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User:Egm6321.f10.team2.oztekin/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)
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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,

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

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

(Eqs 4.4)
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= 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,

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$$ \displaystyle

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

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

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$$ \displaystyle

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

(Eq 5.2)
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The integrating factor is,

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$$ \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)
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Hence,

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$$ \displaystyle

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

(Eq 5.4)
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Then we can derive $$\displaystyle y$$,

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

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$$ \displaystyle

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

(Eq 5.6)
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On the other hand,

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$$ \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.oztekin/HW7


text: File:CV-multi-inlet-outlet.svg

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

title: File:CV-multi-inlet-outlet.svg


text: File:Streamtube-mass-balance.svg

title: File:Streamtube-mass-balance.svg


text: Motivation and emotion/Textbook/Emotion/Sport/Title


text:
Introduction to Sport Biomechanics/Resources

A range of resources to help with understanding the topics, completing the assignments, and submitting for assessment. Resources here are open educational resources, meaning the copyrights are free for reuse in any context.


text:
Talk:Motivation and emotion/Book/2010/Emotional development in children

-- Jtneill - Talk - c 07:49, 14 December 2010 (UTC)


text:
Motivation and emotion/Book/2010/Sex offender motivation/Jeffrey Dahmer videos

The following YouTube videos are a sequence of interviews with Jeffrey Dahmer conducted by Stone Phillips a Dateline NBC reporter. The first 3 parts (out of 6) are linked below.

Jeffrey Dahmer Interview Part I

Jeffrey Dahmer Interview Part II

Jeffrey Dahmer Interview Part III


text: Unconscious motivation

title: Unconscious motivation


text:
Leukotrienes

Leukotrienes are products of the enzyme arachidonate 5-lipoxygenase (5-LO). They are fatty molecules that contribute to inflammation in asthma and allergic rhinitis. Leukotrienes are typically biosynthesized in concert with histamine. As the name suggests, arachidonic acid (AA) is a substrate of arachidonate 5-lipoxygenase. AA is an omega-6 fatty acid. Eicosapentaenoic acid (EPA) is also a 5-LO substrate. EPA is an omega-3 fatty acid. The 4-series leukotrienes (LTB4, LTC4, LTD4, LTE4) are synthesized from AA. They are generally proinflammatory. The 5-series leukotrienes (LTB5, LTC5, LTD5, LTE5) are synthesized from EPA. They are anti-inflammatory. Leukotrienes are released from mast cells as part of the inflammation response. They attract white blood cells to sites of tissue damage and cause smooth muscles to contract.

quiz
{Leukotrienes are products of the enzyme ____________. - arachidonate 5-lipoxygenase (5-LO) + histamine - omega-6 fatty acid - Eicosapentaenoic acid (EPA)
 * type="[]"}
 * Correct! This enzyme is responsible for the production of leukotrienes.
 * Incorrect. Histamine is biosynthesized in concert with leukotrienes, but it is not the enzyme responsible for their production.
 * Incorrect. Omega-6 fatty acid is a substrate, not the enzyme.
 * Incorrect. EPA is a substrate of arachidonate 5-lipoxygenase, not the enzyme itself.

{As the name suggests, ____________ is a substrate of arachidonate 5-lipoxygenase. - Eicosapentaenoic acid (EPA) + arachidonic acid (AA) - histamine - 4-series leukotrienes
 * type="[]"}
 * Incorrect. While EPA is a substrate, it's not the one suggested by the name.
 * Correct! The name "arachidonate 5-lipoxygenase" suggests arachidonic acid as the substrate.
 * Incorrect. Histamine is not a substrate of arachidonate 5-lipoxygenase.
 * Incorrect. The 4-series leukotrienes are products, not substrates.

{AA is an ____________. + omega-6 fatty acid - omega-3 fatty acid - enzyme - product
 * type="[]"}
 * Correct! Arachidonic acid (AA) is an omega-6 fatty acid.
 * Incorrect. AA is not an omega-3 fatty acid.
 * Incorrect. AA is a substrate, not an enzyme.
 * Incorrect. AA is a substrate, not a product.

{EPA is an ____________. - omega-6 fatty acid + omega-3 fatty acid - enzyme - product
 * type="[]"}
 * Incorrect. EPA is not an omega-6 fatty acid.
 * Correct! Eicosapentaenoic acid (EPA) is an omega-3 fatty acid.
 * Incorrect. EPA is a substrate, not an enzyme.
 * Incorrect. EPA is a substrate, not a product.

