How can I use "e" (Euler's number) and power operation?


How can I write 1-e^(-value1^2/2*value2^2) in Python?

I don’t know how to use power operator and e.

Asked By: Thanos Smar



Power is ** and e^ is math.exp:

x.append(1 - math.exp(-0.5 * (value1*value2)**2))
Answered By: Daniel

Python’s power operator is ** and Euler’s number is math.e, so:

 from math import e
Answered By: karlosss

You can use exp(x) function of math library, which is same as e^x. Hence you may write your code as:

import math
x.append(1 - math.exp( -0.5 * (value1*value2)**2))

I have modified the equation by replacing 1/2 as 0.5. Else for Python <2.7, we’ll have to explicitly type cast the division value to float because Python round of the result of division of two int as integer. For example: 1/2 gives 0 in python 2.7 and below.

Answered By: Moinuddin Quadri

math.e or from math import e (= 2.718281…)

The two expressions math.exp(x) and e**x are equivalent
Return e raised to the power x, where e = 2.718281… is the base of natural logarithms. This is usually more accurate than math.e ** x or pow(math.e, x). docs.python

for power use ** (3**2 = 9), not " ^ "
" ^ " is a bitwise XOR operator (& and, | or), it works logicaly with bits.
So for example 10^4=14 (maybe unexpectedly) → consider the bitwise depiction:

(0000 1010 ^ 0000 0100 = 0000 1110) programiz

Answered By: Alexey K.

Just saying: numpy has this too. So no need to import math if you already did import numpy as np:

>>> np.exp(1)
Answered By: gosuto

Just to add, numpy also has np.e

Answered By: Rik Mulder

In my case, the exponent happens to be complex number with angle expressed in radians. So my approach was:

import cmath 
theta = cmath.pi/4
output = cmath.exp(theta*1j)   # LaTeX: $e^{itheta}$ 
print(output)                  # (0.7071067811865476+0.7071067811865476j)

Note: Use 1j instead of j since python throws NameError for j. And used cmath instead of math.

Answered By: Rajesh Swarnkar