Question.
Find the first 5 terms of the Maclaurin Series of $\frac{\sin x}{e^x}$.
I'm trying to find it using division. Is there possibly a shorter way because every time I try to do it, I get the wrong answer: $$ x-2x^3/3+\cdots$$
I don't really like polynomial long division.
This is how I set it up: $$ \frac{x - x^3/3! + x^5/5! - x^7/7!+\cdots}{ 1 + x + x^2/2! + x^3/3! + x^4/4!+\cdots} $$
I'm still getting the same answer.
You can avoid division by writing $$\frac{\sin x}{e^x}=e^{-x} \sin x.$$ Multiply the series for $e^{-x}$ and $\sin x$ together to get the result.
Here are the first two terms:
$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$
$=x\left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+ \dfrac{x^5}{5!}+ \dfrac{x^6}{6!}\cdots\right) -x^2 -\dfrac{2x^3}{3}-\dfrac{x^4}{6}-\dfrac{x^5}{30}-\dfrac{x^6}{120}-\dfrac{x^7}{5040}-\cdots$
$= (x-x^2)\left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+ \dfrac{x^5}{5!}+ \dfrac{x^6}{6!}\cdots\right) +\dfrac{x^3}{3}+\dfrac{x^4}{3}+\dfrac{2x^5}{15}+\dfrac{x^6}{30}+\dfrac{17x^7}{2520}+\cdots$
and you can continue from there, taking the first term of the residual of the second part as the next term to put into the multiplication in the first part.
First off, $\sin(x)/e^x = Im(e^{(i-1)x})$ so the desired series is:
$$0+(-1)x^1/1!+(-2)x^2/2!+(-1+3)x^3/3!+(-4+4)x^4/4! + ...$$
So it looks to me like you're going to have to go to 6th order to get the first five nonzero terms, depending on what they mean by "terms". Now as to doing the division,
$$(x - x^3/3! + x^5/5! - x^7/7!) / (1 + x + x^2/2! + x^3/3! + x^4/4!)$$
The first result is $x$. So you multiply the $e^x$ by $x$ and subtract. The remainder is:
$$-x^2-(1/3!+1/2!)x^3 - x^4/4!+(1/5!-1/4!)x^5...$$
So the next term is $-x^2$. Continue on with this...
HINT $\ $ It's the imaginary part of an expression, and every coefficient of $\rm\:x^{4\:n}\:$ is zero by multisection / symmetry, so you need only compute it up to $\rm\:x^3\:.$
By the Euler formula $$ \textrm{e}^{\textrm{i}x}=\cos x+\textrm{}\sin x, $$ we find the relation $$ \sin x=\frac{\textrm{e}^{\textrm{i}x}-\textrm{e}^{-\textrm{i}x}}{2\textrm{i}}. $$ Hence \begin{equation*} \frac{\sin x}{\textrm{e}^x}=\frac{\textrm{e}^{(\textrm{i}-1)x}-\textrm{e}^{-(\textrm{i}+1)x}}{2\textrm{i}} =\sum_{k=0}^\infty\frac{(\textrm{i}-1)^k-(-1)^k(\textrm{i}+1)^k}{2\textrm{i}}\frac{x^k}{k!}. \end{equation*} Then one can derive any terms in the series expansion at $x=0$ of the function $\frac{\sin x}{\textrm{e}^x}$.