{The 4-series leukotrienes are synthesized from ____________ and are generally ____________. + AA | proinflammatory - EPA | anti-inflammatory - AA | anti-inflammatory - EPA | proinflammatory
 * type="[]"}
 * Correct! The 4-series leukotrienes are synthesized from AA and are proinflammatory.
 * Incorrect. The 5-series leukotrienes are synthesized from EPA and are anti-inflammatory.
 * Incorrect. The 4-series leukotrienes synthesized from AA are proinflammatory.
 * Incorrect. The 5-series leukotrienes synthesized from EPA are anti-inflammatory.

{Leukotrienes are released from ____________ as part of the inflammation response. - white blood cells + mast cells - smooth muscles - sites of tissue damage
 * type="[]"}
 * Incorrect. While leukotrienes attract white blood cells, they are not released from them.
 * Correct! Leukotrienes are released from mast cells during the inflammation response.
 * Incorrect. Leukotrienes cause smooth muscles to contract, but they are not released from them.
 * Incorrect. Leukotrienes attract white blood cells to sites of tissue damage, but they are not released from these sites.

title: Leukotrienes


text:
User talk:Momemeca

 Hello Momemeca, 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) 04:24, 8 December 2010 (UTC)

title: User talk:Momemeca


text:
Talk:Motivation and emotion/Book/2010/Sex offender motivation

-- Jtneill - Talk - c 04:44, 14 December 2010 (UTC)


text:
Atherosclerosis

Atherosclerosis is a disease characterized by the buildup of plaque inside the arteries. Plaque is made up of fat, cholesterol, calcium, and other substances found in the blood. Coronary artery disease (CAD), is atherosclerosis of the coronary arteries. The cause of atherosclerosis is not known. Smoking, diet, physical activity, and genes influence the risk of atherosclerosis. Atherogenic infections have been associated with atherosclerosis in animals.

Additional information

 * LysoSENS - Reversing Heart Disease (video)
 * LysoSENS is a project by SENS Foundation (nonprofit) to clear out intracellular junk so as to potentially reduce the occurrence of atherosclerosis. The project is part of Strategies for Engineered Negligible Senescence.

Readings

 * LysoSENS: Clearing waste accumulations out of cells - a description of theories related to cellular waste accumulations that relates to atherosclerosis.

title: Atherosclerosis


text:
User talk:Norton~enwikiversity/Wikistution

It seems that a more "Wiki" style of editing would be to simply delete portions of the Wikistution that have been superseded, rather than use bracketed italics (e.g., [Prohibition])--Keith1 15:05, 3 April 2011 (UTC)

Ideas for improving this project
This project must be deleted for a number of reasons:
 * 1) It is "dead" in that almost nobody reads it and nobody contributes to it.
 * 2) Posting a paraphrased version of the US constitution harms the reputation of Wikipedia.
 * 3) Readers might be misinformed into thinking this is the actual US constitution

There is a possibility (but not a certainty) that the Wikiversity community MIGHT accept a short stub describing the idea of having students collaboratively create a document like this (though I personally see no need for such a document --Guy vandegrift (discuss • contribs) 20:46, 6 December 2022 (UTC))

XXIX
This doesn't seem feasible. Keith1 04:35, 8 December 2010 (UTC)

XXX
This provision seeks to eliminate the phenomenon of uninformed citizens voting strictly by party affiliation, name recognition, or other superficial reasons. Keith1 04:46, 8 December 2010 (UTC)

This seems like two totally different amendments - maybe it should be split in the future. Keith1 04:46, 8 December 2010 (UTC)


text:
User talk:Egm6936.f09/Kolmogorov scales

= Old discussion page =

= Issues to resolve (reverse chronological order) =

Lecture notes on probability
[[media:Probability.djvu|Probability concepts and notation (djvu)]]

Uncertainty quantification (wiki)

Probability distributions
While working on the | probability density function (PDF) and the | mean value of a random variable, we were interested in different probability distributions. Here, we study some of the | common probability distributions and focus on the formulas, the history, the applications, and the person who discovered it. Some key points and references are:

AI Access: Web site that includes a glossary, definitions, and animations

NIST Dataplot: It is a free, public-domain, multi-platform software system for plotting, statistical analysis, and non-linear modeling.

