Therefore

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

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

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

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

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

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

(1) Consider when n is odd, then

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

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

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

(2) Consider when n is even, then

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

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

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

Conclusion

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

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

Problem Statement
In the problem that formed the basis of the Legendre equation the distance between the centers of two attracting spheres was used to define the Newtonian Potential. The square of the distance between points is given as :

Required
Show equation 7.1.

Solution
First, recall that both points P and Q can be represented in both Cartesian and spherical coordinates:

Or putting equations 7.2 and 7.3 in a form that is more compatible with equation 7.1 (note: this is for the astronomical convention as specified in lecture):

To make the solution more tractable we will examine each component of the summation in equation 7.1:

Now the second term:

Finally the third term:

Now, if we combine all terms we obtain the following:

Equation 7.8 is relatively confusing so we will split it up again and examine terms independently. First we examine all terms with a common $$r_{Q}^{2}$$:

Now we will do the same for terms with $$r_{P}^{2}$$:

Now for the remaining terms:

Combining all simplified terms we obtain:

Which can be rearranged to the following form:

Problem Statement
The binomial theorem describes the algebraic expansion of powers of binomials. It can be represented generally as: :

Required
Derive the following variant of the binomial theorem:

Solution
First, let us re-examine equation 8.2 and consider the relation of the various terms with respect to the generalized binomial theorem in equation 8.1:

From equation we can clearly see the values to be used in equation 8.1, which are as follows:

We will now insert the above values back into equation 8.1, temporarily ignoring r choose k, and simplify a little:

Now since we want $${{x}^{i}}$$ and not $${{\left( -x \right)}^{i}}$$ so let’s factor out $${{\left( -1 \right)}^{i}}$$:

Now we turn our attention to r choose k:

Now, since there are i terms in the equation we must multiply by a factor of $${{\left( -2 \right)}^{i}}$$ in order to obtain a form similar to that given in equation 8.2:

We now recombine r choose k in equation 8.8 with the result of equation 8.6:

We now rearrange terms and cancel to obtain:

Now let’s expand the denominator to see how we can obtain the original equation:

Since there are i terms in the factorial and 2 is to the ith power we can push $${{2}^{i}}$$ into the factorial to obtain:

This is identical to 8.2.

Given
Reccurence relation of Legendre polynomials:

Legendre polynomials for $$n=0,1$$:

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

Solution
Eq. (9.1) can be written as:

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

Let n=1 in  (1), we get

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

Let n=2 in eq. (9.1), we get

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

Let n=3 in eq. (9.1), we get

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

Let n=4 in eq. (9.1), we get

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

Let n=5 in eq. (9.1), we get

Conclusions

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

= Problem 10 - Find Legendre Polynomials from $$\begin{align}P_3 \end{align}$$ to $$\begin{align}P_6 \end{align}$$ = From class note lecture 41_2 and class note lecture 40_3

Problem Statement
We have recurrence relations that

Binomial Theorem :

Thus,

Required
Continue power series expansion to find $$\begin{align}{P_3, P_4 , P_5 , P_6 } \end{align}$$, and compare the result to that obtained by a) (7) & (8) class note lecture 36_2

b) the result of HW 7.9

a)
We need the first 7 terms of the expansion

Using (10.5), we get

Therefore, we have confirmed two results are equivalent.

b)
Problem 9 results:

SAME!

title: User:EGM6321.f10.team6.cook/hw7


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User:Ramillav

Wikiversity user Ramillav has a bachelor degree in Public Administration and Governance, and is a present pursuing a degree in Law. He works as a reporter for an online business news publication.

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

title: User:Ramillav


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User:Thuvack/Mind-Logs/2010 December

December-05-2010:Wikiversity Summer
I've been in contact with a new user here at wikiversity. Our conversation built on my mute intentions to resume editing in the school of electrical engineering this summer (December is summer for us in South Africa). Any ways I realised how rusted my wiki editing techniques have become after such a long time. Baby steps for me all over again.

My day was made when I logged on to the wikiversity main page, only to be gret by Circuit Analysis as a featured page. Awesome!

Anyways this is my last week at work before the break. After that I intend to do more than lurking arround here. Thuvack | talk

title: User:Thuvack/Mind-Logs/2010 December


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User:Egm6321.f10.team2.dube/HW7

=Problem 7.1=

From the lecture slide Mtg 37-1

Given
The first four legendre polynomials are given as
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$$  \displaystyle P_{0}(x)=1 $$     (Eq 1.1)
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$$  \displaystyle P_{1}(x)=x $$     (Eq 1.2)
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$$  \displaystyle P_{2}(x)=\frac{1}{2}\left ( 3x^2-1 \right ) $$     (Eq 1.3)
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$$  \displaystyle P_{3}(x)=\frac{1}{2}\left ( 5x^3-3x \right ) $$     (Eq 1.4)
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$$  \displaystyle P_{4}(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} $$     (Eq 1.5)
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Find
Plot$$\displaystyle \left \{ P_{0},P_{1},P_{2},P_{3} \right \} $$,$$\displaystyle\left \{ Q_{0},Q_{1},Q_{2},Q_{3} \right \}$$,$$\displaystyle \left \{ P_{0},Q_{0} \right \}$$....$$\displaystyle\left \{ P_{4},Q_{4} \right \}$$ and observe evenness and oddness.

Solution
Example
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$$  \displaystyle sin(-\varphi )=-sin(\varphi ) $$     (Eq 1.6) The sine function is an odd function. Notice how the graph of the sine function is rotation symmetric about the origin as it goes into itself under a rotation of 180 degrees. 'Odd functions possess this type of antisymmetry.'
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$$  \displaystyle cos(-\varphi )=cos(\varphi ) $$     (Eq 1.7) The cosine function is an even function. Notice how the graph of the cosine function is reflection symmetric about the vertical axis throught 0 degrees. 'All even functions possess this type of symmetry.'
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Legendre Polynomials

As we see from the graph $$\displaystyle P_{0}(x)$$,$$\displaystyle P_{2}(x)$$,$$\displaystyle P_{4}(x)$$ are symmetric about the vertical axis. They all possess the characteristic of even functions. On tht other hand $$\displaystyle P_{1}(x)$$ and $$\displaystyle P_{3}(x)$$ are symmetric about the origin. They posess the characteristic of odd functions.

