1. ## Is this maclaurin series answer wrong?

I dont think the answer is correct. I am getting something like x- x^4/6...

(just looking for someone to confirm that the answer is wrong)

2. ## Re: Is this maclaurin series answer wrong?

Originally Posted by lc99

I dont think the answer is correct. I am getting something like x- x^4/6...

(just looking for someone to confirm that the answer is wrong)
The listed answer is correct. Can you show us how you are getting your result?

-Dan

3. ## Re: Is this maclaurin series answer wrong?

e^-x = (1-x+x^2/2! -x^3/6!+...)
sinx = (x-x^3/3! + x^5/5! -....)

Then i multiply term by term

I think i may be distributing the terms wrong?

4. ## Re: Is this maclaurin series answer wrong?

Originally Posted by lc99
e^-x = (1-x+x^2/2! -x^3/6!+...)
sinx = (x-x^3/3! + x^5/5! -....)

Then i multiply term by term

I think i may be distributing the terms wrong?
that's a really hard way to go...

I'd write

$e^{-x}\sin(x) = Im(e^{(i-1)x})$

and use the Imaginary part of the Maclaurin series for $e^z,~z=(i-1)x$

it is a mess but there is a pattern

5. ## Re: Is this maclaurin series answer wrong?

I think my calc class didn't learn that yet or wont learn it. Is this part of complex numbers unit? Cause my calc department took that away for our final exam since we had snow days on the east coast.

I was able to find the same answer by just distributing it and removing the high degrees

6. ## Re: Is this maclaurin series answer wrong?

Since you just want the first four nonzero terms, I would do it with a chart and multiply one term of the sine expansion by every term of $e^{-x}$, then line them up by powers of $x$:

$\begin{matrix}\text{multiplication} & \vert & 1 & x & x^2 & x^3 & x^4 & x^5 & x^6 \\ \hline xe^{-x} & \vert & & x & -x^2 & +\dfrac{x^3}{2!} & -\dfrac{x^4}{3!} & +\dfrac{x^5}{4!} & -\dfrac{x^6}{5!} \\ -\dfrac{x^3}{3!}e^{-x} & \vert & & & & -\dfrac{x^3}{3!} & +\dfrac{x^4}{3!} & -\dfrac{x^5}{3!2!} & +\dfrac{x^6}{3!3!} \\ \dfrac{x^5}{5!}e^{-x} & \vert & & & & & & +\dfrac{x^5}{5!} & -\dfrac{x^6}{5!} \\ \hline \text{Total} & \vert & & x & -x^2 & +\dfrac{x^3}{3} & & -\dfrac{x^5}{30} & +\dfrac{x^6}{90}\end{matrix}$

Then you just take the first four terms. There was no need to do find the multiplication of $-\dfrac{x^7}{7!}$ unless I was going out to at least the $x^7$ term.