Simple iteration in python

Question

help in writing the code. What am I doing wrong?

My goal is to find the roots of an equation by an iterative method with precision: ξ = 10^-5f(x) (or ε = 0.00001). Equation: 2.056x^43+3x^31+4x^12+x^8-3,478 = 0.

Code:

# -*- coding: utf-8 -*-
import math
#Definition of function
def phi (x):
    return 2.056*(x**43)+3*(x**31)+4*(x**12)+(x**8)-3.478
#Recursive search function
def findRoot (f, x, q, epsilon):
    fx=f(x)
    #Checking the ending condition
    if (1 / (1-q) * abs (fx-x) <epsilon):
        print 'Root value', fx
        print '1 / (1-q) * abs (fx-x)=', 1 / (1-q) * abs (fx-x)
    else:
        print 'Current approximation', fx
        print '1 / (1-q) * abs (fx-x)=', 1 / (1-q) * abs(fx-x)
        findRoot (f, fx, q, epsilon)
findRoot(phi, 0.5, 0.5, 0.00001)

Execution

Current approximation -3.4731171861
1 / (1-q) * abs (fx-x)= 7.94623437221
Current approximation -3.66403074312e+23
1 / (1-q) * abs (fx-x)= 7.32806148624e+23
Traceback (most recent call last):
  File "Zavd1f.py", line 17, in <module>
    findRoot(phi, 0.5, 0.5, 0.00001)                
  File "Zavd1f.py", line 16, in findRoot
    findRoot (f, fx, q, epsilon)
  File "Zavd1f.py", line 16, in findRoot
    findRoot (f, fx, q, epsilon)
  File "Zavd1f.py", line 8, in findRoot
    fx=f(x)
  File "Zavd1f.py", line 5, in phi
    return 2.056*(x**43)+3*(x**31)+4*(x**12)+(x**8)-3.478
OverflowError: (34, 'Numerical result out of range')

Possible duplicate of [Handling very large numbers in Python](https://stackoverflow.com/questions/538551/handling-very-large-numbers-in-python) — Christopher Schneider, Jun 11 '19 at 15:23
It would help if you'd be specific about *which* "iterative method" you're trying to implement. There are several. — Prune, Jun 11 '19 at 15:41

Prune · Accepted Answer · 2019-06-11T15:59:27.593

This iterative method simply applies the functional value repeatedly as the argument. This works only when your initial guess is within the radius of convergence. Yours is not. You need to implement an algorithm that gets closer with each iteration.

The current algorithm has an initial guess at 0.5; this raised to high powers (8 is high enough) is close to 0, so we get a result very close to the constant term.

f(0.5)   => -3.47...
f(-3.47) => -3.66...e+23    (really big)
f(really_big) => out of bounds

So ... it's either your starting value, or your algorithm, or your implementation of that algorithm.

I played with your code a little; I think you might be trying to implement a bisection algorithm (from q=0.5) or Newton's method. In either case, you've neglected to code the next guess.

You simply use f(x) as the next guess for the root, an x value. This is incorrect; you need to remember that this is a y value; you use it to compute a better guess for your next x value. f(x) is not, itself, that next guess.

Since you didn't post your algorithm, I'm not sure how twice the error (the expression you hard-coded three times) is supposed to related to the iterative process.

Simple iteration in python

1 Answers1