# Evaluation of the derivative

1. Nov 3, 2006

Let be a function so the derivative of any order n exist, my question is if there is a way to evaluate:

$$\frac{d^{n}f(x)}{dx^{n}}$$ as $$n\rightarrow \infty$$

2. Nov 3, 2006

### Office_Shredder

Staff Emeritus
Depends on the function. If it's ex, it's trivial. If it's a polynomial, it's trivial. If it's cos(x), it can't be done (cyclic, so there's no limit). If it's something that's a complete mess of a rational function, you're probably righteously too lazy to figure it out

3. Nov 3, 2006

### arildno

Why should it exist some general way apart from differentiating the function?
The derivatives of the functions f(x)=cos(x) g(x)=1 "evolve" in totally different manners.

Last edited: Nov 3, 2006
4. Nov 4, 2006

well i was thinking about "Cauchy's theorem" so you have that the derivative of any order should satisfy:

$$2\pi i f^{(n)}(a)= n! \oint_{C}dzf(z)(z-a)^{-n-1}$$

and from this integral, if C is a circle of unit radius and centered at z=a then, the contour integral just becomes:

$$2\pi f^{(n)}(a)= n! \int_{-\pi}^{\pi} dxf(e^{ix}+a)e^{-inx}$$

EDIT: another good idea would be perhaps to use the "generalized difference" operator so..

$$f^{(n)}(x)=\frac{\nabla ^{n}}{h^{n}}$$ as h-->0 (small h)

Last edited: Nov 4, 2006
5. Nov 4, 2006

### arildno

Why do you assume that Cauchy's theorem in general gives you a neat way to actually evaluate the derivative??

You are not Eljose, are you?

6. Nov 4, 2006