Taylor polynomial 1/(1-x^2)

  • Thread starter holezch
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
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omg, silly me. The question was 1/(1 + x^2) , not 1 / ( 1- x^2 )
 
  • #3
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so should I just keep differentiating ? :S
 
  • #4
Office_Shredder
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If you know what the taylor series polynomial is for 1/(1-x) it's just a simple substitution
 
  • #5
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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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