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User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework6

= Problem 2 - Solution of L2-ODE-VC=

Given

 * Given two $$\displaystyle L2-ODE-VC $$,

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

 * $$\displaystyle \left( a \right) \left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=f\left( x \right)$$
 * $$\displaystyle (Eq. 2.1)$$
 * }
 * $$\displaystyle (Eq. 2.1)$$
 * }
 * }

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

 * $$\displaystyle \left( b \right) x{{y}^{''}}+2{{y}^{'}}+xy=f\left( x \right)$$
 * $$\displaystyle (Eq. 2.2)$$
 * }
 * $$\displaystyle (Eq. 2.2)$$
 * }
 * }

Find

 * (1) Solve $$\displaystyle Eq.2.1$$ and $$\displaystyle Eq.2.2$$ for the case: $$\displaystyle f\left( x \right)=0$$

 * (2) Solve $$\displaystyle Eq.2.1$$ and $$\displaystyle Eq.2.2$$ for the case: $$\displaystyle f\left( x \right)=\sin x$$

(1.1) Solve $$\displaystyle Eq.2.1$$ for $$\displaystyle f\left( x \right)=0$$ (homogenous)

 * For the case of $$\displaystyle f\left( x \right)=0$$, $$\displaystyle Eq.2.1$$ can be written as:

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

 * $$\left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=0$$
 * $$\displaystyle (Eq. 2.3)$$
 * }
 * }
 * }

 * Consider a trial solution,

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

 * $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$
 * $$\displaystyle (Eq. 2.4)$$
 * }
 * where $$\displaystyle c$$ and $$\displaystyle r$$ are constants to be determined.
 * }
 * where $$\displaystyle c$$ and $$\displaystyle r$$ are constants to be determined.

 * Taking derivative,

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

 * $$\displaystyle{{y}^{'}}=c{{x}^{c-1}}{{e}^{rx}}+r{{x}^{c}}{{e}^{rx}}$$
 * }
 * }
 * }

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

 * $$\displaystyle{{y}^{''}}=c\left( c-1 \right){{x}^{c-2}}{{e}^{rx}}+2cr{{x}^{c-1}}{{e}^{rx}}+{{r}^{2}}{{x}^{c}}{{e}^{rx}}$$
 * }
 * }
 * }

 * Now substitute $$\displaystyle y$$ and whose derivatives in $$\displaystyle (Eq. 2.3)$$,implies:

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

 * $$\left( x-1 \right){{e}^{rx}}\left[ c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c}} \right]-x{{e}^{rx}}\left[ c{{x}^{c-1}}+r{{x}^{c}} \right]+{{x}^{c}}{{e}^{rx}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow c\left( c-1 \right){{x}^{c-1}}-c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c}}-2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c+1}}-{{r}^{2}}{{x}^{c}}-c{{x}^{c}}-r{{x}^{c+1}}+{{x}^{c}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{x}^{c+1}}\left[ {{r}^{2}}-r \right]+{{x}^{c}}\left[ 2cr-{{r}^{2}}-c+1 \right]+{{x}^{c-1}}\left[ {{c}^{2}}-c-2cr \right]-{{x}^{c-2}}\left[ {{c}^{2}}-c \right]=0$$
 * $$\displaystyle (Eq. 2.5)$$
 * }
 * }
 * }

 * Solving $$\displaystyle (Eq. 2.5)$$ using Wolfram Alpha give constants $$\displaystyle c$$ and $$\displaystyle r$$

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

\begin{array}{*{20}l} i) & c=0 & r=1\\  ii) & c=1 & r=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Now homogenous solutions are obtained by substituting $$\displaystyle c$$ and $$\displaystyle r$$ in $$\displaystyle (Eq. 2.4)$$

 * First Homogenous solution is

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

 * $${{y}_}={{x}^{0}}{{e}^{1x}}={{e}^{x}}$$
 * }
 * }
 * }

 * and second homogenous solution is

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

 * $${{y}_}={{x}^{1}}{{e}^{0x}}=x$$
 * }
 * }
 * }

 * Therefore solution of $$\displaystyle (Eq. 2.3)$$ is the linear combination of two homogenous solution

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

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}={{C}_{1}}\cdot {{e}^{x}}+{{C}_{2}}\cdot x$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 2.6)$$
 * }
 * }

.

(1.2) Solve $$\displaystyle Eq.2.2$$ for $$\displaystyle f\left( x \right)=0$$ (homogenous)

 * For $$\displaystyle f\left( x \right)=0$$, $$\displaystyle Eq.2.2$$ reduces to:

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

 * $$\displaystyle x{{y}^{''}}+2{{y}^{'}}+xy=0$$
 * $$\displaystyle (Eq. 2.7)$$
 * }
 * }
 * }

 * Considering a trial solution for y $$\displaystyle (Eq. 2.4)$$, and substituting y and its derivatives in $$\displaystyle (Eq. 2.7)$$,yields:

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

 * $$x{{e}^{rx}}\left[ c\left( c-1 \right){{x}^{c-2}}+2cr{{x}^{c-1}}+{{r}^{2}}{{x}^{c}} \right]+2{{e}^{rx}}\left[ c{{x}^{c-1}}+r{{x}^{c}} \right]+{{x}^{c+1}}{{e}^{rx}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow c\left( c-1 \right){{x}^{c-1}}+2cr{{x}^{c}}+{{r}^{2}}{{x}^{c+1}}+2c{{x}^{c-1}}+2r{{x}^{c}}+{{x}^{c+1}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{x}^{c+1}}\left( {{r}^{2}}+1 \right)+{{x}^{c}}\left( 2cr+2r \right)+{{x}^{c-1}}\left( {{c}^{2}}+c \right)=0$$
 * $$\displaystyle (Eq. 2.8)$$
 * }
 * }
 * }

 * By Solving $$\displaystyle (Eq. 2.8)$$ using Wolfram Alpha yields constants $$\displaystyle c$$ and $$\displaystyle r$$

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

\begin{array}{*{20}l} i) & c=-1 & r=i\\  ii) & c=-1 & r=-i\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * So homogenous solutions are obtained by plugging $$\displaystyle c$$ and $$\displaystyle r$$ in $$\displaystyle (Eq. 2.4)$$

 * First Homogenous solution is

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

 * $${{y}_}={{x}^{-1}}{{e}^{ix}}=\frac{x}$$
 * }
 * }
 * }

 * and second homogenous solution is

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

 * $${{y}_}={{x}^{-1}}{{e}^{-ix}}=\frac{x}$$
 * }
 * }
 * }

 * Hence solution of $$\displaystyle (Eq. 2.7)$$ is the linear combination of two homogenous solutions

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

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}={{C}_{1}}\cdot \frac{x}+{{C}_{2}}\cdot \frac{x}$$ .
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 2.9)$$
 * }
 * }

(2.1) Solve $$\displaystyle Eq.2.1$$ for $$\displaystyle f\left( x \right)=\sin x$$

 * For the case of $$\displaystyle f\left( x \right)=\sin x$$, $$\displaystyle Eq.2.1$$ can be written as:

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

 * $$\left( x-1 \right){{y}^{''}}-x{{y}^{'}}+y=\sin (x)$$
 * $$\displaystyle (Eq. 2.10)$$
 * }
 * }
 * }

 * Above equation can be rewritten as:

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

 * $$\underbrace{1}_\cdot {{y}^{''}}\underbrace{-\frac{x}{\left( x-1 \right)}}_{{y}^{'}}+\underbrace{\frac{1}{\left( x-1 \right)}}_y=\underbrace{\frac{\sin (x)}{\left( x-1 \right)}}_{F\left( x \right)}$$
 * $$\displaystyle (Eq. 2.11)$$
 * }
 * }
 * }

 * Method of variation of parameters could be used to obtain the complete solution of $$\displaystyle (Eq. 2.11)$$.

