Use the fact that sin(x) =Im(e^ix) to calculate a simpler derivation of this result.
Could someone explain how to solve this?
Hello,
As mentioned :
$\displaystyle e^x \sin(x)=\text{Im} \left(e^x e^{ix}\right)=\text{Im} \left(e^{x(1+i)}\right)$
Now use the power series expansion of the exponential :
$\displaystyle =\text{Im}\left(\sum_{n=0}^\infty \frac{\left(x(1+i)\right)^n}{n!}\right)$
But $\displaystyle 1+i=\sqrt{2} e^{i \pi/4}$
So the series is now :
$\displaystyle =\text{Im} \left(\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n e^{in\pi/4}}{n!}\right)=\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n}{n!} \cdot \text{Im}\left(e^{in \pi/4}\right)=\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n}{n!} \cdot \sin \left(\frac{n \pi}{4}\right)$