- #1
wackensack
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My question is presented in the uploaded pdf file.
[For experts] Derivatives of 1/f(x)^2
- Thread starter wackensack
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SUMMARY
The discussion focuses on the derivatives of the function \( \frac{1}{f(x)^2} \) and the application of Faá di Bruno's Formula for calculating higher-order derivatives of composite functions. The key formula presented is \( \frac{d^{n}}{dx^{n}}\left(-\frac{1}{f^{2}(x)}\right) \) involving a summation that incorporates the derivatives of \( f(x) \). Participants emphasize the importance of specific conditions on \( f(x) \), such as being a series of even powers and having certain derivatives equal to zero at a point \( a \). The discussion also highlights the need for symbolic computation tools to derive coefficients in the sequence related to the derivatives.
PREREQUISITES- Understanding of Faá di Bruno's Formula for derivatives
- Knowledge of composite functions and their derivatives
- Familiarity with properties of smooth functions and analytic functions
- Basic proficiency in symbolic computation software
- Study Faá di Bruno's Formula in detail to apply it effectively
- Explore symbolic computation tools like Mathematica or Maple for derivative calculations
- Research the properties of analytic functions versus smooth functions
- Investigate integer partitions and their applications in combinatorial mathematics
Mathematicians, students studying advanced calculus, and researchers working on derivative calculations and function analysis will benefit from this discussion.
- #2
loloPF
- 15
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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)}
But You might want to know that:
\frac{d}{dx}(\frac{1}{f^2(x)})=-2\frac{f'(x)}{f^3(x)}
- #3
Jameson
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I just briefly looked at the problem, but I wanted to say that the file is fine. Just a math problem. :)
- #4
Hurkyl
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I have a nitpick -- smooth functions (i.e. infinitely differentiable) are not required to have a MacLauren series -- you need to be analytic.
- #5
benorin
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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...
http://mathworld.wolfram.com/FaadiBrunosFormula.html
hope you like integer partitions...
- #6
wackensack
- 4
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It may help
Thank you, Mr. Benorin. I'm trying to adapt the Faá di Bruno's formula to my problem.
Bob
Thank you, Mr. Benorin. I'm trying to adapt the Faá di Bruno's formula to my problem.
Bob
- #7
benorin
Science Advisor
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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
\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
\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
wackensack
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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