Evaluate 1/tanh(x) - 1/x for very small x

问题

I need to compute the quantity

1/tanh(x) - 1/x

for x > 0, where x can be both very small and very large.

Asymptotically for small x, we have

1/tanh(x) - 1/x  ->  x / 3

and for large x

1/tanh(x) - 1/x  ->  1

Anyhow, when computing the expression, already from 10^-7 and smaller round-off errors lead to the expression being evaluated as exactly 0:

import numpy
import matplotlib.pyplot as plt


x = numpy.array([2**k for k in range(-30, 30)])
y = 1.0 / numpy.tanh(x) - 1.0 / x

plt.loglog(x, y)
plt.show()

回答1:

For very small x, one could use the Taylor expansion of 1/tanh(x) - 1/x around 0,

y = x/3.0 - x**3 / 45.0 + 2.0/945.0 * x**5

The error is of the order O(x**7), so if 10^-5 is chosen as the breaking point, relative and absolute error will be well below machine precision.

import numpy
import matplotlib.pyplot as plt


x = numpy.array([2**k for k in range(-50, 30)])

y0 = 1.0 / numpy.tanh(x) - 1.0 / x
y1 = x/3.0 - x**3 / 45.0 + 2.0/945.0 * x**5
y = numpy.where(x > 1.0e-5, y0, y1)


plt.loglog(x, y)
plt.show()

回答2:

Use the python package mpmath for arbitrary decimal precision. For example:

import mpmath
from mpmath import mpf

mpmath.mp.dps = 100 # set decimal precision

x = mpf('1e-20')

print (mpf('1') / mpmath.tanh(x)) - (mpf('1') / x)
>>> 0.000000000000000000003333333333333333333333333333333333333333311111111111111111111946629156220629025294373160489201095913

It gets extremely precise.

Look into mpmath plotting. mpmath plays well with matplotlib, which you are using, so this should solve your problem.

Here is an example of how to integrate mpmath into the code you wrote above:

import numpy
import matplotlib.pyplot as plt
import mpmath
from mpmath import mpf

mpmath.mp.dps = 100 # set decimal precision

x = numpy.array([mpf('2')**k for k in range(-30, 30)])
y = mpf('1.0') / numpy.array([mpmath.tanh(e) for e in x]) - mpf('1.0') / x

plt.loglog(x, y)
plt.show()

回答3:

A probably simpler solution to overcome this is changing the data type under which numpy is operating:

import numpy as np
import matplotlib.pyplot as plt

x = np.arange(-30, 30, dtype=np.longdouble)
x = 2**x
y = 1.0 / np.tanh(x) - 1.0 / x

plt.loglog(x, y)
plt.show()

Using longdouble as data type does give the proper solution without rounding errors.

I did sightly modify your example, in your case the only thing you need to modify is:

x = numpy.array([2**k for k in range(-30, 30)])

to:

x = numpy.array([2**k for k in range(-30, 30)], dtype=numpy.longdouble)

来源：https://stackoverflow.com/questions/43279071/evaluate-1-tanhx-1-x-for-very-small-x

标签

python

numpy

math

rounding