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

[wackensack]

My question is presented in the uploaded pdf file.

 

[loloPF]

I am not sure I want to download the file... sorry.
But You might want to know that:
\frac{d}{dx}(\frac{1}{f^2(x)})=-2\frac{f'(x)}{f^3(x)}

 

[Jameson]

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

 

[Hurkyl]

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

 

[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:

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

hope you like integer partitions...

 

[wackensack]

It may help

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

Bob

 

[benorin]

Ok, so I found another formulation of Faa di Bruno's formula for the n^th 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 k^th power of f(x) (not the k^th derivative.)

-Ben

 

[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}​

Now, the task would be:

Find (b_n)​

Any symbolic software may show us that the first elements of this sequence are:

(b_n) = (1, 22, 584, 28384, 2190128, ...)​

I've encountered some difficulties to solve my task...

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

Bob