text:
User:Egm6321.f11.team2/HW7

=R*7.1 - Infinitesimal Length in Spherical Coordinates =

Find
Show that ds2 in spherical coordinates is as given below:

$$\displaystyle ds^2 = dr^2 + r^2d\theta^2 + r^2\cos^2\theta\, d\phi^2$$

Solution
This problem was solved without referring to past solutions.

Differentiating 7.1.1 yields

$$\displaystyle dx_1 = \cos\theta\cos\phi\,dr - r\sin\theta\cos\phi\,d\theta - r\cos\theta\sin\phi\,d\phi$$

Squaring this expression gives

Similarly for 7.1.2:

$$\displaystyle dx_2 = \cos\theta\sin\phi\,dr - r\sin\theta\sin\phi\,d\theta + r\cos\theta\cos\phi\,d\phi$$

And again for 7.1.3:

$$\displaystyle dx_3 = \sin\theta\,dr + r\cos\theta\,d\theta$$

Finally, substituting 7.1.5 — 7.1.7 into 7.1.4 and making much use of the trigonometric identity $$\displaystyle \cos^2 a + \sin^2 a = 1$$, we find

$$\displaystyle \begin{align} ds^2 = &\sum_{i=1}^3 dx_i^{\,2} \\ = & \,( \cos^2\theta \cos^2\phi + \cos^2\theta \sin^2\phi + \sin^2\theta ) dr^2 \\ &+ ( r^2 \sin^2\theta \cos^2\phi + r^2 \sin^2\theta \sin^2\phi + r^2 \cos^2\theta) d\theta \\ &+ ( r^2 \cos^2\theta \sin^2\phi + r^2 \cos^2\theta \cos^2\phi) d\phi^2 \\ &- 2r\cos\theta\sin\theta \cancel{( \cos^2\theta + \sin^2\phi - 1 )} dr\,d\theta \\ \\ = &\,\left[\cos^2\theta ( \cos^2\phi + \sin^2\phi ) + \sin^2\theta \right] dr^2 \\ & + r^2 \left[ \sin^2\theta (\cos^2\phi + \sin^2\phi) + \cos^2\theta \right] d\theta^2 \\ & + r^2\cos^2\theta (\sin^2\phi + \cos^2\phi) d\phi^2 \\ \\ = &\, dr^2 + r^2 d\theta^2 + r^2\cos^2\theta\,d\phi^2 \end{align} $$

=R*7.2 - Heat Conduction on a cylinder =

Given
$$\displaystyle x_1 = r \cos \theta = \xi_1 cos \xi_2 $$ $$\displaystyle x_2 = r \sin \theta = \xi_1 sin \xi_2 $$ $$\displaystyle x_3 = z = \xi_3 $$

Find
1. Find $$\displaystyle \left \{ dx_i \right \}=\left \{dx_1,dx_2,dx_3\right \} $$ in terms of $$\displaystyle \left \{ d\xi_j \right \}=\left \{d\xi_1,d\xi_2,d\xi_3\right \} $$ and $$\displaystyle \left \{ d\xi_k \right \}$$ 2. Find $$\displaystyle ds^2=\sum_{i=1}(dx_i)^2= \sum_{k}(h_k)^2(d\xi_k)^2$$. Identify $$\displaystyle \left \{ dh_i \right \}$$ in terms of $$\displaystyle \left \{ d\xi_j \right \}$$ 3. Find $$\displaystyle \Delta u $$ in cylindrical coordinates

4. Use seperation variable and compare to the Bessel equation.

Solution
This problem was solved without referring to past solutions.

part 1

$$\displaystyle ds^2=\sum_{k=1}^3(h_k)^2(d\xi_k)^2= \sum_{i=1}^{3}(dx_i)^2 $$ $$\displaystyle dx_i=\frac{\partial x_i}{\partial \xi_1} d\xi_1+\frac{\partial x_i}{\partial \xi_2} d\xi_2+\frac{\partial x_i}{\partial \xi_3} d\xi_3 $$ $$\displaystyle dx_1 =\cos \xi_2 d\xi_1-\xi_1 \sin \xi_2 d\xi_2+(0) d\xi_3 $$ $$\displaystyle dx_2 =\sin \xi_2 d\xi_1+\xi_1 \cos \xi_2 d\xi_2+(0) d\xi_3 $$ $$\displaystyle dx_3 =(0) d\xi_1+(0) d\xi_2+(1) d\xi_3 $$

Part 2

$$\displaystyle ds^2=(\cos \xi_2 d\xi_1-\xi_1 \sin \xi_2 d\xi_2)^2 + (\sin \xi_2 d\xi_1+\xi_1 \cos \xi_2 d\xi_2)^2 + (d\xi_3)^2=(d\xi_1)^2+\xi_1^2 (d \xi_2)^2+(d\xi_3)^2 = (h_1)^2(d\xi_1)^2 + (h_2)^2(d\xi_2)^2 + (h_3)^2(d\xi_3)^2 $$ $$\displaystyle ds^2=dr^2+r^2d \theta^2 + dz^2 $$ Thus, $$\displaystyle h_1(\xi)=1 $$ $$\displaystyle h_2(\xi)=r=\xi_1 $$ $$\displaystyle h_3(\xi)=1 $$

Part 3 $$\displaystyle \Delta u = \frac{1}{h_1h_2h_3}\sum_{i=1}^{3} \frac{\partial}{\partial \xi_i}\left [ \frac{h_1h_2h_3}{h_i^2}\frac{\partial u }{\partial \xi_i} \right] $$ Where $$\displaystyle h_1h_2h_3=\xi_1=r $$ So, $$\displaystyle \Delta u = \frac{1}{\xi_1} \left [\frac{\partial}{\partial \xi_1} \left (\frac{\xi_1}{1^2} \frac{\partial u }{\partial \xi_1}\right ) +\frac{\partial}{\partial \xi_2} \left (\frac{\xi_1}{\xi_1^2}\frac{\partial u }{\partial \xi_2}\right )+\frac{\partial}{\partial \xi_3} \left ( \frac{\xi_1}{1^2}\frac{\partial u }{\partial \xi_3} \right ) \right ] =\frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial u }{\partial \xi_1}\right ) +\frac{1}{\xi_1^2}\frac{\partial^2 u }{\partial \xi_2^2}+\frac{\partial^2 u }{\partial \xi_3^2} $$

part 4

$$\displaystyle \Delta u = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial u }{\partial \xi_1}\right ) +\frac{1}{\xi_1^2}\frac{\partial^2 u }{\partial \xi_2^2}+\frac{\partial^2 u }{\partial \xi_3^2}=0 $$ $$\displaystyle u = X(\xi_1)Y(\xi_2)Z(\xi_3) $$ $$\displaystyle \frac{Y Z}{\xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{X Z }{\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+X Y \frac{\partial^2 Z }{\partial \xi_3^2}=0 $$ divide through by  $$\displaystyle XYZ $$ $$\displaystyle \frac{1}{X \xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+\frac{1}{Z} \frac{\partial^2 Z }{\partial \xi_3^2}=0 $$ Assuming that $$\displaystyle \frac{1}{Z} \frac{\partial^2 Z }{\partial \xi_3^2}=A $$ $$\displaystyle \frac{1}{X \xi_1} \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y\xi_1^2}\frac{\partial^2 Y }{\partial \xi_2^2}+A=0 $$ multiplying through by $$\displaystyle \xi_1^2 $$ $$\displaystyle \frac{\xi_1}{X } \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +\frac{1}{Y}\frac{\partial^2 Y }{\partial \xi_2^2}+A \xi_1^2 =0 $$

assumeing for second term $$\displaystyle \frac{1}{Y}\frac{\partial^2 Y }{\partial \xi_2^2}=-B $$ $$\displaystyle \frac{\xi_1}{X } \frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) -B+A \xi_1^2 =0 $$ multiplying through by $$\displaystyle X $$ $$\displaystyle \xi_1\frac{\partial}{\partial \xi_1} \left (\xi_1 \frac{\partial X }{\partial \xi_1}\right ) +(A \xi_1^2-B)X =0 $$ The Original equation is following that $$\displaystyle \xi_1^2 \frac{\partial^2 X }{\partial \xi_1^2}+\xi_1 \frac{\partial X }{\partial \xi_1}+(A \xi_1^2-B)X =0 $$

Lets define $$\displaystyle \xi_1 =x, \,\,\,\,\, X =y $$ Thus, the above equations is following that $$\displaystyle x^2 y''+xy'+(A x^2-B)y =0 $$

=R*7.3 -Laplacian operator in spherical coordinate using Math/Physics convention =

Given
Spherical coordinate using Math/Physics convention:

$$\displaystyle \begin{cases} x_{1}=r\cos\overline{\theta}\cos\varphi=\xi_{1}\cos\xi_{2}\cos\xi_{3}\\ x_{2}=r\cos\overline{\theta}\sin\varphi=\xi_{1}\cos\xi_{2}\sin\xi_{3}\\ x_{3}=r\sin\overline{\theta}=\xi_{1}\sin\xi_{2}\end{cases}$$

(7.3.1) Where $$\displaystyle \overline{\theta}=\frac{\pi}{2}-\theta $$

Find
Find $$\displaystyle \triangle u$$ in this coordinate.

Solution
From 7.3.1, we can obtain

$$\displaystyle \begin{cases} x_{1}=r\sin\theta\cos\varphi=\xi_{1}\cos\xi_{2}\cos\xi_{3}\\ x_{2}=r\sin\theta\sin\varphi=\xi_{1}\cos\xi_{2}\sin\xi_{3}\\ x_{3}=r\cos\theta=\xi_{1}\sin\xi_{2}\end{cases} $$ (7.3.2) Thus $$\displaystyle \begin{cases} dx_{1}=\mathrm{cos}\phi\, r\,\mathrm{cos}\theta\, d\theta+\mathrm{cos}\phi\,\mathrm{sin}\theta\, dr-\mathrm{sin}\phi\, r\,\mathrm{sin}\theta\, d\phi\\ dx_{2}=\mathrm{sin}\phi\, r\,\mathrm{cos}\theta\, d\theta+\mathrm{sin}\phi\,\mathrm{sin}\theta\, dr+\mathrm{cos}\phi\, r\,\mathrm{sin}\theta\, d\phi\\ dx_{3}=\mathrm{cos}\theta\, dr-r\,\mathrm{sin}\theta\, d\theta\end{cases}$$ (7.3.3) Since $$ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}$$, then $$\displaystyle ds^{2}={dr}^{2}+{r}^{2}\,{d\theta}^{2}+{r}^{2}\,{\mathrm{sin}^2\theta}\,{d\phi}^{2}$$ (7.3.4) So $$\displaystyle h_{1}=1,\; h_{2}=r,\; h_{3}=r\sin\theta$$ (7.3.5) The Laplacian operator is as follows: $$\displaystyle \Delta u = \frac{1}{h_1h_2h_3}\sum_{i=1}^{3} \frac{\partial}{\partial \xi_i}\left [ \frac{h_1h_2h_3}{h_i^2}\frac{\partial u }{\partial \xi_i} \right ] $$ (7.3.6) hence $$\displaystyle \Delta u=\frac{1}{r^{2}\sin\theta}\left(\frac{\partial}{\partial r}\left(\frac{r^{2}\sin\theta}{1}\frac{\partial u}{\partial r}\right)+\frac{\partial}{\partial\theta}\left(\frac{r^{2}\sin\theta}{r^{2}}\frac{\partial u}{\partial\theta}\right)+\frac{\partial}{\partial\phi}\left(\frac{r^{2}\sin\theta}{r^{2}\sin^{2}\theta}\frac{\partial u}{\partial\phi}\right)\right)$$ (7.3.7)

$$\displaystyle \Delta u=\frac{1}{r^{2}\sin\theta}\left(\frac{\partial}{\partial r}\left(r^{2}\sin\theta\frac{\partial u}{\partial r}\right)+\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial u}{\partial\theta}\right)+\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\theta}\frac{\partial u}{\partial\phi}\right)\right)$$ (7.3.8)

=R*7.4 - Laplacian in elliptic coordinates =

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

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

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

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

Find
Verify the Laplacian in elliptic coordinates given by

Solution
Elliptical coordinates are

First, we assume

And then, we differentiate (7.4.2) and (7.4.3) with respect to each direction (i.e. x1 and x2).

