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Polynomial approximation

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Obtain the Taylor polynomials Tnf(x) as indicated. In each case, it
    is understood that f(x) is defined for a11 x for which f(x) is meaningful.


    Problem one
    Tn = (a^x) = sigma from k = 0 to n of ((log a)^k)/k! x^k

    Problem two
    Tn = (1/(1+x)) = sigma from k = o to n of (-1)^k x^k


    2. Relevant equations



    3. The attempt at a solution

    Im totally lost and I dont know where to start can anyone help me please?
     
  2. jcsd
  3. Apr 14, 2010 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Did you really have to use "a11" when you have a perfectly good "l" key?

    They Taylor series for function f(x) is [tex]\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n[/tex] where [itex]f^{(n)}(0)[/itex] indicates the nth derivative of f evaluated at 0.

    (Actually, that is the "Taylor series at 0" or "MacLaurin" series.)

    So for [itex]a^x[/itex] you only need to find the derivatives of [itex]a^x[/itex] and evaluate at x= 0. That can be done most efficiently by writing [itex]a^x= e^{ln a^x}= e^{x ln a}[/itex].

    It is also true that any power series equal to a given function is a Taylor series. If you remember that the sum of the geometric series [itex]\sum_{n=0}^\infty r^n[/itex] is equal to 1/(1- r), then the Taylor series for 1/(1+x) should be obvious.
     
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