Maclaurin series coefficient

  • Thread starter jnimagine
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  • #1
jnimagine
178
0

Homework Statement


find coefficient of x^4 in the MAclaurin series for f(x)=e^sinx


Homework Equations


ok... so taking derivatives 4 times for this function...gave me a mess! @.@
can someone help me in simplying the derivatives...?
1. cosxe^sinx
then for 2. is it -sinxe^sinx-sinxcosxe^sinx...?
= . = I'm getting lost with these derivatives...

same for e^3xcos2x... gets sooo complicated...
 
Last edited:

Answers and Replies

  • #2
xepma
525
7
Don't try to find it by determining the derivatives. You already know the expansions of the function sin x and e^y. Just plug them in, and expand up till 4th order (or better to just look what terms contribute to the x^4 coefficient).

E.g.:
[tex]e^{\sin x} = \sum_{k=0}^\infty \frac{(\sin x)^k}{k!}[/tex]

Then plug in the expansion for sin x and collect the terms.
 
  • #3
jnimagine
178
0
Don't try to find it by determining the derivatives. You already know the expansions of the function sin x and e^y. Just plug them in, and expand up till 4th order (or better to just look what terms contribute to the x^4 coefficient).

E.g.:
[tex]e^{\sin x} = \sum_{k=0}^\infty \frac{(\sin x)^k}{k!}[/tex]

Then plug in the expansion for sin x and collect the terms.

hmm I'm still not too sure...
I do get the series u've written above... now for sinx i expand it like x - x^3/3! + x^5/5!... up to the fourth order... and plug in 0??... doesn't that just make everything equal to 0...?
 
  • #4
36,655
8,654
I think what xepma is talking about is this[tex]e^{\sin x} = \sum_{k=0}^\infty \frac{(\sin x)^k}{k!}[/tex]
[tex]= 1 + sin(x)/1 + sin^2(x)/2 + sin^3(x)/3! + ...[/tex]

Now, put in your series for sin(x), sin2(x) and so on. You probably won't need the sin3(x) and might need only a term or two for the sin2(x) part.
 

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