And then,

Since $$\displaystyle ds^2 = \sum_{i} (dx_i)^2= \sum_{k} {(h_k)^2 (d \xi _k)^2} $$,

And the Laplacian equals

Thus,

=R 7.5 - Laplacian in parabolic coordinates. =

Given
Two dimensional parabolic coordinates $$ (\mu,\nu) $$ are defined by the equations:

Find
Verify the Laplacian in parabolic coordiantes as:

Solution
Differentiating first equation: Differentiating second equation: First equation becomes: Second equation becomes: Now: From Equation 7.5.6: Comparing with equation 7.5.6 we analayze that: Laplace operator for two dimensional system in general curvilinear coordinates is given as: Putting the value of $$ h_1, h_2 $$ in equation 7.5.8 we get

=R*7.6 - =

Given
The Legendre differential equation: $$\displaystyle (1-x^2) y'' - 2 x y' + n(n+1) y = 0$$(7.6.1) The 1st homogeneous solutions of the Legendre differential equation for n = 0,1,2,3,4. $$\displaystyle P_0(x) = 1$$(7.6.2) $$\displaystyle P_1(x) = x$$(7.6.3) $$\displaystyle P_2(x) = \frac{1}{2} (3x^2 - 1)$$(7.6.4) $$\displaystyle P_3(x) = \frac{1}{2} (5x^3 - 3x)$$(7.6.5) $$\displaystyle P_4(x) = \frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}$$(7.6.6) Which can be generalized as $$\displaystyle P_n(x) = \sum^{[n/2]}_{i=0} (-1)^i \frac{(2n-2i)! x^{n-2i}}{2^n i! (n-i)! (n-2i)!}$$(7.6.7) Where [n\2] = integer part of $$\frac{n}{2}$$

Find
(A) Verify that equations (7.6.2)-(7.6.6) are homogeneous solutions of the Legendre differential equation. (B) Show that equation (7.6.7) can be rewritten as
 * $$\displaystyle P_n(x) = \sum^{[n/2]}_{i=0} \frac{1 \cdot 3 \cdots (2n-2i-1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$$(7.6.8)

(C) Verify that equations (7.6.2)-(7.6.6) can be obtained from equations (7.6.7) and (7.6.8).

Solution
This problem was solved without referring to past solutions.

Solution Part A
To show that equations (7.6.2)-(7.6.6) are homogeneous solutions of their respective Legendre equations, we must first show these equations. For n = 0,
 * $$\displaystyle (1-x^2) y'' - 2 x y' = 0$$

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

For n = 2,
 * $$\displaystyle (1-x^2) y'' - 2 x y' + 6y = 0$$

For n = 3,
 * $$\displaystyle (1-x^2) y'' - 2 x y' + 12y = 0$$

For n = 4,
 * $$\displaystyle (1-x^2) y'' - 2 x y' + 20y = 0$$

For n=0, $$\displaystyle P_0(x) = 1$$ $$\displaystyle P_0'(x) = 0$$ $$\displaystyle P_0''(x) = 0$$ Substituting $$\displaystyle P_0(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) 0 - 2 x 0 = 0$$

Which proves it is a homogeneous equation of the Legendre equation. For n=1, $$\displaystyle P_1(x) = x$$ $$\displaystyle P_1'(x) = 1$$ $$\displaystyle P_1''(x) = 0$$ Substituting $$\displaystyle P_1(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) 0 - 2 x 1 + 2x = 0$$
 * $$\displaystyle - 2x + 2x = 0$$

Which proves it is a homogeneous equation of the Legendre equation. For n=2, $$\displaystyle P_2(x) = \frac{1}{2} (3x^2 - 1)$$ $$\displaystyle P_2'(x) = 3x$$ $$\displaystyle P_2''(x) = 3$$ Substituting $$\displaystyle P_2(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) 3 - 2 x (3x) + 6(\frac{1}{2} (3x^2 - 1)) = 0$$
 * $$\displaystyle 3 - 3x^2 - 6x^2 + 9x^2 - 3 = 0$$

Which proves it is a homogeneous equation of the Legendre equation. For n=3, $$\displaystyle P_3(x) = \frac{1}{2} (5x^3 - 3x)$$ $$\displaystyle P_3'(x) = \frac{1}{2} (15x^2 - 3)$$ $$\displaystyle P_3''(x) = 15x$$ Substituting $$\displaystyle P_3(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) (15x) - 2 x (\frac{1}{2} (15x^2 - 3)) + 12(\frac{1}{2} (5x^3 - 3x)) = 0$$
 * $$\displaystyle 15x - 15x^3 -  15x^3 + 3x + 30x^3 - 18x = 0$$

Which proves it is a homogeneous equation of the Legendre equation. For n=4, $$\displaystyle P_4(x) = \frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}$$ $$\displaystyle P_4'(x) = \frac{35}{2}x^3 - \frac{15}{2}x$$ $$\displaystyle P_4''(x) = \frac{105}{2}x^2 - \frac{15}{2}$$ Substituting $$\displaystyle P_4(x)$$ for $$\displaystyle y(x)$$ we obtain,
 * $$\displaystyle (1-x^2) (\frac{105}{2}x^2 - \frac{15}{2}) - 2 x (\frac{35}{2}x^3 - \frac{15}{2}x) + 20(\frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}) = 0$$

 * $$\displaystyle \frac{105}{2}x^2 - \frac{15}{2} - \frac{105}{2}x^4 + \frac{15}{2}x^2 - \frac{70}{2}x^4 + \frac{30}{2}x^2 + \frac{175}{2}x^4 - \frac{150}{2} x^2 + \frac{15}{2} = 0$$

Which proves it is a homogeneous equation of the Legendre equation.

Solution Part B
Looking at equations (7.6.7) and (7.6.8) we can see that the terms that must be equal for the equations to be equal is
 * $$\displaystyle \frac{(2n-2i)!}{2^n (n-i)!}$$(7.6.9)

and
 * $$\displaystyle \frac{1 \cdot 3 \cdots (2n-2i-1)}{2^i}$$(7.6.9)

These can be shown to be equal by experimentation. For n = 1,2,3,4 and i = 0,1,2 we see that equation (7.6.9) becomes For n = 1 i=0 :$$\displaystyle \frac{(2-0)!}{2^1 (1-0)!} = 1$$ For n = 2 i=0 :$$\displaystyle \frac{(4-0)!}{2^2 (2-0)!} = 3$$ i=1 :$$\displaystyle \frac{(4-2)!}{2^2 (2-1)!} = \frac{1}{2}$$ For n = 3 i=0 :$$\displaystyle \frac{(6-0)!}{2^3 (3-0)!} = 15$$ i=1 :$$\displaystyle \frac{(6-2)!}{2^3 (3-1)!} = \frac{3}{2}$$ For n = 4 i=0 :$$\displaystyle \frac{(8-0)!}{2^4 (4-0)!} = 105$$ i=1 :$$\displaystyle \frac{(8-2)!}{2^4 (4-1)!} = \frac{15}{2}$$ i=2 :$$\displaystyle \frac{(8-4)!}{2^4 (4-2)!} = \frac{3}{4}$$ We see that equation (7.6.10) becomes For n = 1 i=0 :$$\displaystyle \frac{1}{2^0} = 1$$ For n = 2 i=0 :$$\displaystyle \frac{1 \cdot 3}{2^0} = 3$$ i=1 :$$\displaystyle \frac{1}{2^1} = \frac{1}{2}$$ For n = 3 i=0 :$$\displaystyle \frac{1 \cdot 3 \cdot 5}{2^0} = 15$$ i=1 :$$\displaystyle \frac{1 \cdot 3}{2^1} = \frac{3}{2}$$ For n = 4 i=0 :$$\displaystyle \frac{1 \cdot 3 \cdot 5 \cdot 7}{2^0} = 105$$ i=1 :$$\displaystyle \frac{1 \cdot 3 \cdot 5}{2^1} = \frac{15}{2}$$ i=2 :$$\displaystyle \frac{1 \cdot 3}{2^2} = \frac{3}{4}$$ This shows that equations (7.6.9) and (7.6.10) are equivalent, which proves equation (7.6.8) is equivalent to equation (7.6.7).

Solution Part C
Equations (7.6.2)-(7.6.6) can be obtained from equation (7.6.7) and (7.6.8). For n = 0, $$\displaystyle P_0(x) = \sum^{0}_{i=0} (-1)^i \frac{(2(0)-2i)! x^{(0)-2i}}{2^{(0)} i! ((0)-i)! ((0)-2i)!}$$ $$\displaystyle P_0(x) = (-1)^0 \frac{(2(0)-2(0))! x^{(0)-2(0)}}{2^{(0)} (0)! ((0)-(0))! ((0)-2(0))!}$$ Since (0)! = 1, $$\displaystyle P_0(x) = 1 \frac{1 (1)}{1 (1) (1) (1)} = 1$$ For n = 1, $$\displaystyle P_1(x) = \sum^{0}_{i=0} (-1)^i \frac{(2(1)-2i)! x^{(1)-2i}}{2^{(1)} i! ((1)-i)! ((1)-2i)!}$$ $$\displaystyle P_1(x) = (-1)^0 \frac{(2-2(0))! x^{(1)-2(0)}}{2 (0)! ((1)-(0))! ((1)-2(0))!}$$ $$\displaystyle P_1(x) = (1) \frac{2 x}{2 (1) (1) (1)} = x$$ For n = 2, $$\displaystyle P_2(x) = \sum^{1}_{i=0} (-1)^i \frac{(2(2)-2i)! x^{(2)-2i}}{2^{(2)} i! ((2)-i)! ((2)-2i)!}$$ $$\displaystyle P_2(x) = (-1)^0 \frac{(2(2)-2(0))! x^{(2)-2(0)}}{2^{(2)} (0)! ((2)-(0))! ((2)-2(0))!} + (-1)^1 \frac{(2(2)-2(1))! x^{(2)-2(1)}}{2^{(2)} (1)! ((2)-(1))! ((2)-2(1))!}$$ $$\displaystyle P_2(x) = 1 \frac{(4)! x^2}{4 (1) (2) (2)} - 1 \frac{2 x^0}{4}$$ $$\displaystyle P_2(x) = \frac{3 x^2}{2} - \frac{1}{2}$$ $$\displaystyle P_2(x) = \frac{1}{2} (3 x^2 - 1)$$ For n = 3, $$\displaystyle P_3(x) = \sum^{1}_{i=0} (-1)^i \frac{(2(3)-2i)! x^{(3)-2i}}{2^{(3)} i! ((3)-i)! ((3)-2i)!}$$ $$\displaystyle P_3(x) = (-1)^0 \frac{(2(3)-2(0))! x^{(3)-2(0)}}{2^{(3)} (0)! ((3)-(0))! ((3)-2(0))!} + (-1)^1 \frac{(2(3)-2(1))! x^{(3)-2(1)}}{2^{(3)} (1)! ((3)-(1))! ((3)-2(1))!}$$ $$\displaystyle P_3(x) = \frac{6! x^3}{8 (3!) (3!)} - \frac{6! x}{8 (2!)}$$ $$\displaystyle P_3(x) = \frac{5 x^3}{2} - \frac{3 x}{2}$$ $$\displaystyle P_3(x) = \frac{1}{2} (5 x^3 - 3x)$$ For n = 4, $$\displaystyle P_4(x) = \sum^{2}_{i=0} (-1)^i \frac{(2(4)-2i)! x^{(4)-2i}}{2^{(4)} i! ((4)-i)! ((4)-2i)!}$$ $$\displaystyle P_4(x) = \frac{8! x^4}{16 (4!) (4!)} - \frac{6! x^2}{16 (3!) (2!)} + \frac{4! x^0}{16 (2!) (2!)}$$ $$\displaystyle P_4(x) = \frac{35 x^4}{8} - \frac{15 x^2}{4} + \frac{3}{8}$$

=R*7.7 - Separating Laplace's Equation in Elliptical and Parabolic 2D Coordinates =

Given
Laplace's Equation in Elliptical Coordinates is

$$\displaystyle \nabla^2 u = \frac{1}{a^2 \left( \sinh^2\mu + \sin^2\nu \right)}\left( \frac{\partial^2 u}{\partial \mu^2} + \frac{\partial^2 u}{\partial \nu^2}\right) = 0$$

Laplace's Equation in Parabolic Coordinates is

$$\displaystyle \nabla^2 u = \frac{1}{\mu^2 + \nu^2}\left( \frac{\partial^2 u}{\partial \mu^2} + \frac{\partial^2 u}{\partial \nu^2}\right) = 0$$

Find
Separate the Laplace equations in both elliptical and parabolic coordinates.

