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taylor polynomial 1/(1-x^2) |
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| Apr3-10, 08:48 PM | #1 |
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taylor polynomial 1/(1-x^2)
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 |
| Apr3-10, 08:59 PM | #2 |
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omg, silly me. The question was 1/(1 + x^2) , not 1 / ( 1- x^2 )
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| Apr3-10, 09:05 PM | #3 |
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so should I just keep differentiating ? :S
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| Apr3-10, 09:07 PM | #4 |
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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
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| Apr3-10, 09:13 PM | #5 |
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