Use a higher number than the maximum float for the special function lambertw

Question

I am using the special function lambertw (k=-1) in Python 3 and I need to use it with numbers higher/lower than the maximum/minimum float number (1.7976931348623157e+308).

What can I do?

Also I tried with "decimal", but it did not work, i. e.,

from decimal import Decimal
from scipy.special import lambertw

lambertw(Decimal('3.1E+600'))

obtained this,

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/share/apps/sistema/Python-3.5.1/lib/python3.5/site-packages/scipy/special/lambertw.py", line 107, in lambertw
return _lambertw(z, k, tol)
TypeError: ufunc '_lambertw' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''

The first thing you can do is post what you tried along with an explanation of why it did not work; telling people here that *"I tried with "decimal", but it did not work".* isn't helpful. — Dimitris Fasarakis Hilliard, May 24 '17 at 17:29
I would be surprised if `scipy.special.lambertw` works with `decimal.Decimal`... — juanpa.arrivillaga, May 24 '17 at 17:42

score 2 · Answer 1 · answered May 24 '17 at 17:35

The decimal module should be able to solve your problem. The issue you ran into may be that you did not set the precision higher than the default 28, as mentioned in the docs. To do that, just call getcontext().prec = 100 or whatever degree of precision you need.

For instance, using your example number, I just ran this interactive session:

>>> decimal.getcontext().prec = 1000
>>> d = decimal.Decimal(1.7976931348623157e+308)
>>> d
Decimal('179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368')

Ah, thanks for your edit to the question, it was pretty vague at first. I did a bit of research and it looks like you just can't; scipy seems to only support native numeric types: https://docs.scipy.org/doc/numpy-1.10.1/user/basics.types.html — Personman, May 24 '17 at 18:10

Warren Weckesser · Accepted Answer · 2017-05-24T18:46:50.040

The mpmath library in SymPy includes an implementation of lambertw. mpmath implements arbitrary precision floating point arithmetic.

Here's an example. First, import mpmath from sympy, and set the digits of precision to 100 (chosen arbitrarily--change to fit your needs):

In [96]: from sympy import mpmath

In [97]: mpmath.mp.dps = 100

Verify that the mpmath function gives the same results as scipy.special.lambertw:

In [98]: from scipy.special import lambertw

In [99]: lambertw(123.45)
Out[99]: (3.5491328966138256+0j)

In [100]: mpmath.lambertw(123.45)
Out[100]: mpf('3.549132896613825444243187580460572741065183903716765715536934583554830913412258511917029758623080475405')

Compute lambertw(3.1e600). The argument is entered as a string, because we can't represent 3.1e600 as a regular floating point value. mpmath will convert the string to a high-precision floating point value using the precision that we set earlier.

In [101]: mpmath.lambertw('3.1e600')
Out[101]: mpf('1375.455917376503282959382815269413629072666427317318260231463057587794635136887591876065911283365916388')

We can also create a variable x to hold the input value, and then call mpmath.lambertw(x):

In [102]: x = mpmath.mpf('3.1e600')

In [103]: x
Out[103]: mpf('3.099999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999e+600')

In [104]: mpmath.lambertw(x)
Out[104]: mpf('1375.455917376503282959382815269413629072666427317318260231463057587794635136887591876065911283365916388')

The result can be represented as a regular floating point value, so we pass it to the builtin function float() to do the conversion:

In [105]: float(mpmath.lambertw(x))
Out[105]: 1375.455917376503

Use a higher number than the maximum float for the special function lambertw

2 Answers2