 * Integrating factor for the above equation [[media:2010_10_26_15_07_13.djvu|Page3, Mtg 29 (b)]]is,

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

 * $$h\left( x \right)={{u}_{1}}^{2}\left( x \right)\cdot \exp \left[ \int{{{a}_{1}}\left( x \right)dx} \right]$$
 * $$\displaystyle (Eq. 2.12)$$
 * }
 * }
 * }

 * From 1.1, one homogenous solution is,

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

$${{u}_{1}}\left( x \right)={{y}_}={{e}^{x}}$$
 * $$\displaystyle (Eq. 2.13)$$
 * }
 * }

 * $$\displaystyle (Eq. 2.12)$$ becomes:

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

$$h\left( x \right)={{e}^{2x}}\cdot \exp \left[ \int{\left( -\frac{x}{\left( x-1 \right)} \right)dx} \right]$$
 * }
 * }

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

$$h\left( x \right)={{e}^{x-\log \left( x-1 \right)}}$$
 * $$\displaystyle (Eq. 2.14)$$
 * }
 * }

 * Particular solution is given by[[media:2010_10_28_14_00_19.djvu|Page1, Mtg 30 (b)]],

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

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

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

$${{y}_{P}}={{e}^{x}}\int\left( \int{{{e}^{x-\log \left( x-1 \right)}}\cdot \frac{\sin (x)}{\left( x-1 \right)}dx} \right)dx$$
 *
$$\displaystyle (Eq. 2.15)$$
 * }
 * }

 * Now complete solution is obtained by combining $$\displaystyle (Eq. 2.15)$$ and $$\displaystyle (Eq. 2.6)$$,

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

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}+{{y}_{P}}$$
 * }
 * }

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

$$y={{C}_{1}}\cdot {{e}^{x}}+{{C}_{2}}\cdot x+{{e}^{x}}\int\left( \int{{{e}^{x-\log \left( x-1 \right)}}\cdot \frac{\sin (x)}{\left( x-1 \right)}dx} \right)dx$$ .
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.16)$$
 * }
 * }

(2.2) Solve $$\displaystyle Eq.2.2$$ for $$\displaystyle f\left( x \right)=\sin x$$

 * For $$\displaystyle f\left( x \right)=\sin x$$, $$\displaystyle Eq.2.2$$ becomes:

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

$$x{{y}^{''}}+2{{y}^{'}}+xy=\sin \left( x \right)$$
 *
$$\displaystyle (Eq. 2.17)$$
 * }
 * }

 * $$\displaystyle Eq. 2.17$$can be rewritten as:

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

$$\underbrace{1}_\cdot {{y}^{''}}+\underbrace{\frac{2}{x}}_{{y}^{'}}+\underbrace{y}_=\underbrace{\frac{\sin (x)}{x}}_{F\left( x \right)}$$
 *
$$\displaystyle (Eq. 2.18)$$
 * }
 * }

 * As in the previous case, method of variation of parameters could be used to obtain the complete solution of $$\displaystyle (Eq. 2.17)$$.

 * From 1.2, one homogenous solution is,

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

$${{u}_{1}}\left( x \right)={{y}_}=\frac{x}$$
 *
$$\displaystyle (Eq. 2.19)$$
 * }
 * }

 * Integrating factor for $$\displaystyle Eq. 2.18$$ is,

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

 * $$h\left( x \right)={{u}_{1}}^{2}\left( x \right)\cdot \exp \left[ \int{{{a}_{1}}\left( x \right)dx} \right]$$
 *
$$\displaystyle (Eq. 2.20)$$
 * }
 * }
 * }

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

$$h\left( x \right)=\frac\cdot \exp \left[ \int{\left( \frac{2}{x} \right)dx} \right]$$
 * }
 * }

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

$$h\left( x \right)={{e}^{2ix}}$$
 *
$$\displaystyle (Eq. 2.21)$$
 * }
 * }

 * Particular solution is given by,

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

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

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

$${{y}_{P}}=\frac{x}\int\left( \int{{{e}^{2ix}}\frac{\sin (x)}{x}dx} \right)dx$$
 * }
 * }

 * Simplifying using Wolfram Alpha,

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

$${{y}_{P}}=\frac{x}\int\left( i\left( \frac{Ei\left( ix \right)-Ei\left( 3ix \right)}{2} \right) \right)dx$$
 * }
 * }

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

$${{y}_{P}}=\frac{4x}\left( {{e}^{2ix}}Ei\left( -ix \right)-\left( 1+{{e}^{2ix}} \right)Ei\left( ix \right)+Ei\left( 3ix \right) \right)$$
 *
$$\displaystyle (Eq. 2.22)$$
 * }
 * }

 * where $$\displaystyle Ei(x)$$ is the exponential integral

 * Therefore complete solution is the combination of $$\displaystyle (Eq. 2.22)$$ and $$\displaystyle (Eq. 2.9)$$,

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

$$y={{C}_{1}}\cdot {{y}_}+{{C}_{2}}\cdot {{y}_}+{{y}_{P}}$$
 * }
 * }

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

$$y={{C}_{1}}\cdot \frac{x}+{{C}_{2}}\cdot \frac{x}+\frac{4x}\left( {{e}^{2ix}}Ei\left( -ix \right)-\left( 1+{{e}^{2ix}} \right)Ei\left( ix \right)+Ei\left( 3ix \right) \right)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.23)$$
 * }
 * }

= Problem 9 - Verification of First Homogenous Solutions of Legendre Equation=

Given

 * Legendre Equation,

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

$$\displaystyle \left( 1-{{x}^{2}} \right){{y}^{''}}-2x{{y}^{'}}+n\left( n+1 \right)y=0$$
 *
$$\displaystyle (Eq. 9.1)$$
 * }
 * }

 * Legendre polynomials:

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

$$\displaystyle{{P}_{0}}(x)=1$$
 *
$$\displaystyle (Eq. 9.2)$$
 * }
 * }

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

$$\displaystyle{{P}_{1}}(x)=x$$
 *
$$\displaystyle (Eq. 9.3)$$
 * }
 * }

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

$${{P}_{2}}(x)=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$
 *
$$\displaystyle (Eq. 9.4)$$
 * }
 * }

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

$${{P}_{3}}(x)=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$
 *
$$\displaystyle (Eq. 9.5)$$
 * }
 * }

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

$${{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$
 *
$$\displaystyle (Eq. 9.6)$$
 * }
 * }

Find

 * Verify that $$\displaystyle Eqs. 9.2-9.6 $$ are solutions of Legendre equation.

(1) n=0,$$\displaystyle {{P}_{0}}(x)=1$$

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

\begin{array}{*{20}l} y=1 & {{y}^{'}}=0 & {{y}^{''}}=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

$$\left( 1-{{x}^{2}} \right)\times 0-2x\times 0+0\times 1=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.7)$$
 * }
 * }

(2) n=1,$$\displaystyle{{P}_{1}}(x)=x$$

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

\begin{array}{*{20}l} y=x & {{y}^{'}}=1 & {{y}^{''}}=0\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

$$\left( 1-{{x}^{2}} \right)\times 0-2x\times 1+2\times x=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.8)$$
 * }
 * }

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

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

\begin{array}{*{20}l} y=\frac{1}{2}\left( 3{{x}^{2}}-1 \right) & {{y}^{'}}=3x & {{y}^{''}}=3\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

$$\left( 1-{{x}^{2}} \right)\times 3-2x\times 3x+6\times \frac{1}{2}\left( 3{{x}^{2}}-1 \right)=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.9)$$
 * }
 * }

(4) n=3,$${{P}_{3}}(x)=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$

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

\begin{array}{*{20}l} y=\frac{1}{2}\left( 5{{x}^{3}}-3x \right) & {{y}^{'}}=\frac{1}{2}\left( 15{{x}^{2}}-3 \right) & {{y}^{''}}=15x\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

$$\left( 1-{{x}^{2}} \right)\times 15x-2x\times \frac{1}{2}\left( 15{{x}^{2}}-3 \right)+12\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.10)$$
 * }
 * }

(5) n=4,$${{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$

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

\begin{array}{*{20}l} y=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} & {{y}^{'}}=\frac{35}{2}{{x}^{3}}-\frac{15}{2}x & {{y}^{''}}=\frac{105}{2}{{x}^{2}}-\frac{15}{2}\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above values in LHS of $$\displaystyle Eq.9.1$$ yields:

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

 * $$\left( 1-{{x}^{2}} \right)\times 15x-2x\times \frac{1}{2}\left( 15{{x}^{2}}-3 \right)+12\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)=0$$
 * }
 * }
 * }

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

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

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

$$=\frac{2}\left( 175-105-70 \right)+\frac{2}\left( 105+15+30-150 \right)+\frac{15}{2}-\frac{15}{2}=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 9.11)$$
 * }
 * }

 * Hence $$\displaystyle Eqs. 9.2-9.6 $$ are solutions of Legendre equation.

