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Finding Maclaurin Series Expansions of Functions

  1. Dec 2, 2008 #1
    Hey, here's is my problem as the exam states it:

    A) Write out the first four non-zero terms of the Maclaurin Series for [tex]F(x) = (1+X^{7})^{-4}[/tex] Give all of the coefficients in exact form, simplified as much as possible.

    B) Find the exact value of the 21st order derivative of [tex]F(x) = (1+X^{7})^{-4}[/tex] evaluated at x=0

    I can not figure out an efficient(or simple) way to find the expansion of the given function. At first I tried using the known series 1/(1-X) = [tex]\sum X^{n}[/tex]

    I replaced X in the equality with -X^7 to get [tex]\sum (-1)^{n}X^{7n}[/tex]

    After this though, I'm lost. Apparently I'm not allowed to just multiply the general term by 1/(1+X^7)^3. So I've no idea what to do. :(

    Hopefully the solution to A simplifies the solution to B
    Last edited: Dec 2, 2008
  2. jcsd
  3. Dec 2, 2008 #2
    Did you find the first 4 terms of the MacLaurin series for it yet? The idea is to look for a pattern in those to find the general form of the series.
  4. Dec 2, 2008 #3
    I tried doing that, the issue I ran into was that the question asks for the first four non-zero terms of the series. I attempted to take a few derivatives, but by the 3rd one I was already dealing with 4 separate sums of functions(that in turn have product rule applied and each become 2 more sums). Not only that, at that point all the terms would have been zeros if they had been evaluated. Seemingly it would take forever to do this the direct method.
  5. Dec 2, 2008 #4
    I also have another question but don't feel like making another thread for it, I've already typed it up on mathbin:

  6. Dec 2, 2008 #5


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    Take u=X^7. Then 1/(1+u)=(1-u+u^2-u^3+...). To the accuracy you need it 1/(1+u)^4=(1-u+u^2-u^3)^4. You can just brute force multiply it out. Drop any term with a power over u^3. You don't need it. You can also be clever and use counting arguments. For example to get the coefficient of the u^2 term you can either choose one power of u from two of the factors, C(4,2)=6 ways to do that, or one power of u^2 from one of the four factors, C(4,1)=4 ways to do that. So the coefficient of u^2 is 4+6=10.
    Last edited: Dec 2, 2008
  7. Dec 2, 2008 #6
    Yea it does take awhile; it looks like the first 4 non-zero terms are (it takes 21 terms to get them)


    That looks like [tex]\sum a_n(-1)^nx^{7n}[/tex]

    where the [itex]a_n[/tex] is some sequence that makes the {1,4,10,20,...}; I can't think of what would work for that at the moment though.
  8. Dec 2, 2008 #7


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    It's a multinomial expansion. There's probably a concise expression for it somewhere.
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