Legendre Functions

Controversely from polynomials, as we see from the graph $$\displaystyle Q_{0}(x)$$,$$\displaystyle Q_{2}(x)$$,$$\displaystyle Q_{4}(x)$$ are symmetric about the origin. So they are odd functions. And same with same approach $$\displaystyle Q_{1}(x)$$ and $$\displaystyle Q_{3}(x)$$ are symmetric about the vertical axis. They are even functions.

Functions Vs. Polynomials

The graph shows that whenever for the same indices$$\displaystyle P_{i}(x)$$ is even then $$\displaystyle Q_{i}(x)$$ is odd and controversely whenever $$\displaystyle P_{i}(x)$$ is odd then $$\displaystyle Q_{i}(x)$$ is even. So if we look at the multiplication of two functions,

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$$  \displaystyle F(x)=P_{i}(x)*Q_{i}(x) $$     (Eq 1.8)
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is always be odd.For i=1,2,3,4,.....

For i =2k
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$$  \displaystyle F(-x)=P_{i}(-x)*Q_{i}(-x)=P_{i}(x)*[-Q_{i}(x)] $$     (Eq 1.9) For i=2k+1
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$$  \displaystyle F(-x)=P_{i}(-x)*Q_{i}(-x)=[-P_{i}(x)]*Q_{i}(x) $$     (Eq 1.10)
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$$  \displaystyle \int P_{i}(x)Q_{i}(x)\Rightarrow even $$     (Eq 1.11)
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From the integral of multiplication of two functions we have even functions. And for integral boundries +1 and -1 These integrals will be same and their difference will be zero.
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$$  \displaystyle \int_{-1}^{1}P_{i}(x)Q_{i}(x)=R(1)-R(-1)=0 $$     (Eq 1.12) Where R is an even function.
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=Problem 7.2=

From the lecture slide Mtg 38-1

Given
In meeting 38-1 the equation is given as

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$$  \displaystyle f=\sum_{i}g_{i } $$     (Eq 2.1) Where f is a linear combination of g(x).
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Find
Show that Eq(2.1) is odd when gi is odd and Eg(2.1) is even when gi is even.

Solution
For even functions if we change the independent variable x into -x the relationship between two functions can be written by definition

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$$  \displaystyle f(x)=f(-x) $$     (Eq 2.2)
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And for odd functions if we change the independent variable x into -x the relationship for odd functions by definition

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$$  \displaystyle f(-x)=-f(x) $$     (Eq 2.3)
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So now we can expand the Eq(2.1) as

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$$  \displaystyle f(x)=g_{1}(x)+g_{2}(x)+g_{3}(x)+............. $$     (Eq 2.4)
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Then we can change the independent variable with its minus sign
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$$  \displaystyle f(-x)=g_{1}(-x)+g_{2}(-x)+g_{3}(-x)+............. $$     (Eq 2.5)
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Case 1
If $$\displaystyle g(x)$$ is even then by definition $$\displaystyle g(x)=g(-x)$$ so we can write
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$$  \displaystyle f(-x)=g_{1}(x)+g_{2}(x)+g_{3}(x)+............. $$     (Eq 2.6) Where the RHS of the function is the same as Eq(2.4). Hence f is an even function.
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Case 2
And If $$\displaystyle g(x)$$ is an odd function then $$\displaystyle g(x)=-g(x)$$. Substituting into Eq(2.2) yields;
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$$  \displaystyle f(-x)=-g_{1}(x)-g_{2}(x)-g_{3}(x)+............. $$     (Eq 2.7)
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Where the RHS of the functions is minus f. So $$\displaystyle f(x)=-f(x)$$ when g(x) is odd then f(x) is odd.

=Problem 7.3=

From the lecture slide Mtg 38-1

Given

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$$  \displaystyle P_{n}(\mu )=\sum_{i=0}^{[n/2]}(-1)^i\frac{(2n-2i)!\mu^{n-2i}}{2^ni!(n-i)!(n-2i)!} $$     (Eq 3.1) or
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$$  \displaystyle P_{n}(\mu )=\sum_{i=0}^{[n/2]}\frac{1,3,...(2n-2i-1)}{2^ii!(n-2i)!}(-1)^i\mu^{n-2i} $$     (Eq 3.2) where $$ \displaystyle [n/2] $$ = integer part of $$ \displaystyle n/2 $$ and
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$$  \displaystyle q(x)=\sum_{i=0}^{4}c_ix^i $$     (Eq 3.3) where $$ \displaystyle c_0=2, c_1=-5, c_2=-3, c_3=11, c_4=7, c_5=6 $$
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Find
Show that
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$$  \displaystyle P_{2k}(\mu) $$     (Eq 3.4) is even ($$ \displaystyle k=0,1,2... $$) and
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$$  \displaystyle P_{2k+1}(\mu) $$     (Eq 3.5) is odd.
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Find $$ \displaystyle {a_i} $$ such that $$ \displaystyle q(x)=\sum_{i=0}^{5}a_iP_i(x) $$

and plot $$ \displaystyle q=\sum_{i}^{}c_ix^i=\sum_{i}^{}a_iP_i $$

Solution
An equation is even when $$ \displaystyle f(x)=f(-x) $$.

Taking Equation 3.1 and plugging in $$ \displaystyle n=2k $$ we get:
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$$  \displaystyle P_{2k}(\mu )=\sum_{i=0}^{[2k/2]}(-1)^i\frac{(2(2k)-2i)!\mu^{2k-2i}}{2^{2k}i!(2k-i)!(2k-2i)!} $$     (Eq 3.6) Now let us plug in $$ \displaystyle \mu=x $$ and $$ \displaystyle \mu=-x $$ and compare $$ \displaystyle P_{2k}(x) $$ and $$ \displaystyle P_{2k}(-x) $$.
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$$  \displaystyle P_{2k}(x)=\sum_{i=0}^{[2k/2]}(-1)^i\frac{(2(2k)-2i)!x^{2k-2i}}{2^{2k}i!(2k-i)!(2k-2i)!} $$     (Eq 3.7) and
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$$  \displaystyle P_{2k}(-x)=\sum_{i=0}^{[2k/2]}(-1)^i\frac{(2(2k)-2i)!(-x)^{2k-2i}}{2^{2k}i!(2k-i)!(2k-2i)!} $$     (Eq 3.8) As you can see, $$ \displaystyle x $$ and $$ \displaystyle -x $$ are both getting raised to an even power, so they will both be equal and satisify the even clause, $$ \displaystyle f(x)=f(-x) $$.
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An equation is odd when $$ \displaystyle f(-x)=-f(x) $$.