=problem3=

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

$$ {Z}^{\prime}+ \left[ \frac{2 {{u}_{1}}^{\prime}}{{u}_{1}}- \frac{2x}{ \left(1- {x}^{2} \right)} \right]Z= \frac{1}{ {u}_{1} {\left(1- {x}^{2} \right)}^{2}} $$ The integrating factor is
 *
$$\displaystyle (Eq. 3.8) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$h\left( x \right)=exp\left[ \int{\left( \frac{2{{u}_{1}}^{\prime }}-\frac{2x}{\left( 1-{{x}^{2}} \right)} \right)}dx \right]$$ Substituting in our homogeneous solution and integrating gives
 *
$$\displaystyle (Eq. 3.9) $$
 * }
 * }

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

$$h\left( x \right)=exp\left[ log\left( 1-{{x}^{2}} \right)+2log\left( x \right) \right]$$
 * }
 * }

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

$$h\left( x \right)={{x}^{2}}\left( 1-{{x}^{2}} \right)$$
 *
$$\displaystyle (Eq. 3.10) $$
 * }
 * }

The solution to equation 3.8 can be written as

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

$$Z\left( x \right)=\frac{1}{h\left( x \right)}\left[ {{k}_{2}}+\int{\left[ h\left( x \right)\frac{f\left( x \right)}{{{u}_{1}}(x)} \right]}dx \right]$$ Substituting in
 *
$$\displaystyle (Eq. 3.11) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}+\int{\left[ \frac{{{x}^{2}}\left( 1-{{x}^{2}} \right)}{x{{\left( 1-{{x}^{2}} \right)}^{2}}} \right]}dx \right]$$
 * }
 * }

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

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}+\int{\left[ \frac{x}{\left( 1-{{x}^{2}} \right)} \right]}dx \right]$$
 * }
 * }

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

$$Z\left( x \right)=\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}\left[ {{k}_{2}}-\frac{1}{2}\log \left( 1-{{x}^{2}} \right) \right]$$
 *
$$\displaystyle (Eq. 3.12) $$
 * }
 * }

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

$$U\left( x \right)={{k}_{1}}+\int{Z}\left( x \right)dx$$
 * }
 * }

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

$$U\left( x \right)={{k}_{1}}+\int{\left( \frac{{{x}^{2}}\left( 1-{{x}^{2}} \right)}-\frac{1}{2}\frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 *
$$\displaystyle (Eq. 3.13) $$
 * }
 * }

Therefore,

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

$$y\left( x \right)=U\cdot {{u}_{1}}\left( x \right)$$
 * }
 * }

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

$$y\left( x \right)={{k}_{1}}x+{{k}_{2}}x\int{\frac{1}{{{x}^{2}}\left( 1-{{x}^{2}} \right)}dx}-\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 * }
 * }

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

$$y\left( x \right)={{k}_{1}}x+{{k}_{2}}x\cdot \frac{1}{2}\left( -\frac{2}{x}-\log (1-x)+\log 1+x \right)-\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx$$
 * }
 * }

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

$$y\left( x \right)={{k}_{1}}\underbrace{x}_+{{k}_{2}}\left( \underbrace{\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1}_ \right)-\underbrace{\frac{x}{2}\int{\left( \frac{\log \left( 1-{{x}^{2}} \right)}{{{x}^{2}}\left( 1-{{x}^{2}} \right)} \right)}dx}_$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 3.14) $$
 * }
 * }


text:
Creating a video lecture

How to use a web cam to create a video recorded lecture, publish it to Youtube, and embed that Youtube video on another website - such as a Learning Management System.

Considerations

 * As a rule of thumb, keep your lectures short. 5-10 minutes short. If you must take the full hour, break it up into 6 or 12 chapters
 * Video recorded lectures made available online are a very different format compared to a live lecture given in a theatre. Your audience online is more easily distracted, can stop, pause and replay, and is usually not in a social setting. Many advise that short videos delivering concise information, presented close up, personally, and with an authentic voice will help to maintain interest and engagement.
 * Video data is relatively expensive to download, therefore keep your videos short, concise and on message. Consider the option of offering audio recording only, and offering your slides to download as well. Audio only is less expensive to download, and gives the added flexibility of listening while doing other things.

Recording a video
You have 3 options:
 * 1) Record with a webcam and computer, using software on a website (fastest option)
 * 2) Record with a webcam and computer, using software on the computer (Requires software installation, but affords better vision quality)
 * 3) Record with a video camera and computer (Requires more expensive camera, and software installation, but gives best quality in both vision and sound)

Webcam and website software

 * Many popular video sharing websites, such as Youtube, offer the feature of recording video through a webcam using software on their website. This method is certainly the fastest way to get a video online, but the quality is low, and the reliability of the recording is determined by the reliability of your interent connection at during the time of recording.
 * Step by step instructions: Using webcam to record video on Youtube
 * Video demonstration: Adoptafeature: Quick Capture

Webcam and computer software

 * Most webcams come with software you can use to record with. If you are recording with such software, make a test recording and find out where the video is being saved. It will be either on your desktop, in your system folder called "videos", or in a folder with a similar name to the software you're using.
 * 1) Find the software for your webcam. Look for an icon for the webcam software on your desktop or task bar. If you can't find it, try unplugging your webcam and plugging it back in. The option to open the software should appear - depending on your computer. Otherwise, look through the software that is available on the computer.
 * 2) Record a 10 minute video. The video will save either on your computer's desktop, "Videos" folder, or in a folder with a similar name to the webcam software.
 * 3) Preview the video.
 * 4) Upload to You Tube.
 * Step by step instructions: How to Make Movies on Your Windows XP Computer
 * Video demonstration: How To Record From Your Webcam

Publish the video
You should aim to upload your video to at least 3 locations. Think of this as a backup. One location should be a popular site where your video is most likely to be found and seen, like Youtube. You're next location should be somewhere it is most likely going to remain available in a public archive, like Archive.org. Finally, you might like your video to be available on a very popular and non commercial education service, like Wikimedia Commons.

Upload to Youtube

 * 1) Go http://youtube.com (type this in your web browser address bar)
 * 2) Check that you are logged in (see top tight of Youtube. If you see your name you are logged in)
 * 3) Click “Upload” then click the yellow button “upload video”
 * 4) Find the file you recorded previously on your computer
 * 5) Start filling in the details on the form. Name and describe the video to a wide audience. Be sure to use key word tags that link your video to subjects, projects and groups
 * 6) When video has finished uploading, you will see a message in a green box at the top, with a link the video page. Click that to see the video's page on Youtube.

Embeding Youtube videos on your own website
In this step, we display a Youtube video on our own website.
 * 1) On the video's Youtube page, find the "embed" code button and copy that text.
 * 2) Go to your website, in this case we are using Moodle.
 * 3) Turn editing on and click HTML view.
 * 4) Paste the embed code, then save.

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

title: Creating a video lecture


text:
User:KBlott

Categories that interest me (Wikiversity may call these “seminars”)

 * math
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Pages I've created

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Articles that I am currently writing.

 * Mathics
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Pages I've modified
Public health strategies for HIV

How to cite stuff
template filler

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I’m going to Florida, whether or not the weather is clement.

title: User:KBlott


text:
User:Spock of Vulcan

Hi! I'm Lianne, and I'm new here (joined on November 14, 2010). I am not so new over on the English Wikipedia, where I am quite an active editor.

I'm quickly learning that some parts of Wikiversity have sort of been abandoned, which is unfortunate. Therefore, I'm placing high value on community here. So, if you need any help with computer science or philosophy, want to talk about editing, school organization, or whatever, or just want to say hi, post on my talk page. Maybe we can get some life back in those abandoned schools.

What I can give to Wikiversity
I have a B.C.Sc. (bachelor of computer science). As such, I probably know most of the material covered by a lot of the computer science courses offered here. So if you need help with any of that, feel free to leave me a message on my talk page, and I'll see what I can do!

I also have a minor in philosophy. That certainly doesn't make me any kind of expert, but I'm pretty well versed in the basics, especially on the topic of ethics. I'm currently working on organizing the resources in the School of Philosophy.

My Learning Goals

 * I am currently learning Esperanto and French. These are my top priority learning goals.
 * I'd like to take the courses offered in the School of Computer Science that are on topics I didn't get a chance to study while doing my B.C.Sc..

News
This section is mostly here so that you can tell I'm still here. There are so many Wikiversity users that seem to have disappeared. So, as long as there is a recent news item here, you'll know I'm still here!

 * July 5, 2011 - I haven't been here in a while, largely due to major life changes such as graduating, finding a job, moving out of my parents' house, etc. Also, I've been devoting the bulk of my free time to my language studies. However, I do still have goals here at Wikiversity, so hopefully in the near future I'll have more time to work on them.
 * December 31, 2010 - I got back to editing after a Wikibreak necessitated by university. I'm done school now!
 * November 16, 2010 - I found the School of Comparative Mythology, and started improving it.
 * November 15, 2010 - I found the School of Philosophy, and started improving it.
 * November 14, 2010 - I joined Wikiversity!

title: User:Spock of Vulcan


text:
User:Egm6321.f10.team4.Yoon/HW6

= Relationship Between Artanh(x) function And Legendre Functions $$\displaystyle {{Q}_{0}}(x),{{Q}_{1}}(x)$$ =

Solution for part.a
To find out the relationship between Legendre equation and artanh function, let's derivate artanh function from tanh fuction first.

let

Then, moving the denominator of the right hand side to the left hand side.

