User:Silly rabbit/Sandbox

~~User:Silly rabbit/Sandbox/Spinors~~

~~User:Silly rabbit/Sandbox/Cartan affine connection~~

~~User:Silly rabbit/Sandbox/List of missing differential geometry topics~~

~~User:Silly rabbit/Sandbox/Differentiable manifold~~

~~User:Silly rabbit/Sandbox/Exterior algebra~~

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Proof of (1)

{\begin{matrix}V&\to &C(q)\\\downarrow &\swarrow &\\A&&\end{matrix}}

{{a\times n=} \atop {\ }}{{\underbrace {a+\cdots +a} } \atop n}

Doesn't work, ℝⁿ ℝℝn ℝℜℛ

Proof of (1)

$({\vec {a}}\cdot {\vec {\sigma }})({\vec {b}}\cdot {\vec {\sigma }})\,$	$=a_{i}\sigma _{i}b_{j}\sigma _{j}\,$
	$=a_{i}b_{j}\sigma _{i}\sigma _{j}\,$
	$=a_{i}b_{j}\left(\delta _{ij}+i\varepsilon _{ijk}\sigma _{k}\right)\,$
	$=a_{i}b_{j}\delta _{ij}+i\sigma _{k}\varepsilon _{ijk}a_{i}b_{j}\,$
	$={\vec {a}}\cdot {\vec {b}}+i{\vec {\sigma }}\cdot ({\vec {a}}\times {\vec {b}})\,$

Proof of (2)

First notice that for even powers

({\hat {n}}\cdot {\vec {\sigma }})^{2n}=I\,

but for odd powers

({\hat {n}}\cdot {\vec {\sigma }})^{2n+1}={\hat {n}}\cdot {\vec {\sigma }}\,

Combine these two facts with the knowledge of the relation of the exponential to sine and cosine:

$e^{ix}\,$	$=\sum _{n=0}^{\infty }{\frac {i^{n}x^{n}}{n!}}\,$
	$=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{2n!}}+i\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}\,$

Which, when we use $x=a({\hat {n}}\cdot \sigma )\,$ gives us

=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(a{\hat {n}}\cdot \sigma )^{2n}}{2n!}}+i\sum _{n=0}^{\infty }{\frac {(-1)^{n}(a{\hat {n}}\cdot \sigma )^{2n+1}}{(2n+1)!}}\,

=\sum _{n=0}^{\infty }{\frac {(-1)^{n}a^{2n}}{2n!}}+i({\hat {n}}\cdot \sigma )\sum _{n=0}^{\infty }{\frac {(-1)^{n}a^{2n+1}}{(2n+1)!}}\,

The sum on the left is cosine, and the sum on the right is sine so finally,

e^{ia({\hat {n}}\cdot {\vec {\sigma }})}=\cos {a}+i({\hat {n}}\cdot {\vec {\sigma }})\sin {a}\,

-{\frac {d}{dx}}\left[p(x){\frac {dy}{dx}}\right]+q(x)y=\lambda w(x)y.

(1)

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A ration of food. (Silly rabbit 15:17, 20 June 2007 (UTC))