Sean Meyn

Jeff Miller: Web page that includes Earliest Known Uses of Some of the Words of Mathematics among other various very interesting pages

Richard Tweedie, The Probability Web (Berkeley), Probability distributions

Exponential: When a specific event repeats itself at some amount of time such as cars crossing an intersection, lifetime of electronic components, number of customers at the grocery check-out, amount of money customers spend in a grocery store, an exponential distribution is used. Exponential distribution is widely used in the field o reliability, i.e., lifetime of a product, such as a car battery. Exponential distribution does not exhibit a memory of the event. For example, if the probability of the lifetime of a first battery to be at least t years is P then the probability of a second battery to last at least t years is again P. Some real data related to the faults found in a software system can be found in the book by Cook and Lawless (2007, Appendix D, Table D3, p.368) ,

Cauchy

Gamma

Log-normal

Weibull distribution
Moved to the issues resolved section.

Also as a new article at Weibull distribution.

Cauchy distribution
It is a special case of | student’s t-distribution. Prof. Loc Vu-Quoc from University of Florida explains the Cauchy distribution in detail in his Numerical Methods course (see Lecture 23 p.2 and Lecture 24). The single-slit diffraction application of the Cauchy distribution is analyzed in detail in Lecture 24.

Definition
The PDF of the Cauchy distribution located at $$\displaystyle x_0 $$ with half width at half maximum height $$\displaystyle \gamma $$ is given as (Pope, 2000, p.53, Eq.3.79, Prof. Loc Vu-Quoc, Numerical Methods Spring 2011 Lecture 23, p.2)

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

$$  \displaystyle f(x; x_0,\gamma) = \frac{1}{\pi} \frac{\gamma} { \left(x - x_0 \right)^2 + \gamma^2} $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

A simulation of the Cauchy distribution can be performed using AI Access. An application of the simulation to the single-slit diffraction is shown in Lecture 24.

The blindfolded archer’s score
Mandelbrot and Hudson (2004, p.37) explains Cauchy distribution with a story called the blindfolded archer’s score. An archer stands in front of a target on an infinitely long wall. He starts shooting in random directions.

On Cauchy distribution in finance
Mandelbrot wrote a book on fractals in finance, in which there was long discussion on the Cauchy distribution, which has extensive applications in finance. The Cauchy distribution is an example of an "L-stable" distribution, introduced by Paul Le'vy, a math professor at the Ecole Polytechnique, France.

 NOTE:  In a quotation, to have the source listed in a separate line, and right-justified, see the Template:Quotation of mediawiki.

discussion 2011.07.26: turbulence in pipe flow
Kerem Uguz attended the Bifurcations and Instabilities in Fluid Dynamics (BIFD2011) Fourth International Symposium at Barcelona, Spain (July 18-21, 2011). Two of the invited talks were about turbulence in pipe flow. It was interesting to see that there is still a scientific curiosity to pressure-driven pipe flow 128 years after the original papers by Reynolds (1883a, 1883b). In fact there is a very recent paper published in Science (2011) by Björn Hof who was one of the invited speakers.

Why are people still interested in pressure-driven pipe flow?
First of all, after Reynolds’ papers  in pipe flow, both theoretically and experimentally, there are still uncertainties in determining the critical Reynolds number for which the flow becomes turbulent (See Darbyshire and Mullin (1995) ) and also setting efficient theoretical and experimental ways to understand turbulence. The question to answer is “At what Reynolds number the flow is persistently turbulent (sustained turbulence) or it is ultimately laminar?”. It is important to resolve this problem. It was shown that the transition to turbulence depends on the type and magnitude of the perturbations but a detailed understanding is still missing. Second, understanding turbulence might open doors to new theoretical approaches. Third, understanding turbulence gives the chance to control it. Consequently, a turbulent flow can be relaminarized, which in turn reduces the control costs.

The contents of the talks
The title of Björn Hof’s talk was “The Onset of Turbulence in Pipe Flow”. The key points of Hof were

 * Pipe flow is linearly stable
 * Flow is intermittent
 * Puffs occur and decay in the flow

His approach is to separate relevant processes: decay of turbulent puffs and spreading of turbulence. When the Reynolds number reaches a critical value the decay rate of puffs are of similar magnitude to the branching of the puffs. So at that critical Reynolds number, even though the puffs that occur in the pipe decay with time some new puffs occur in the pipe. So, sustained turbulence is obtained. A similar argument is also present in the paper by Darbyshire and Mullin (1995, check the quotation below). Isolated regions of turbulence (which is represented with puffs) are transient. They experimentally determined that critical Reynolds number as $$ \displaystyle 2040 $$. They showed that this critical Reynolds number is independent of the disturbance introduced into the system. The details are given in their Science paper (2011).