Multiplying e^y on the both side of the equation

Collecting the term of x yields

Deviding by (x-1)

Taking log on the both side of the equation,

Deviding by 2

In conclusion, artanh fuction is identical with Legendre series(Q0).

Solution for part.b
Multiplying x and substract 1 on both side of equation yield,

= Problem 2. Non-homog. L2_ODE_VC using Variable of Paramerters ==

From Meeting 33, p 33-1.

Given
Solve the Non-homog L2_ODE_VC with the condition $$\displaystyle f(x)=0 $$. K.p.28, Pb 1.1 ab '''Part. a'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (x-1)y''-xy'+y=f(x) $$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

'''Part. b'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle xy''+2y'+xy=f(x) $$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

General trial solution is given on Lecture note p.32-1 as,
 * {| style="width:100%" border="0" align="left"

 * style="width:10%; padding:10px; border:2px solid #ff0000" |
 * style="width:10%; padding:10px; border:2px solid #ff0000" |

$$ \displaystyle y={x}^{c}\cdot {e}^{rx} $$ $$
 * $$\displaystyle (Eq. 2-3)
 * }
 * }

Find
Find solutions of the equations.

'''Part. a'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (x-1)y''-xy'+y=f(x) $$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

'''Part. b'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle xy''+2y'+xy=f(x) $$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

Solution
Taking 1st and 2nd derivatives of the trial solution is

Solution for Part. a
Plugging the derivatives of trial solutions into the equation given (eq 2.1) yields,

Collecting x terms by reduction order,

To make above equation into the format of trial solution, only one of the x term should be survived. Thus, the values for c and r by (Eq 2-7) are,

Substituting the values (Eq.2-1) into the trial solution, the homogeneous solutions are determined as

Solution for Part. b
Plugging the derivatives(from (eq.2-4) to (eq.2-6)) of trial solutions into the equation given (eq 2.2) yields,

Collecting x terms by reduction order,

To make above equation into the format of trial solution, only one of the x term should be survived. Thus, the values for c and r by (Eq.2-11) are,

Substituting the values (Eq.2-2) into the trial solution, the homogeneous solutions are determined as

Egm6321.f10.team4.Yoon 00:00, 14 Nov 2010 (UTC) - Primary Author

= Problem 1. L2_ODE_VC using Var. of Paramerters = From Meeting 33, p 33-1.

Given
Legendre Equation

Legendre Polynomials

where, $$\displaystyle m = \frac{n}{2}$$

When a particular solution P2(n=2,r=1) is given as below,

Find
Legendre Equation will be given

And then show that the second solution is as below.

Solution
When n=2, the given Legendre Equation is as

Firstly, writing the equation in standard format as

On the assumption that we have one solution, say $$\displaystyle y(x) = u_1(x)$$. We will find the other solution by differentiating $$\displaystyle y(x) = U(x)\cdot {u}_{1}(x)$$ gives

Plugging the above deriviates into the (Eq. 1-7) gives

Since u1(x) is a solution of (Eq. 1-7), the term is zero. We will use $$\displaystyle Z:=U'$$ to apply missing dependent variable method,

After deviding the equation in terms of Z by the highest coefficient of Z, now we can find intergrating factor h(x),

By the definition we set in advance, $$\displaystyle Z=U'=\frac{k2}{h(x)}$$, we can go back to U(x) by integrating Z

Since we have U(x) now, we can solve y(x)

By plugging what we have, h(x), u1(x), into above equation, we can get the desired solution Q2(x)

=hw7 p1=

Given
The general form of the Legendre Equation:

And the solutions have the following format.

Definitions of the Legendre Polynomials:

 * where $$\displaystyle m = \frac{n}{2}$$.

title: User:Egm6321.f10.team4.Yoon/HW6


text:
Category:Resources needing sources

__HIDDENCAT__

title: Category:Resources needing sources


text:
User talk:Spock of Vulcan

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --La comadreja formerly AFriedman RESEARCH (talk) 02:13, 15 November 2010 (UTC)

 Hello , you can call me Jacob Frank. I'd like to be your mascot and welcome you to Wikiversity. Even though I was a plain and untutored man back in 18th-century Europe, that still didn't stop me from contributing my ideas to the world. Neither should it stop you. As a religious leader who developed the doctrine of "purification through transgression," my advice to you is to be bold and to ignore all rules—yes, these are two of Wikiversity's official policies! Happy editing.

title: User talk:Spock of Vulcan


text:
User talk:KBlott

Glad to see you having fun
Let me suggest something for your work/play here that is possible on Wikiversity and not on Wikipedia, which is the organization of "articles" into "courses," or into topic coverage that is deep without the individual pages becoming overwhelming.

Basically, subpages can be used in mainspace here, unlike Wikipedia, so hierarchical hypertext can be used, and deep coverage of subtopics, but all organized under a master topic, which can itself be organized under a larger topic.

Pages can also be created as what I've been calling "seminars," where various users can discuss, research, and report, sometimes on very detailed points.

Wikiversity has a neutrality policy, but it's radically inclusive, and minor points of view, or points of view that would be different from those of a majority of (possibly uninformed) Wikipedia editors, can be covered in depth here. There are no barriers, really, only behavioral guidelines or some sense that people should try to cooperate and help each other. I'm engaged, at the moment, in massive debate on pages under Cold fusion, but all this will be refactored into seminars that cover very specific subtopics in depth, that, eventually, will present, in an organized fashion, all advocated points of view -- even silly ones, though usually those advocating silly points of view know they are considered silly by most and won't mind them being placed lower in a hierarchy, as long as they are treated fairly.

This was actually Wikipedia theory, but dominant factions arose among the administrative core that believed fringe theory -- as they understood it -- should be excluded, and the notability guidelines there set up conditions for constant battles over page space. Here, there is no notability guideline. We even have, sometimes, a little tolerance for nonsense. We might actually keep a modern Jabberwocky, and wouldn't it have been a shame if the original Jabberwocky had been deleted. "Patent nonsense! Out of scope!"

By providing a place for Wikipedia refugees, we may be getting the better part of the deal, in the long run. Articles can be built here that meet Wikipedia standards, and then something becomes possible that otherwise is very difficult: direct comparison of alternate articles, with an RfC on Wikipedia, where non-involved editors would read more than one article, check out sources, etc., and make a choice. The one-edit-at-a-time, with varied eyes seeing each edit, and factions coming and going, is guaranteed to create messes and stuck situations. In the end, we may help save Wikipedia. Right now, of course, it doesn't look like that. But sometimes I can see what's coming, though I never know for sure.

I'm just trying to help it happen, which is why I helped when you first arrived, and why I've continued to help others when they run into problems here. Little by little, I believe, there will be less trouble, as those who understand the potential of Wikiversity return to activity here, it was difficult for a while.

