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Maclaurin Series

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data

    The Maclaurin series for a function f is given by [tex]\sum\frac{x^n}{2n}[/tex]. What is the value of f(4)(0), the fourth derivative of at x = 0?

    a.) 1
    b.) 2
    c.) 3
    d.) 4
    e.) 5

    2. Relevant equations

    The Maclaurin Series is the infinite series centered at x = 0 with the following formula:

    f(x) = f(0) + f'(0)x + [tex]\frac{f''(0)x^2}{2!}[/tex] + [tex]\frac{f^n (0)x^n}{n!}[/tex]

    3. The attempt at a solution

    I tried differentiating the series starting from f(x) to f4(x) Here is my work:

    f(x) = [tex]\sum[/tex][tex]\frac{x^n}{2n}[/tex]


    f'(x) = [tex]\sum[/tex][tex]\frac{x^{n-1}}{2}[/tex]


    f''(x) = [tex]\sum[/tex][tex]\frac{(n-1)x^{n-2}}{2}[/tex]



    f'''(x) = [tex]\sum[/tex][tex]\frac{(n-1)(n-2)x^{n-3}}{2}[/tex]


    f4(x) = [tex]\sum[/tex][tex]\frac{(n-1)(n-2)(n-3)x^{n-4}}{2}[/tex]



    Therefore:

    f4(0) = [tex]\sum[/tex][tex]\frac{(n-1)(n-2)(n-3)(0)}{2}[/tex] = 0.

    0 is not in the answer choice. I would like hints please, not answers... what am I doing wrong?
     
    Last edited: Mar 28, 2009
  2. jcsd
  3. Mar 28, 2009 #2
    Depends on the limit of the sum. Say I gave you [itex]x^5+x^4+x^3+x^2+x+1=\sum_{n=0}^5 x^n[/itex] and asked you what the derivative at zero was, you wouldn't say its zero would you?
     
  4. Mar 28, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I think you are making this much too complicated!

    The nth term of a MacLaurin series is, as you say,
    [tex]\frac{f^{(n)}(0)}{n!}x^n[/tex].

    If a power series is [itex]\sum a_nx^n[/itex], then
    [tex]\frac{f^{(n)}(0)}{n!}x^n= a_n x^n[/tex]
    so
    [tex]f^{(n)}(0)= n!a_n[/tex]

    In your example, [itex]a_n= 1/2n[/itex].
     
  5. Mar 28, 2009 #4
    Thank you. I see what I did wrong now then. I didn't know we can be able to equate the nth term of the MacLaurin with the series.

    Thanks again!
     
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