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Finding quadratic maclaurin polynomial

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    the question asks to find the quadratic maclaurin polynomial for f(x)
    Given f(x) = x sin(x)


    3. The attempt at a solution
    i know that a maclaurin series is when a=0 in a taylor series.

    i did the 1st-5th derivatives of f(x) and then used the formula for taylor polynomial and set a=0.

    f'=sin(x) + x cos(x)
    f''=2cos(x) -x sin(x)
    f'''=-3sin(x) -xcos(x)
    f''''=-4cos(x) +xsin(x)
    f'''''=5sin(x) +xcos(x)

    with the formula for taylor polynomial

    Pn(x) = n
    Ʃ [f^k (a) (x-a)^k] / k!
    k=0

    by setting a=0 i got Pn(x)=0+0+x^2 -0- (x^4)/6............


    however the answer seems to be P(x)=x^2
    anyone able to explain to me how that is so?

    i think i got it... is the quadratic maclaurin polynomial just the 4th order maclaurin polynomial??
     
  2. jcsd
  3. Nov 1, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, the "quadratic" MacLaurin polynomial is NOT fourth degree. It is, as is any quadratic polynomial, second degree. You are correct that the function and first derivative are both 0 at x= 0 so the first non-zero coefficient is the second degree term. The "quadratic MacLaurin polynomial" for this function is just that second degree term, [itex]x^2[/itex].

    Graph both y= x sin(x) and [itex]y=x^2[/itex] on the same coordinates. You should find that they remarkably similar close to the origin.
     

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    Last edited: Nov 1, 2011
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