Sun Mar 29 09:50:18 2026

newton_test():
  numpy version:  1.26.4
  python version: 3.10.12
  newton() implements a Newton solver for nonlinear equations.

newton1_test():
  newton1() is a simple implementation of Newton's method
  for estimating a solution of F(X)=0.

  F1 = sin ( x ) - 0.5 * x
  X0 = 1.5, F(X0) = 0.247495
  X* = 1.89549, F(X*) = -7.73058e-09
  Number of steps was 4

  F2 = 2.0 * X - EXP ( - X )
  X0 = 0.1, F(X0) = -0.704837
  X* = 0.351734, F(X*) = -4.09699e-11
  Number of steps was 3

  F3 = X * EXP ( - X )
  X0 = 0.5, F(X0) = 0.303265
  X* = -9.38962e-14, F(X*) = -9.38962e-14
  Number of steps was 6

  F4 = EXP ( X ) - 1 / ( 100 * X^2 )
  X0 = 0.03, F(X0) = -10.0807
  X* = 0.0953446, F(X*) = -1.60398e-10
  Number of steps was 7

newton2_test():
  newton2() is a more robust implementation of Newton's method
  for estimating a solution of F(X)=0.

  F1 = sin ( x ) - 0.5 * x
  X0 = 1.5, F(X0) = 0.247495
  X* = 1.89549, F(X*) = -7.73058e-09
  Number of steps was 4

  F2 = 2.0 * X - EXP ( - X )
  Root X =  0.35173371123404124
    0: f(0.1) = -0.704837
    1: f(0.342643) = -0.0246066
    2: f(0.351723) = -2.91774e-05
    3: f(0.351734) = -4.09699e-11

  F3 = X * EXP ( - X )
  Root X =  -9.389621148813321e-14
    0: f(0.5) = 0.303265
    1: f(-0.5) = -0.824361
    2: f(-0.166667) = -0.196893
    3: f(-0.0238095) = -0.0243832
    4: f(-0.00055371) = -0.000554017
    5: f(-3.06425e-07) = -3.06425e-07
    6: f(-9.38962e-14) = -9.38962e-14

  F4 = EXP ( X ) - 1 / ( 100 * X^2 )
  Root X =  0.09534461719362387
    0: f(0.03) = -10.0807
    1: f(0.04359) = -4.21836
    2: f(0.060984) = -1.62598
    3: f(0.0792032) = -0.511673
    4: f(0.0915816) = -0.0963876
    5: f(0.0951339) = -0.00510924
    6: f(0.095344) = -1.60817e-05
    7: f(0.0953446) = -1.60398e-10

newton_test():
  Normal end of execution.
Sun Mar 29 09:50:18 2026