CHOOSE ONLY 1! 4A or 4B, up to you. We are programming in python. We are able to
ID: 3872690 • Letter: C
Question
CHOOSE ONLY 1! 4A or 4B, up to you.
We are programming in python. We are able to use numpy and scipy
CHOOSE ONLY ONE 4a) Consider f(x)-(1-x)/x, which has a root at x-1. Write a code to find the root using the secant method with xo=01 and X1-10. Comment on the convergence. Try some other starting intervals. Consider the function glx)-(x-a)/(bx-c), where a,b, and c are free parameters. Use your 3 data points (x i,f(x_i) to construct an approximation for f and use it to numerically determine the root 4b) Let a root finding method be given by: Xn+1-Xn +d , where d is a positive integer and fíx) is a smooth function, and (g)d) means d derivatives of the function g. Give a proof of the convergence rate of the iteration near a simple root r (f(r) 0). Code the method up for d-2 efficiently and demonstrate the convergence rate for a given test function(you pick, no trivial cases).Explanation / Answer
from math import *
def scrub(f):
nf=''
nums='0123456789'
for i in range(len(f)):
nf += f[i]
if f[i] in nums:
if i+1 < len(f):
if f[i-1] != '.' and f[i+1] != '.' and f[i-1] not in nums and f[i+1] not in nums:
nf += '.0'
elif i+1 == len(f):
nf +='.0'
return nf
def funct(f):
f=scrub(f)
def aval(x):
try:
x=float(x)
return eval(f)
except:
return f
return aval
def newton(f,d,n,guess,exact=None):
F=funct(f)
D=funct(d)
p=[guess]
for i in range(1,n+1):
prv = p[i-1]
p.append(prv-F(prv)/D(prv))
if not exact==None: exact=(len(p)-1)*[exact]
else: exact=(len(p)-1)*[p[len(p)-1]]
error = [abs(p_i-exact_i) for p_i,exact_i in zip(p,exact)]
return [p,exact,error]
#==============================================================================
f='x**2-329'
d='2*x'
n=16
guess=2
exact=sqrt(329)
[p,exact,error]=newton(f,d,n,guess,exact)
import matplotlib
matplotlib.use("TKAgg")
import matplotlib.pyplot as plt
import numpy as np
plt.ion()
plt.plot(p,'r')
plt.plot(exact,'y')
#plt.plot(np.log(error),'b--')
#