Thanks again, for blessing us with your presence and work, and your little joyful song made me happy, one of my high points here. Makes it all worthwhile. --Abd 03:54, 9 January 2011 (UTC)
 * Thanks. There is no saving Wikipedia.  When I complained to the powers that be about being blocked, they responded by blocking more IP addresses.  More than a thousand, I estimate, just to keep me from criticising Anthony Fauci.  Wikipedia is blocking millions of IP addresses just to keep people like me off its talk pages.  Volunteers are simply responding by going elsewhere.  At the rate that it is haemorrhaging sane people,  its days are numbered.  And, since the ratio of fanatics to ordinary volunteers is rising, the rate of volunteer loss  is accelerating.
 * Just so you know, I will be biting WP in the a** around the end of the month. The backlash may spill over onto WV.  Could you mention to the WV ‘crates that WMF may get its panties in a bunch?  I don’t think Jimbo is a Faucist but some of his people are.
 * I recall, back in the days when I worked for a living, when a co-worker (with no background in physics) came to me (excited) and asking me about cold fusion. I recall thinking
 * D + D --> 4He + gamma
 * and responding with scepticism. I had look at some of the evidence you provided.  It had not occurred to me to consider
 * 2D + 2D + 2D --> 12C*
 * 12C* --> 3 4He + heat.
 * Although the reaction is very unlikely, it is not forbidden by quantum mechanics. Such a reaction can be catalyzed, in principle, by trapping deuterium in a face centered cubic lattice.  I checked the literature and, indeed, there are reports that palladium can form such a lattice.   Are you a physicist? --KBlott 05:22, 9 January 2011 (UTC)
 * No. But I studied physics with Richard P. Feynman when I was at Cal Tech. I was in his class the two years when they put together his phyisics text by filming him.
 * The CF reaction is almost certainly not d+d > 4He. However, helium has in fact been measured and confirmed, by multiple groups, to be the ash, and heat/helium is 25 +/- 5 MeV (Storms, 2007 and 2010), compared to 23.8 MeV for deuterium fusion. See Status of cold fusion (2010), or the preprint at . I'm mentioned, I'm happy as a clam! (I helped edit the paper.)
 * Storms and most people in the field are assuming it's some kind of cluster fusion. There has been particular study by Takahashi of 4D -> Be-8 -> 2* 4He + energy, which is much simpler, it is really just two deuterium molecules. Takahashi did the math to predict that if four deuterons, with their electrons, form a tetrahedron, they will collapse (to a Bose-Einstein condensate) and fuse within about a femtosecond. Storms doesn't like this theory, because he thinks that it would take too much energy to form the required configuration in the required confinement. I don't know that anyone has done the work to predict rate, however. There is some desire to start looking for the occurrence of D2 in confinement, it could be a clue. It's devilishly difficult to see what is happening inside some metal!
 * This is a fascinating case: chemists discovered something that physicists could not explain, so the physicists, largely, assumed that the chemists, who were doing what was within their field of expertise, must have been wrong. Only a minority of physicists actually tried to find explanations. (Like Takahashi, who is a hot fusion physicist, or like some Nobel Prize winners who worked on it for a time.) Storms thinks that no theory explains all the confirmed observations, and he may be quite right. But fusion it is, from the heat/helium value. Mechanism unknown. And the chemists aren't likely to find the mechanism, it's not their field! Still, some are working on ways to make the effect more reliable and stronger, and that might provide clues. They were experts at calorimetry and at what chemistry could do, and they said, "This isn't chemistry!" It was a tad rude for the physicists to say, "It must be chemistry!" simply because they had never seen such a reaction before. Generally, they applied theory only to the brute-force deuterium fusion idea, assuming that, if it was fusion, the reaction must be d+d fusion, which it almost certainly was not, no neutrons, only a little tritium, and far fewer neutrons -- and no gamma rays.
 * To the extent that anyone considered multibody fusion at all, they immediately rejected it, as if the reaction would be as in a plasma, requiring the chance meeting of three or four deuterons in one place at one time. As you have correctly suspected, lattice confinement changes that radically. Takahashi, early on, discovered that bombardment of loaded deuterium foils with energetic deuterons, but below fusion energy, enhanced the expected fusion rate by 10^26. This led him to look at what might be happening. He's still working on it, publishing fairly frequently.
 * I'm working, myself, on a replication of the SPAWAR work, recently reported after the Navy declassified it, that found a few -- but significant, well above background -- energetic neutrons from a "co-deposition" experiment. My plan is to confirm this and sell cheap kits to confirm it. I'll be selling science, not energy. It's possible nobody will ever sell energy from this, it is one fragile effect, really easy to miss, and very difficult to scale up. I have reason to believe I'll be able to confirm, for this experiment doesn't require much to happen, I'll probably not see any excess heat at all, or something ambiguous at best. But, hey, a few neutrons? I think it's cool! And if I'm able to get the level of effect that they showed, my detector results will be conclusive as to neutrons. I'm using two layers of LR-115 solid state nuclear track detector film, sandwiched, detector surface in, only put together when the experiment starts, so that background, from before the experiment, doesn't show up on both layers at the same places.... This is really, really cheap. --Abd 00:37, 10 January 2011 (UTC)
 * A student of Feynman? I am in awe.   I think lattice confinement is the only possible explanation.   Any other explanation that I could think of would allow ordinary matter to collapse at STP.  Since that hasn’t happened, I think we can safely rule those mechanisms out.  I am curious.  What happens when one bombards carbon-12 with neutrons?  Does one get
 * n + 12C --> 3 4He + n
 * or
 * n + 12C --> 3 4He + p + e-?
 * Is there a resonant energy that makes the reaction more likely?
 * I don't know about resonant energy, but see and . SPAWAR thinks they are seeing neutrons at about 13.5 MeV, probably from secondary DT fusion. The "triple tracks" they see are from C-12 breakup in the CR-39 detectors they use. The reaction they describe is the first. Knock-on protons are the main charged particle radiation from neutron bombardment of hydrocarbons, I expect to see mostly proton tracks. The detectors I'm using are pretty different from CR-39, the detector layers are quite thin. I don't know what a triple track would look like if I get any, but only neutron-caused tracks should show up in my detector stack interior (coincident on both adjacent detector layers). Even if some radon gets into the middle, each alpha would then only cause a track on one layer. --Abd 05:10, 10 January 2011 (UTC)
 * Thanks. Based on the information you have provided, it should be easy enough to compute the activation energy of the intermediate quantum state.  However, it will take me some time to digest the information.  I think one can rule out the reaction  2D + 2D --> 8Be* --> 2 4He + energy from the outset.  The simplest Feynman diagram that I can think of for such a reaction would require a colourless glueon pair.  Colourless glueons are forbidden.   The simplest possible Feynman diagram for the reaction 2D + 2D + 2D --> 12C* --> 3 4He + energy requires three gluon pairs, one corresponding to each primary glueon colour.  So, the reaction is permitted but is unlikely—unless the reacting particles are confined to a space which is sufficiently small that the ground state energy of the reactants is greater than the activation energy.  Do you happen to know the unit cell size of palladium?
 * You can find papers by Takahashi that probably have the information you are interested in. What Takahashi does is to assume confinement of four deuterons in a single lattice cell. The most efficient packing would be with them in a tetrahedron, I believe. This is high confinement. There is speculation that there is a phase of palladium deuteride with two deuterons per cell. So four is quite high. However, if there are four, and, while he doesn't state it, I think, they must have low relative temperature, (possible transiently within the Boltzmann distribution), he calculates that they will collapse into a Bose-Einstein Condensate. To do this, I think, the electrons must be present, and he includes them in his model. This condensate is extremely small. I think he considers an inertia of collapse, his papers are sometimes titled with "Collapse Motion." He predicts that once the BEC forms, there is nuclear fusion within a femtosecond. Detecting the BEC, then, will probably be utterly impossible. Nature is hiding a huge secret here. The only sign of this occuring could be the helium and the heat. What happens to the energy? Well, he considers that the Be-8, before it decays, radiates energy in a series of nuclear transitions, his energy is as photons that will ba absorbed by the lattice. However, the ground state will still produce, I'd think, charged particle radiation, above the Hagelstein limit (see Hagelstein, Naturwissenschaften, 2010). Bottom line, I think, is that we have no idea what happens when fusion takes place inside a BEC. I don't even know what it means. After all, there is a single wave equation for the whole BEC. Normal deuterium BECs don't fuse. Etc.
 * Kim has also (Naturwissenschaften, 2010? or 2009) studied more general BECs as involved in cold fusion. Storms thinks that larger clusters are involved, and a lot of work is being done assuming whole-nanoparticle quantum states where some of the deuterium might fuse, but the whole nanoparticle shares in the energy. This field is awakening from a twenty-year torpor, a socially-induced paralysis of a kind. Lots of people kept working, but this is really a huge problem that is going to take a whole lot more attention, from many physicists, before there is much progress, I believe. It's one difficult problem, as I think you might appreciate. --Abd 15:31, 10 January 2011 (UTC)

Invitation to Cold fusion
Since you have indicated an interest in "luke warm fusion" on your User page, please consider watching and commenting or contributing as you find time at the Cold fusion resources. I have a sense, from your user Talk comments, that you've got quite enough physics under your belt to be very helpful. More than me, and more up-to-date.

"cold fusion" is sometimes used to refer to any fusion below normal fusion temperatures, including very high temperature experiments, often where elements are fused through energetic collisions and the goal is the formation of maximally stable very-high-Z elements, since excess energy from the collision forces will break up an unstable nucleus very easily. So they run these at lower than normal fusion energy and simply ride on the incidence of tunneling or the like.

Originally, "cold fusion" referred to muon-catalyzed fusion which was accomplished at near absolute zero, but it came to refer to the room-temperature (or at least not much above boiling temperatures) experiments, starting with Pons and Fleischmann.