Taking Equation 3.1 and plugging in $$ \displaystyle n=2k+1 $$ we get:
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$$  \displaystyle P_{2k+1}(\mu )=\sum_{i=0}^{[2k+1/2]}(-1)^i\frac{(2(2k+1)-2i)!\mu^{(2k+1)-2i}}{2^{(2k+1)}i!((2k+1)-i)!((2k+1)-2i)!} $$     (Eq 3.9) Now let us plug in $$ \displaystyle \mu=x $$ and $$ \displaystyle \mu=-x $$ and compare $$ \displaystyle P_{2k+1}(x) $$ and $$ \displaystyle P_{2k+1}(-x) $$.
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$$  \displaystyle P_{2k+1}(x)=\sum_{i=0}^{[2k+1/2]}(-1)^i\frac{(2(2k+1)-2i)!x^{(2k+1)-2i}}{2^{(2k+1)}i!((2k+1)-i)!((2k+1)-2i)!} $$     (Eq 3.10) and
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$$  \displaystyle P_{2k+1}(-x)=\sum_{i=0}^{[2k+1/2]}(-1)^i\frac{(2(2k+1)-2i)!(-x)^{(2k+1)-2i}}{2^{(2k+1)}i!((2k+1)-i)!((2k+1)-2i)!} $$     (Eq 3.11) As you can see, $$ \displaystyle P_{2k+1}(x) $$ and $$ \displaystyle P_{2k+1}(-x) $$ satisfy the odd clause, $$ \displaystyle f(-x)=-f(x) $$, therefore they are odd.
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We can see from (36-2) that
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$$ \displaystyle P_0(x)=1 $$ (Eq 3.12)
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$$ \displaystyle P_1(x)=x $$ (Eq 3.13)
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$$ \displaystyle P_2(x)=\frac{3x^2}{2}-\frac{1}{2} $$ (Eq 3.14)
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$$ \displaystyle P_3(x)=\frac{5x^3}{2}-\frac{3}{2}x $$ (Eq 3.15)
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$$ \displaystyle P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} $$    (Eq 3.16)
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$$ \displaystyle P_5(x)=\frac{63}{8}x^5-\frac{35}{4}x^3+\frac{15}{8}x $$    (Eq 3.17)
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We know that $$ \displaystyle P_i(x)=A*x_i $$ so:
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$$ \displaystyle \begin{bmatrix}P_0 \\P_1 \\ P_2 \\ P_3 \\ P_4 \\ P_5 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 & 0\\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} & 0\\ 0 & \frac{15}{8} & 0 & -\frac{35}{4} & 0 & \frac{63}{8} \end{bmatrix} \begin{bmatrix}1 \\ x \\ x^2 \\ x^3 \\ x^4 \\ x^5
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\end{bmatrix} $$    (Eq 3.18)
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We also know that $$ \displaystyle q(x)=a_iP(x)=(c_i)^Tx^i $$.

If $$ \displaystyle P_i(x)=Ax^i $$,

then

$$ \displaystyle (a_i)^TAx_i=(c_i)^Tx_i $$

and

$$ \displaystyle (a_i)^T=A^{-1}(c_i)^T $$.

We know from (Mtg38-3) that:
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$$ \displaystyle c_i=\begin{bmatrix}2 \\ -5 \\ -3 \\ 11 \\ 7 \\ 6 \end{bmatrix} $$ (Eq 3.19)
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and
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$$ \displaystyle A^{-1}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 & 0\\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 & 0\\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} & 0\\ 0 & \frac{27}{63} & 0 & \frac{28}{63} & 0 & \frac{8}{63} \end{bmatrix} $$    (Eq 3.20) Therefore:
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$$ \displaystyle (a_i)^T=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 & 0\\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 & 0\\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} & 0\\ 0 & \frac{27}{63} & 0 & \frac{28}{63} & 0 & \frac{8}{63} \end{bmatrix} \begin{bmatrix}2 \\ -5 \\ -3 \\ 11 \\ 7 \\ 6 \end{bmatrix}^T $$    (Eq 3.21)
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$$ \displaystyle a_i=\begin{bmatrix}\frac{12}{5} \\ \frac{146}{35} \\ 2 \\ \frac{106}{15} \\ \frac{8}{5} \\ \frac{16}{21} \end{bmatrix} $$ (Eq 3.22)
 This problem was solved by referencing [| Problem #11].
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= Problem 7.4 =

From the lecture slide Mtg 38-3

Given

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$$ \displaystyle

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

(Eq 4.1)
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$$ \displaystyle

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

(Eq 4.2)
 *

 * }

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

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

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

ii) Evaluate non-zero coefficients.

Part a

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

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

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

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

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

Then for evaluating non-zero coefficients.

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

Therefore we derive coefficients using WolframAlpha ,

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

(Eqs 4.3)
 *

 * }

Part b

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

Then,

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

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

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

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

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

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

(Eqs 4.4)
 *

 * }

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

From the lecture slide Mtg39-1

Given
The Legendre equation is,

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

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

 * $$\displaystyle \begin{align}

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

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

 * $$\displaystyle \begin{align}

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

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

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

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

$$ \displaystyle

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

(Eq 5.1)
 *

 * }

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

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

$$ \displaystyle

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

(Eq 5.2)
 *

 * }

The integrating factor is,

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

$$ \displaystyle

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

(Eq 5.3)
 *

 * }

Hence,

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

$$ \displaystyle

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

(Eq 5.4)
 *

 * }

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

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

$$ \displaystyle

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

(Eq 5.5)
 *

 * }

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

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

$$ \displaystyle

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

(Eq 5.6)
 *

 * }

On the other hand,

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

$$ \displaystyle

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

(Eq 5.7)
 *

 * }

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

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

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

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

(Eq 5.8)
 *

 * }

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

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

or

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

$$ \displaystyle

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

(Eq 5.9)
 *

 * }

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

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

$$ \displaystyle

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

(Eq 5.10)
 *

 * }

and then,

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

$$ \displaystyle

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

(Eq 5.11)
 *

 * }

By Eq 5.7,

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

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

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

(Eq 5.12)
 *

 * }

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

From the lecture slide Mtg 39-1

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

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

$$ \displaystyle

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

(Eq 6.1)
 *

 * }

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

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

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

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

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

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

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

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

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

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

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

the first tern of Eq 6.1 is even;

$$\displaystyle J$$ is always odd;

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

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

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

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

= Problem 7.7 Distance Between Two Spheres =

From the lecture slide Mtg 40-3

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

In Astronomical Convention we have,

 * $$\displaystyle \begin{align}

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

 * $$\displaystyle \begin{align}

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

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

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

Find
Show that,

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

Solution
Since,

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

$$ \displaystyle

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

(Eq 7.1)
 *

 * }

so we need to calculate each terms.

