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

1. Sep 28, 2005

wackensack

My question is presented in the uploaded pdf file.

:surprised

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2. Sep 29, 2005

loloPF

But You might want to know that:
$$\frac{d}{dx}(\frac{1}{f^2(x)})=-2\frac{f'(x)}{f^3(x)}$$

3. Sep 29, 2005

Jameson

I just briefly looked at the problem, but I wanted to say that the file is fine. Just a math problem. :)

4. Sep 29, 2005

Hurkyl

Staff Emeritus
I have a nitpick -- smooth functions (i.e. infinitely differentiable) are not required to have a MacLauren series -- you need to be analytic.

5. Oct 5, 2005

benorin

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:

hope you like integer partitions...

6. Oct 9, 2005

wackensack

It may help

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

Bob

7. Jan 23, 2006

benorin

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

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

where $$f^{k}(x)$$ is the kth power of f(x) (not the kth derivative.)

-Ben

8. Jan 28, 2006

wackensack

Mr. Benorin, you see, this is a local problem: the final result is evaluated at $$x = a$$. Besides that, $$f$$ satisfies some particular conditions, which must be considered:
(a) $$f(x) \neq 0$$, over some open interval $$A$$;
(b) $$f$$ is a series of even powers;
(c) $$f^{(2n+1)}(a) = 0$$ and $$f^{(2n)}(a) \neq 0$$, $$n = 0, 1, 2, ...$$;
The final result is a function of $$a$$, and the sum symbol, $$\Sigma$$, will not appear in the final answer.
As I've pointed,
$$g^{(2n)}(a)=-b_n f(a)^{-3n-2}$$​
Find $$(b_n)$$​
$$(b_n) = (1, 22, 584, 28384, 2190128, ...)$$​