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> 
> > 
> > Given the following theorem $f$ and $g$ 
> >  $$f(x) = \sum\_{n=0}^{\infty} a\_n (x-x\_0)^n, |x-x\_0| < R\_1 \\ g(x) = \sum\_{n=0}^{\infty} b\_n (x-x\_0)^n, |x-x\_0| < R\_2$$
> >  then
> >  $$f(x)g(x) = \sum\_{n=0}^{\infty} c\_n (x-x\_0)^n, |x-x\_0| < R$$
> >  Where
> >  $$c\_n = \sum\_{r=0}^{n} a\_rb\_{n-r} = \sum\_{r=0}^{n} a\_{n-r}b\_{r} \\ R \geq \min \{R\_1, R\_2\}$$
> > 
> > 
> > 
> 
> 
> Use the theorem above to find the first four nonzero terms in the Maclaurin of $$e^x \sin x$$
> 
> 
> 

My understanding **RE-EDIT based on Answers**

$$f(x) \rightarrow 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 +\frac{1}{5!} x^5 ...
\\ g(x) \rightarrow 0+ x + 0 - \frac{1}{3!} x^3 + 0 + \frac{1}{5!} x^5 ... $$

$$c\_0 = \sum\_{r=0}^{0} a\_r b\_{n-r} = a\_0b\_0 = 1 \cdot 0 = 0$$
$$c\_1 = \sum\_{r=0}^{1} a\_r b\_{n-r} = a\_0b\_1 +a\_1b\_0 = 1 \cdot 1 + 1 \cdot 0 =1$$
$$c\_2 = \sum\_{r=0}^{2} a\_r b\_{n-r} = a\_0b\_2 + a\_1b\_1 + a\_2b\_0 = 1$$
$$c\_3 = ... = \frac{1}{3}$$
$$c\_4 = ... = 0$$
$$c\_5 = ... = - \frac{1}{30}$$

The first four nonzero terms of $e^x \sin x$:
$$ \implies c\_1 x^1+ c\_2 x^2 +c\_3 x^3 + c\_5 x^5 $$
$$ \implies x+ x^2 +\frac{1}{3} x^3 - \frac{1}{30} x^5$$

Would this be correct?

Thx for your input