[수치해석] Parabolic Interpolation 파이썬

import math
 
def getX3(x0, x1, x2, f0, f1, f2):
  x3 = ((f0*(x1**2 - x2**2)+f1*(x2**2-x0**2)+f2*(x0**2-x1**2))/(f0*(x1-x2)+f1*(x2-x0)+f2*(x0-x1)))/2
  return x3
  
def parabolic(x0, x1, x2, maxit, es):
  # See x1 as intermediate point 
  # you can substitute it for (x0 + x2)/2
  iter = 1
  f0 = f(x0)
  f1 = f(x1)
  f2 = f(x2)
  x3 = getX3(x0, x1, x2, f0, f1, f2)
  f3 = f(x3)
  xopt = x1
 
  print(" i    x0    f(x0)   x1     f(x1)    x2   f(x2)   x3    f(x3)")
  print("%2d %.4f  %.4f %.4f %.4f %.4f %.4f %.4f %.4f" % (iter, x0, f0, x1, f1, x2, f2, x3, f3))
 
  # Suppose We're finding maximum
 
  while True:
    x3 = getX3(x0, x1, x2, f0, f1, f2)
    f3 = f(x3)
    
    # new point's f value is larger than old intermediate one's 
    if f3 > f1:
      xold = xopt
      xopt = x3
 
      if x3 > x1:
        x0 = x1
        f0 = f1
        x1 = x3
        f1 = f3
      else:
        x2 = x1
        f2 = f1
        x1 = x3
        f1 = f3
 
    iter += 1
 
    ea = abs((xopt - xold)/xopt)*0.01
 
    print("%2d %.4f  %.4f %.4f %.4f %.4f %.4f %.4f %.4f" % (iter, x0, f0, x1, f1, x2, f2, x3, f3))
 
    if iter >= maxit or ea <= es:
      print("%.4f is optimal point" % xopt)
      break
 
 
def f(x):
  # Min = 3 : 
  # return (x**2) -  (6*x) + 15
  # Max = 1.4274
  return 2*math.sin(x) - (x**2)/10
 
parabolic(0,1,4, 1000,0.00001)