Solving Integral Equations

[Albert Chan]

Hi, Eddie

My guess is the integral calculations will take the most time.
Instead of always do integral from 0 to x, why not re-use old integral calculations ?

∫(f(u), u=0 to x) = ∫(f(u), u=0 to t) + ∫(f(u), u=t to x)

(08-24-2019 03:34 PM)Eddie W. Shore Wrote:  ∫( sin^2 t dt, 0, x) = 1.4897
Guess = 1
(Radians Mode)
Root ≈ 2.4999

This is my Mathematica code, only for proof of concept.

Quote:f[x_] := Sin[x]^2;
F[a_,b_] := Integrate[f[x], {x,a,b}];
a = 1.4897

newton[{x_,t_,Ft}] := Module[{Fx = Ft + F[t,x]}, {x - (Fx-a)/f[x], x, Fx}];
newton[x_] := newton[{x,0,0}]

> NestList[newton, 1.0, 5]

{1.0,
 {2.71878, 1., 0.272676},
 {2.38150, 2.71878, 1.54649},
 {2.48529, 2.38150, 1.44043},
 {2.49964, 2.48529, 1.48436},
 {2.49991, 2.49964, 1.4896}}

Note: x=2.49991 already converged to 1.4897. The value 1.4896 were previous integral, 0 to 2.49964