Solution
This problem was solved without referring to past solutions. After the problem was solved, it was checked against F10 Team6's solution for the elliptic case.

We begin with elliptical coordinates. We'll try a solution of the form

$$\displaystyle u(\mu,\nu) = \Mu(\mu)\Nu(\nu)$$

Plugging this into Laplace's equation yields

$$\displaystyle \nabla^2 u = \frac{1}{a^2 \left( \sinh^2\mu + \sin^2\nu \right)}\left( \Nu \frac{\partial^2 \Mu}{\partial \mu^2} + \Mu \frac{\partial^2 \Nu}{\partial \nu^2}\right) = 0 $$

Multiplying both sides of the equation by $$\displaystyle a^2\left(\sinh^2\mu + \sin^2\nu\right)$$ and dividing by $$\displaystyle \Mu(\mu)\Nu(\nu)$$ yields

$$\displaystyle \frac{1}{\Mu}\frac{\partial^2\Mu}{\partial\mu^2} = - \frac{1}{\Nu}\frac{\partial^2\Nu}{\partial\nu^2}$$

The only way this equation can hold is if both sides are equal to a constant k. Thus, the separated equations are

$$\displaystyle \frac{\partial^2\Mu}{\partial\mu^2} = k\Mu$$

and

$$\displaystyle \frac{\partial^2\Nu}{\partial\nu^2} = -k\Nu$$

For parabolic coordinates, the procedure is much the same. We use the same form for u, but this time we multiply both sides of the equation by $$\displaystyle \left(\mu^2 + \nu^2\right)$$ before dividing by $$\displaystyle \Mu(\mu)\Nu(\nu)$$:

$$\displaystyle \begin{align} 0 &= \frac{1}{\mu^2 + \nu^2}\left( \Nu \frac{\partial^2 \Mu}{\partial \mu^2} + \Mu \frac{\partial^2 \Nu}{\partial \nu^2}\right) \\ \frac{1}{\Mu}\frac{\partial^2 \Mu}{\partial \mu^2} &= -\frac{1}{\Nu}\frac{\partial^2 \Nu}{\partial \nu^2} \end{align} $$

As before, both sides must be equal to a constant k, so the separated equations are again

$$\displaystyle \frac{\partial^2\Mu}{\partial\mu^2} = k\Mu$$

and

$$\displaystyle \frac{\partial^2\Nu}{\partial\nu^2} = -k\Nu$$

=R*7.8 - Plot legendre polynomial function and observe the limit of P and Q=

Given
$$\displaystyle y=A \cdot P_n(x)+B \cdot{Q_n(x)} $$

Legendre Polynomials are following that $$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)! $$ where $$\displaystyle m = \frac{n}{2}$$. $$\displaystyle {P}_{0}(x) = 1 \,\,\,\, {P}_{1}(x) = x \,\,\,\,{P}_{2}(x) = \frac{1}{2}(3x^2-1)\,\,\,\,{P}_{3}(x) = \frac{1}{2}(5x^3-3x,) $$ $$\displaystyle {Q}_{0}(x) = \frac{1}{2}\mathrm{log}\left(\frac{1+x}{1-x}\right) \,\,\,\,{Q}_{1}(x) = \frac{1}{2}x \mathrm{log}\left(\frac{1+x}{1-x}\right)-1 \,\,\,\, $$ $$\displaystyle {Q}_{2}(x) = \frac{1}{4}(3x^2-1)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{3}{2}x \,\,\,\,$$ $$\displaystyle {Q}_{3}(x) = \frac{1}{4}(5x^3-3x)\mathrm{log}\left(\frac{1+x}{1-x}\right)-\frac{5}{2}x^2+\frac{2}{3} $$

Find
Show that for n=0; $$\displaystyle \mu \to \pm1 => Q_0(\mu) \to oo$$ Plot $$\displaystyle \begin{align} &\left[P_0, P_1, P_2, P_3\right]\\ &\left[Q_0, Q_1, Q_2, Q_3\right]\\ \end{align} $$ Oberve the limit of $$\displaystyle P_n(\mu) \,\,\, and\,\,\, Q_n(\mu) \,\,\,as\,\,\, \mu \to \pm 1 $$

Solution
This problem was solved by referring to past solutions. Egm6321.f10.team4/HW7.

From the given equations, legendre polynomial equation is following that This is the legendre polynomial graph. The above graph is a function of Q When n=oo, $$\displaystyle P_n(\mu) \,\,\, and\,\,\, Q_n(\mu) \,\,\,as\,\,\, \mu \to \pm 1 $$ It shows that a higher order number is shown the Runge's phenomenon at the edge of graph.

=R7.9 - Finding $$ v_i$$ =

Given
Consider the non-orthonormal basis $$ \{\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}\}$$, expressed in the orthonomal basis $$ \{\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\}$$ as folllows: $$\displaystyle \mathbf{b}_{i}=A_{ij}\mathbf{e}_{j}$$ (7.9.2) here $$\displaystyle \mathbf{A}=\left[A_{ij}\right]=\left[\begin{array}{ccc} 5 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 5\end{array}\right]$$ and $$\displaystyle \mathbf{v}=-2\mathbf{e}_{1}+4\mathbf{e}_{2}-5\mathbf{e}_{3}$$ (7.9.3)

Find
$$\displaystyle{\mathbf v_i}\in\mathbb{R}^{3\times 1}$$ such that $$\displaystyle \mathbf{v}=v_i\mathbf b_i$$

Solution
From 7.9.2 and 7.9.3, we have $$\displaystyle \mathbf{v}=v_{i}A_{ij}\mathbf{e}_{j}$$ (7.9.4) $$\displaystyle \mathbf{\mathbf{A}}^{T}\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.9.5) $$\displaystyle \left[A_{ij}\right]^{T}\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.9.6) Since $$\displaystyle det(\mathbf{A})=-80 \neq 0$$

we have $$\displaystyle \left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[A_{ij}\right]^{-T}\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]=\left[\begin{array}{ccc} \frac{23}{80} & -\frac{11}{40} & \frac{3}{80}\\ -\frac{7}{40} & -\frac{1}{20} & \frac{13}{40}\\ \frac{3}{80} & \frac{9}{40} & -\frac{17}{80}\end{array}\right]\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.9.7) thus $$\displaystyle \left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -\frac{149}{80}\\ -\frac{59}{40}\\ \frac{151}{80}\end{array}\right]$$ (7.9.8)

=Contributing Members= Egm6321.f11.team2.Xia I solved R*7.3 and R7.9, proofread R*7.2 and R*7.5 Egm6321.f11.team2.johri.y 03:08, 7 December 2011 (UTC) I solved R 7.5 and proofread R*7.1 and R*7.4 Egm6321.f11.team2.epps 03:38, 7 December 2011 (UTC) I solved R*7.1 and R*7.7.

Egm6321.f11.team2.kim 15:33, 7 December 2011 (UTC) I solved R*7.2 and R*7.8, proofread R*7.1, R*7.3, R7*.4 and R7.9 Egm6321.f11.team2.rho17:47, 7 December 2011 (UTC) I solved R*7.4, proofread R*7.5, R*7.7, and R7.8. Egm6321.f11.team2.wise 20:06, 7 December 2011 (UTC) I solved R*7.6, and proofread R7.9

title: User:Egm6321.f11.team2/HW7


text:
User talk:Syedahinaosfur

 Hello Syedahinaosfur, 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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title: User talk:Syedahinaosfur


text:
User talk:Kraŭs

 Hello Kraŭs, 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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title: User talk:Kraŭs


text:
User:Wagino 20100516

   Hello, I am from Indonesia and now live on Pekanbaru City

title: User:Wagino 20100516


text:
User talk:Malithyapa

 Hello Malithyapa, 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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title: User talk:Malithyapa


text:
Wikiversity talk:Request custodian action

/Archive 1 /Archive 2

Category:Requests for own page deletion
I deprecated this category, and Template:Delete My Page because "own page deletion" is routinely handled by the ordinary deletion template, giving the reason "author request." There is no need for a special template or category. The Template:Delete My Page required editing Please delete my page, a totally unnecessary complication.

I also edited that Wikiversity page to provide instructions on deleting your pages. The page had been used only very rarely, adding no value over a simple request through a deletion template.

We have become very efficient at cleaning up the various deletion categories. Own page deletion can sometimes be controversial, as can any speedy deletion reason. I see no need for a separate category.

I am asking here before removing Own Page Deletion from the action box displayed on WV:RCA. --Abd (discuss • contribs) 03:07, 14 September 2015 (UTC)

 * Seems reasonable to me. I have put my own pages up for deletion but I'd never heard of these before. --Marshallsumter (discuss • contribs) 18:28, 14 September 2015 (UTC)


text:
Wikiversity:Request custodian action/Archive/13

Poetlister1
It will probably be taken care of before it is noticed here, but in case not: User:Poetlister1 appears to be a sock/alternate account of the globally banned User:Poetlister, so presumably needs to be blocked accordingly. - Bilby 13:33, 4 November 2011 (UTC)

 * I would be uncomfortable enforcing the global ban discussion at meta for several reasons.
 * It was not advertised to our community, so it is hard to feel that it represents a community consensus here.
 * There is no policy, at here or meta, for who/what/when/where/why/how to handle global bans.
 * Poetlister himself was not part of the discussion so he could not "defend himself".
 * Etc, it is all just a bit too informal. That being said he is clearly stepping around a global lock.  We have allowed this (against my personal objectsions) with thekohser, but in that cases community consensus was sought before editing occurred, and the original account was detached so it could edit (if memory serves).

 * All that being said, I have been out of the loop for a while. If there was a discussion to block/ban him here, just point a diff and I will act swiftly. Thenub314 14:21, 4 November 2011 (UTC)`

 * The closest thing to a policy that I'm aware of is at meta - perhaps I'm reading it wrong, but it seems to call for blocking based on a violation of the WMF Terms of Use. I assume these cases are rare enough that they haven't put a lot of work into the policy side, and I gather that the process may have been put in place specifically because of Poetlister's editing here. I'm not sure how the project independence fits into this model, but my assumption is that this is meant to be where project independence ends. :) - Bilby 14:34, 4 November 2011 (UTC)

 * A follow up for Colloquium/archives/July 2011 with local consensus for enforcing the global ban is likely needed to help guide what actions custodians should or shouldn't take. -- dark lama  14:42, 4 November 2011 (UTC)
 * That seems reasonable. :) Although I'm curious about where the WMF sits on this - it wasn't clear from what I read if they saw allowing a globally banned editor access is an exclusively community decision or not. The draft global ban page seems to suggest it isn't a community decision, but that, of course, is both just a draft and an interpretation. - Bilby 14:47, 4 November 2011 (UTC)
 * Sadly, to what extent a policy is global, or whether a Meta policy is intended to address issues local to the Meta project itself are just two areas I've seen debated and a cause for confusion at other projects. The global ban proposal itself can be seen as intended to guide the actions of people who are members of the group that add global locks, and not intended to override any local consensus that may differ. -- dark lama  15:13, 4 November 2011 (UTC)
 * I agree. An apparent defacto consensus settled out of the global lock for Thekohser. By whatever process, a "global ban" may be declared and implemented with a global lock. This only affects SUL accounts. At one point in the sequence with Thekohser, the global lock was lifted and a steward went around to all projects with Thekohser accounts, locally blocking them. He was explicit that local wikis could unblock. Then a steward, without any public discussion, reinstated the global lock. This lock was not lifted, because local wikis with bureaucrats realized that they could delink the account, and a number of WMF wikis did so for Thekohser. There has been no resulting disruption, except that the steward who had globally locked again strongly opposed it on one wiki, acting locally (he was also an admin on that wiki), and he resigned when the community consensus overruled him.
 * So, if a user is considered "broadly disruptive," or something like that, the SUL account can be locked. It's been reasonably well established that local communities may overturn this to allow the user to edit there, and this is then a safeguard. There is no developed policy on what to do if the user creates a new, non-SUL account. Global locks do affect single accounts, I think, but such usage has not been broadly approved. A steward who blocked a non-disruptive local account, not SUL'd, would clearly be attempting to interfere in local business, however. I don't expect it, for it is effectively an over-ride, since local admins can't unlock or set lock exemptions (as they may with global blocks). I have seen local accounts locked, in spite of no cross-wiki disruption, based on a request from a Wikiversity user. No notice was provided on Wikiversity of the request. I considered that disruptive. The user has stopped editing; the last time he requested such, it was denied. The original grant was probably based on inadequate review of evidence and lack of opposition. --Abd 16:33, 4 November 2011 (UTC)

