py newton-raphson secant

October 4, 2023

direction

Mr. Satoru is stuck in a train station with a bunch of monsters (which he needs to kill) and humans (which he needs to save). He can expand a “sure-kill” domain that kills anyone within its perimeters, with him as the center, but he must consider the optimal amount of monsters killed and humans saved.
The monsters are distributed around him in a randomized manner, but can be represented with the equation $$\tag{1} M(x) = 20x - x^2. $$
The humans distributed around him can also be represented with the equation $$\tag{2} H(x) = \sin 2x. $$
Mr. Satoru’s judgement on the radius in which he would expand him domain can be represented with the following equation $$\tag{3} R(x) = M(x) + H(x). $$
Optimize the radius of Mr. Satoru’s domain using the Newton-Raphson method, and Secant Method until the error $\varepsilon < 10^{-3}$! (Hint: To optimize a function, differentiate it first and find its root).

derivative

differentiate Eqn (3), with the help of Eqns (1) and (2) wil give $$\tag{4} R'(x) = M'(x) + H'(x) = 20 - 2x + 2\cos 2x. $$
Optimum radius is obtained when $$\tag{5} R'(x_{\rm optimum}) = 0. $$
Find root of $x'(x)$ to get $x_{\rm optimum}$.

iterative method

Newton-Raphson method $$\tag{6} x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. $$
Secant method $$\tag{7} x_{n+2} = x_{n+1} - \frac{(x_{n+1} - x_n) f(x_{n+1})}{f(x_{n+1}) - f(x_n)}. $$
Newton-Raphson method requires one initial value $x_0$ and secant method requires two initial values $x_0$ and $x_1$.

solution

Code

import math

# test find optimum or y = (x - 2)(x - 6)
# it will give x = 4
# and ok
TEST = False

# break
ITERMAX = 100

# show iteration
SHOWITER = False

# Method
METHOD = 'Newton-Raphson'

def R(x):
  if TEST:
    y = (x - 2) * (x - 6)
  else:
    y = 20 * x - x**2 + math.sin(2 * x)
  return y

def R_(x):
  if TEST:
    y = (x - 2) + (x - 6)
  else:
    y = 20 - 2*x + 2 * math.cos(2 * x)
  return y

def R__(x):
  if TEST:
    y = 2
  else:
    y = -2 - 4 * math.sin(2 * x)
  return y

# method 1
print("Newton-Raphson method")
if SHOWITER:
  print("i", "err", "R_(x)", "x", sep='\t')

x0 = 19
err = 1
i = 1
while err > 1E-3:
  # finding root for R_(x) not for R(x)
  x1 = x0 - R_(x0) / R__(x0)
  err = abs((x1 - x0))

  if SHOWITER:
    errs = f'{err:.2f}'
    R_xs = f'{R_(x1):.2f}'
    xs = f'{x1:.2f}'
    print(i, errs, R_xs, xs, sep='\t')

  x0 = x1
  i += 1

  if i > ITERMAX:
    break

i -= 1
x = x1

print()
print("Results")
print("iteration =", i)
print("err =", err)
print("x =", x)
print("R(x) =", R(x))
print("R_(x) =", R_(x))
print("R__(x) =", R__(x))
print()

# method 2
print("secant method")
if SHOWITER:
  print("i", "err", "R_(x)", "x", sep='\t')

x0 = 1
x1 = 50
err = 1
i = 1
while err > 1E-3:
  # finding root for R_(x) not for R(x)
  x2 = x1 - R_(x1) / (R_(x1) - R_(x0)) * (x1 - x0)
  err = abs((x2 - x1))

  if SHOWITER:
    errs = f'{err:.2f}'
    R_xs = f'{R_(x2):.2f}'
    xs = f'{x2:.2f}'
    print(i, errs, R_xs, xs, sep='\t')

  x0 = x1
  x1 = x2
  i += 1

  if i > ITERMAX:
    break

i -= 1
x = x2

print()
print("Results")
print("iteration =", i)
print("err =", err)
print("x =", x)
print("R(x) =", R(x))
print("R_(x) =", R_(x))
print("R__(x) =", R__(x))

Result

Newton-Raphson method

Results
iteration = 8
err = 7.699729229315722e-06
x = 10.139963731429795
R(x) = 100.97056678487193
R_(x) = -3.319355901254539e-11
R__(x) = -5.960626523950763

secant method

Results
iteration = 5
err = 0.0002986050511903926
x = 10.139963647548765
R(x) = 100.97056678487193
R_(x) = 4.9995029555161e-07
R__(x) = -5.960626430028292

url https://onecompiler.com/python/3zpg9vgqa

notes

It seem that Newton-Raphson method is very sensitive to initial value to obtain optimum of Eqn (3).
Furhter study is required to investigate the good function to optimize and not too sensitive to initial value for Newton-Raphson method and to initial values for secant method.