Numerical result out of range in Python

Question

I am a beginner to Python. I have written a function which calculates numbers that are of the order 10^-15.

While generating the output of the code, an error says "Numerical result out of range". What are possible reasons for which this error occurs?

Here is my code.

import numpy as np
import matplotlib.pyplot as plt
T=4.32*10**19
i=input("Number of iterations ")
h=T/i
a=[0.01]*6000
t=[0.0]*6000
data = np.loadtxt('/home/user/t.txt')
t=data[:]
def f (a):
    if a!=0:    
        c=((0.75/a + 0.044/(a**2) + 0.74*(a**2))**0.5)  
        return c
    else :
        return 0

for n in range(0,5998):
    k1=f ( a[n] );
    k2=f ( a[n]+(h/2)*k1 );
    k3=f ( a[n]+(h/2)*k2 );
    k4=f ( a[n]+h*k3 ); 
    a[n+1]=(a[n]+(h/6)*(k1 + 2*k2 + 2*k3 + k4))

fo=open("a_of_t.txt", "w")
for item in a:
    fo.write("%e\n" % item)
fo.close()

plt.plot(t,a, 'k')
plt.show()

The error reads

line 23, in <module>
    k3=f ( a[n]+(h/2)*k2 );
line 15, in f
    c=((0.75/a + 0.044/(a**2) + 0.74*(a**2))**0.5)  
OverflowError: (34, 'Numerical result out of range')

Please show your code and the specific error message (stack trace) that you're getting. — yole, Jun 30 '15 at 22:09
At some order of magnitude floats will stop working, but I forgot where. Perhaps https://docs.python.org/2/library/decimal.html `decimal` will interest you? — NightShadeQueen, Jun 30 '15 at 22:11
It's similar to [link] (http://stackoverflow.com/questions/29488213/handling-very-small-numbers-in-python) — Adolfo Correa, Jun 30 '15 at 22:21
Can you write out the entire error? It tells you where the computer messed up — rassa45, Jun 30 '15 at 23:02
Can you please tell me exactly what is the value of a[n] + (h/2) * k2. If it is inonvenient I can definitely try to find out myself but it might take a few seconds more — rassa45, Jun 30 '15 at 23:08
This is a code for numerically solving a differential equation using 4th order Runge-Kutta method. a[n+1] represents the value of a for next step. — user40330, Jun 30 '15 at 23:12
Actually, I think I have your answer. Try not using Decimal() and see how it goes — rassa45, Jun 30 '15 at 23:12
If you really didn't know, you could have printed n in the second to last for loop and found out where it was crashing. — rassa45, Jun 30 '15 at 23:14

score 0 · Accepted Answer · answered Jun 30 '15 at 22:13

0

Python decimals are stored as 64-bit floats, and as such, have finite precision. 10^-15 cannot be represented using a float. Consider using the decimal module.

answered Jun 30 '15 at 22:13

James

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Python uses double (64-bit floats) which have a magnitude range of roughly 1e308 to 1e-308, plenty good enough for the OP's stated range of 1e-15. – wallyk Jun 30 '15 at 22:39
@James Can you suggest me how to incorporate decimal module in this code? I am not getting it. – user40330 Jun 30 '15 at 23:09
@user40330 just convert all your variables to `Decimal` by doing `Decimal(var)`. For example, `T=Decimal(4.32*10**19)`, `i=Decimal(input("Number of iterations "))`. If you want a `float` at the end, just cast using `float(someDecimal)`. – James Jun 30 '15 at 23:36
@James Thanks! that helped – user40330 Jul 01 '15 at 11:14

Numerical result out of range in Python

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