 * (edit conflict) I'd suggest reviewing the meta ban discussion. The claims of improper behavior were all old. There was no current disruption, other than the global ban discussion itself, which more or less invited every enemy of Poetlister to show up (and which originally asserted "identity theft" which was inflammatory and not true. The proposer later redacted that, but meanwhile many !votes had been added.) The implementing steward was not opposed to local action bypassing the "global ban." I had proposed, in the ban discussion, that Wikiversity be exempted by the steward doing the renaming which delinks. He declined, suggesting that this would require local action. 'Crats here have been asked to accomplish the delinking, they have not done so, apparently either opposing it or not being ready or willing to take a stand in this controversy.
 * Meta has never had policy-making authority over the individual wikis. Global locks were designed to be uncontroversial, serving the wikis, not constraining them.
 * In the case of Thekohser, the global ban was not enforced. A local account was created so he'd have a new local talk page he could use. Because of substantial and active local sentiment for keeping him blocked, that user was blocked immediately (by me) -- with his consent. He then edited project pages by IP, listing them locally on his Talk page. He used self-reversion to reflect respect for the local blocks which existed. All his edits in that phase were reverted back in by local users. When a custodian decided "the experiment was over," and blocked (though no harm was being done at all), the matter then went to the community, which decided by 3:1 to allow him to edit. It was considered cleaner to use the renaming device to detach his local account from the global account. Our tradition is strong: users are not disruptive here, they are welcome. This is not an excuse to use this project to host attacks on other wikis and users at other wikis. Neither Thekohser (after the delinking) nor Poetlister (ever) have done that. I would (and have) supported steward intervention should this happen and local custodians not handle the matter.
 * Poetlister is a highly experienced user, he's been a 'crat and checkuser, and was never shown to have abused those privileges. What he did in the past was irregular (and has been highly dramatized to maximize the apparent outrageousness), but I put a lot of effort into reviewing his history on the Talk page of the global ban discussion, and I found that the only "disruption" was, in fact, over his identity, for years. The meta RfC was started as a direct descendant of a foundation-l posting which was lamenting the independence of Wikiversity. Since the only known active account of Poetlister was the Wikiversity account, since there was no disruption at all from this account, the global ban must be seen as a direct attempt to interference in the affairs of Wikiversity, and to exert control over this wiki by users who are not dedicated to its success.
 * As near as I can tell, the close of the RfC was based more on numbers than on arguments, for no present danger was asserted, and my own analysis was that, if Poetlister were inclined to be disruptive, the ban would make it more difficult to detect and interdict, not easier. What I've seen on Wikipedia is an increased reliance on quick noticeboard bans, moving away from the original requirement for an RfC and thus patient and reasoned argument. It becomes a question of what size of mob can quickly be assembled. And where. --Abd 16:20, 4 November 2011 (UTC)
 * About the comment Poetlister is a highly experienced user, he's been a 'crat and checkuser, and was never shown to have abused those privileges., given that checkuser logs are private and not publicly discussed (excluding WR leaks), how would you suppose that it would be "shown"? Perhaps it would be better stated that he has not been publically accused of abusing either of these tools.  It is mainly the admin tools that he stands accused of abusing (unblocking his socks, for example) and accused of violating policy in ways that do not involve tools (such as vote stacking). Thenub314 19:46, 4 November 2011 (UTC)
 * That's correct. "Never accused" is actually stronger than "never shown," because the latter might require proof. The "vote stacking" that I reviewed was strange. All of it was of no effect, as far as I could tell, it did not shift outcomes. Let me put it this way, I can go on Wikipedia any day and see tolerated user behavior that is worse in its effect on the project and the community than what I saw in reviewing the evidence in the meta RfC. Poetlister definitely made mistakes. So do others whom we don't block and ban. We fix the errors, and we do what we can to make sure they don't happen again.
 * If Poetlister were inclined to do what he did before, the global lock would not prevent it. For example, the latest offenses alleged at meta involved using email. I asked for evidence of this, none was provided. However, global locks don't prevent email. If anyone here wants to communicate with Poetlister, it's easy, just email him through the interface and if he chooses to reply, he can directly, but he can also send emails through the user interface here.
 * The biggest concern seemed to be socking. Locking his known accounts won't prevent socking, it may encourage it. If he has an open account, openly used, it is then easy to pick up current IP, which can then be compared with that of a suspected sock. So what was the point of the global ban? It's fairly obvious to me, and it is certainly not about protecting Wikiversity, nor is it about protecting the other wikis, it does nothing to do that. It's about anger and punishment, an attitude that has infected the wikis, which is far from the original vision. --Abd 20:11, 4 November 2011 (UTC)
 * I think your being a bit unfair to the people commenting at meta. I do not believe it is about anger and punishment, but rather the belief that he will disrupt communities.  Are you arguing creating multiple socks, giving them the appearence of being different people and having several of them vote on issues didn't cause any disruption?  Someone had to initially go through the votes of all of these socks to look for vote stacking, are you sure they never changed the outcome?  How can you be sure? For example, since !voting is not voting, I myself don't always comment if I cam clearly hold a minority point of view that will not get consensus.  How do you measure that number of people who did not comment because they came and saw a plethora of people supporting the same position?  More importantly, if the individual communities involved say he did cause disruption within their community then how can you claim he did not?  Your correct that locking/blocking doesn't prevent socking for a skilled user, but neither does doing nothing.  I think everyone involved agrees with that.  The point is to make clear what should happen if a steward discovers a sock, I don't think this was an emotional issue for the people who were involved.  As far as Poetlister personally goes, I have never seen him cause a problem here so he may go on editing here, I am not going to block him.  I am just commenting your logic seems rather shaky to me. Thenub314 21:30, 4 November 2011 (UTC)
 * Thanks, Thenub, I'm glad we agree on not blocking. My comment is based on extensive research and reflection. I actually reviewed many of those AfDs and documented my observations. Have you read this? I reviewed the history of some of the users commenting with regard to Poetlister. I have only written about a fraction of what I've seen. You are making observations, Thenub, which might seem reasonable, but which appear to be void of review of evidence. In any case, this is not the place to judge the "guilt or innocence" of Poetlister. My point is that discussions on other wikis can suffer from participation bias and reliance on comments from users with a "history" with the subject (as can discussions here). There are three central points I'm making, the rest may be dicta.
 * Poetlister has not been disruptive here, except as others have attempted to harass him here. We should guard against that.
 * There were no credible allegations of recent abuse from him, on any wiki. The actual evidence from his last session, on WikiSource, as Longfellow, showed the reverse, a history of contributions so solid that he was about to be approved as an admin, and the only problem was that his prior identity had not been disclosed, and we can see why, in fact, and, note, this was not secret from the 'crats, who had explicitly permitted it. I see some sign that they may have been embarrassed by the subsequent disclosure, and later blamed him for what they, in fact, allowed. Which in no way harmed the community, when compared with his positive contributions. Nevertheless, my opinion is that they were foolish to allow an admin candidacy for a user with a concealed past, and I argued the same here in a recent case. "Returning user," former identity not disclosed, with the past being, very likely, problematic, not a problem. Probationary custodian, here, yes, not really a problem. Permanent custodianship, with a concealed past, a problem. Those 'crats (in hindsight, to be sure), should have required Longfellow to disclose his former identity in order to become an administrator. We don't need that for probationary custodians, because the work is supervised and may be easily restrained (with the Standard stop agreement, truly easy).
 * There was, mentioned in the discussion, an origin for the initiative, and it was explicitly aimed at Wikiversity and at our well-established policy of allowing users with negative records elsewhere to openly edit here. This was, then, not a discussion aimed at protecting unsuspecting wikis from allowing Poetlister to edit. It was aimed at us. That's just what happened, Thenub. There is a faction of users who dislike our policy and would stop it, who would deny us the benefit of these experienced users. That, sometimes, bans elsewhere might be unfair or unjust is something that we don't attempt to decide, and we even avoid detailed discussion of it, if possible, because that will then invite more fruitless debate over matters that we, by our own policy, don't need to decide. If users abuse Wikiversity for an external agenda, that's to be stopped. There is a fine line, sometimes difficult to discern, over "wikistudies," vs "cross-wiki disruption," and we need to develop ethical standards for these studies, but mostly users have been behaving and not abusing Wikiversity.
 * The general argument seems to be that Poetlister causes disruption. However, at WikiSource, he was a net positive, even considering the relatively mild disruption over his admin candidacy and what ensued. He's been a net positive here, as well, worth this low level of discussion or "disruption." --Abd 23:06, 4 November 2011 (UTC)
 * I've raised a question as to the status of global bans, and whether they can be overridden locally, on Meta. As that page isn't particularly active, if that doesn't work I'll handball the query more directly to the WMF - the new Terms of Use are likely to make the question moot if they pass, but clearly they're still a draft. Having read the various discussions, it does seem that the intent of the global ban procedure was to override local consensus for seriously problematic users, and, more particularly, to prevent Poetlister from editing here. Whether or not this is possible is, I guess, the question that needs clarification, although my assumption is that the decision on meta was intended to be binding. - Bilby 21:17, 4 November 2011 (UTC)
 * Abd has pointed out that Meta was the wrong place to ask,, so I've sent an email to Steven Walling, as he seemed to be the WMF member most active in the global ban discussion. Either he's the right person to send the query to, or I figure he'll let me know who is. :) I should mention that I'm not asking if Wikiversity should allow Poetlister to edit, as that is rightly a question for the Wikiversity community. My only question is whether or not Wikiversity can let Poetlister edit under a global ban from Meta. That, it seems to me, is a WMF question. Obviously, if this is a decision we can make, then the issue returns to us, so some clarity would help. - Bilby 00:20, 5 November 2011 (UTC)
 * If the intent of the global ban was to prevent Poetlister from editing here, that was never explicit in the meta discussion, it's inferred from the source. Thus the meta ban, for many commenting, may be based on a general perception of danger. I see that Phillipe has now commented from his personal experience. It appears to be a repeat of the very old complaint about Poetlister, but is presented as a current danger, in spite of lack of evidence that whatever happened then is not being repeated, and hasn't been repeated for a long time. If there were evidence of immediate disruption, of the usage of Wikiversity as a haven from which to damage other wikis, there would be a coordinating issue, i.e., justification for steward action. But there has been none of this. Phillipe is asserting secret evidence. If there is evidence that must remain secret, the WMF may indeed act according to its own counsel. But Phillipe is not only not speaking for the WMF, he's written he would recuse from that board discussion, which I find unexplained.
 * The closing steward for the global lock was Fr33kman. I discussed this issue extensively with him.. The steward does not see the global ban as preventing local wikis from bypassing it (though he doesn't like the idea). This is not a WMF action. And precedent has been that the local wikis may indeed undo the effect of a global lock, it's been done quite a few times by the local wikis, and the principle that meta and other wikis don't set policy for the individual wikis is well-established. The WMF, the legal owner, may intervene, but has almost entirely kept away from that, for obvious reasons. It is no-win and substantial risk, for them. --Abd 21:15, 5 November 2011 (UTC)
 * I've had a reponse, both on wiki and via email, from Steven Walling. In essence, this is an unusual case, and not comparable to TheKosher or previous situations, as they weren't the result of a wide community discussion for a global ban. The Wikiversity community was informed of the ban discussion and the result,  as was Poetlister, and were able to be involved - certainly Abd was, extensively. The decision, though, was intended to be cross-wiki, and the decision was intended to mean that PoetLister, as a user not an account, was no longer permitted to edit on any WMF wiki.
 * On those grounds, I feel strongly we should respect the decision reached on Meta, rather than second-guessing the community discussion there. If that can't be decided here, I guess we need to take it to the Colloquium. - Bilby 23:53, 5 November 2011 (UTC)

 * Poetlister on any name should be immediately blocked, and any admin who unblocks any of his socks should be immediately desysopped. As for claims he never abused the CU tool, when he was a CU he used his access to gain access to the global CU mailing list and then used some of the personal information that was part of the global list in outing at least one other user. There are people above spreading disinformation in support of Poetlister and that shouldn't be tolerated. Ottava Rima (talk) 23:49, 5 November 2011 (UTC)
 * Ottava, can you provide evidence about the checkuser abuse claim?

