Euler'ss formula, Rodrigues' rotation formula

Euler_formulas.txt

Video in which this page is discussed:

https://youtu.be/jckyLXSWgZM

===== Euler formlua =====

$z,t\in{\mathbb C}$. Recall

$\cosh(z):=\sum_{k=0}^\infty\frac{z^{2k}}{(2k)!}$

$\cos(z):=\sum_{k=0}^\infty\frac{z^{2k}}{(2k)!}(-1)^k$

$\sinh(z):=\sum_{k=0}^\infty\frac{z^{2k+1}}{(2k+1)!}$

$\sin(z):=\sum_{k=0}^\infty\frac{z^{2k+1}}{(2k+1)!}(-1)^k$

$\exp(z):=\sum_{k=0}^\infty\frac{z^k}{k!}=\cosh(z)+\sinh(z)$

With the above we now have:

$i^2=-1 \implies$

$\sin(t)=-i\sinh(it)$

$\cos(t)=\cosh(it)$

$\exp(it)=\cos(t)+i\sin(t)$

Note:

$i^0=1$ (as used in power series)

----
Real $2\times 2$-matrix representation of $\mathbb C:$

$a+ib\mapsto a\cdot E_2+b\cdot I_2$

where

$I_2=\begin{bmatrix}
0 & -1 \\
1 &  0
\end{bmatrix}, E_2=\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}$

as

$I_2^2=\begin{bmatrix}
0 & -1 \\
1 &  0
\end{bmatrix} * \begin{bmatrix}
0 & -1 \\
1 &  0
\end{bmatrix} = \begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}=-E_2=-I_2^0$

$\implies \forall d.\ I_2^{d+2}=-I_2^{d}$

Oscillates between anti-symmetric and symmetric ever step and between positive and negative
(w.r.t. to the previous one) every two steps.

Trivial 3-dimensional extension this case:

$I_3=\begin{bmatrix}
0 & -1 & 0 \\
1 &  0 & 0 \\
0 &  0 & 0
\end{bmatrix}, F_3=\begin{bmatrix}
1 &  0 & 0 \\
0 &  1 & 0 \\
0 &  0 & 0
\end{bmatrix}, I_3^2 = -F_3$

$\implies \forall d>0.\ I_3^{d+2} = -I_3^d$ (Note: $d>0$!)

Difference: $F_3$ is not the identity, $I_3^2\neq -I_3^0$.

Generalizing this:
Define

$\vert \vert \bullet\vert \vert: {\mathbb R}^3\to{\mathbb R},\ \ \vert \vert X\vert \vert := (X_1^2+X_2^2+X_3^2)^{1/2}$

$N_\bullet: {\mathbb R}^3\to{\mathbb R}^3,\ \ N_X:=X\,/\,\vert \vert X\vert \vert$

$[\bullet\times]: {\mathbb R}^3\to{\mathbb R}^{3\times 3},\ \ [X\times]:=\begin{bmatrix}
0 & -X_3 & X_2  \\
X_3 &  0 & -X_1 \\
-X_2 & X_1 &  0
\end{bmatrix}$ (anti-symmetric)

$\bullet\otimes\bullet: {\mathbb R}^3\times{\mathbb R}^3\to{\mathbb R}^{3\times 3},\ \ X\otimes X:=\begin{bmatrix}
X_1\cdot X_1 & X_1\cdot X_2 & X_1\cdot X_3  \\
X_2\cdot X_1 & X_2\cdot X_2 & X_2\cdot X_3  \\
X_3\cdot X_1 & X_3\cdot X_2 & X_3\cdot X_3  \\
\end{bmatrix}$ (symmetric)

Related as

$[X\times]^2 = -(\vert \vert X\vert \vert^2\cdot E - X\otimes X)$

(Reduces to $-F_3$ when $X=(0,0,1)$)

$[X\times]^3 = -\vert \vert X\vert \vert^2\cdot [X\times]$

$\implies \forall d>0.\ [X\times]^{2+d} = -\vert \vert X\vert \vert^2\cdot [X\times]^d$

In particular, for non-zero $X$:

$[N_X\times]^3 = i^2\cdot [N_X\times]^1$

----

Let's consider

$J^P J^{B} = j^P J^{B}$, with $j$ a non-zero field element, say.
This is realized, in particular, in the trivial case $J=j$.
But the above is what we have in mind.

Generally, here the lowest $J$-power that can't be broken down is $P-1+B$.

Will make use of the partitioning bijection $f(p, i) = i\,P + p$

$f(0,0)= 0, f(1,0)=1, f(2,0)=2, \dots, f(P-1,0)=P-1$

$f(0, 1)=P, f(1, 1)=P+1, \dots, f(P-1,1)=2P-1$

$f(0,2)=2P,\dots$.

Thus

$\sum_{k=B}^\infty a_k J^k = \left(\sum_{p=0}^{P-1}\left(\sum_{i=0}^\infty a_{i\,P+p+B} \,j^{i\,P} \right)J^{p}\right)J^B $

Consider

$P=2$, i.e. $J^2 J^B = j^2 J^B$

$\sum_{k=B}^\infty a_k J^k = \sum_{i=0}^\infty \left(a_{2i+B}\, E +  a_{2i+B+1}\, J\right)\,(j^{2})^iJ^B $

Consider moreover

$B=1$, i.e. $J^2 J = j^2 J$

$\sum_{k=0}^\infty a_k J^k = a_0\,E + \sum_{i=0}^\infty \left(a_{2i+1}\, E +  a_{2i+2}\, J\right)\,(j^{2})^iJ $

$a_k=\frac{z^k}{k!}$

$\exp(z\,J) = E + \dfrac{\sinh(z\,j)}{j} J + \dfrac{\cosh(z\,j)-1}{j^2} J^2 $

To approach Eulers formula more closely, after formally rescaling of $z$ by $i/j$, we may also express this as

$\exp(i\,z\,J/j) = \left(E-(J/j)^2\right) + \cos(z)\,(J/j)^2 + i\,\sin(z)\,J/j$

Consider

$j^2=-1$, which when also taking $J^2 J = j^2 J$ is just $J^3=-J$, E.g., $j=i$.
Then

$\exp(z\,J) = (E + J^2) + \cos(z)\,(-J^2) + \sin(z)\,J$

Consider

$J^2=-E$

$\exp(z\,J) = \cos(z) E + \sin(z) J$

----

Call $x\equiv \vert\vert X\vert\vert$ and recall

$[N_X\times]^2 = -E + N_X\otimes N_X$

$[N_X\times]^3 = -[N_X\times]$

So

$\exp(x\,[N_X\times]) = N_X\otimes N_X + \cos(x) (E - N_X\otimes N_X) + \sin(x) [N_X\times]$

Consider

$X\cdot R = 0$ and call a third vector $I \equiv X\times R$, then

$\exp(x[N_X\times])N_R = \cos(x) N_R + \sin(x)\, N_I$