Python Calculate log(1+x)/x for x near 0

Question

Is there a way to correctly calculate the value of log(1+x)/x in python for values of x close to 0? When I do it normally using np.log1p(x)/x, I get 1. I somehow seem to get the correct values when I use np.log(x). Isn't log1p supposed to be more stable?

for `log(1+x)` you want `np.log1p(x)` not `np.log1p(1+x)`. – hobbs Sep 13 '16 at 18:59 — hobbs, Sep 13 '16 at 18:59

score 1 · Answer 1 · answered Sep 13 '16 at 19:01

1

np.log1p(1+x)

That gives you log(2+x). Change it to np.log1p(x).

answered Sep 13 '16 at 19:01

wim

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score 0 · Accepted Answer · answered Oct 05 '16 at 23:15

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So I found one answer to this. I used a library called decimal.

from decimal import Decimal
x = Decimal('1e-13')
xp1 = Decimal(1) + x
print(xp1.ln()/x)

This library seems to be much more stable than numpy.

answered Oct 05 '16 at 23:15

Aditya369

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Python Calculate log(1+x)/x for x near 0

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