How Can We Approximate the n-th Derivative Using the n-th Difference?

[Klaus_Hoffmann]

hi, my problem is to evaluate the n-th difference to be able to approximate the n-th derivative of a function f(x) at the point x=0

as for small 'h' then $$\frac{\Delta ^{n}}{h^{n} \sim \frac{d^{n}f(0)}{dx^{n}}$$

i think that the n-th difference (with step h) has the representation:

$$\Delta ^{n} f(0) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty}ds \frac{f(s)}{(s(s-h)(s-2h)(s-3h)...(s-nh)}x^{s}$$

and that evaluating this we could obtain an (approximate) asymptotic expansion for n-->infinity.

the intention is given a generating function

$$f(x)= a(0)+a(1)x+a(2)x^{2}+.....$$

then to calculate a(n) you need to know the n-th derivative, we can approximate this derivative by the n-th forward difference at x=0 with an small step 'h'

 

[lurflurf]

http://en.wikipedia.org/wiki/Newton_series
Don't start with that
exp[aD]=1+delta
D=(1/h)log(1+delta)
Analogously for different differences
Beware numerical differentiation by differences is very unstable

 

[tacman]

and then use the formula above to evaluate it.


Thank you for sharing your approach to evaluating the n-th difference. It seems like you are using a complex integral to approximate the n-th derivative at x=0. This can be a useful technique, especially for functions with a known generating function. However, it may not always provide an accurate approximation, especially for functions that are not analytic or have singularities at x=0. It is important to carefully consider the choice of c in the integral and the step size h to ensure accurate results. Additionally, for functions with known derivatives, it may be more efficient to simply calculate the derivatives at x=0 directly rather than using this integral method. Overall, your approach is an interesting way to approximate the n-th derivative and could be useful in certain situations.