Bruno Eckhardt was the second invited speaker. He is the author of the introduction to the Theme Issue ‘Turbulence transition in pipe flow: 125th anniversary of the publication of Reynolds' published in the Philosophical Transactions of the Royal Society A. 2009 was the 125th anniversary of the publication of Reynolds' papers.

The title of Bruno Eckhardt’s talk was “Turbulence transition in shear flows: what can we learn from pipe flow?”. Eckhardt groups different critical Reynolds numbers:
 * Reynolds number near zero: flow is laminar
 * $$ \displaystyle Re_E = 81 $$: Energy stability of pipe flow. Joseph and Carmi (1969) determined that up to Reynolds number of $$ \displaystyle 81 $$ the energy content of perturbations decay monotonically.
 * $$ \displaystyle Re_{ECS} = 773 $$: Turbulence needs three dimensional structures as a scaffold that supports the spatio-temporal turbulent dynamics. These are called Exact Coherent Structures (ECS), which require prominent downstream vortices.
 * $$ \displaystyle Re_{turb} \approx 1600 $$: Darbyshire and Mullin (1995) determined experimentally a Reynolds number around $$ \displaystyle 1600 $$. They found that this number depends on how you put your disturbance. You need to be very careful near transition region. It is very difficult to reproduce the experiments in that region even with same initial conditions.

 * $$ \displaystyle Re_{space}=2040$$: Turbulence is transient. Spatio-temporal coupling matters.

Questions

 * What does “energy content of perturbations decay monotonically” mean?
 * How to trigger turbulence? This is a paper for which both Bruno Eckhardt and Hof contributed.
 * Localized coherent structures (edge state)

discussion 2011.06.28
[[media:vql.random.process.djvu|Random processes]]

See also Category:Egm6936.f10.

discussion on 2011.03.01
The aim is to solve Exercise 3.16 in (Pope, 2000, p.59), which is the proof of Eq.(7).

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

$$  \displaystyle \left(     \frac{u_1}   {\langle u_1^2   \rangle^{1/2}} - \frac{u_2}   {\langle u_2^2   \rangle^{1/2}} \right)^2 = \frac{u_1^2} {\langle u_1^2 \rangle} - \frac{2u_1 u_2} { \langle u_1^2 \rangle^{1/2} \langle u_2^2 \rangle^{1/2} } + \frac{u_2^2} {\langle u_2^2 \rangle} $$.     (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

As the left hand side of Eq.(1) is greater or equal to zero, so is the right hand side. Therefore,

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

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

\frac{2u_1 u_2} { \underbrace{ \langle u_1^2 \rangle^{1/2} }_{{u_1}^\prime } \underbrace{\langle u_2^2 \rangle^{1/2}}_{{u_2}^\prime }

} \le \frac{u_1^2} {({u_1}^\prime)^2} + \frac{u_2^2} {({u_2}^\prime)^2 }

$$.     (2)
 *
 * }

Taking the mean of Eq.(2) yields

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

$$  \displaystyle \underbrace{ \frac{ \langle u_1 u_2 \rangle }  { {u_1}^\prime {u_2}^\prime } } _ {\rm{auto \; correlation}} \le \frac{1}{2} \left [
 * style="width:95%" |
 * style="width:95%" |

\frac{ ({u_1}^\prime)^2 } { ({u_1}^\prime)^2 } + \frac{ ({u_2}^\prime)^2 } { ({u_2}^\prime)^2 } \right ] = 1 $$.     (3)
 *
 * }

Now, for the plus sign of the term, i.e.,

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

$$  \displaystyle \left(     \frac{u_1}   {\langle u_1^2   \rangle^{1/2}} + \frac{u_2}   {\langle u_2^2   \rangle^{1/2}} \right)^2 = \frac{u_1^2} {\langle u_1^2 \rangle} + \frac{2u_1 u_2} { \langle u_1^2 \rangle^{1/2} \langle u_2^2 \rangle^{1/2} } + \frac{u_2^2} {\langle u_2^2 \rangle} $$.     (4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Similar to the treatment in Eq.(2) yields

