Approximating function derivatives

[Namir]

(01-28-2024 01:37 AM)Albert Chan Wrote:  Hi, Namir

You could still do 2 function calls per Newton step, yet getting much better correction!

\(\displaystyle
\frac{f(x)}{f'(x)}
= \frac
{\frac{f(x+h) + f(x-h)}{2} + \mathcal{O}(h^2)}
{\frac{f(x+h) - f(x-h)}{2h} + \mathcal{O}(h^2)}
≈ \frac{f(x+h) + f(x-h)}{f(x+h) - f(x-h)} ×h
\)

see thread (HP71B) Newton's method

I found your approach interesting. I tested it with f(x)=exp(x)-3*x^2 an used two versions of Newton's and Halley's methods--one using calculated f(x) and one approximating f(x) with f(x+h) and f(x-h). I came to the conclusion that your approach attempts to find the root of some (f(x+h)-f(x-h))/2 and NOT f(x) itself. Your approach works only if f(x) represents a straight line. The results of your averaging f(x) yields roots with fewer accurate decimals.

Namir