I want to find the series of Sinh(x) around x=ln(2) . I solved it using 2 different method .
First method: $$\sinh(x)=\frac{e^{x}-e^{-x}}{2}=\frac{e^{(x-\ln(2))+\ln(2)}-e^{-(x-\ln(2))-\ln(2)}}{2}=e^{(x-\ln(2)}-0.25e^{-(x-\ln(2))}$$ hence
$$\sinh(x)=\sum_{x=0}^\infty \frac{(x-\ln(2))^n}{n!}-0.25\frac{(-1)^n(x-\ln(2))^n}{n!}=\sum_{x=0}^\infty\frac{1}{n!}[1+0.25(-1)^{n-1}](x-\ln(2))^n$$ Second method : $$\sinh(x)=\sinh((x-\ln(2))+\ln(2))=\sinh(x-\ln(2))\cosh(\ln(2))+\sinh(\ln(2))\cosh(x-\ln(2))$$ $$\sinh(x)=\sinh((x-\ln(2))+\ln(2))=\frac{5}{4}\sinh(x-\ln(2))+\frac{3}{4}\cosh(x-\ln(2))$$ $$\sinh(x)=\frac{5}{4}\sum_{n=0}^\infty\frac{(x-\ln(2))^{2n+1}}{(2n+1)!}+\frac{3}{4}\sum_{n=0}^\infty\frac{(x-ln(2))^{2n}}{(2n)!}$$
My question is :
I want to apply some operations on the solution of the second method to have the same form of the solution gotten from first method. How can I deal with ($2n$) and ($2n+1$) in the summation? Can I just make some substitutions? for example : $m=2n+1$ and $r=2n$? as following? $$\sinh(x)=\frac{5}{4}\sum_{m=0}^\infty\frac{(x-\ln(2))^{m}}{m!}+\frac{3}{4}\sum_{r=0}^\infty\frac{(x-\ln(2))^{r}}{r!}$$ (I think no because when I computed some terms, I found it is wrong! But how can I deal with this!?)
