# Series expansion

• January 18th 2009, 10:41 AM
Haris
Series expansion
Use the fact that sin(x) =Im(e^ix) to calculate a simpler derivation of this result.

Could someone explain how to solve this?
• January 18th 2009, 10:55 AM
o_O
Quote:

Originally Posted by Haris
Use the fact that sin(x) =Im(e^ix) to calculate a simpler derivation of this result.

Could someone explain how to solve this?

What result are you referring to?
• January 18th 2009, 11:00 AM
Haris
Sorry, the series expansion of $e^x.sinx$.

$e^x.sinx=1+x^2+\frac{1}{2}x^2+\frac{1}{24}x^4-\frac{1}{36}x^6...$
• January 19th 2009, 04:10 AM
Moo
Hello,

As mentioned :

$e^x \sin(x)=\text{Im} \left(e^x e^{ix}\right)=\text{Im} \left(e^{x(1+i)}\right)$

Now use the power series expansion of the exponential :
$=\text{Im}\left(\sum_{n=0}^\infty \frac{\left(x(1+i)\right)^n}{n!}\right)$

But $1+i=\sqrt{2} e^{i \pi/4}$

So the series is now :

$=\text{Im} \left(\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n e^{in\pi/4}}{n!}\right)=\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n}{n!} \cdot \text{Im}\left(e^{in \pi/4}\right)=\sum_{n=0}^\infty \frac{(\sqrt{2} x)^n}{n!} \cdot \sin \left(\frac{n \pi}{4}\right)$