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

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

\frac{2u_1 u_2} { {u_1}^\prime {u_2}^\prime

} \ge - \frac{u_1^2} {({u_1}^\prime)^2} + \frac{u_2^2} {({u_2}^\prime)^2 }

$$.     (5)
 *
 * }

Taking the mean of Eq.(5) yields

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

$$  \displaystyle \underbrace{ \frac{ \langle u_1 u_2 \rangle }  { {u_1}^\prime {u_2}^\prime } } _ {\rm{auto \; correlation}} \ge -\frac{1}{2} \left [
 * style="width:95%" |
 * style="width:95%" |

\frac{ ({u_1}^\prime)^2 } { ({u_1}^\prime)^2 } + \frac{ ({u_2}^\prime)^2 } { ({u_2}^\prime)^2 } \right ] = -1 $$.     (6)
 *
 * }

Combining Eq.(3) and Eq.(6) gives

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

$$  \displaystyle -1 \le \underbrace{ \frac{ \langle u_1 u_2 \rangle }  { {u_1}^\prime {u_2}^\prime } } _ {\rm{auto \; correlation}} \le 1 $$ (7)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

With Eq.(7) ends the derivation of Eq.(3.94) in (Pope, 2000, p.57) and also the solution to the exercise (3.16) on page59.

For the proof of Eq.(7), Pope suggests to use the Cauchy-Schwarz Equation, which is

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

$$  \displaystyle = \|x\| \|y\| \underbrace{ | \cos \angle (x,y) |}_{\le 1} $$.     (8)
 * style="width:95%" |
 * style="width:95%" |
 * \langle x,y\rangle|
 *

 * }

or

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

$$  \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\le \|x\| \|y\|
 * \langle x,y\rangle|

$$ (9)
 *

 * }

As the proof of Eq.(7) does not include the Cauchy-Schwarz inequality, probably Pope misused the name, Cauchy-Schwarz (Pope, 2000, p.59).

= Issues resolved (chronological order) =

Weibull distribution
Also as a new article at Weibull distribution.

Introduction
Here, we present some real experimental data that fit the Weibull distribution, which is a more general form of the exponential distribution. The data can be found in a spreadsheet format, here. The probability density function (pdf) of the Weibull distribution is given as

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

$$  \displaystyle F_X (x) = \Big[  1 - exp \Big( - \frac{x}{p_2} \Big)^{p_1} \Big] $$.     (A1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To fit to the experimental data, a generalization of the Weibull distribution is applied by introducing a third coefficient as

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

$$  \displaystyle F_X (x) = p_3 \Big[ 1 - exp \Big( - \frac{x}{p_2} \Big)^{p_1} \Big] $$.     (A2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

When a nonlinear fitting is applied, the parameters are found as

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

$$  \displaystyle p = [1.6407, 740.2473, 933.7323] $$.     (A3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

With the above parameters, the distribution looks like

FIG 1: The distribution and the data points

When we take $$ \displaystyle p(3) = 1 $$ as in the usual Weibull distribution, we obtain the pdf as

FIG 2. The Weibull distribution for $$ \displaystyle p(3) = 1 $$

The story about the experimental data
The data was first given in the paper by Dalal and McIntosh (1994) and reproduced in the book by Cook and Lawless (2007, Appendix D, Table D3, p.368). The example comes from a debugging process of a software system with around seven million noncommentary source lines. Cumulative staff days, $$ \displaystyle t $$, the cumulative number of faults detected, $$ \displaystyle N(t) $$, and the cumulative number of lines of code added, $$ \displaystyle C(t) $$, were recorded. Here is a table of selected data (1 data/10 data).

As seen from the table, over $$ \displaystyle 1,300 $$ total staff days were spent for the debugging process, $$ \displaystyle 870 $$ faults were detected, and more than $$ \displaystyle 342,000 $$ lines were added. The reason to fit a model to the data is to guess when to stop testing the code. The selected data and the full set is given in the following figure.

FIG 3 Selected and the full set data

At the beginning, the total number of faults found increases very slowly, then it rises quickly, and finally reaches a plateau with time. As the data represents an event that repeats itself at some amount of time and also from the shape of the curve, one may think of fitting an exponential distribution or a more general distribution, e.g., Weibull distribution that would also fit the early times.

Curve fitting with actual data
The general form of the Weibull distribution with 3 parameters is given in Eq.(A2). To fit this curve to the existing data, one may use a nonlinear least square analysis. First, an initial guess for the parameters is required. Then, the estimation of the coefficients is done with a nonlinear fit. If the initial guess is very far from the actual values, it is not possible to obtain converged values for the parameters. For example, an initial guess of

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

$$  \displaystyle p = [100, 10, 1] $$.     (A4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

does not yield converged estimates of the parameters. The output of Matlab is:

Warning: The Jacobian at the solution is ill-conditioned, and some model parameters may not be estimated well (they are not identifiable). Use caution in making predictions.

