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[For experts] Derivatives of 1/f(x)^2

  1. Sep 28, 2005 #1
    My question is presented in the uploaded pdf file.

    :surprised
     

    Attached Files:

  2. jcsd
  3. Sep 29, 2005 #2
    I am not sure I want to download the file... sorry.
    But You might want to know that:
    [tex]\frac{d}{dx}(\frac{1}{f^2(x)})=-2\frac{f'(x)}{f^3(x)}[/tex]
     
  4. Sep 29, 2005 #3
    I just briefly looked at the problem, but I wanted to say that the file is fine. Just a math problem. :)
     
  5. Sep 29, 2005 #4

    Hurkyl

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    Staff Emeritus
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    Gold Member

    I have a nitpick -- smooth functions (i.e. infinitely differentiable) are not required to have a MacLauren series -- you need to be analytic.
     
  6. Oct 5, 2005 #5

    benorin

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    Homework Helper

    Use Faá di Bruno's Formula for the Nth derivative of a composition of functions (since 1/f(x)^2=h(f(x)) where h(x)=1/x^2). Here is a link:

    http://mathworld.wolfram.com/FaadiBrunosFormula.html

    hope you like integer partitions...
     
  7. Oct 9, 2005 #6
    It may help

    Thank you, Mr. Benorin. I'm trying to adapt the Faá di Bruno's formula to my problem. :rolleyes:

    Bob
     
  8. Jan 23, 2006 #7

    benorin

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    Ok, so I found another formulation of Faa di Bruno's formula for the nth derivative of a composition of functions: here's your answer

    [tex]\frac{d^{n}}{dx^{n}}\left(-\frac{1}{f^{2}(x)}\right) = \sum_{m=1}^{n}\left\{\frac{1}{m!}\left[\sum_{j=0}^{m-1}(-1)^{j}\frac{m!}{j!(m-j)!}f^{j}(x)\frac{d^{n}}{dx^{n}}\left( f^{m-j}(x)\right)\right]\frac{(-1)^{m+1}(m+1)!}{f^{m+2}(x)}\right\}[/tex]

    where [tex]f^{k}(x)[/tex] is the kth power of f(x) (not the kth derivative.)

    -Ben
     
  9. Jan 28, 2006 #8

    Mr. Benorin, you see, this is a local problem: the final result is evaluated at [tex]x = a[/tex]. Besides that, [tex]f[/tex] satisfies some particular conditions, which must be considered:
    (a) [tex]f(x) \neq 0[/tex], over some open interval [tex]A[/tex];
    (b) [tex]f[/tex] is a series of even powers;
    (c) [tex]f^{(2n+1)}(a) = 0[/tex] and [tex]f^{(2n)}(a) \neq 0[/tex], [tex]n = 0, 1, 2, ...[/tex];
    The final result is a function of [tex]a[/tex], and the sum symbol, [tex]\Sigma[/tex], will not appear in the final answer.
    As I've pointed,
    [tex]g^{(2n)}(a)=-b_n f(a)^{-3n-2}[/tex]​
    Now, the task would be:
    Find [tex](b_n)[/tex]​
    Any symbolic software may show us that the first elements of this sequence are:
    [tex](b_n) = (1, 22, 584, 28384, 2190128, ...)[/tex]​
    I've encountered some difficulties to solve my task... :confused:

    Mr. Benorin, your result may come in handy, thank you.

    Bob
     
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