First Term:

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

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

Second Term:

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

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

Third Term:

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

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

Add all three terms together, then we get

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

$$ \displaystyle

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

(Eq 7.2)
 *

 * }

From trigonometric equations, we know that

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

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

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

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

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

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

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

$$ \displaystyle

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

(Eq 7.3)
 *

 * }

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

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

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

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

(Eq 7.4)
 *

 * }

= Problem 7.8: Derivation of a Specialized Case of the Binomial Series =

From the lecture slide Mtg 40-4

Given
Given the following binomial series equation:

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

$$ \displaystyle \left( x +y \right)^r=\sum_{k=0}^{\infty}\dbinom{r}{k}x^{r-k}y^{k} $$ (Eq 8.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where the binomial coefficient, "r choose k", is defined as follows:

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

$$ \displaystyle \dbinom{r}{k}=\frac{r(r-1)\cdots (r-k+1)}{k!} $$ (Eq 8.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

and r is a real number, $$\displaystyle r \in \mathbb{R}$$

Derive the following equation:

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

$$ \displaystyle \left(1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\alpha_{i}x^{i}
 * style="width:95%" |
 * style="width:95%" |

$$ (Eq 8.3)
 *

 * }

where

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

$$ \displaystyle \alpha_i = \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2i - 1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2i)} $$ (Eq 8.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
In comparing the left hand side (LHS) of Eq 8.3 to the LHS of Eq 8.1, the following values for r, x, and y are obtained:

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

$$ \displaystyle r=-\frac{1}{2}, \qquad x=1, \qquad y=(-x) $$ (Eq 8.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Plugging these values into the right hand side (RHS) of Eq. 8.1 yields the following:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left( 1-x \right)^{-\frac{1}{2}}=\sum_{k=0}^{\infty}\dbinom{-\frac{1}{2}}{k}1^{\left(-\frac{1}{2}-k\right)}(-x)^{k}

$$ (Eq 8.6)
 *

 * }

Note that for every k

$$ \displaystyle 1^{\left(-\frac{1}{2}-k\right)}=1,\quad \forall k $$

Next, the binomial coefficient in Eq 8.6 is expanded by using Eq 8.2 and the values in Eq 8.5 to yield the following:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left( 1-x \right)^{-\frac{1}{2}}=\sum_{k=0}^{\infty}\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)\cdots (-\frac{1}{2}-k+1)}{k!} (-x)^{k}=\sum_{k=0}^{\infty}\frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})\cdots (-k+\frac{1}{2})}{k!} (-x)^{k}

$$ (Eq 8.7)
 *

 * }

Next, factor out -1/2 from each term in the numerator and change the summation index k to i as follows:

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

$$ \displaystyle \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\frac{\left(-\frac{1}{2}\right)[1]\left(-\frac{1}{2}\right)[3]\left(-\frac{1}{2}\right)[5]\cdots \left(-\frac{1}{2}\right)[2i-1]}{i!} (-x)^{i}=\sum_{i=0}^{\infty}\left(\left(-\frac{1}{2}\right)(-x)\right)^i\frac{(1)(3)(5)\cdots (2i-1)}{i!} $$ (Eq 8.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq 8.8 can be simplified as follows:

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

$$ \displaystyle \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\frac{x^i}{2^i}\cdot \frac{(1)(3)(5)\cdots (2i-1)}{i!} $$ (Eq 8.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Note that the terms in the denominator can be rewritten as:
 * {| style="width:100%" border="0"

$$ \displaystyle (2^i)(i!)=\underbrace{(2\cdot 2\cdot 2\cdot \ldots)}_{i\ times}(1\cdot 2\cdot 3\cdot \ldots \cdot i)=2\cdot 4\cdot 6\cdot \ldots \cdot 2i $$ (Eq 8.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The RHS of Eq 8.10 can replace the denominator of Eq 8.9 resulting in the equation found in Eq 8.3:

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

$$ \displaystyle \begin{align} \left( 1-x \right)^{-\frac{1}{2}}=\sum_{i=0}^{\infty}\cdot \underbrace{\frac{(1)(3)(5)\cdots (2i-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2i}}_{\alpha_i}x^i
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\end{align}$$ (Eq 8.11)
 *

 * }

 * The following link was consulted after this problem was solved in an attempt to provide the most clear and comprehensive presentation :

http://en.wikiversity.org/w/index.php?title=User:Egm6321.f09.Team1/HW5&oldid=510466#Problem_6:_Binomial_Series

= Problem 7.9 Deriving Legendre polynominals using recurrence relation =

From the lecture slide Mtg 41-2

Given
The second recurrence relation of Legendre polynominals is,

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

$$ \displaystyle

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

(Eq 9.1)
 *

 * }

And Legendre polynominal of order 0 and 1 are,

 * $$\displaystyle \begin{align}

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

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

Solution
From Eq 9.1 we have,

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

$$ \displaystyle

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

(Eq 9.2)
 *

 * }

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

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

$$ \displaystyle

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

(Eqs 9.3)
 *

 * }

Solve Eqs 9.3 by sequence we have,

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

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

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

(Eqs 9.4)
 *

 * }

=Problem 7.10=

From the lecture slide Mtg 41-3

Given
To derive $$ \displaystyle P_{3}(x),P_{4}(x),P_{5}(x),P_{6}(x)$$ from the expansion of binomial expansion we have to start with giving generation function of legendre series. Generating function was derived in meeting 40 form Newtonian potential.

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

$$  \displaystyle \left ( 1-2\mu \rho +\rho ^{2} \right )^{-\frac{1}{2}}=\sum_{n=0}^{\infty }P_{n}(\mu )\rho ^{n} $$     (Eq 10.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle -2\mu \rho +\rho ^{2}=-x $$     (Eq 10.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (1-x)^{-\frac{1}{2}}=\sum_{n=0}^{\infty }\alpha _{i}x^{i} $$     (Eq 10.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now we have to use binmial expansion. But our power is not an integer. So we use 'r choose k' for general derivation.

