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

i need compute quantity

1/tanh(x) - 1/x 

for x > 0, x can both small , large.

asymptotically small x, have

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

and large x

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

anyhow, when computing expression, 10^-7 , smaller round-off errors lead expression being evaluated 0:

import numpy import matplotlib.pyplot plt   x = numpy.array([2**k k in range(-30, 30)]) y = 1.0 / numpy.tanh(x) - 1.0 / x  plt.loglog(x, y) plt.show() 

use python package mpmath arbitrary decimal precision. example:

import mpmath 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 mpmath plotting. mpmath plays matplotlib, using, should solve problem.

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here example of how integrate mpmath code wrote above:

import numpy import matplotlib.pyplot plt import mpmath mpmath import mpf  mpmath.mp.dps = 100 # set decimal precision  x = numpy.array([mpf('2')**k k in range(-30, 30)]) y = mpf('1.0') / numpy.array([mpmath.tanh(e) e in x]) - mpf('1.0') / x  plt.loglog(x, y) plt.show()