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Can someone explain this?(Taylor polynomials)

  1. Nov 10, 2004 #1
    For function f(x)=1/(1+x^2), calculate Taylor polynomials for the 2nd and 4th degree about the point a=0.

    The answer was:

    P2 = 1-x^2;
    P4 = 1-x^2+x.^4

    for 2nd degree I got -2x/[(1+x^2)^2]
    for 4th degree I got 12x/[(1+x^2)^4]
  2. jcsd
  3. Nov 10, 2004 #2
    But neither of your solutions are polynomials?
  4. Nov 10, 2004 #3
    Why aren't they?
  5. Nov 11, 2004 #4
    Last edited: Nov 11, 2004
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