Find fixed point for a function

Question

A fixed point for a function is the point where f(x)=x.

For a specific function I'm supposed to find the fixed point by starting with a random guess and then calucalting f again and again, i.e.: calculating f(x), f(f(x)), f(f(f(x))),... until the value doesn't change over epsilon.

the function I'm supposed to write gets as an input: a function, a random guess, epsilon, and n the number of iterations. the function should calculate an approximation for a fixed point. It will stop when the difference between the two numbers is smaller than epsilon, or when n iterations have been done. input and output examples:

    >>> fixed_point(lambda x:x**2, 1, n=5) 
1 #0 iterations needed, initial guess suffices 

>>> fixed_point(lambda x:x**2, 0.5, n=5) 
5.421010862427522e-20 #after 5 iterations 

>>> fixed_point(lambda x:x**2, 0.5, n=4) 
>>> #returns None 

>>> fixed_point(lambda x:x**2, 2, n=5) 
>>> #returns None, guesses were: 2, 4, 16, 256, 65536, 4294967296

My code gives a correct answer only for the first example, What should I fix in it?

def fixed_point(f, guess, epsilon=10**(-8), n=10):
itr=0
test=f(guess)
if (abs(test-guess)<epsilon):
    return(test)


while ((n>itr) and (abs(test-guess)>=epsilon)):
    itr+=1
    test=f(test)

    if ((abs(test-guess))<epsilon):

        return(test)
return(None)

Victory · Accepted Answer · 2014-03-29 18:23:36Z

2

You almost had it! You were just failing to "remember" your last guess, check it out:

 def fixed_point(f, guess, epsilon=10**(-8), n=10):
      itr=0
      print "Guess:", guess
      test=f(guess)
      if (abs(test-guess)<epsilon):
          return(test)


      while ((n>itr) and (abs(test-guess)>=epsilon)):
          itr+=1
          guess = test
          test = f(test)
          print "Guess:",guess

          if ((abs(test-guess))<epsilon):
              return(test)

      return(None)


  print fixed_point(lambda x:x**2, 1, n=5)
  print fixed_point(lambda x:x**2, 0.5, n=5)
  print fixed_point(lambda x:x**2, 0.5, n=4)
  print fixed_point(lambda x:x**2, 2, n=5)

answered Mar 29, 2014 at 18:23

Victory

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1 Comment

Steve Jessop Over a year ago

Heh, I was wrong. The code can be fixed, but the problem is still the approach :-)

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