Is My Taylor Polynomial for 1/(1-x^2) Correct?

[holezch]

Homework Statement

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

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

 

[holezch]

omg, silly me. The question was 1/(1 + x^2) , not 1 / ( 1- x^2 )

 

[holezch]

so should I just keep differentiating ? :S

 

[Office_Shredder]

If you know what the taylor series polynomial is for 1/(1-x) it's just a simple substitution

 

[holezch]

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 that's all they want the function to be equal up to