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

$$  \displaystyle (x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k} $$     (Eq 10.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \binom{n}{k}=\frac{n(n-1)....(n-k+1)}{k!} $$     (Eq 10.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where our n designates real number.

Solution
Now we can use Eq(10.4) and Eq (10.5) to obtain expansion of generating function.

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

$$  \displaystyle \begin{align} &(1-(2\mu \rho +\rho ^{2}))^{-\frac{1}{2}}=1+(-\frac{1}{2})(\rho ^{2}-2\mu \rho )+ \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(\rho ^{2}-2\mu \rho ) ^{2}+\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{^3!}(\rho ^{2}-2\mu \rho )^{3}+\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2})}{4!}(\rho ^{2}-2\mu \rho )^{4}\\
 * style="width:95%" |
 * style="width:95%" |

&+\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2})(-\frac{9}{2})}{5!}(\rho ^{2}-2\mu \rho )^{5}+\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-\frac{7}{2})(-\frac{9}{2})(-\frac{13}{2})}{6!}(\rho ^{2}-2\mu \rho )^{6}\\ \end{align} $$     (Eq 10.6) Since we need sixth power of $$\displaystyle \rho $$ ,6th binomial expansion is enough for observing sixth P.
 *

 * }

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

$$  \displaystyle 1+\mu \rho -\frac{1}{2}\rho ^{2}+\frac{3}{2}\mu ^{2}\rho ^{2}+\frac{5}{2}\mu ^{3}\rho ^{3}-\frac{3}{2}\mu \rho ^{3}+\frac{3}{8}\rho ^{4}-\frac{15}{4}\mu ^{2}\rho ^{4}+\frac{35}{8}\mu ^{4}\rho ^{4}+\frac{15}{8}\mu \rho ^{5}-\frac{35}{4}\mu ^{3}\rho ^{5}+\frac{63}{8}\mu ^{5}\rho ^{5}-\frac{5}{16}\rho ^{6}+\frac{105}{16}\mu ^{2}\rho ^{6}-\frac{315}{16}\mu ^{4}\rho ^{4}+\frac{273}{16}\mu ^{6}\rho ^{6}.....
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 10.7)
 *

 * }

Eq(10.7) has also other higher order terms. Since the definition is given in Eq(10.1) we have to collect same orders togerher. Then we can observe six legendre polynomials.

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

$$  \displaystyle 1+\mu \rho +\frac{1}{2}(3\mu ^{2}-1)\rho ^{2}+\frac{1}{2}(5\mu ^{3}-3\mu) \rho ^{3}+(\frac{35}{8}\mu ^{4}-\frac{15}{4}\mu ^{2}+\frac{3}{8})\rho ^{4}+(\frac{63}{8}\mu ^{5} -\frac{35}{4}\mu ^{3}+\frac{15}{8}\mu)\rho ^{5}+(\frac{273}{16}\mu ^{6}-\frac{315}{16}\mu ^{4}+\frac{105}{16}\mu ^{2}-\frac{5}{16})\rho ^{6} $$     (Eq 10.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Manipulating Eq(10.1) abit more ,
 * {| style="width:100%" border="0"

$$  \displaystyle (1-2\mu \rho +\rho ^{2})^{-\frac{1}{2}}=P_{0}(\mu )\rho ^{0}+P_{1}(\mu )\rho ^{1}+P_{2}(\mu )\rho ^{2}+P_{3}(\mu )\rho ^{3}+P_{4}(\mu )\rho ^{4}+P_{5}(\mu )\rho ^{5}+P_{6}(\mu )\rho ^{6}.... $$     (Eq 10.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If we compare with Eq (10.8)and Eq(10.9) we can see that coefficients of different powers of $$ \displaystyle\rho$$'s are our legendre polynomials. These polynomials are the same as what we have found from Eq.(7)and(8) from the meeting 36-2. Where
 * {| style="width:100%" border="0"

$$  \displaystyle P_3(x) = \sum_{i=0}^1 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} =\frac{5}{2} x^3 -\frac{3}{2} x $$ (Eq 10.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle P_4(x) = \sum_{i=0}^2 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}=\frac{35}{8} x^4-\frac{15}{4} x^2+\frac{3}{8} $$     (Eq 10.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle P_5(x) = \sum_{i=0}^2 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}=\frac{63}{8} x^5 - \frac{35}{4} x^3 +\frac{15}{8} x $$ (Eq 10.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle P_6(x) = \sum_{i=0}^3 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}=\frac{273}{16} x^6 - \frac{315}{16} x^4 + \frac{105}{16} x^2 - \frac{5}{16} $$     (Eq 10.13) =Contributions Of Team Members=
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * Problem 1:    Solved by: Oztekin

Author: Oztekin

Proofread by: Dube,Zou

 * Problem 2:    Solved by: Oztekin

Author: Oztekin

Proofread by: Ismail,Zou

 * Problem 3:    Solved by: Dube

Author: Dube

Proofread by: Abudaram,Zou

 * Problem 4:    Solved by: Abudaram

Author: Zou

Proofread by: Ismail,Oztekin

 * Problem 5:    Solved by: Zou

Author: Zou

Proofread by: Dube,Oztekin

 * Problem 6:    Solved by: Zou

Author: Zou

Proofread by: Ismail,Abudaram

 * Problem 7:    Solved by: Ismail

Author: Zou

Proofread by: Zou,Abudaram

 * Problem 8:    Solved by: Zou

Author: Zou

Proofread by: Ismail

 * Problem 9:    Solved by: Patterson

Author: Patterson

Proofread by: Oztekin,Patterson

 * Problem 10:    Solved by: Oztekin

Author: Oztekin

Proofread by: Zou,Patterson

= References =

title: User:Egm6321.f10.team2.dube/HW7


text:
User talk:Ramillav

 Hello Ramillav, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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) 14:26, 5 December 2010 (UTC)

title: User talk:Ramillav


text:
Ideas in Geometry/Analytic Geometry

Analytic Geometry is covered in section 1.2.3 of our textbook.

When addressing this topic I asked three questions:
 * 1) What is analytic geometry?
 * 2) What is it used for?
 * 3) What do I need to know and be able to do for the homework and the exam?

Below are the responses I came up with to answer my questions.

