How can i avoid getting result 0 from exp() for very small float numbers?

Question

I'm trying to get the exact value from an exponential that gets a very small floating point as input and gives me the result 0.

You can reproduce the problem with the following code:

import numpy as np
from math import sqrt, log, exp, pi

k = np.array([[-746.9292399]])
z = exp(k)
print(z)

The result of this will be 0 , for my pc the result of anything smaller than -743.0 will be 0

I have tried using mpmath to solve this as follows:

import numpy as np
from math import sqrt, log, exp, pi
import mpmath as mp

k = np.array([[-746.9292399]])
z = mp.exp(float(k))
print(z)

det = np.linalg.det([[z,2,3],[2,2,z],[3,6,2]])
print(det)

Which is giving a good answer for the exp() , however i will further need to put that result in a numpy array , and extract the determinant from it , which throws an error in the code i attached because numpy doesn't like mpf numbers in its arrays.

Anybody knows how could i get the result from exp() in a way that i can use it with numpy and its functions?

For reference, mathematically, `exp(-746.9292399)` is ‭4.0997...e-325‬. The minimum value of [binary64](https://en.wikipedia.org/wiki/Double-precision_floating-point_format) is 4.940656...E-324 — chux - Reinstate Monica, Mar 28 '20 at 09:06

score 2 · Answer 1 · answered Mar 26 '20 at 18:22

This is a common problem with exp and log functions, as the output tends to disappear to zero rather quickly. A good observation is that you typically want expr(x) - 1 in most scientific computations, and most implementations provide a custom function for that to avoid numerical stability issues. So, if it fits your use case, you should definitely consider expm1. Similar considerations also apply to log, and hence we have log1p, expm1, log1pexp, log1mexp etc. See here for details: https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.expm1.html

im trying to implement a probability density function for an expectation-maximization algorithm , i have tried your apprach however the final estimations were more accurate before (at the moment im just doing a simple if(x < -743): x = -743 in an attemp to make the program not throw an error) . @alias — george joker, Mar 27 '20 at 09:19

score 0 · Answer 2 · answered Mar 26 '20 at 10:52

0

it seems to me that around exp(-745) the value is the smallest that a float can hold.

notice that

>>> exp(-744)
1e-323

The floating point has moved almost entirely to the right. I believe that this is the limit of python's float.

answered Mar 26 '20 at 10:52

Tom Gil

9
2

Yes , i was wondering if there is a way to overcome this limit. – george joker Mar 26 '20 at 10:53
I see. I'm not an expert or anything so I probably am no use in this case, but I hope you'll overcome this issue. – Tom Gil Mar 27 '20 at 11:13

score 0 · Answer 3 · answered Mar 26 '20 at 12:03

Okay I've been working on this for about 30 mins now and I got this

import decimal
pow(decimal.Decimal(math.exp(1)),decimal.Decimal(-748.9292399))

Also, check this Exponential of large negative numbers

there is one small issue. I calculated your problem with windows calculator and it gave me this 3.819636355e-325

How can i avoid getting result 0 from exp() for very small float numbers?

3 Answers3