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Maclaurin Series help

  1. Nov 28, 2007 #1
    Find the MacLaurin polynomial of degree 4 for f(x)
    f(x)= (integral from 0 to x) sin(3t^2)




    [f^(n)(0)/n!]*x^n



    3. The attempt at a solution

    - I took the 4th derivative of sin(3t^2) and got:

    f''''(x)= -108sin(3t^2)-1296t^2cos(3t^2)+1296t^4sin(3t^2)

    Not real sure what to do from there. I plugged 0 in for t to find my f^(n)(0) and got 0,1,1,-1,-1 ...but I'm not sure if that's right. Can somebody please check my derivative and point me in the right direction for finding the series?
     
  2. jcsd
  3. Nov 29, 2007 #2

    Avodyne

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    This problem is easier than you're making it out to be.

    Since f(x) is given by the integral, to get the degree 4 and lower terms in f(x), we need only the degree 3 and lower terms in the integrand. Then, since sin(z) = z + O(z^3), we have sin(3t^2) = 3t^2 + O(t^6), and so we need only keep the first term.
     
  4. Nov 29, 2007 #3
    - I understand what you're saying about only needing to go to the degree 3, but I'm not quite sure I get the sin(z) = z +O(t^6). Can you please explain that a little more?
     
  5. Nov 29, 2007 #4

    Avodyne

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    sin(z) = z - z^3/3! + ... is the standard series for the sine function. If we substitute 3t^2 for z, we get sin(3t^2) = 3t^2 - 9t^6/3! + ... The t^6 and higher terms will not be needed, as they integrate to x^7 and higher, and we only want terms up to x^4.
     
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