 * As to an admin unblocking, we have ample precedent here that blocking and unblocking follows local policy and local consensus, not something else. We have very sound precedent that says, as SBJ once put it, to maintain a block requires consensus, not unblock. If any user is blocked here in violation of our policy, I would expect any custodian to respond responsibly and appropriately. This community has a voluntary relationship with the WMF, involving mutual service, and meta is a coordinating wiki, not a government, nor is the meta community the owner of this site, nor does the meta community create and maintain this site, we do. Stewards and global sysops serve us, not the other way around. (And that support is routine and deeply appreciated.) I am concerned that some might be inviting stewards to intervene without necessity. If that is happening, it's cross-wiki disruption. I'm confident that this community has both the courage and the maturity to respond as needed. --Abd 01:56, 6 November 2011 (UTC)

 * I presume that means the decision won't be happening here. I'll give it some more time to see if things change, then I guess we will need to bring it to the community via the Colloquium for discussion, per your and DarkLama's suggestion. - Bilby 07:28, 6 November 2011 (UTC)
 * Evidence has been provided over and over and over. You were banned at Meta partly because people kept providing you with it then a short while later you would feign forgetting and demand it over again. Such actions are disruptive, especially when you were a prior participant in the discussions. There is also no precedence for mass disruption that unblocking Poetlister would be. Such is completely inappropriate. Ottava Rima (talk) 15:03, 6 November 2011 (UTC)

 * Can anyone point to evidence that Poetlister abused checkuser? I just reviewed the RfC at meta, and found no evidence of it there. The claim of abuse seems to be inconsistent with testimony presented in Requests for comment/Poetlister and Cato, which would surely have covered such abuse if it had been known. An article in the Signpost seems to deny that any abuse was even possible.
 * "Cato's checkuser actions were monitored by another Wikiquote checkuser; throughout his time as a checkuser, only one action, the check of one of his own sockpuppets, appeared illegitimate. While Cato's OTRS access was terminated, his checkuser access was not yet removed, because it was determined that abuse was unlikely as long as the investigation remained private."
 * (Cato = Poetlister.) If evidence has been provided "over and over," then it should be easy to find and point to it. Thanks for any help with this. --Abd 15:32, 6 November 2011 (UTC)
 * "Can anyone point to evidence that Poetlister abused checkuser?" He was removed as a CU because he used the CU email list to get personal information on people which he posted at places like WR. This is common knowledge and has been discussed forever. Ottava Rima (talk) 03:23, 7 November 2011 (UTC)

 * As to my request above, as I'm closing this, and unless this is re-opened, please provide any evidence on my Talk page, or in subsequent discussion. That evidence would not be relevant to a present custodian action. If another discussion is opened, a note should be placed here linking to it. --Abd 17:18, 6 November 2011 (UTC)

I request that only a custodian willing to act to block the user, or personally needing more immediate community comment to make such a decision, because there is consensus that this belongs on the Colloquium or possibly a Community Review, if it is to be pursued. --Abd 17:18, 6 November 2011 (UTC)}

 * This certainly needs discussion, yes. --SB_Johnny talk 22:01, 6 November 2011 (UTC)

I'm blocking the account now as the "default position". I think we should have community consensus locally before ignoring community consensus in the interwiki space. --SB_Johnny talk 21:56, 6 November 2011 (UTC)

Request unblock of Poetlister1

I will now assert that SB_Johnny should have recused from this, but I will not provide evidence on this point at this time. Recusal failure, in fact, is not a reason to unblock.

Rather, the reason for unblock is simple. WV:Blocking policy is clear.
 * Blocks are used to stop behavior inconsistent with the Wikiversity mission and scope ...
 * Only people who knowingly act and are likely to continue to act in a way that is inconsistent with the Wikiversity mission and scope should be blocked. In practice, this means that only blatant vandals and spammers can be blocked without prior warning about their behavior.
 * Blocks for behavior that has ceased or may happen in the future are inappropriate. Blocks are to deter continuing recent behavior.
 * Blocks that divide the Wikiversity community or cause community tension should be reversed quickly.

SBJ has resurrected an old, discredited reason for blocking here: global ban. This has been asserted before and rejected by consensus, and not just on Wikiversity, on many wikis. Indeed, SBJ himself rejected it. A global ban existed in that former case, the only difference was that it was a "defacto ban," i.e., no steward was willing to unlock, whereas the Poetlister ban is by a community, all right, but not the Wikiversity community, only a narrow slice of users who inhabit meta, and who know little about Wikiversity's function, purposes, and traditions.

Because the record of Poetlister's edits shows extensive positive contribution, with no sign of disruption or policy violation, because policy does not, then, allow him to be blocked, because Poetlister1 is an openly declared alternate account, which we permit, I am therefore requesting that another custodian unblock Poetlister1, pending further discussion if needed. --Abd 23:41, 6 November 2011 (UTC)

 * Declined. On the one hand, I would never enact an unblock request began by someone other than the blocked user. It is up to him to request an unblock on his talk page if he desires.  Secondly, there is no clear policy violation, the blocking policy is a proposal that has never been passed.  Finally, the facts which kept me from acting (Poetlister and WV not being invited into the discussion) turned out to be false, thus making the whole process much more reasonable.  I do not personally see this as an attack on WV, or SBJ serving some sort of "other master".  As far as I understand, SBJ and Poetlister have no particular history that should cause him to recuse.  Also we should keep in mind, if memory serves, SBJ has been the victim of "outside intervention" when Jimbo removed his tools.  It seems a bit odd to now declare he is the agent of some sort of outside intervention. Thenub314 02:37, 7 November 2011 (UTC)
 * True, there has been no unblock request. However, it's obvious that Poetlister wants to be able to edit, he wouldn't have created the alternate account if he didn't want to.
 * Poetlister was informed, all right, but was unable to participate in the discussion. As to Wikiversity, there is a problem with holding a discussion on meta about the future of a Wikiversity user. Our users participate and have watchlists here. They may or may not even have accounts at meta.
 * There is a history with Poetlister and SBJ. Trust me. It will come out in subsequent process. And that's not relevant to unblock, remember? As to an attack on WV, that I can establish with plenty of evidence, but, again, the point is not that, it's that Poetlister has not been disruptive here, yet has extensive positive contributions. Wikiversity has a strong tradition on this, and this has shown up in many discussions. It's consensus here, whether or not a particular policy has ever formally been adopted.
 * SBJ was standing, during that Jimbo incident, for Wikiversity vs outside intervention. Yes, it's odd, indeed. SBJ resigned for a long time, and came back to break the Gordian knot over the global ban of Thekohser, though I'd paved the way. There, he stood for local autonomy as well. Something shifted, it shows in his contributions, but, again, SBJ isn't the point. He's not the "agent" of some conspiracy, he's carrying water for outside forces, perhaps for his own reasons. --Abd 03:08, 7 November 2011 (UTC)

 * Oppose and block proposer - There is no way to assume this proposal is done in good faith and Abd was banned at Meta for disruption associated with Poetlister, Abigor, etc. Abd has a history of using Wikiversity for disrupting other Wikis and that is embarrassing to the community as a whole. Ottava Rima (talk) 03:23, 7 November 2011 (UTC)
 * I'm not "banned" at meta, I'm blocked, and it had little to do with Poetlister, if at all. Unblock has not been denied, and there is a steward who is investigating. Slowly.

 * Ottava, above, makes some serious and disruptive charges, violating declared restrictions. Is that going to be tolerated? Obviously, I'm not the one to answer that.

 * As to the meta events, let me put it this way: if that block stands at meta, meta is lost, all semblance of respect for due process and for the freedom to criticize within the bounds of civility is gone. And if the right of a meta discussion, involving a relatively small number of users at meta, not here, to ban users from participating at Wikiversity, prohibiting us from allowing it, with no offenses taking place at wikiversity, and no cross-wiki disruption using Wikiversity as a base, if that stands, Wikiversity as a safe haven for positive contributions is lost and it's dangerous to continue investing in this community. It's become a tail for the Wikipedia community to wag. --Abd 14:58, 7 November 2011 (UTC)
 * You were banned there, not blocked. The ban was determined by many, many admin on Meta. Wikiversity would do good to ban you here too. You have provided nothing to Wikiversity but continue your crusade to try and get obviously dangerous people unblocked elsewhere. That is not what Wikiversity is for. Ottava Rima (talk) 15:45, 7 November 2011 (UTC)

 * @Abd.
 * In response to your first point, one thing your arguments have abundantly clear is that I have no way to interpret Poetlister's motives.  Why CU one of his own socks? Why double vote in something that will obviously pass?  Why create an account with no intention of editing.  I don't know, but if he wants to be unblocked I am sure he will request it.
 * Well, I have not seen any such evidence about SBJ and Poetlister, so accepting your word on it makes as much sense as accepting other peoples word about Poetlister's deeds. In Thekohser's case a unlock was sought as a first step, has Poetlister requested the crat's delink his accounts here?  Unlike Thekohser's case we were invited to discuss the matter before the lock was instituted.  So the two cases are in no way comparable.
 * I have read the foundation-l list. I contend there is nothing about WV that was at issue.  Had it been any WMF project people would have reacted similarly.  It is an attack on Poetlister directly.
 * I wasn't part of the global block request, which is a shame, but for various reasons I was taking a wikibreak then and I was not aware it was going on. But overall my opinion is that if he want to unblocked, delinked, allowed to edit in some way would then we should take things one step at a time.  First he can request it at his talk page, then we can seek a community consensus on the matter.  Figure out what to do from there. Thenub314 03:57, 7 November 2011 (UTC)

 * Oppose - Poetlister is globally banned. We have a duty to enforce the global ban. There really can be no local unblock decision without a reversal of the global ban.
 * I share my concerns with Ottava Rima about Abd's behaviour here. This is not appropriate behaviour by a custodian. Custodians should not be supporting users which have been banned by the whole Wikimedia community and helping them to edit past their restrictions. -- Simone  08:09, 7 November 2011 (UTC)
 * Custodians have tools to serve the community. But we are also members of the community and may advocate our own positions. I have not used tools to "support" Poetlister, except as I support every participant here. I have argued for positions that serve the Wikiversity community and that maintain protection against harassment, both of our users from outside and of users at the other wikis by users here. Simone is not a supporter of Wikiversity, see . Rather, she came here to enforce outside standards, see her record, and, by the way, the reinstatement of her Wikipedia block. Poetlister has not been "banned by the whole Wikimedia community," and certainly not by this community. The issue here is whether or not meta serves Wikiversity or Wikiversity serves meta. The design was the former, not the latter. If the contract is being changed, we should know. --Abd 13:12, 7 November 2011 (UTC)

