Taylor series for $e^z\sin(z)$

Question

How can I write the Taylor series for $e^z\sin(z)$ at $z=0$ without making the procedure too complicated?

Isn't there an easier way than to compute it's derivatives and find a pattern?

Answer 1

Yes, there are simpler methods based on the power series expansion of $\exp$. I indicate two equivalent methods, the first one being more elementary than the second one.

Method 1: Use Euler's formula $\sin(z) = \dfrac{e^{iz} - e^{-iz}}{2i}$.

Method 2: For all $x \in \Bbb R$,

$$ e^{x}\sin(x) = \Im(e^{(1+i)x}) = \sum_{n=0}^\infty\frac{\Im((1+i)^n)}{n!}x^n= \sum_{n=0}^\infty \frac{2^{n/2}\sin(\frac{n\pi}{4})}{n!}x^n. $$ By analytic continuation, the identity still holds for $z \in \Bbb C$.

Answer 2

to find the $nth$ derivative of $$e^{az}\sin (bz+c)$$ differentiating this expression once $$D^1=e^{az}(a\sin (bz+c)+b\cos (bz+c))$$ let $a=r\cos \theta$ and $b=r\sin \theta$. you get $$D^1= e^{az}r\sin (bz+c+\theta)$$ if you observe you see $$D^n= e^{az}r^n\sin (bz+c+n\theta)$$ where $r=\sqrt{a^2+b^2}$ and $\theta=\arctan(\frac{b}{a})$ now substitute for $a,b,c$.

Answer 3

I think the easiest procedure is the multiplication of power series. Let \begin{align} g(z)= & a_0+b_1z+a_2z^2+a_3z^3+\ldots+a_nz^n+\ldots\\ f(z)= & b_0+b_1z+b_2z^2+b_3z^3+\ldots+b_nz^n+\ldots \end{align} Then \begin{align} g(z)f(z) = & (a_0b_0) \\ + & (a_1b_0+a_0b_1)z \\ + & (a_0b_2+a_1b_1+a_2b_0)z^2 \\ + & (a_0b_3+a_1b_2+a_2b_1+a_3b_0)z^3 \\ + & (a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0)z^4 \\ \vdots & \\ \vdots & \\ + & \sum_{i+j=k}a_ib_jz^k \end{align} implies $$ g(z)f(z)=\sum_{k=1}^\infty\color{red}{\sum_{i+j=k}a_ib_j}z^k. $$ In your case, \begin{align} e^z\sin(z) = & \left(\sum_{\alpha=0}^{\infty}\frac{1}{\alpha!}z^\alpha \right)\left(\sum_{\beta=0}^{\infty}\frac{(-1)^{\beta}}{(2\beta+1)!}z^{2\beta+1} \right) \\ = & \sum_{k=0}^{\infty}\;\; \color{red}{\sum_{\alpha+(2\beta+1)=k} \left[\frac{1}{\alpha!}\cdot\frac{(-1)^{\beta}}{(2\beta+1)!} \right]} z^{k} \end{align}