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I'm trying to find the first five terms of a Maclaurin series using division.

Is there possibly a shorter way because every time I try to do it for $\frac{\sin x}{e^x}$

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:

x - x^3/3! + x^5/5! - x^7/7! ÷ 1 + x + x^2/2! + x^3/3! + x^4/4!

I'm still getting the same answer.

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    How did you get your answer Mariano Suárez-Alvarez?2011-03-28

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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.

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    For general $f(x)/g(x)$, this probably isn't much savings. But since we happen to know the series for $1/g(x)$ here, yeah, I think it helps.2011-03-28
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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.

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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...

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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 $\rm\:x^3\:.$