 * Comment: While I've historically opposed global bans imposed without community discussion, this particular case did come out of a community discussion. As Bilby pointed out, there seems to be changes going on with how these things are handled in general, see meta:Talk:Global bans, and in particular meta:Talk:Global bans/Archive 1. --SB_Johnny talk 13:17, 7 November 2011 (UTC)
 * There are changes happening at meta, for sure. What I see is the rise there of users who don't understand the original wiki vision, and the early community norms, which were highly libertarian. I've seen stuff going down at meta that would have been unthinkable even a year ago. Consider my current block there. I'd been civilly critical of certain specific actions. I was blocked at the demand of a steward, who provided piles of "evidence," and the obvious cause was that I'd asked him what he was concerned about, he'd claimed I'd made misleading comments. In other words, I'd asked him some clue as to what he proceeded to assert as evidence for my block, so that I could correct any errors. He'd refused, and I wrote "suit yourself," and dropped it. I was then blocked, per his request, without warning, without violating any warning, and with a clean block record, and with no immediate risk or ongoing disruption. I was blocked infinite with email cut off. And with no general discussion in any place designed for that. One of the "offenses," about which there had been admin complaint, was being a non-admin discussing requests for admin action, there were meta sysops who thought non-admins should not discuss proposed actions on that page. In other words, that page could not be a venue for a community decision. It's Orwellian. (A parallel is community bans decided on AN/I on Wikipedia, bypassing the original requirement for a preceeding user RfC. General users don't watch AN/I. It floods the watchlist.)
 * It used to require a user RfC to manage a block like that (i.e, for some sort of "general disruption," as distinct from immediate disruption, which was not alleged) And even then, RfCs were often ignored, there are plenty of old ones never closed. Meta is a mess.
 * I was actually shocked. I knew the work of that steward. He'd seemed sober and responsive (as you commented, SBJ). Something has changed. I suspect burnout. I've seen it happen to many once-good admins on Wikipedia. And then the unblock template sits with no response. I know what used to happen with things like this. It isn't happening any more. I see this as a sign that the meta community is dying, it's becoming senescent, there is a loss of vision. It used to be very difficult to get blocked at meta, and to stay blocked, you had to continue and repeat offenses, and even then it was difficult.
 * The ones who are taking over want meta to be a government, for control is what they want. It appeals to them. --Abd 14:48, 7 November 2011 (UTC)

 * Decline. I think Abd and SB_Johnny both agree community consensus is needed. I think the disagreement is over what the status of a global locked account should be in the mean time. I think RCA is intended to request action for obvious cases where community discussion and consensus are unlikely to be needed. I think whether the right or wrong action was taken, and what should happen now are outside the intended uses of RCA. -- dark lama  17:01, 7 November 2011 (UTC)
 * The position is reasonable. Because I believe our policy is clear, that we have a standing consensus that users who are not disruptive here are not to be blocked here, I'll leave this request open, myself. However, the whole discussion here should be closed, based on your argument, but I'd ask that it not be moved to the archive yet. A custodian might see it and decide to act according to existing policy. If SBJ had not acted without first finding consensus, there would be no issue for me, for this page, so it remains possible that another custodian may act to reverse that. --Abd 17:30, 7 November 2011 (UTC)

Unblock request
Poetlister1 has requested unblock. Except for my involvement, I would unblock, because there is no policy that Poetlister1 (or Poetlister) have violated here, and our traditions (and policies and guidelines) are clear that a user is not to be blocked who has not violated policies, unless there is an established community consensus otherwise. Our policy is explicit that behavior elsewhere isn't relevant. Prior opinion, even by SBJ, who blocked Poetlister1, has confirmed this clearly. That can be overridden by a ban consensus here, regardless of "violations," but there is no such consensus, nor has it been sought. Arbitrary, ad hoc blocks like this have ripped Wikiversity apart in the past. How about some sane, sober, and straightforward adherence to policy, and if the policy is wrong, we change it by consensus? --Abd 13:22, 8 November 2011 (UTC)
 * A community consensus to unblock will need to be established. Someone should start a Community Review. (I have to travel to speak today, so I couldn't do a fair job.) Thenub314 14:26, 8 November 2011 (UTC)
 * Thanks, Thenub. Ordinarily, it does not require the formation of a specific consensus to follow established policy. The prior case of Thekohser confirmed that, in the close, that block should require consensus, if not based clearly in policy, not unblock. It's an important issue that all custodians should understand, I'll claim. There is no allegation on the table that Poetlister violated *any* Wikiversity policy, and precedent and our explicit established policy is that behavior elsewhere is irrelevant. Hence custodians do have guidance, based on community consensus, already. The burden of establishing a CR should be on those who want to make an exception to what's established. It's always been that way, in fact. What's established is that users are not blocked except for violations of our policy, aside from emergencies, and an emergency should be based on actual and present damage, likely to continue, not mere fear of same. What was the emergency? --Abd 16:13, 8 November 2011 (UTC)
 * To be clear: Thekohser's case established consensus for one thing and one thing only, how to handle Thekohser's case. No policy or general consensus was established.  There is a global community involved, and so the question is not simply what has he done here ignoring what happened elsewhere, but his actions on the whole should be considered.
 * Even accepting that Thekohser's case were to establish some consensus about larger matters then Thekohser, the facts of this case are too different to apply an "unblock because that is what we do" mentality. There was no real discussion in Thekohser's case, if there was WV was not invited to participate in that discussion.  Here there was a discussion at meta, it involved our community as well as others and the consensus was to ban.  So in my view there is no "exception" as to what has been established.  Thenub314 16:56, 8 November 2011 (UTC)
 * Closing comment in Thekohser's case was explicit on our policy and traditions. It was not simply about Thekohser, personally. Indeed, that was hardly an issue. The argument against Thekohser at the time was quite the same: he'd been "globally banned," and, in fact, he was globally banned -- and still is -- just not by a meta discussion. The dangerous precedent here is that a meta discussion (which could easily have been obtained re Thekohser, closed as ban, if it were thought necessary, see the Jimbo RfC which was basically about the same issue: running 2:1 to support interference in Wikiversity, until similar interference at Commons, which has a much larger user based, led to 4:1 against interference) is being presumed to have overriding authority over our local policies and consensus. Yes, our community was invited to participate in the meta discussion -- through a non-neutral notice, by the way, later refactored -- but that's something that would appear only transiently on watchlists of our users, and a discussion of this import would, here, probably go into the site notice. In any case, meta discussions will be dominated by meta users. That's not our community, though certainly we have users who do watch meta. Some users, such as Poetlister, are banned at meta, so Poetlister could not defend himself. The fundamental issue here is the same as before: are Wikiversity sysops responsible to meta, or to the local community? If any sysop here considers the answer to that to be "meta," then I suggest they be explicit about it, so that we, as a community, may make informed decisions.
 * This is not about the ability of stewards to act to protect other wikis. They already have that power, and we may not locally, as individual sysops, interfere with that. But no harm to other wikis has been alleged here, and certainly no ongoing harm, no present danger, no actual disruption, and our policy on blocking is quite clear, supported by precedent. Any sysop here may enforce that policy. As usual, no sysop is obligated to take any specific action. A CR, as matters stand, will review the blocking action, given that it was contrary to policy. I'd rather make that unnecessary. --Abd 17:32, 8 November 2011 (UTC)
 * In one sentence or less, which policy was this contrary to? Thenub314 18:06, 8 November 2011 (UTC)
 * WV:Blocking policy is stated as "proposed," but the relevant language has stood without objection for years, and it is actual practice. --Abd 19:11, 8 November 2011 (UTC)
 * My memory is quite different, I recall much discontent over the blocking policy as a whole, particularly when Moulton was active and Adambro had placed a large range block to stop him editing. Some felt it should be more strongly worded, others felt it should be weaker, overall there was no sense of consensus, and I haven't seen anything to suggest that it ever did have a consensus.  Just because something is untouched (or a particular portion) does not constitute any sort of approval or consensus. In partial reply to your email, your vote in my candidacy for custodianship has no barring on my opinions.  This is not politics. Thenub314 03:58, 9 November 2011 (UTC)
 * There was, at times, support for enforcing global bans, but most of those users left. The community settled on a position which was already reflected in the policy, so the policy was not changed. That something is left on a wiki, like WV:Blocking policy, does constitute a kind of consensus. Consensus can change. Is Thenub claiming that we have no blocking policy, that custodians are free to do block whomever, that it's a free-for-all? The position is "interesting," and that's why Thenub's candid remarks (and the candor is appreciated) will likely come up in the discussion of his candidacy. Freedom from harassment here due to outside conflicts is a fundamental Wikiversity tradition, and I don't want to see any more custodians who support outside interference (absent clear and present damage and danger) and who might be willing to enable and reinforce it.
 * The Moulton case, though it involved a global ban, was different because Moulton had been locally blocked and was block evading, and his editing was often disruptive in itself. I attempted to engage Moulton, as I had Thekohser, and Moulton was not cooperative. He was unblocked anyway, and the community accepted it to give him another chance. He abused it. I criticized Adambro for his personally placing the range blocks, because he'd become a personal enforcer (which inflames conflicts), but the range blocks were needed. He did block Thekohser's alternative account, the same kind of problem, because it wasn't disruptive, and that triggered the discussion that led to delinking Thekohser. --Abd 13:14, 9 November 2011 (UTC)

 * Oppose unblock Ottava Rima (talk) 18:06, 8 November 2011 (UTC)
 * Oppose unblock. Poetlister is globally banned. Ban evasion cannot be tolerated. If Poetlister should be allowed to edit Wikiversity, the global ban needs to be undone first. Simone  18:12, 8 November 2011 (UTC)
 * This is the same argument as was presented in 2010 for User:Thekohser, and it was not only rejected by numerous wikis, Wikiversity included, by consensus, the effective global ban for Thekohser still stands, and would probably be impossible to lift at meta. Meta is its own wiki, and it doesn't operate or control this wiki, period. The last time that outside interference was attempted, here and at Commons, Jimbo lost most of his Founder tools over it. This is an attempt to create an end run around that. The existence of a global ban discussion at meta -- which could easily have been done with Thekohser, meta normally follows a certain line -- changes nothing about the fundamental issue: local control or meta control?
 * We had a local sysop in 2010 who carried water for the outside interference, it was one who has commented here and who lost his tools here, though not specifically over that. In 2010, SBJ stood for local control and was desysopped by Jimbo for it, but the tools were returned when SBJ essentially agreed not to wheel-war with Jimbo. And then he resigned, because he was unwilling to work under that kind of coercion. I thought that quite noble, it's part of what influenced me to participate here. SBJ came back to close the unblock/delinking of Thekohser, thus taking a clear stand for our independence. I do not know why he has changed his position here. There is no practical difference in the situations.
 * The status quo is this: meta can lock accounts globally, thus implementing a defacto global ban, as a decision to protect all the WMF wikis by default. Local wikis may then undo this, as to themselves, in a number of ways, and all have been used, I believe. A local, alternate, non-SUL account can be created and used, ideally with proper supervision. And a 'crat may rename the locked account and name it back, thus delinking the account, preventing its use for global editing and removing it from the linkage needed for a lock. This protects other wikis, that account cannot then be SUL'd, if I'm correct. These have all been accepted, and no process has yet established the contrary. But these discussions can be quite disruptive, and I've attempted to avoid this. The local block of Poetlister, contrary to our precedent and policy, may be forcing that discussion to take place. Maybe that's a good thing. Maybe not. The claim that the "global ban" must be lifted is pure invention. --Abd 19:11, 8 November 2011 (UTC)
 * Thekohser was never globally banned. He was merely locked by Jimbo. There is a difference. Ottava Rima (talk) 16:26, 9 November 2011 (UTC)
 * Thekohser global status:. Global ban was declared by Jimbo, yes, and many steward actions refer to it. Jimbo also blocked Thekohser locally. Jimbo did not issue any global lock. That was done by stewards, to implement the ban. Apparently because it was realized that this created a problem, a possible abuse of the global lock tool, a steward then went around and locally blocked, to allow local admins to unblock, and the lock was lifted. Some local wikis unblocked. Then another steward, about a month later, re-instated the lock, per global ban, "per discussion," I asked him about it. He clearly claims that Thekohser is globally banned. The lock was bypassed and the account was unblocked through discussion and 'crat action, on several wikis. In many discussions, Thekohser has been referred to, by many users, as globally banned. Sometimes the term "globally blocked" is used. That's the status quo. Thekohser is still globally banned, just not by a specific meta discussion. The account is still SUL locked. If Thekohser is not globally banned, then why has the lock been maintained? He is banned, but local wikis may decide to allow him to edit. Strong precedent. I fail to see a major difference, in principle, as to what we are considering, between an "open discussion" on meta -- not open to all members in good standing of the individual communities --, and a private discussion with an obvious operating consensus between stewards. From a prior meta discussion of Thekohser (where most of those commenting were in favor of unblocking Thekohser on meta), we can see how the reality is that meta administrators toe the meta line, and they, and stewards, who are also elected at meta, control conclusions. That's just the way it is. I'm not complaining, but it's preposterous to imagine that global bans only exist if there is a formally closed "ban discussion" at meta. It's just like Wikipedia, in fact. The meta ban for Poetlister was properly closed, as to procedure. The issue, however, is one of jurisdiction. Meta discussions can properly control use of meta tools, such as the global lock, and there are sound reasons for this. But if they are going to force communities to act or refrain from acting, according to local consensus, a major basis for the wiki system largely collapses. The people who do the work, as a community, control the community, i.e., it controls itself. I should not be able to control whether or not a user can edit here, by going to another wiki and voting for a global ban. Yet that is exactly what Wikipedians (and others) have attempted to do. --Abd 17:57, 9 November 2011 (UTC)
 * Jimbo was the source of the lock. There was no global ban. The Stewards only implemented what Jimbo declared. Why you bothered to post so much of the above when it is obvious that Thekohser was never globally banned is not apparent. Ottava Rima (talk) 18:16, 9 November 2011 (UTC)

