taylor polynomial 1/(1-x^2)


by holezch
Tags: 1 or 1x2, polynomial, taylor
holezch
holezch is offline
#1
Apr3-10, 08:48 PM
P: 253
1. The problem statement, all variables and given/known data

The question asks me to write out a taylor polynomial for 1/(1-x^2)
of degree 2n+1 at 0.



3. The attempt at a solution

My answer was 1 + x^2 + x^4 + x^6 + ..... + (x^4)/(1-x^2) which I just got from using hte geometric series formula. The textbook answer however is this:

1 - x^2 + x^4 - ............. + (-1)^n x^2n

Am I wrong? I thought so, but I took out a graphing calculator, and the larger I make my polynomial degree, the closer it looks to 1/(1-x^2). So my answer must be correct?

thanks
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holezch
holezch is offline
#2
Apr3-10, 08:59 PM
P: 253
omg, silly me. The question was 1/(1 + x^2) , not 1 / ( 1- x^2 )
holezch
holezch is offline
#3
Apr3-10, 09:05 PM
P: 253
so should I just keep differentiating ? :S

Office_Shredder
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#4
Apr3-10, 09:07 PM
Mentor
P: 4,499

taylor polynomial 1/(1-x^2)


If you know what the taylor series polynomial is for 1/(1-x) it's just a simple substitution
holezch
holezch is offline
#5
Apr3-10, 09:13 PM
P: 253
Quote Quote by Office_Shredder View Post
If you know what the taylor series polynomial is for 1/(1-x) it's just a simple substitution
right. that was very blind of me.. thanks! I also don't need to include the remainder term in my answer right? I just stop at 2n, since thats all they want the function to be equal up to


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