Approximating function derivatives

[Albert Chan]

(01-29-2024 01:25 PM)Namir Wrote:  I used an initial guess of 6 for exp(x)-3*x^2 and 1e-16 as the tolerance for x.

Ah, I missed the big root.

Since I expected quadratic convergence, I set eps = 1e-9, initial h = 1e-3
Again, h is decreasing along with dx, not a fixed number.

lua> S = require'solver'
lua> f = fn'x:exp(x) - 3*x^2'
lua> S.newton(f, 6, 1e-3, 1e-9, true)
6
5.195955942003588
4.52948873087134
4.053975908526381
3.80408413181784
3.737403183742807
3.7330961934908533
3.7330790288874782
3.7330790286328144      17
lua> f(_)
7.105427357601002e-15

This is very similar to true Newton's method

lua> df = fn'x:exp(x) - 6*x'
lua> x = 6
lua> for i=1,9 do print(x); x = x-f(x)/df(x) end
6
5.195956335690451
4.529489328402691
4.053976528531787
3.8040845309684834
3.7374032842820784
3.733096198370094
3.733079028904745
3.7330790286328144