It's also been used to refer to bubble fusion, but that's a misnomer, bubble fusion is hot fusion, if it's real. Thus, as well, it was a bit confusing when piezoelectric fusion was called "cold fusion" in some reports, merely because the device could sit on your bench and not melt, it merely generated a few neutrons. From hot fusion. I've seen reports of that where someone scoffed at the device being real, since, after all, wasn't cold fusion debunked twenty years ago? Confused, people can get.

Really, CF is "luke warm fusion" because the rate apparently does increase with temperature, and the limit would be the destruction of the necessary confinement. I'm not sure how high in temperature the reports go, but the temperature dependence is pretty well known, some researchers try to operate close to the boiling point of the electrolyte, in electrolytic experiments, for that reason. Complicated issue, eh? It makes the calorimetry more difficult. --Abd 20:15, 10 January 2011 (UTC)
 * Thanks for the invitation. I would be more than pleased if you would be willing to co-author a short communication proposing a mechanism for the observed neutron generation in the Fleishmann-Pons reaction.  --KBlott 17:13, 11 January 2011 (UTC)

 * The Pons-Fleischmann reaction doesn't generate neutrons. Rather, with great variability depending on accessory conditions, very low levels of neutrons are seen; these are clearly from secondary reactions, possibly connected with rare branches. Tritium is much more commonly generated, but, still, it can be said that the P-F reaction produces (normally) no radioactive byproducts that can be externally observed, tritium is likewise found at low levels, compared to what would be expected if it was a part of the main reaction. That is, indeed, part of why the original discovery was so roundly rejected. The "ash" expected was missing. The helium actually produced, from known reactions, would be accompanied by a 23.8 MeV gamma, which wasn't seen, instead the energy, effectively all of it, ended up as heat.

 * To give you an example of the level of mystery involved, my own work is to replicate the SPAWAR work finding a few neutrons, in codeposition experiments. Cold fusion is believed to be a surface effect, partly from where the helium is found. However, the generation of neutrons in the SPAWAR experiments seems to be heavily dependent upon the substrate for codeposition, there are few neutrons with a silver wire cathode, more with platinum, and relatively copious neutrons with gold. Nobody knows why, and SPAWAR doesn't even speculate on it. (For perspective, "relatively copious" may be a rate of neutron detection of not much more than one per hour. Electronic detectors, relying upon a bulky detector device, can't distinguish that from background unless there are bursts, and prior neutron work involved bursts. An SSNTD can distinguish this by spatial location, by the concentration of tracks near the cathode wire, whereas background would generally come in from cosmic rays, all over the detector surface.

 * I came here today because of the interest on your user page in "luke warm fusion." Cold fusion refers to any fusion below the astronomical temperatures involved with hot fusion (or as the tunneling version of that, where "inadequate" temperature still allows for some fusion rate.) "Cold fusion" is used in nuclear physics for the fusion of high-Z elements with bombardment by other elements, to create very high-Z nucleotides. Popularly, of course, it refers to any relatively low-temperature process, even up to what we might think of as "hot," such as plasma electrolysis. If it can be contained in a container, it's "cold"! And even more specifically, it refers to what P and F discovered and reported, at room temperatures, and up to the boiling point of heavy water, and would be -- and is -- applied to even hotter processes, perhaps up to the melting point of metals.

 * Bubble fusion, on the other hand, is really hot fusion, if it works. The fusion is obtained through the extremely high temperatures in collapsing bubbles.... so .... what do you mean by "luke warm fusion?" --Abd 17:18, 4 February 2011 (UTC)

 * I agree that neutron production in the F-P reaction is probably a tertiary by-product of the main reaction. My original model contains several errors that I can identify.  Don’t make the mistake of assuming I have a broad familiarity with the F-P literature.  I never gave the reaction serious consideration until you introduced me to the literature.  I was busy working for a living when the F-P reaction was initially reported.

 * My current model of the F-P reaction predicts that neutron production should be correlated with iodine synthesis when gold is used as the cathode. The side reaction requires roughly eight steps before the Iodine + neutron is produced.  Any one of these steps could potentially be forbidden by the standard model.  It will take me some to compute the mass excess and spin quantum numbers associated with each step and verify that each step is permitted.  If any step is forbidden by the standard model it would, of course, be necessary to reject the model.    I will also look at what the model predicts when silver and platinum is used.  I would be happy to co-author a paper with you if the mechanism pans out.  The model is quite conservative and (if it is viable) should not disputed by anyone who is not vested in a competing model.

 * When palladium is used, the model predicts that relatively large amounts of Cadmium should be produced (helium-4 >> Cd >> neutrons). The model can be used to predict the relative abundance of each Cd isotope.  However, I set the computations aside due to a lack of empirical data.  Do you happen to know the relative abundance of each Cd isotope produced in the F-P(palladium) reaction?

 * My current model (may) also predict the production of relatively large amounts of deuteron radiation (helium-4 >> deuterons >> neutrons) in some cases. The deuteron radiation may be in the form of high energy hydrogen-2 nucleons + beta particles or (more likely) low energy atomic deuterium.  Alternatively, (and even more likely) the cathode may be transmuted to a different isotope, depending on the cathode used.  These computations are also simple but time consuming and important as they are a major prediction of the model.

 * I don’t know why tritium is produced in the F-P reaction. Does the addition of light water increase the production of hydrogen-3 relative to helium-4?   --KBlott 08:10, 5 February 2011 (UTC)

 * Make sure you have read Storms' 2010 review. He lays out the requirements for a theory, and describes the experimental behavior to be explained. Lots of theories have been proposed that explain one or two phenomena, Storms deprecates that, but, we should remember, it's possible that this strange, previously unexplored territory contains more than one new species. Storms' monograph on Low Energy Nuclear Reaction (World Scientific, 2007) is more thorough.
 * What is fatal to a theory, at least provisionally, is that if makes predictions that don't match the experimental data. In some cases, the data has not been completely reported. I can ask questions of researchers, I have good access to them, generally.
 * As an example, Moulton has proposed two sources of artifact: misting, where tiny droplets of electrolyte are entrained in outflow gas, causing loss of water as liquid instead of as water vapor. If the model used in that experiment considers the heat of vaporization of lost water, and counts the mist leaving the cell as vapor, the correction will produce an appearance of excess heat, adequate, it's proposed, to explain some excess heat results. (Moulton claims "all," in effect, but he has another theory which would also operate under the same conditions.) There are many reasons why misting is unlikely, but one would be that if mist is escaping with the gases (which are slowly evolved, bubbling is not terribly rapid), the droplets would attach to the apparatus around the vent and evaporate there, and since the electrolyte contains dissolved salts to make it conductive, deposits of salt would be seen. I asked. These have not been observed. From the apparatus construction and operation, misting is not to be expected.
 * The other source of the heat anomaly that Moulton proposes is power supply noise, caused by the variations in resistance due to formation and release of bubbles. This is a real noise source, a known effect. However, CF researchers typically use a constant current power supply, that adjusts the voltage to make the current constant. Moulton has never analyzed this quantitatively using realistic numbers. However, bottom line, if the noise is entirely manifest as voltage noise, the method of estimating power used by McKubre et al will be accurate: McKubre's equipement takes many samples of the voltage and averages them and records the average, perhaps once per minute. He assumes constant current. Moulton notes, correctly, that there must be some current noise, and conflates that into an assumption that the noise is significant. He then suggests that researchers failed to check for "AC power" in the input to the cell. That's not the case, for AC power is correctly handled by McKubre's method if the current is constant, which is what McKubre himself notes. McKubre is an electrochemist, this kind of work is in his field. It turns out that McKubre did, in fact, look at the current with an oscilloscope, and, while he has not reported numbers, significant current noise was absent. Another researcher told me that he'd measured input power with three methods: a wattmeter (high bandwidth), a digital storage oscilloscope (multiplying instant voltage by instant current, high data acquisition rate), and using the GPIB data from the power supply. Since all three methods came up with the same result, he now only uses the power supply, no additional equipment is needed.
 * Further, dead cells, cells with, apparently, palladium that isn't ready to show the effect, control cells with platinum cathodes, and hydrogen cells, all show zero excess heat even if the bubbling is the same.
 * Thus Moulton looked at work that showed that excess heat was correlated with current (it is, but not at all rigidly), current is correlated with bubbling (it is, for sure) and jumped to a conclusion. However, as Storms has pointed out, Excess heat also occurs when there is no current. There appears to be some homeostatic effect: once the reaction sets in, it seems, sometimes to "want" to continue. And it's been known to do that for days, ultimately generating more energy than the entire input to the cell, sometimes by a large factor. but not usually!
 * You are now looking at codeposition. At this point, there is a large body of work. It's fine, now, to come up with explanations for, say, the drastic variation between gold and platinum and silver cathodes, and then consider predictions from these theories. Just understand that the effect being observed with this variation is a tiny, tiny one. My guess is that it may be horrifically difficult to explain the tiny side-effects absent an explanation for the main show. But if you assume deuterium fusion as the main show, say through the Be-8 pathway inside a BEC, you might then be able to come up with plausible secondary reactions, and, indeed, SPAWAR does posit a pathway through tritium, perhaps generated by a rare interaction with light hydrogen, which then, even more rarely, fuses with some hot product from the main show. Producing a few neutrons.
 * Theories which involve a long series of reactions are probably quite unlikely unless the intermediate reactions proceed 100% to the next step; that's why Widom-Larsen theory seems implausible to me. No theory is complete without study of reaction rate.
 * And do understand the morphology here: the cathode wire is a substrate, and there is, in the initial electrolyte, some palladium chloride; the palladium is plated onto the cathode wire, in the first stages of the experiment. The initial current is quite low, so that the deposit made is relatively uniform and will adhere well. (It doesn't do if the plating falls off because the cell is jarred!) This is called "codeposition," but it's quite possible that this is incorrect. I don't have the experimental data, but one researcher claimed that the initial voltage was too low to evolve deuterium at all, so what is being done is to prepare a palladium-plated cathode in situ, and then to load it with deuterium.
 * I'm starting with an almost-pure replication. I'm using the same materials as SPAWAR (though not the same production batches, and my heavy water is Canadian, theirs came from a supplier in the U.S.), but I'm using different SSNTDs, I think I can get more precise and reproducible results with LR-115. I'm not, in the first run at least, varying the electrolysis protocol and the cell materials. No matter what great ideas come along! That desire to amplify the effect, for fun or profit, has plagued the field, exact replications were unusual.
 * I see from your to-do list that you might be interested in doing actual experimental work. I can supply all materials needed for a SPAWAR replication, except for the power supply and current meter. I intend to sell a complete kit, one cell, everything fabricated, gold cathode, ready to pour in a vial of prepared electrolyte and hook up a power supply, for about $100. Just raw materials might cost about the same or more, but you'd get materials for more than one cell, I assume. This doesn't include the sodium hydroxide for etching the films, and it doesn't include the bells and whistles I'm adding, such as piezoelectric sensors, looking for shock waves, and thermocouples. A light water control cell would be about $70. Heavy water is the most expensive thing in the experimental cells! --Abd 15:45, 5 February 2011 (UTC)