 * A point that I'm unclear about: I'd noticed that socks of Thekohser had been globally locked. However, the one mentioned above, m:User:Thekohser-2 was an SUL account, with two wikis attached. User:Ethical Accountability, also SUL, was locally blocked here (without any disruption here, similar issue), leading to a mess, ultimately concluded by delinking the main account. EA SUL was locked much later. Can a steward lock a non-SUL account, like User:Poetlister1? I think not, CentralAuth doesn't even list the account, and if not, then this account can only be prevented from editing by a local admin action, like that of User:SB_Johnny. (Or by a steward or global sysop using tools locally, and I've only seen that happen in emergencies.) If a matter has the attention of local sysops, and there is no emergency (such as outing), it's broadly understood that this is not done.
 * Meta (properly) has control over SUL accounts, to prevent widespread disruption, not over individual unlinked, local accounts. A global ban is implemented with a global lock, just as a local site ban is implemented with a local block. Bans are generally of two kinds. The Wikipedia w:WP:Banning policy used to describe both: a formal ban, issued by any of a number of different means (which is what the present version of the policy refers to, note no mention of "global bans") and a "defacto ban," which now has some lipstick on it, implying that it's based on a discussion; however, if we look at actual cases, that discussion may merely be a request to administrators to unblock, with no administrator being willing. Wikipedia seems to allow this to take place in 24 hours! One of the basic kinds of community bans is that an editor has been indef blocked, and, after some notice, no administrator is willing to unblock. The equivalent with a global ban, implemented by global lock, is that no steward is willing to unlock. So Thekohser is clearly globally banned. He can only edit, under that account name, where permitted by a local community. Or with socks, and it's clear that if a sock is created that is SUL, it will be locked when noticed. Still. Per ban. All bans can be reversed. A defacto local ban is reversed by any administrator willing to unblock, or a defacto global ban by any steward willing to unlock. --Abd 19:06, 9 November 2011 (UTC)

Ban Abd
I am formally requesting a ban of Abd for disruption in his attempt to try and get Poetlister unblocked among many, many other problems. He has directly made up claims about Poetlister, ignored evidence that has been discussed many times, and puts up walls of text to try and wear others down. There is also no legitimate reason to even think about unblocking Poetlister yet he continued. This is a very long pattern of disruption, and it will never stop as long as Abd is allowed to edit here.

 * Support
 * 1) Ottava Rima (talk)  15:47, 7 November 2011 (UTC)

 * Oppose
 * 1) This is an extreme solution to a genuine problem. Abd's support of Poetlister is disruptive in my view, needs discussing, and may be a reason to reconsider Abd's sysop status. However, Abd is generally a productive user at Wikiversity, and I don't think a ban is in order at the moment. Perhaps it would be good to have a community review of the subject. Abd's use of "walls of text" is indeed a genuine issue, and needs a solution apart from this. -- Simone  16:18, 7 November 2011 (UTC)
 * 2) Ottava has done this before, start a ban discussion in an inappropriate place. When it was pointed out that it was the wrong place, he then selectively copied part of it to the proper place, WV:CR, to create an impression of "momentum." There is no allegation of tool abuse, so this is not for WV:Custodian feedback. I do not recommend allowing a Community Review started by Ottava, his have proven extraordinarily disruptive in the past. A ban on an active sysop is a matter that would require site message. This is a page for asking custodians to act, not for seeking community consensus. If the day has come that a user cannot request reversal of a custodian action on this page, without being accused of disruption, Wikiversity is lost. I'd welcome a Community Review if not for the disruption and wasted time it would create. Procedurally, by the way, I should have asked SBJ on his talk page to reverse his action first. I overlooked that step, my apologies. Nevertheless the request I made here is solidly grounded in WV:Blocking policy and our traditions. --Abd 16:48, 7 November 2011 (UTC)
 * Comments
 * @Simone. Thanks. I have not used my sysop tools in this matter. I use the tools to serve this community, and very few of my actions have been controversial (and many of the controversial ones have been ultimately confirmed by the community -- a recent exception is standing on RfD right now, I certainly won't repeat what I did then, I just want the community to express itself.) As to "walls of text," I have never rejected reasonable refactoring, by collapsing or moving to a Talk page, with neutral summary -- or no summary. Collapse can be used with request for summarization. I hope that people understand that sometimes issues are complex and require more than sound bites for response. I'll note that I was last topic banned on Wikipedia because of alleged excessive text on meta, requesting delisting of a web site, which arguments proved necessary, the request would have clearly been denied if the issues were not explored in detail. In my view this page should not be a discussion page at all, it should be requests for custodian review, with practically nothing else. We do not want this to become the Wikiversity w:WP:ANI, a continuous train wreck. --Abd 16:35, 7 November 2011 (UTC)

 * "Abd is generally a productive user at Wikiversity" Simone, there is no evidence for this. Instead, Abd has chased off many academics, encouraged disruptive users, and has done whatever he can to try and destroy Wikiversity. Ottava Rima (talk) 17:16, 7 November 2011 (UTC)

Crosswiki spam
In addition to this request: please delete User:Scotthancocksc as well. Thanks in advance. Trijnstel 17:33, 7 November 2011 (UTC)
 * ✅. Thanks, Trijnstel. --Abd 17:41, 7 November 2011 (UTC)
 * Please, check also user:Grabthecheapautoinsurance. Not 'cross-wiki' spam, but should be blocked too IMO and edits deleted due to 'blatant' spam... a great opportunity to use Special:Nuke. If possible, please pay a little more attention to new pages on userspace; spamming is daily. Thanks a lot.” TeleS (T M @ C G) 17:45, 7 November 2011 (UTC)

 * ✅. Indeed. Thanks, Teles. Nuked contributions, indeffed account. Never used Nuke before. Cool. At one point I was watching RC every day, but I'm a bit distracted lately, eh? I'll check that page, though, at least briefly. I did look at Newpages, and found some pages, all of which were already tagged for speedy deletion. I cleaned that up, there were only three files. I blocked some more spam-only accounts. --Abd 18:19, 7 November 2011 (UTC)

Spam
Please, check edits done by this IP. A lot spam edits have been done. Thanks.” TeleS (T M @ C G) 07:31, 20 November 2011 (UTC)
 * ✅ - Thankyou - blocked 1 week. -- Jtneill - Talk - c 08:11, 20 November 2011 (UTC)

Crosswiki spam (2)
Please delete User:Jasondixonjaso and User:Michaelcampbel for crosswiki spam by using multiple accounts as well as anonymous. Thanks in advance. Trijnstel 23:03, 22 November 2011 (UTC)
 * ✅ Thank you, and keep up the good work! Thenub314 23:07, 22 November 2011 (UTC)


text:
User:Egm6321.f10.team2.oztekin/hw7.1

R*7.1 - Finding ds and Laplace Operator Equivalent in Spherical Coordinates
From the lecture slide Mtg 39-1

Given
Infinitesimal length in Cartesian coordinates given as in Mtg 38-5 ;

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

$$\begin{align} ds=dx_{j}.e_{j} \end{align}$$ (7.1.1)
 *
 * }

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

$$\begin{align} ds^{2}=ds.ds=\left ( dx_{j}.e_{j} \right )\left ( dx_{i}e_{i} \right )=dx_{i}.dx_{j}(e_{i}e_{j}) \end{align}$$ (7.1.2)
 *
 * }

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

$$\begin{align} e_{i}e_{j}=\delta _{ij} \end{align}$$ (7.1.3)
 *
 * }

Delta function defined as ;

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

$$\begin{align} \delta _{ij}=\left \{ \begin{matrix} 1 & for & i=j\\ 0& for  & i\neq j \end{matrix} \right. \end{align}$$ (7.1.4)
 *
 * }

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

$$\begin{align} ds^{2}=dx_{i}.dx_{i}=\sum_{i=1}^{3}\left ( dx_{i} \right )^{2} \end{align}$$ (7.1.5)
 *
 * }

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

$$\begin{align} &x_{1}=x=rcos(\theta )cos(\phi )=\xi _{1}cos(\xi _{2})cos(\xi _{3}) \\ &x_{2}=y=rcos(\theta )sin(\phi )=\xi _{1}cos(\xi _{2})sin(\xi _{3}) \\ &x_{3}=z=rsin(\theta )=\xi _{1}sin(\xi _{2}) \\

\end{align}$$ (7.1.6)
 *
 * }

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

$$\begin{align}

\Delta u =\left [ \frac{1}{h_{1}h_{2}h_{3}} \right ]\sum_{i=1}^{3}\frac{\partial }{\partial \xi _{i}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{i}^{2}} \frac{\partial u }{\partial \xi _{i}}\right ]

\end{align}$$ (7.1.7)
 *
 * }

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

$$\begin{align} \begin{matrix} \hat{\xi }=(\hat{\xi _{1}},\hat{\xi _{2}},\hat{\xi _{3}}) \end{matrix} \end{align}$$ (7.1.8)
 *
 * }

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

$$\begin{align} \begin{matrix} h_{1}h_{2}h_{3}=r^{2}cos(\theta ) \end{matrix} \end{align}$$ (7.1.9)
 *
 * }

Find

 * Show that infinitesimal length 'ds' can be written as (5) in meeting 39-1 in spherical coordinates.
 * Derive Laplace operator in spherical coordinates.

Solution

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

$$\begin{align} dx_{1}=\left [ \frac{\partial x_{1}}{\partial r} \right ]dr+\left [ \frac{\partial x_{1}}{\partial \theta } \right ]d\theta +\left [ \frac{\partial x_{1}}{\partial \phi } \right ]d\phi

\end{align}$$ (7.1.10)
 *
 * }
 * {| style="width:100%" border="0"

$$\begin{align} dx_{2}=\left [ \frac{\partial x_{2}}{\partial r} \right ]dr+\left [ \frac{\partial x_{2}}{\partial \theta } \right ]d\theta +\left [ \frac{\partial x_{2}}{\partial \phi } \right ]d\phi \end{align}$$ (7.1.11)
 *
 * }

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

$$\begin{align} dx_{3}=\left [ \frac{\partial x_{3}}{\partial r} \right ]dr+\left [ \frac{\partial x_{3}}{\partial \theta } \right ]d\theta +\left [ \frac{\partial x_{3}}{\partial \phi } \right ]d\phi \end{align}$$ (7.1.12)
 *
 * }

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

$$\begin{align} dx_{1}=\frac{\partial r}{\partial r}cos(\theta )cos(\phi )dr+r\frac{\partial cos(\theta )}{\partial \theta }cos(\phi )d\theta +rcos(\theta )\frac{\partial cos(\phi )}{\partial \phi }d\phi

\end{align}$$ (7.1.13)
 *
 * }

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

$$\begin{align} dx_{2}=\frac{\partial r}{\partial r}cos(\theta )sin(\phi )dr+r\frac{\partial cos(\theta )}{\partial \theta }sin(\phi )d\theta +rcos(\theta )\frac{\partial sin(\phi )}{\partial \phi }d\phi \end{align}$$ (7.1.14)
 *
 * }