Parameter estimation
So, it is important to understand the role of each parameter. Here, we present two graphs where $$ \displaystyle p_3 = 1 $$ for both cases.

FIG 4. Effect of parameters when $$ \displaystyle p_3 = 1 $$, a) Study of $$ \displaystyle p_1 $$, b) Study of $$ \displaystyle p_2 $$.

The first figure aims to study the effect of the first parameter, $$ \displaystyle p_1 $$. Note that the exponential distribution is recovered when $$ \displaystyle p_1 = 1 $$. For $$ \displaystyle p_1>1 $$, the curve increases slowly at the beginning, which is the case in our problem (Figs 1-3) and $$ \displaystyle p_1 $$ turned out to be $$ \displaystyle 1.64 $$. The parameter $$ \displaystyle p_2 $$ shows how fast the curve increases after it passes through this slow increase region. Finally, a good initial guess for the parameter $$ \displaystyle p_3 $$ is the maximum of the data $$ \displaystyle N(t) $$, i.e. $$ \displaystyle 870 $$.

An alternative method for parameter estimation
Another way to estimate the parameters $$ \displaystyle p_1 $$, $$ \displaystyle p_2 $$, and $$ \displaystyle p_3 $$ is to take two times the logarithm of the generalized Weibull distribution as

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

$$  \displaystyle ln\bigg(-ln\big(1-\frac{F_X }{p_3} \big) \bigg) = p_1 ln(x) - p_1 ln(p_2) $$.     (A5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

As explained in the previous section, a good guess for $$ \displaystyle p_3 $$ would be around the maximum value of $$ \displaystyle N(t) $$, which is $$ \displaystyle 870 $$. A plot of the left hand side versus $$ \displaystyle ln(x) $$ gives p1 as the slope and $$ \displaystyle - p_1 ln(p_2) $$ as the intercept. Using a spreadsheet, if we insert $$ \displaystyle p_3=880 $$ so that the natural logarithm of the last couple of data exist, the slope becomes $$ \displaystyle 1.7177 $$ and the intercept $$ \displaystyle -11.195 $$. Hence, $$ \displaystyle p_1 $$ is $$ \displaystyle 1.728 $$ and $$ \displaystyle p_2 $$ is $$ \displaystyle 678 $$. Recall that the result of the nonlinear fit was $$ \displaystyle p = [1.6407, 740.2473, 933.7323] $$. So an initial guess of $$ \displaystyle p = [1.728, 678, 880] $$ is a good starting point. If there were only two parameters, this analysis would have given the end result without the need for a nonlinear fit.

Detailed Octave Procedure
Copy the first two columns in Exponential Data into soffice (or excel), then save as .csv file, e.g., junk.csv. Use vim to replace blank space with comma, and save the result into the file exponential.cdf.csv:
 cat junk.csv | sed 's/ /,/g' >! exponential.cdf.csv

Load the 1st column into matrix x and the 2nd column into matrix y with the octave commands:
 x = csvread ('exponential.cdf.csv','A1:A158')

 y = csvread ('exponential.cdf.csv','b1:b158')

Define the inline function to do the curve fitting:
 F = inline ("p(3) * ( 1 - exp ( - ( x/p(2) ).^p(1) ) ) ","x","p")

Define the initial parameters in the matrix "pin":
 pin = [2 ; 600 ; 870]

Then give the nonlinear least square curve fitting command:
 [f,p,kvg,iter,corp,covp,covr,stdresid,Z,r2]=leasqr(x,y,pin,F);

Now plot the data (x,y) and the fitted function (x,"f"):
 plot (x,y, "+;y;", x, f , "-;f;")

Save the figure in png format for uploading to google doc:
 print -dpng weibull.fit.png

Detailed Matlab Procedure
First, create the model
 modelFun = @(p,x)p(3) * ( 1 - exp ( - ( x/p(2) ).^p(1) ) )

Then, assign an initial guess for the parameters p(i)
 startingVals = [2 600 870];

The command nlinfit gives the coefficients based on the model and the starting values
 coefEsts = nlinfit(x, y, modelFun, startingVals)

We give an initial guess
 startingVals = [2 600 870];

The parameters are found as
 coefEsts = [1.6404 740.2474  933.7324]
 compared to the optimal result from Octave:
 p (result) = [1.6407, 740.2473, 933.7323]
 which is very close to what was found with Octave leasqr command.