 * Hagelstein has (Naturwissenschaften, 2010, I think) set an upper limit on charged particle radiation produced in the main reaction, of about 20 KeV. Above that level, effects would be predicted that are not observed. There is data on isotopic abundances in reported transmutation. Just remember, the transmutations are quite rare, compared to helium. The first place to look is Storms, and there is more detail in his book. You might also look at the Earthtech replication, which ascribed some reported transmutation to cell material contamination. In an electrolytic cell, the cathode will sweep up any cations in the solution, concentrating them. Iwamura uses a different approach, with gas-loading, and shows transmutation as the primary effect being observed (I think he doesn't measure heat at all.) Vyosotskii looks at apparent biological transmutation, shown by isotopic distribution of product radically different from the norm. Nobody has replicated Vyosotskii, to my knowledge, and, AFAIK, nobody has tried. Too weird! But Vyosotskii is an experienced scientist, widely published before he was so crazy as to involve himself with cold fusion and stuff that is Not Supposed to Happen. --Abd 16:02, 5 February 2011 (UTC)

User:KBlott/TGF-beta/TGF beta 1
was moved from a mainspace page that appeared to be a Wikipedia article, though it was different. It was replaced with a soft redirect to the relevant Wikipedia article. You may develop draft wikipedia articles in your user space, please don't do it in mainspace top level. It may be done as a resource project as a subpage, just as we may have discussion of or critiques of Wikipedia articles (my opinion, and I've been doing this for years). --Abd (discuss • contribs) 23:06, 12 September 2015 (UTC)

title: User talk:KBlott


text:
User talk:RadiX

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

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Motivation and Emotion (Gambling) Book
Hi Ruy,

I'm Alex MC (the author of the chapter on gambling). I was actually working on the page rather than experimenting:) As this is all new to me (well, to most of us in the class actually) I don't doubt I was making some errors. The reason I was still messing with it when your message appeared was because I was trying to get my edits recognised as contributions - it had been working OK but for some reason the IP address started to appear rather than Alex MC (even though I was using the four tildes).  Got me confused, but I guess it's all just part of the challenge, huh.

Alex

 * I've offered this user some explanation on User talk:AlexMC. Alex, you had logged out, apparently. This edit was also logged out, you can look at the history tab and see. --Abd 20:35, 16 October 2011 (UTC)
 * Thanks, Abd! Ruy Pugliesi ◥ 11:52, 17 October 2011 (UTC)

Custodianship
Hi RP,

Thanks for your message on my talk page. I'm familiar with your work and I think you could make a strong contribution to the administrative side of Wikiversity. I'm planning to take something of a wikibreak shortly, so I'm not in the best position to mentor you, but as you are already a meta sysop and a global rollbacker I'd be comfortable mentoring you jointly with any other active custodian, and nominating you when you're ready after I return. Alternatively, you're more than welcome to select an alternative mentor, and you can always ask me for feedback on a more specific basis. Please let me know your thoughts and don't hesitate to ask around and discuss with the other custodians.

Draicone (talk) 08:10, 20 October 2011 (UTC)

 * Apologies for delay, but welcome aboard - you are now an en.wv probationary custodian. Please let me know if I can do anything to help out. Sincerely, James -- Jtneill - Talk - c 01:58, 30 October 2011 (UTC)

Hi Ruy. It's been a week or so... any questions? Your logs look fine to me, so no issues there.

BTW, if you haven't done so yet, please make sure WV:RCA is on your watchlist. There's currently a discussion going on there about how a particular decision on meta should relate to WV, so your input would certainly be helpful there (since you presumably have a better feel for meta than any of the local 'crats, myself included). --SB_Johnny talk 18:22, 6 November 2011 (UTC)
 * I've closed that discussion, but it was an involved close. If you feel, Ruy, that more discussion should take place in that location, you could re-open it, but consensus seemed to be that what was needed was either a community discussion on the Colloquium, or a Community Review. This wasn't an ordinary Request for custodian action, based on Wikiversity disruption, the only disruption visible was being created by that RCA discussion! SBJ's suggestion that you watch RCA, if you aren't already, is quite sound. Thanks for your work. --Abd 19:29, 6 November 2011 (UTC)

 * Sorry, guys! :( I was a little busy IRL these days. I'm totally agreed with that. What do you think about starting a RfC regarding such proposal? :) Ruy Pugliesi ◥ 01:22, 10 November 2011 (UTC)
 * There are a number of processes here. We have WV:Community Review, which is pretty much the equivalent of RfC elsehwere. My position here is that our consensus is already clear, and it's stated in WV:Blocking policy. While that's titled as a proposed policy, it's been that way for donkey's ages, lots of our operating policies are like that. Doesn't mean that it's not policy, the principles have been tested again and again, and the critical text has been standing for years. I'm chary of starting a Community Review myself on a matter which I consider settled. Instead, what I see is a custodial action which is contrary to standing policy, and the issue of "global bans" has been addressed before, so I've started by asking the custodian to reverse his action, and the next step is WV:Custodian feedback, if I or someone else is not satisfied. Community review for actions and users is down the line, only if necessary. On the other hand, anyone is free to start up discussion of changing our policy, which should start on Wikiversity talk:Blocking policy, and if there is dispute that cannot be resolved there, then there would be the possibility of a CR on our policy. We should have a passed blocking policy, to avoid confusion and for the guidance of custodians. The discussion would go in the site notice.
 * There is also a long term process that I've created to support finding deep consensus. It's slow, it's not intended, at least not at this point, for rapid decision-making, it's intended for producing coherent reports on topics, for use by the community as it sees fit. That's the Assembly. Any user may join the Assembly, instructions are on the page. As a deliberative process, it needs defined membership, which makes it different from the usual wiki usual. --Abd 01:53, 10 November 2011 (UTC)

