<s>[INST] Maclaurin series of $f(x) = e^x \sin x$ $$f(x) = e^x \sin x$$
I tried applying the given formula in my book but it didn't work.
The maclaurin for $e^x$ is given as $\displaystyle \sum \frac{x^n}{n!}$ and $\sin x$ $\displaystyle \sum \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
I attempted to multiply them together, failed teh books answer. I tried inputting values for them and then multiplying and that fails as well. Why?
$\displaystyle  \frac{x^n}{n!}  \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
Fails
 [/INST] It’s just like multiplying polynomials. 
Just as 
$$(a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+ce+cf$$
is the sum of all possible products of one term from the first factor and one term from the second factor, so also is the product
$$\left(1+x+\frac{x^2}2+\frac{x^3}6+\ldots\right)\left(x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots\right)\;.$$
Thus, it must be
$$\begin{align*}
x&-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots\\
&+x^2-\frac{x^4}6+\frac{x^6}{120}-\frac{x^8}{5040}+\ldots\\
&+\frac{x^3}2-\frac{x^5}{12}+\frac{x^7}{240}-\frac{x^9}{10080}+\ldots\\
&+\frac{x^4}6-\frac{x^6}{36}+\frac{x^8}{720}-\frac{x^{10}}{30240}+\ldots\\
&+\ldots\;.
\end{align*}$$
Of course some of these terms can be combined, since they involve the same power of $x$, but we can already see that the first four powers of $x$ that will appear in this product are $x,x^2,x^3$, and $x^4$: the constant term is $0$. What products of one term of
$$1+x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{24}+\ldots$$
and one term of $$x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^7}{5040}+\ldots$$ are of the form $ax^k$ with $k=1,2,3$, or $4$? Obviously the second factor can’t be an $x^5$ term or higher. The second factor is always at least $x$, so the first factor can’t be an $x^4$ term or higher. Thus, the only partial products that contribute to the first four terms of the product are ones found in the polynomial product
$$\left(1+x+\frac{x^2}2+\frac{x^3}6\right)\left(x-\frac{x^3}6\right)\;,$$
and not all of those will be needed. Specifically, we need only
$$1\cdot x-1\cdot\frac{x^3}6+x\cdot x-x\cdot\frac{x^3}6+\frac{x^2}2\cdot x+\frac{x^3}6\cdot x\;;$$
every other partial product yields a power of $x$ higher than the fourth power and therefore does not contribute to the first four terms of the desired series.
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