Ytukyg

RE-EDITED:

I've come to conclusion that $e^{At}e^{Bt}=e^{Bt}e^{At}$ if:

$sum_{k=0}^{infty}sum_{j=0}^{infty}frac{t^{k+j}}{k!j!}A^kB^j=sum_{j=0}^{infty}sum_{k=0}^{infty}frac{t^{k+j}}{k!j!}B^jA^k$

Does anyone have any idea how I can verify that it only is true if $AB=BA$?

Using induction we have $AB=BA$. Also if $A^mB^n=B^nA^m$ we obtain$$A^{m+1}B^n=A.A^mB^n=AB^nA^m=B^nA^{m+1}$$similarly$$A^mB^{n+1}=B^{n+1}A^m$$and $$A^{m+1}B^{n+1}=B^{n+1}A^{m+1}$$which means that for all natural $l,k$ we have $$A^lB^k=B^kA^l$$

We have
begin{align}
A^x B^y &= underbrace{Acdots (A}_{x text{ times}}underbrace{B)cdots B}_{y text{ times}}\
&= underbrace{Acdots (B}_{x text{ times}}underbrace{A)cdots B}_{y text{ times}}\
&= underbrace{Acdots (A}_{x-1 text{ times}}B)Aunderbrace{Bcdots B}_{y-1 text{ times}}\
&= underbrace{Acdots (A}_{x-2 text{ times}}B)AAunderbrace{Bcdots B}_{y-1 text{ times}}\
&= ldots\
&=Bunderbrace{Acdots A}_{x text{ times}}underbrace{Bcdots B}_{y-1 text{ times}}\ &= BA^xB^{y-1}.
end{align}
Now repeat this $y-1$ more times.