Hi again. I'll be recommending you for full custodian in a few days... ready? --SB_Johnny talk 19:08, 24 November 2011 (UTC)
 * Hey, SB Johny. Sorry for not responding yesterday, I was a kinda busy IRL. Oh, yes, if you trust me, I have no reason to refuse ;) Thank you. Ruy Pugliesi ◥ 20:30, 27 November 2011 (UTC)
 * Hi Ruy Pugliesi - I think you may now the record now for the longest probationary custodianship. Are you still willing and are you ready to be nominated for full custodianship? Sincerely, James -- Jtneill - Talk - c 22:19, 20 September 2013 (UTC)
 * Ruy, congratulations on your election as a steward! You are still a probationary custodian here, for far longer than policy contemplates; your mentor dropped the ball over three years ago, and you must have missed Jtneill's question. Do you wish to continue? If so, we can get the permanent discussion going. Otherwise, you are a steward and nobody will complain if you remove the tools directly. In any case, thanks for your service to Wikiversity. --Abd (discuss • contribs) 19:21, 7 March 2015 (UTC)
 * Dear Abd,
 * Thank you!
 * I've really missed Jtneill's comment. Sorry for that.
 * I'd rather keep helping Wikiversity as a administrator. There is always something I could do that is not solely within the scope of stewards duties, which is mostly intended for assisting with emergency actions. However, if you think I am not suitable for full custodianship, please, tell me. :)
 * Best regards, Ruy Pugliesi ◥ 04:47, 8 March 2015 (UTC)
 * That, for me, is not the issue. Candidates for Custodianship/Ruy Pugliesi is still open. This should be completed. There are some practical consequences. For example, as a permanent custodian, you may mentor probationary custodians. Probationary custodianship was conceived of as a one-month process. That is possibly short, there is no harm in extensions, as such. However, I will treat the above as indicating your interest in continuing as a custodian, and I will facilitate, then, the PC discussion, allowing the request to be closed. SB_Johnny should have the opportunity to make a recommendation, so first thing I'll do is to ask him. Thanks. --Abd (discuss • contribs) 18:12, 8 March 2015 (UTC)

FYI, your nomination for full custodian has been closed as successful. Congratulations. Thanks for your work in helping out the community. Sincerely, James -- Jtneill - Talk - c 06:03, 1 April 2015 (UTC)
 * Yes, congratulations. --Abd (discuss • contribs) 15:47, 1 April 2015 (UTC)
 * Thank you! Ruy Pugliesi ◥ 03:35, 2 April 2015 (UTC)

How to be a Wikimedia sysop
You may be interested in participating in and/or contributing to this learning project, and see also: How to be a Wikimedia sysop/Wikiversity. Sincerely, James. -- Jtneill - Talk - c 02:03, 30 October 2011 (UTC)
 * I'll read them carefully. Thanks, :) Ruy Pugliesi ◥ 02:05, 30 October 2011 (UTC)

Welcome
Have fun. :) Ottava Rima (talk) 03:45, 30 October 2011 (UTC)
 * Thank you! Ruy Pugliesi ◥ 20:45, 30 October 2011 (UTC)

Mail
- Ottava Rima (talk) 18:11, 7 November 2011 (UTC)

Your block of 150.176.249.100
Thanks! I've extended the block to two weeks, standard escalation for persistent vandalism from an apparently stable IP, with no positive contributions ever, as far as I could see. I opened up the talk page just in case some legit user gets caught in this.

I've developed a low tolerance for accounts and IPs with nothing but vandalism in their record. If a user wants to contribute usefully, they would never want that as the first edits recorded in their account, so I save them and us the trouble and indef registered users, and use escalating blocks for IP. Just my thinking.... --Abd (discuss • contribs) 23:29, 14 November 2011 (UTC)

Test template
I have not made much use of the test template myself because it doesn't seem to have much of an effect. Wondering whether you have ever received feedback from IPs which make test edits? --HappyCamper 07:56, 19 January 2012 (UTC)

Nice to see you come back
Welcome back :-) --Goldenburg111 00:44, 29 January 2015 (UTC)
 * Thanks, Goldenburg111. I returned in November. I will be around. :) Ruy Pugliesi ◥ 00:55, 29 January 2015 (UTC)
 * Why did you leave anyways? You got bored? That's why I'm a bit inactive as of now, I used to be such a Wikipedia Addict but now I just lost that jig. --Goldenburg111 00:59, 29 January 2015 (UTC)
 * I was a bit busy IRL due to academic reasons from February to mid-October, Goldenburg111. I also had some heatlh issues :/ But now everything is okay. Why are you bored with Wikipedia? Ruy Pugliesi ◥ 01:17, 29 January 2015 (UTC)
 * Sorry to hear that with the health issues Ruy. Nice to hear it's taken care of :). Anyways, with the bored with Wikipedia thing. An 11 year old having interest in editing Wikipedia is abnormal, for my age (don't worry a lot of people know I'm 11). So yeah you expect 11 year old's to be playing games. But the reason why i was not interested in Wikipedia as much is because it tends to get boring, I can't really describe it well. But I know I'll get back on my feet again, and that'll be in a few days (hopefully). And yeah, school brings you down far well. --Goldenburg111 14:28, 29 January 2015 (UTC)

Please reconsider deletion
You deleted DCG/Community_Lab as an (Abandoned resource). This was under discussion at, please see that. There, I suggested a move to the user space of the creator. It was part of a larger resource that had been so moved. I don't know why this piece was isolated; as I recall, the user had moved this one page back to mainspace, perhaps planning to do something with it that never happened.

Guy is new as a probationary custodian and was (very properly) asking. Dave suggested using proposed deletion, which is also done for pages like this.

Generally, we do not delete things like this, move to user space is a very normal response, unless the page is totally and completely obvious as being of no use. The benefit is that this practice does no harm, it causes no controversy or hurt feelings, ever. I had thought of just leaving this alone, since the value of that page is marginal, but today decided to request you undelete and move it. Thanks. --Abd (discuss • contribs) 14:36, 20 March 2015 (UTC)
 * ✅. Regards, Ruy Pugliesi ◥ 19:26, 21 March 2015 (UTC)
 * Thanks. --Abd (discuss • contribs) 23:24, 21 March 2015 (UTC)

Staff Listing
Welcome back! Should we change the listing on the STAFF page to RadiX? -- Dave Braunschweig (discuss • contribs) 00:56, 19 February 2016 (UTC)
 * Hello! Thank you. And thanks for showing me this page. I am going to update de list to match my current username. :) R adi X  ∞  01:31, 19 February 2016 (UTC)

Invariant energy - why is deleted?
The page Invariant energy was deleted without any explanation. What is the problem with the page or may be it was a mistake? Fedosin (discuss • contribs)
 * Sorry, I deleted the page by accident. It had been blanked by I vandal so that I didn't noticed that. Restored :) R adi X  ∞  14:05, 3 October 2019 (UTC)

Now it is OK, thank you.Fedosin (discuss • contribs)

How we will see unregistered users
Hi!

You get this message because you are an admin on a Wikimedia wiki.

When someone edits a Wikimedia wiki without being logged in today, we show their IP address. As you may already know, we will not be able to do this in the future. This is a decision by the Wikimedia Foundation Legal department, because norms and regulations for privacy online have changed.

Instead of the IP we will show a masked identity. You as an admin will still be able to access the IP. There will also be a new user right for those who need to see the full IPs of unregistered users to fight vandalism, harassment and spam without being admins. Patrollers will also see part of the IP even without this user right. We are also working on better tools to help.

If you have not seen it before, you can read more on Meta. If you want to make sure you don’t miss technical changes on the Wikimedia wikis, you can subscribe to the weekly technical newsletter.

We have two suggested ways this identity could work. We would appreciate your feedback on which way you think would work best for you and your wiki, now and in the future. You can let us know on the talk page. You can write in your language. The suggestions were posted in October and we will decide after 17 January.

Thank you. /Johan (WMF)

18:14, 4 January 2022 (UTC)

Activity Review
Wikiversity has adopted a policy that expects users with administrative rights to be active participants and to regularly use those rights (at least five edits and five actions within the last 12 months). It appears that you are no longer active at Wikiversity and do not currently have a need for administrative rights.

If you would prefer to retain your administrative role, please rejoin us with your contributions and support. If your account remains inactive, we will need to request that stewards remove your administrative rights. Let me know if you have any questions, and thank you for your previous efforts on our behalf.

Dave Braunschweig (discuss • contribs) 13:55, 25 July 2022 (UTC)
 * Hi. Sorry for the lack of activity over here. I've been in a wikibreak lately due to issues in my job, but I just returned to Wikimedia projects. I wish to retain adminship since I'm going to more involved with wikivesity projects as I am taking part of some educational wiki projects right now. Thank you! R adi X  ∞  13:53, 28 July 2022 (UTC)

title: User talk:RadiX