Here is the code:
exp = 1.79
def calc(t):
return pow(t - 1, exp)
The input values of t range from 0 to 1 (e.g. 0.04). This code throws a "math domain exception," but I'm not sure why.
How can I solve this?
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Karl Knechtel
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Joan Venge
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What version of Python is this? I tried `pow(-.6, 1.79)` in Python 3.2 and got a complex result. – Free Monica Cellio Feb 01 '12 at 03:13
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@FreeMonicaCellio the built-in `pow` does that, as does `**` (with appropriate parentheses), but `math.pow` from the standard library reports an exception. Confusingly, both `pow`s describe themselves as `
`, but they are not the same. – Karl Knechtel Jan 11 '23 at 03:14
3 Answers
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If t ranges from 0 to 1, then t - 1 ranges from -1 to 0. Negative numbers cannot be raised to a fractional power, neither by pow builtin nor math.pow.
wim
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1@Joan: No. `pow()` supports 3 arguments, for modulus, whereas `math.pow()` does not. – Ignacio Vazquez-Abrams Feb 01 '12 at 03:14
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Thanks didn't know that. But when I do `from math import pow`, which pow is that? – Joan Venge Feb 01 '12 at 03:18
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That's the same as using `import math` and then `math.pow`, except that the name will now shadow the builtin `pow` in your namespace. Note you can also use `x**y` for exponentiation. – wim Feb 01 '12 at 03:19
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Negative numbers raised to a fractional exponent do not result in real numbers. You will have to use cmath if you insist on calculating and using them, but note that you will need some experience in complex numbers in order to make use of the result.
>>> cmath.exp(cmath.log(0.04 - 1) * 1.79)
(0.7344763337664206-0.5697182434534497j)
Ignacio Vazquez-Abrams
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exp = 1.79
def calc(t):
return pow(t - 1, exp)
print calc(1.00) # t-1 is 0, there will be no error.
print calc(0.99) # t-1 is negative, will raise an error.
Jason Sundram
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cola
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