Definition
Initially, I looked at this section and found no clear definition as to what Analytic Goemetry was actually about. I saw a nice little story about Sir Isaac Newton, a plane with a line and a curve, and a formula that seemed a little complicated, but no idea what we were talking about.

After a little research, my definition of Analytic Geometry is this: The study of geometric objects within the coordinate plane, or you could call it Coordinate Geometry.

This means that any geometric object which can be drawn on a coordinate plane is part of this realm. This includes all of your basic shapes - circles, quadrilaterals, triangles, etc.

Why Do We Use This Stuff?
Analytic Geometry can be used for many things in math:

Coordinates
The most basic form of this would be the coordinate plane. On which we can plot and manipulate points on a plane using a values of x and y.

Distance and Angles
By using our coordinate planes, we can determine distance between points along with the angles created by line in relationship to other lines.

This linked picture show us the relationship between angles, distance, and coordinate planes.

Equations of Curves
Analytic Geometry can also be used to plot curves of equations that use polynomials and other equations that require an input of x and which results in an output of y (x and y act would be coordinates for a point on a plane).

This picture shows up a curve in black with a line tangent to the curve:

Other Themes
Below are some other important themes of analytical geometry, some of which we need not worry about right now:
 * vector space
 * definition of the plane
 * the dot product, to get the angle of two vectors
 * the cross product, to get a perpendicular vector of two known vectors (and also their spatial volume)
 * intersection problems

Many of these problems involve a good deal of linear algebra and Lesson Four: Algebraic Geometry.

What Do We Need To Know For This Class?
As you can see, much of what we see in Analytic Geometry will be used throughout this class, but for the purposes of this particular chapter and section, we will focus on the concept of slope of a tangent line at a given point on a curve (In Calculus it is known as the concept of a derivative).

For our example in the book we are given a picture similar to this:

Where:

h is a small number which approaches 0 (For Example: 0.1, 0.01, 0.001...).

x is our input number.

f(x) is our function f with the x value plugged into it (represented by the green curve).

Is this making sense? If not think of the picture in terms of basic slope:

Slope (m) = Rise (Change in y)/Run (Change in x)

So this pictures really shows that our formula to solve this is: m = f(x+h) - f(h)/h Because: f(x+h) - f(h) is the change in the y value and h is the change in x.

Solving for the Slope of a Tangent Line Given an Equation and a Point
On page 18 of the .pdf textbook, you see the process for solving this kind of problem. To sum it up:
 * Set up the formula for slope of a tangent line with the appropriate values inserted for the variables.
 * Gather a selection of numbers approaching 0 (0.1, 0.01, 0.001...)
 * Substitute these numbers in for h.
 * All of the numbers should approach the slope of the tangent line.

The Calculus Derivative Trick
There is actually an easier process for to find the slope of a tangent line at a point on a curve.

So, let's look at the first example shown on page 18.

On this particular problem, we have taken the derivative of f(x). The derivative of this equation gives us another equation which will tell us the slope of the tangent line at any value of x. The derivative is found by multiplying the power of the exponent of a polynomial by the number in front of the varialbe, then reducing the exponent by 1. Here is the general formula.

Derivatives of powers: if

 * $$ f(x) = x^r\,$$,

where r is any real number, then

 * $$ f'(x) = rx^{r-1}\,$$,

wherever this function is defined.

So by this idea we arrive at :$$ f'(x) = 2x\,$$.

And after substituting 2 for x, our answer for this example the slope is 4.

Why does this work?
Let's look at an example when we are solving for an unknown x value (xo) for the same f(x) as in our example.

When this formula is broken down algebraically it becomes clear how this works out. If you were to look at the original method for our example and reduce the formula before substituting smaller values of h, you would notice that the process is similar to the one above.

DISCLAIMER
However, now that we have a nice method for solving for the slope of the tangent line, I must remind you that this is a shortcut. The key to this section for this class is to be able to explain why all of this makes sense in terms of the coordinate plan and values of x and h. Once you move on to Calculus, you will do much more work with this shortcut, but the key to doing well in MATH 119 is to understand these concepts in geometric terms.

title: Ideas in Geometry/Analytic Geometry


text: File:Egm6321.f10.team2.legendre.poly.png


text:
Talk:Replacement of phosphorus by arsenic in living organisms

I have created an XtraNormal cartoon animation of the script.

 * Arsenic-Loving Bacteria

Caprice 17:22, 5 December 2010 (UTC)

 * I love the way the computer says some of the worlds such as "eukaryotic". My reading of the text is at File:Bacteria.ogg. --JWSchmidt 20:09, 5 December 2010 (UTC)

The article content is biased if not trollish as it stands today. The NASA team didn't even purify the bacterial DNA correctly; attempts to reproduce the results showed that all of the field-standard purification procedures through the previous half century reduced the arsenic content of the sample to concentrations below the assay detection threshold. This is bad biochemistry and does not belong on Wikiversity.


text: File:Egm6321.f10.team2.Legendre.func.png


text:
Topic talk:Information security

This page appears unfocused. How about including something along the lines of the confidentiality/integrity/availability model? This is pretty much a classic and basic model in every area of information security that I've seen...

title: Topic talk:Information security


title: File:Egm6321.f10.team2.funcvspoly.png


text:
File:Proofs Involving Boxy Things.pdf

Licensing
for school project

title: File:Proofs Involving Boxy Things.pdf


text:
Ideas in Geometry/Pythagorean Theorem 2

By: Katy Anderson

Click on the picture below to make it bigger. Also, there are two pages, so once you click on it press "next page" to see the second page.

title: Ideas in Geometry/Pythagorean Theorem 2


text:
User:Egm6321.f10.team03.sigillo/Hwk7

=Problem 2: Summation of functions =

Given
Given is the definition of a function, which is the sum of two or more functions:
 * {| style="width:100%" border="0"

$$f(x)=\sum\limits_{i}{{{g}_{i}}\left( x \right)}$$
 * $$\,(Eq. 2.1) $$
 * style= |
 * }

Find
It has to be shown that if each gi(x) is an even function also f(x) is an even function and that if gi(x) is odd also f(x) is an odd function.

Solution
As first we need to look at how even and odd functions are defined.