= Reading = == The Lady Tasting Tea: How statistics revolutionized science in the 20th century ==

title: User talk:Egm6936.f09/Kolmogorov scales


text:
User talk:Keith1

 Hello Keith1, 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) 05:22, 8 December 2010 (UTC)

To :

The project Wikistution has been moved to User:Norton~enwikiversity/Wikistution

Yours truly, --Guy vandegrift (discuss • contribs) 19:22, 6 December 2022 (UTC)

This project had subpages:

title: User talk:Keith1


text:
User talk:Whargoul~enwikiversity

 Hello Whargoul, 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) 05:22, 8 December 2010 (UTC)

Your account will be renamed
Hello,

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

Unfortunately, your account clashes with another account also called Whargoul. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Whargoul~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 23:47, 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) 06:25, 19 April 2015 (UTC)

title: User talk:Whargoul~enwikiversity


text:
Talk:Motivation and emotion/Textbook/Motivation and emotion/Textbook/Emotion/Antisocial personality disorder

-- Jtneill - Talk - c 08:21, 8 December 2010 (UTC)

-- Jtneill - Talk - c 11:42, 14 December 2010 (UTC)


text:
Talk:Motivation and emotion/Textbook/Motivation/Flow

-- Jtneill - Talk - c 10:36, 8 December 2010 (UTC)

-- Jtneill - Talk - c 09:00, 14 December 2010 (UTC)


text:
Talk:Motivation and emotion/Textbook/Motivation/Positive thinking

-- Jtneill - Talk - c 03:20, 14 December 2010 (UTC)


text:
User talk:Frostly (usurped)

 Hello Frostly, 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.

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 * Visit a (kind of) random project.
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 * Give feedback about your initial observations
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 15:11, 8 December 2010 (UTC)

Request for usurpation
Hello,. A request has been made via SRUC (metawiki) to usurp, or "take over", your username because another user would like to use it to edit.

If you do not object to being renamed to a new username in order for another user to use the name you currently have, please log in and post a reply here saying so (you may also tell us what username you would like to be renamed to, or we will provide you with a generic one) or use.

If you do nothing, the request may be filled shortly, and your account will be moved to a generic username. You may request that it be moved to a new username of your choice at any time.

If you object to being renamed, please log in and make an edit to this page clearly stating that you object to usurpation. Any objection on your part will prevent usurpation.

Please note that even if your current username is usurped, you can still edit and your data will not be lost; your preferences, watchlist, and other user settings will be transferred to a new username.

Thank you for your time. Superpes15 (discuss • contribs) 21:12, 21 March 2023 (UTC)

title: User talk:Frostly (usurped)


text:
User talk:MagodBaru

 Hello MagodBaru, 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

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 * Visit a (kind of) random project.
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 * Find out about research activities on Wikiversity.

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 17:00, 8 December 2010 (UTC)

title: User talk:MagodBaru


text:
User talk:Rflejeune

 Hello Rflejeune, 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

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 * Visit a (kind of) random project.
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 * 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) 17:09, 8 December 2010 (UTC)

title: User talk:Rflejeune


text:
User talk:Aaqiby1

 Hello Aaqiby1, 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) 17:36, 8 December 2010 (UTC)

Please stop
Please do not recreate pages like Aaqib as such content is not acceptable for Wikiversity. Ottava Rima (talk) 19:57, 17 December 2010 (UTC)

Continue, but please be careful!
You may continue, but please be careful. If you are User:Mabuabsdd, please don't use multiple accounts without acknowledging them, and using a second account to recreate a deleted page can be considered a problem, as noted above. Don't worry about what has already happened, we care much more about the future than the past.

However, for the moment, anyway, the page itself is okay, it simply should not have been in mainspace, i.e., having a page name that doesn't have some prefix. I have created a special user space under my user name, and arranged for the page to be undeleted, and I've moved it there, though there is a redirect at this point so that you can still find the page by going to Aaqib Azeez and you will be redirected to the present location of the page. You and your friends may edit it there, just please be nice.

Think of this as a library building with a playground attached. Inside the building, which is like mainspace, we'll ask you to be "quiet," i.e., to allow people to focus on study, reading, etc. Outside the building but still part of the library, is a playground. You can laugh and play, try things out, and have fun. If a problem develops, we will let you know. Mostly, just think about the needs of others and Wikiversity, keep the place reasonably clean as to what you make, and you will be fine. Enjoy!