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

$$\begin{align} dx_{3}=\frac{\partial r}{\partial r}sin\theta dr+r\frac{\partial sin(\theta )}{\partial \theta }d\theta \end{align}$$ (7.1.15)
 *
 * }

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

$$\begin{align} dx_{1}^{2}=&(cos\theta)^{2} (cos\phi)^{2} dr^{2}+r^{2}(sin\theta)^{2} (cos\phi )^{2}d\theta ^{2}+r^{2}(cos\theta)^{2} (sin\phi)^{2} d\phi ^{2}+2r^{2}sin\phi cos\phi (sin\theta) (cos\theta) d\theta d\phi \\ &-2r(cos\theta) (cos\phi)^{2} (sin\theta) drd\theta -2r(cos\theta) (cos\phi) (sin\phi) drd\phi

\end{align}$$ (7.1.16)
 *
 * }
 * {| style="width:100%" border="0"

$$\begin{align} dx_{2}^{2}=&(cos\theta)^{2} (sin\phi)^{2} dr^{2}+r^{2}(sin\theta)^{2} (sin\phi )^{2}d\theta^{2} +r^{2}(cos\theta)^{2} (cos\phi)^{2} d\phi ^{2}+2r(cos\theta )^{2}(sin\phi) (cos\phi) drd\phi \\ &-2r^{2}(sin\theta) (sin\phi) (cos\theta) (cos\phi )d\theta d\phi -2r(cos\theta) (sin\theta) sin^{2}\phi drd\theta \end{align}$$ (7.1.17)
 *

 * }

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

$$\begin{align} dx_{3}^{2}=(sin\theta)^{2} dr^{2}+r^{2}(cos\theta )^{2}d\theta ^{2}+2r(sin\theta ) (cos\theta) drd\theta \end{align}$$ (7.1.18)
 *

 * }

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

$$\begin{align} ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2} \end{align}$$ (7.1.19)
 *

 * }

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

$$\begin{align} ds^{2}=&{\color{Green} (cos\theta)^{2} (cos\phi)^{2} dr^{2}}+{\color{Blue} r^{2}(sin\theta)^{2} (cos\phi)^{2} d\theta ^{2}}+{\color{Cyan} r^{2}(cos\theta)^{2} (sin\phi)^{2} d\phi ^{2}}+{\color{Green} (cos\theta)^{2} (sin\phi)^{2} dr^{2}}+{\color{Blue} r^{2}(sin\theta)^{2} (sin\phi)^{2} d\theta^{2}}\\ & +{\color{Cyan} r^{2}(cos\theta)^{2} (cos\phi)^{2} d\phi ^{2}}+(sin\theta)^{2} dr^{2}+r^{2}(cos\theta)^{2} d\theta ^{2}+{\color{Red} 2r^{2}sin\phi cos\phi sin\theta cos\theta d\theta d\phi} -2rcos\theta (cos\phi)^{2} sin\theta drd\theta \\ & -{\color{Orange} 2r(cos\theta)^{2} cos\phi sin\phi drd\phi}+{\color{Orange} 2r(cos\theta)^{2} sin\phi cos\phi drd\phi }-{\color{Red} 2r^{2}sin\theta sin\phi cos\theta cos\phi d\theta d\phi} -2rcos\theta sin\theta sin^{2}\phi drd\theta \\ &+2rsin\theta cos\theta drd\theta \end{align}$$ (7.1.20)
 *

 * }

If we group things together we will get ;

$$  \displaystyle ds^{2}=1.dr^{2}+r^{2}d\theta ^{2}+r^{2}(cos\theta)^{2} d\phi ^{2} $$

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

$$\begin{align} &i=1\\ &\frac{\partial }{\partial \xi _{1}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{1}^{2}} \frac{\partial u }{\partial \xi _{1}}\right ]=\frac{\partial }{\partial r}\left [ \frac{r^{2}cos(\theta )}{(1)^{2}} \frac{\partial u }{\partial r}\right ]

\end{align}$$

(7.1.21)
 *

 * }

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

$$\begin{align} &i=2\\ &\frac{\partial }{\partial \xi _{2}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{2}^{2}} \frac{\partial u }{\partial \xi _{2}}\right ]=\frac{\partial }{\partial \theta }\left [ \frac{r^{2}cos(\theta )}{(r)^{2}} \frac{\partial u }{\partial \theta }\right ] \end{align}$$ (7.1.22)
 *

 * }

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

$$\begin{align} &i=3\\ &\frac{\partial }{\partial \xi _{3}}\left [ \frac{h_{1}h_{2}h_{3}}{h_{3}^{2}} \frac{\partial u }{\partial \xi _{3}}\right ]=\frac{\partial }{\partial \phi }\left [ \frac{r^{2}cos(\theta )}{(r)^{2}(cos\theta )^{2}} \frac{\partial u }{\partial \phi }\right ] \end{align}$$ (7.1.23)
 *

 * }

If we substitute in

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

$$\begin{align} \Delta u =\frac{1}{r^{2}cos(\theta )}\left [ \frac{\partial }{\partial r}\left [ \frac{r^{2}cos(\theta )}{(1)^{2}} \frac{\partial u }{\partial r}\right ]+\frac{\partial }{\partial \theta }\left [ \frac{r^{2}cos(\theta )}{(r)^{2}} \frac{\partial u }{\partial \theta }\right ]+\frac{\partial }{\partial \phi }\left [ \frac{r^{2}cos(\theta )}{(r)^{2}(cos\theta )^{2}} \frac{\partial u }{\partial \phi }\right ] \right ] \end{align}$$ (7.1.24)
 *

 * }

$$  \displaystyle \Delta u =\frac{1}{r^{2}}\left ( r^{2}\frac{\partial u }{\partial r} \right )+\frac{1}{r^{2}cos\theta }\frac{\partial }{\partial \theta }\left ( cos\theta \frac{\partial u }{\partial \theta } \right )+\frac{1}{r^{2}(cos\theta)^{2} }\frac{\partial ^2 u }{\partial \phi ^2} $$

= References =

title: User:Egm6321.f10.team2.oztekin/hw7.1


text:
User:Egm6321.f11.team1.colsonfe/HW7

= R*7.4 Laplacian In Elliptic Coordinates =

Given
See wikipdia reference of ellipitic coordinates.

Laplacian operator:

 * $$\Delta u=\frac{1}{h_1h_2h_3}\sum^3_{i=3}\frac{\partial}{\partial\xi_i}\left[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}\right]$$

== Find ==

Verify the Laplacian in elliptic coordinates:

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

Solution
Using the reference listed above, we find that

$$\displaystyle \xi_1=\mu$$

$$\displaystyle\xi_2=\nu$$

$$\displaystyle h_1=h_2=a\sqrt{\sinh^2\mu+\sin^2\nu}$$

Therefore, using the definition of Laplacian operator in the above section,

$$\nabla^2u=\Delta u=\frac{1}{h_1h_2}\left[\frac{\partial}{\partial\xi_1}\left(\frac{h_1h_2}{(h_1)^2}\frac{\partial u}{\partial\xi_1}\right)+\frac{\partial}{\partial\xi_2}\left(\frac{h_1h_2}{(h_2)^2}\frac{\partial u}{\partial\xi_2}\right)\right]$$

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

Therefore we arrive at our solution.

= R7.5 Laplacian In Parabolic Coordinates =

Given
See wikipdia reference of parabolic coordinates.

Laplacian operator:

 * $$\Delta u=\frac{1}{h_1h_2h_3}\sum^3_{i=3}\frac{\partial}{\partial\xi_i}\left[\frac{h_1h_2h_3}{(h_i)^2}\frac{\partial u}{\partial\xi_i}\right]$$

== Find ==

Verify the Laplacian in parabolic coordinates:

 * $$\nabla^2u=\frac{1}{\mu^2+\nu^2}\left(\frac{\partial^2u}{\partial\mu^2}+\frac{\partial^2u}{\partial\nu^2}\right)$$

Solution
Using the reference listed above, we find that

$$\displaystyle \xi_1=\sigma$$

$$\displaystyle\xi_2=\tau$$

$$\displaystyle h_1=h_2=\sqrt{\sigma^2+\tau^2}$$

I will use the following definitions to match the syntax given in the problem.

$$\displaystyle \mu:=\sigma$$

$$\displaystyle \nu:=\tau$$

Therefore, using the definition of Laplacian operator in the above section,

$$\nabla^2u=\Delta u=\frac{1}{h_1h_2}\left[\frac{\partial}{\partial\xi_1}\left(\frac{h_1h_2}{(h_1)^2}\frac{\partial u}{\partial\xi_1}\right)+\frac{\partial}{\partial\xi_2}\left(\frac{h_1h_2}{(h_2)^2}\frac{\partial u}{\partial\xi_2}\right)\right]$$

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

Therefore we arrive at our solution.

= R7.9 Express Vector As Linear Combination of Basis Vectors =

Given
Non-orthonormal basis:

 * $$\displaystyle{\mathbf b_1,\mathbf b_2,\mathbf b_3}$$

Orthonormal basis:

 * $$\displaystyle{\mathbf e_1,\mathbf e_2,\mathbf e_3}$$

Expressing the non-orthonormal in the orthonormal basis:

 * $$\displaystyle\mathbf b_i=A_{ij}\mathbf e_j$$

 * where

 * $$\mathbf A=[A_{ij}]=\left[\begin{matrix}5&2&3\\4&5&6\\7&8&5\end{matrix}\right]$$

Vector $$\displaystyle\mathbf v$$ in $$\displaystyle\mathbb{R}^3$$:

 * $$\displaystyle-2\mathbf e_1+4\mathbf e_2-5\mathbf e_3$$

Definition of Gram Matrix:

 * $$\boldsymbol\Gamma=[\Gamma_{ij}]=[(\mathbf b_i\cdot\mathbf b_j)]\in\mathbb R^{n\times n}$$

Finding components of $$\displaystyle\mathbf v$$

 * $$\displaystyle{v_j}=\mathbf\Gamma^{-1}\{\mathbf b_i\mathbf v\}$$

== Find ==

 * $$\displaystyle{\mathbf v_i}\in\mathbb{R}^{3\times 1}$$ such that $$\displaystyle\mathbf=v_i\mathbf b_i$$

Solution
To begin, find each entry of the Gram Matrix:

$$\Gamma_{11}=\mathbf b_1\cdot\mathbf b_1=25+4+9=38$$

$$\Gamma_{12}=\mathbf b_1\cdot\mathbf b_2=20+10+18=48$$

$$\Gamma_{13}=\mathbf b_1\cdot\mathbf b_3=35+16+25=76$$

$$\Gamma_{21}=\mathbf b_2\cdot\mathbf b_1=48$$

$$\Gamma_{22}=\mathbf b_2\cdot\mathbf b_2=16+25+36=77$$

$$\Gamma_{23}=\mathbf b_2\cdot\mathbf b_3=28+40+30=98$$

$$\Gamma_{31}=\mathbf b_3\cdot\mathbf b_1=76$$

$$\Gamma_{32}=\mathbf b_3\cdot\mathbf b_2=98$$

$$\Gamma_{33}=\mathbf b_2\cdot\mathbf b_3=49+64+25=128$$

Thus

$$\boldsymbol\Gamma=\left[\begin{matrix}38&48&76\\48&77&98\\76&98&128\end{matrix}\right]$$

And

$$\boldsymbol\Gamma^{-1}=\left[\begin{matrix}-0.017&-0.086&0.076\\-0.086&0.060&0.005\\0.076&0.005&0.041\end{matrix}\right]$$

Also we can see that

$$\displaystyle\mathbf b_1\cdot\mathbf v=-10+8-15=-17$$

$$\displaystyle\mathbf b_2\cdot\mathbf v=-8+20-30=-18$$

$$\displaystyle\mathbf b_3\cdot\mathbf v=-14+32-25=-7$$

So then

$$\displaystyle\mathbf b_i\cdot\mathbf v=\left[\begin{matrix}-17\\-18\\-7\end{matrix}\right]$$

Therefore by matrix multiplication

$$\displaystyle\Gamma^{-1}\{\mathbf b_i\cdot\mathbf v\}=\left[\begin{matrix}1.305\\0.347\\-1.095\end{matrix}\right]$$

This implies our solution is

= References =

title: User:Egm6321.f11.team1.colsonfe/HW7