Even functions: A function gi(x) is even if:
 * {| style="width:100%" border="0"

$${{g}_{i}}\left( x \right)={{g}_{i}}\left( -x \right)$$
 * $$\,(Eq. 2.2) $$
 * style= |
 * }

Odd functions: A function gi(x) is odd if:
 * {| style="width:100%" border="0"

$${{g}_{i}}\left( x \right)={-{g}_{i}}\left( -x \right)$$
 * $$\,(Eq. 2.3) $$
 * style= |
 * }

Now let's split the function g(x) into 3 parts: The part for all x<0, the part for x=0 and the part for all x>0, denoted by g-(x), g0(x) and g+(x) respectively. So that the function g(x) can now be written as:
 * {| style="width:100%" border="0"

$$g\left( x \right)={{g}_{-}}\left( x \right)+{{g}_{0}}\left( x \right)+{{g}_{+}}\left( x \right)$$
 * $$\,(Eq. 2.4) $$
 * style= |
 * }

The properties of the single functions are:
 * {| style="width:100%" border="0"

$$\begin{align} & {{g}_{-}}\left( x \right)=\left\{ _{0\quad for\quad x\ge 0}^{\mathbb{Z}\quad for\quad x<0} \right. \\ & {{g}_{0}}\left( x \right)=\left\{ _{0\quad for\quad x\ne 0}^{\mathbb{Z}\quad for\quad x=0} \right. \\ & {{g}_{+}}\left( x \right)=\left\{ _{0\quad for\quad x\le 0}^{\mathbb{Z}\quad for\quad x>0} \right. \\ \end{align}$$ THE EVEN CASE:
 * $$\,(Eqs. 2.5) $$
 * style= |
 * }

Now, accordingly to Eq. 2.2 and with the properties of Eqs. 2.5 for an even function gi(x) we can set g-(x) = g+(x) so that the function gi(x) can be rewritten as:
 * {| style="width:100%" border="0"

$${{g}_{i}}\left( x \right)={{g}_{i}}_{+}\left( x \right)+{{g}_{_{i}0}}\left( x \right)+{{g}_{i}}_{+}\left( x \right)$$ Summing n even functions together to build the function f(x) yields to:
 * $$\,(Eq. 2.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right) \\ & {{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right) \\ & \quad \vdots \quad \ \ +\quad \vdots \quad \ \ +\quad \vdots \quad \ \ \\ & \underline{{{g}_}\left( x \right)+{{g}_}\left( x \right)+{{g}_}\left( x \right)} \\ & {{f}_{+}}\left( x \right)+{{f}_{0}}\left( x \right)+{{f}_{+}}\left( x \right)\quad =f\left( x \right) \\ \end{align}$$ 'f(x)'' has the same form as Eq. 2.6 which we found to be even, therefore also the sum of all gi(x), which is f(x), has to be even!'''
 * $$\,(Eqs. 2.7) $$
 * style= |
 * }

THE ODD CASE:

Accordingly to Eq. 2.3 and with the properties of Eqs. 2.5 for an odd function gi(x) we can set g-(x) = -g-(x) so that the function gi(x) can be rewritten as:
 * {| style="width:100%" border="0"

$${{g}_{i}}\left( x \right)={{g}_{i}}_{-}\left( x \right)+{{g}_{_{i}0}}\left( x \right)-{{g}_{i}}_{-}\left( x \right)$$ Summing n odd functions together to build the function f(x) yields to:
 * $$\,(Eq. 2.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right) \\ & {{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right) \\ & \quad \vdots \quad \ \ +\quad \vdots \quad \ \ +\quad \vdots \quad \ \ \\ & \underline{{{g}_}\left( x \right)+{{g}_}\left( x \right)-{{g}_}\left( x \right)} \\ & {{f}_{-}}\left( x \right)+{{f}_{0}}\left( x \right)-{{f}_{-}}\left( x \right)\quad =f\left( x \right) \\ \end{align}$$ 'f(x)'' has the same form as Eq. 2.8 which we found to be odd, therefore also the sum of all gi(x), which is f(x), has to be odd!'''
 * $$\,(Eqs. 2.9) $$
 * style= |
 * }

Contribution

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

The solution showed above was derived from scratch by the author without consulting last years Homework!
 * style="width:45%; padding:1px; border:2px solid #CCFF33" |
 * style="width:45%; padding:1px; border:2px solid #CCFF33" |
 *
 * style= |
 * }

Solved and posted by Egm6321.f10.team03.sigillo 21:02, 5 December 2010 (UTC)

title: User:Egm6321.f10.team03.sigillo/Hwk7


text: File:Bacteria.ogg

title: File:Bacteria.ogg


title: File:P18ExampleDiverivative.jpg


title: File:P18ExampleDiverivativeProof.jpg


text:
Ideas in Geometry/Constructible Polygons and Impossibilities

Lesson Twenty-Nine: Constructible Polygons and Impossibilities

While a compass and a straightedge allow for almost any constructible polygons, there are several impossible constructions. There are three main problems that are known as the “impossibilities” which cannot be solved with a compass and a straightedge.

The first impossibility is doubling the cube.

The goal of doubling the constructed cube is to double its volume. To do this, we must construct the roots of the equation.

To demonstrate this, we can begin with a cube that has a volume of 1. In order to “double the cube” we must take v=1*2 (v=2). This means we must construct a cube with a volume of 2. For a cube with this volume, each side length must equal $$\sqrt[3]{2}.$$ We know that the cube root of any number is not constructible so therefore we cannot double the cube. This is true for all volumes of a cube because each will have $$x^3$$ in its equation.

The second impossibility is squaring the circle.

The goal of squaring a constructed circle is to construct a circle with a given radius and then construct a square with the same area as the original circle.

To demonstrate this, we can begin with a circle that has a radius of 1 unit. Area of a circle is A=πr^2 so we know that the area of this circle is A=π(1^2) or A=π. In order to square the circle, we must construct a square with an area of π. We know that π is not a constructible number so this square cannot be constructed. This is true for all constructed circles because each will have π in its radius.

The third impossibility is trisecting an angle.

The goal of this construction is to trisect an angle. Though bisecting an angle is simple enough, to trisect one is impossible. Trisecting an angle is defined as an impossibility because there is no possible way to do it with just a compass and a straightedge. Though the concept of trisecting an angle cannot be constructed, there are certain angles in which it is possible to trisect.