If you would rather have a file in your own user space, you may do so, and ask for assistance if you don't know how to do that.

In the future, please do not create pages in mainspace that are not reasonably educational content and visible as such. It causes others to waste time cleaning them up. You are welcome to create pages, usually, in your own user space; for example, this page could have been User:Aaqiby1/Aaqib Azeez, no problem.

Welcome to Wikiversity! If you have any questions or you need assistance, please ask me on my Talk page, though you can reply here and I'd probably see it. --Abd 18:47, 7 January 2011 (UTC)

Hello!
Hello Aaqiby1. Nice time because also on your user page. You can e-mail me any questions. Mabuabsdd. GGG--Mabuabsdd 14:43, 7 February 2011 (UTC)

title: User talk:Aaqiby1


text:
User talk:Larsoze

 Hello Larsoze, 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

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 * Visit a (kind of) random project.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 17:36, 8 December 2010 (UTC)

title: User talk:Larsoze


text:
User talk:Forbesbuck

 Hello Forbesbuck, 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) 19:43, 8 December 2010 (UTC)

title: User talk:Forbesbuck


text:
User talk:Logan (username usurped)

Computerwiz908 20:17, 8 December 2010 (UTC)

title: User talk:Logan (username usurped)


text:
User:Forbesbuck

=Forbes Buck=

Computer Student and Artist

 * Forbes Alden Buck
 * Born May 6 1978 in Cliton Creek, Yukon, Canada
 * Parents
 * Barbra Joyce (Cocharne) Buck
 * P. Linsday Buck
 * Currently Self Teaching the Following Programing Languages
 * HTML
 * JavaScript
 * AJAX
 * C
 * C++
 * C#
 * Java
 * SQL
 * Python
 * Ruby
 * PHP
 * Experienced in The following programs
 * Adobe
 * Photoshop up to and including CS4
 * Dreamweaver CS4
 * Softimage 2010
 * Mudbox 2010
 * HTML Coding/debugging with notepad

title: User:Forbesbuck


text:
User:MagodBaru

Hi this is my page. I will soon add content here.

title: User:MagodBaru


text:
User talk:Computerwiz908

 Hello Computerwiz908, 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) 21:59, 8 December 2010 (UTC)

title: User talk:Computerwiz908


text:
User talk:Northrupt

 Hello Northrupt, 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

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 * 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) 22:11, 8 December 2010 (UTC)

title: User talk:Northrupt


text:
User talk:TeddieBearFan

 Hello Scott A Herbert, 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

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 * Give feedback about your initial observations
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 22:12, 8 December 2010 (UTC)

title: User talk:TeddieBearFan


text:
User talk:Cojack~enwikiversity

 Hello Cojack, 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.
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 * 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) 22:12, 8 December 2010 (UTC)

Your account will be renamed
Hello,

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

Unfortunately, your account clashes with another account also called Cojack. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Cojack~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 23:33, 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) 06:05, 19 April 2015 (UTC)

title: User talk:Cojack~enwikiversity


text:
User talk:Kalumajhi

 Hello Kalumajhi, 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) 23:06, 8 December 2010 (UTC)

title: User talk:Kalumajhi


text:
User talk:Sehguh94

 Hello Sehguh94, 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) 02:43, 9 December 2010 (UTC)

title: User talk:Sehguh94


text:
User talk:Andy33~enwikiversity

 Hello Andy33, 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) 02:43, 9 December 2010 (UTC)

Your account will be renamed
Hello,

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

Unfortunately, your account clashes with another account also called Andy33. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Andy33~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 23:27, 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) 06:02, 19 April 2015 (UTC)

title: User talk:Andy33~enwikiversity


text:
University of Canberra/RCC/Invite

{| class="infobox" style="width:300px"

Canberra RecentChangesCamp January 2011

 * Rcc2010-icon.svg
 * Read all about RecentChangesCamp 2011
 * at the
 * University of Canberra
 * Planning RecentChangesCamp 2012 for next year has already begun.
 * }
 * University of Canberra
 * Planning RecentChangesCamp 2012 for next year has already begun.
 * }
 * Planning RecentChangesCamp 2012 for next year has already begun.
 * }
 * }

title: University of Canberra/RCC/Invite


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


text: User:Ha98574/Railroad/Korea/Korean Metropolian Subway Gyeongbuk Line/Template


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


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


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


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