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- Thread starter wackensack
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But You might want to know that:

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

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Hurkyl

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benorin

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

hope you like integer partitions...

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

Bob

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benorin

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[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 k

-Ben

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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

(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

As I've pointed,

[tex]g^{(2n)}(a)=-b_n f(a)^{-3n-2}[/tex]

Now, the task would be:[tex](b_n) = (1, 22, 584, 28384, 2190128, ...)[/tex]

I've encountered some difficulties to solve my task... Mr. Benorin, your result may come in handy, thank you.

Bob

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