One of these angles is a 45° angle. Given an angle ABC that equals 45°, we can construct an equilateral triangle at point B. This gives us a new angle DBE that is equal to 60°. Now all we need to do is bisect the 60° angle, giving us one 15° angle (formed between the 45° angle and the 60° angle) and a 30° angle. The last step is to bisect the 30° angle giving us three 15° angles, thus successfully trisecting a 45° angle.


text:
1 Timothy (CCL)/Week 2

Assignments

 * Read 1 Timothy 1:3-11

title: 1 Timothy (CCL)/Week 2


text:
1 Timothy (CCL)/Week 3

Assignments

 * Read 1 Timothy 1:12-17

title: 1 Timothy (CCL)/Week 3


text:
1 Timothy (CCL)/Week 4

Assignments

 * Read 1 Timothy 1:18-20

title: 1 Timothy (CCL)/Week 4


text:
User:Djantonacci

Hello!

I am using Wikiversity to supplment my own independent study of French! I have just recently applied for the French Mentor program so I am looking forward to that :)

title: User:Djantonacci


text:
User talk:Djantonacci

 Hello Djantonacci, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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) 22:09, 5 December 2010 (UTC)

title: User talk:Djantonacci


text:
A crisis for institutions, opportunities for teachers

Based on the blog post of the same name, this presentation is for the Course Convener's Meeting at the University of Canberra, 6 December 2010.

A crisis?
"The Browne Report released on 12 October, and effectively rubber stamped in the savage public sector cuts announced yesterday, was simply the final nail in the coffin. Under the beguiling but misleading title Securing a Sustainable Future for Higher Education it effectively announced that university degrees are no longer considered a public good but a private investment."

- 30px

"The Higher Education Base Funding Review will build on the Government’s higher education reforms by establishing the principles for public investment in Australian higher education, the funding levels required for Australia to remain internationally competitive and the appropriate level of public and private contribution."

- 30px

"While a great deal of virtual ink has been spilled over the need to reform our schools and universities, I think we need to question how we manage education altogether. For it is manifest that the institution, the form in which we have managed education and society in general, has ultimately come to failure."

- 30px

"In the model of coercive capitalism proposed by Naomi Klein, the impact of crisis is used to justify a tightening and a quickening of the dominant neoliberal ideology. This ideology highlights the transfer control of the economy and state or public assets from public to the private sector under the belief that it will produce a more efficient [smaller, less regulatory] government and improve economic outputs. "

- 30px

"‘Unfreezing’ (Lewin, 1951) is an organizational term that has come to mean many things. First, it means that for change to take place members of the organization must see not only a need for change but also an urgent reason to change. Slaughter and Leslie have made the case for urgency by showing us that, out of financial necessity, higher education is already undergoing a quiet revolution that is having some unintended consequences. Second, Lewin’s concept of unfreezing warns us that attempts to change without addressing an organization’s cultures and values will fail in the long run."

- 30px

An opportunity?
"In particular we might now revisit the critical work on the neoliberal university, the student as consumer and the marketisation of HE, in order to critique and negate the path that we are pushed towards. This work identifies the types of controlled, economically-driven, anti-humanist organisations that will possibly emerge, and the ways in which oppositional, alternative, meaningful social change might be realised."

- 30px

I've been trying to think inside out from the institution for 4 or 5 years now. How might those who presently work inside the institutions, work in such a way that makes ready, or feeds opportunities that are developing outside the institutions? I reckon HE teachers ought to beat the policy makers and bosses at their own game, and start setting up for independent, contract work. This means taking full ownership of the units you teach, and start running then independently from the institution, contracted back into the institutions, while they remain.

 * 1) Start with your unit outline.
 * 2) Copy it to Wikiversity and chuck out all that guff, make it fun, engaging, and readable. Link to the guffy version if legal requires it. If your Institution has draconian IP policy, work to change it, or change the unit outline enough to qualify as original work.
 * 3) Start networking online around the subjects of your units, update the wiki as new resources and ideas come your way.
 * 4) Refine your Wikiversity entry and start editing all related Wikipedia articles so that links are prevalent from there to your Wikiversity course.
 * 5) Set up a blog/website for your units, or create a page on your existing blog/website for the units you are developing to teach and assess.
 * 6) Ditch the learning management system and any other platform the institution prescribes (such as email or lecture recording facilities), get it all out in the open, on commonly available and tried and true services, establish your own online identity, use the links, RSS and embed codes to quickly populate an institution's LMS if you must.
 * 7) Set up a Google docs spreadsheet with all the data the institution needs for auditing, such as attendance, contact details, demographics, participation and completion rates, feedback, etc.
 * 8) Work out how much it costs your target universities to run the unit, then work out how you can teach and assess that unit for less.
 * 9) Find other teachers going independent, try to build a network who together might be able to offer units for the better half of a degree, using this approach.
 * 10) Offer all the universities in the world your service, outlining the cost benefit analysis you've done for them (like Google Docs did).  All you need from them is assurance they will give your graduates the rubber stamp on your assessed and moderated say so.
 * 11) Negotiate a 3-4 year contract with each university, to make sure you have time to refine and develop, and so that degree hunting students have consistency in you.
 * 12) Run your unit open access, inviting non enrolled students to participate, offering them post study assessment (Recognition of Prior Learning) should they ever be wealthy enough to pay for the paper.
 * 13) Author a textbook for the unit on Wikibooks, and desktop publish it to Lulu.com, charge a small royalty for the printed version.
 * 14) Create quirky merchandise for your unit, again charging small royalties.
 * 15) Set up a donation widget on your unit website to take donations from anyone who shows their love.
 * 16) Make sure your running costs include time for research and development. I reckon 5-10 hours for every hour of teaching and assessment as a rough guide.
 * 17) Get smart with your work, think about ways to use community projects like Wikipedia's Featured Article initiative and other ideas, or consider pay it forward assignments, to help with assessment and other workload challenges.

None of this necessarily preserves the work of teaching subjects that are in low demand unfortunately. In this regard, I would look to associations and other outside bodies to subsidise the running costs of your units with sponsorship. Whack a heritage order on ancient Greek studies etc. Try to find sustainable online markets for your niche service.

Examples
"There is a chance that Wikiversity will become the Internet’s free university just as Wikipedia is the free encyclopedia on the Internet."

- 30px

 * Introduction to Open Education
 * Composing Free and Open Online Educational Resources
 * Facilitating Online

 * Otago Polytechnic's 2009 open education showcase
 * The Khan Academy
 * Michael Sandel's teachings on Justice


text: Motivation and emotion/Textbook/Motivation/Sexual motivation/Want to know your sex IQ?


text: Ideas in Geometry/Fibonacci Numbers

title: Ideas in Geometry/Fibonacci Numbers