Question

Find first few terms, up to 4th degree, of the function f(x)=e^-x*sin(x) as a power series, centered at 0.

Answer 1

@raffle_snaffle We have
\[f(x)=e^{-x}\times \sin x\]
Do you know taylor's series?

Answer 2

yeah

Answer 3

I am not very familiar with taking the product of two power series

Answer 4

I think you can't take the product of the two unless they bother converge?

Answer 5

both*

Answer 6

We won't find the product here, but we'll use this formula.
F(x) expanded in power series around x=0
\[f(x)=f(0)+\frac{f'(0)}{1!} \times x+\frac{f''(0)}{2!}\times x^2....\]

Answer 7

It's taylor's series about a=0

Answer 8

Okay, I can probably figure it out. It was a bonus question on one of my exams. We went over the product of two power series really quick so the question wasn't clicking at the time I was taking the exam.

Answer 9

Thank you for your help and time

Answer 10

oops, I'm not sure if taking product would work. That would be really difficult.

Answer 11

Welcome @raffle_snaffle

Answer 12

Well, my professor told me it I was suppose to take the product of the two power series. I will probably have to study up on the material.

Answer 13

Maybe I am over analyzing it. Making it more difficult then it really is.

Answer 14

oh professor would obviously know more. thanks, I also learned something.

Answer 15

Wait hold on here a minute... I am not guaranteed what I said is true. Lol

Answer 16

Well we know that the taylor series for \(\bf e^{x}\) is given as:\[\bf e^x=\sum_{n=0}^{k}\frac{ x^n }{ n! }=1+x+\frac{ x^2 }{ 2! }+...+\frac{x^n}{n!}\]Note that by taking both sides to power of -1 we get:\[\bf \implies e^{-x}=\frac{ 1 }{1+x+\frac{ x^2 }{ 2! }+...+\frac{x^n}{n!} }\]Now note that the taylor series for \(\bf sin(x)\) is:\[\bf \sin(x)=x-\frac{ x^3 }{ 3! }+\frac{ x^5 }{ 5! }+...+(-1)^n\frac{ x^{2n+1} }{ (2n+1)! }\]

Answer 17

Now we multiply them together:\[\bf e^{-x}\sin(x)=\frac{\sum_{n=0}^{k}(-1)^n \frac{x^{2n+1}}{(2n+1)!}}{\sum_{n=0}^{k}\frac{x^n}{n!}}= \frac{ x-\frac{x^3}{3!}+\frac{x^5}{5!}... }{ 1+x+\frac{x^2}{2!}... }\]You can simplify further I guess but shouldn't be necessary...

@raffle_snaffle

Answer 18

Okay thanks.