How can I use "e" (Euler's number) and power operation?
Question:
How can I write 1-e^(-value1^2/2*value2^2) in Python?
I don’t know how to use power operator and e.
Answers:
Power is ** and e^ is math.exp:
x.append(1 - math.exp(-0.5 * (value1*value2)**2))
Python’s power operator is ** and Euler’s number is math.e, so:
from math import e
x.append(1-e**(-value1**2/2*value2**2))
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.
math.e or from math import e (= 2.718281…)
The two expressions math.exp(x) and e**x are equivalent
however:
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
Just saying: numpy has this too. So no need to import math if you already did import numpy as np:
>>> np.exp(1)
2.718281828459045
Just to add, numpy also has np.e
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.
How can I write 1-e^(-value1^2/2*value2^2) in Python?
I don’t know how to use power operator and e.
Power is ** and e^ is math.exp:
x.append(1 - math.exp(-0.5 * (value1*value2)**2))
Python’s power operator is ** and Euler’s number is math.e, so:
from math import e
x.append(1-e**(-value1**2/2*value2**2))
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.
math.e or from math import e (= 2.718281…)
The two expressions math.exp(x) and e**x are equivalent
however:
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
Just saying: numpy has this too. So no need to import math if you already did import numpy as np:
>>> np.exp(1)
2.718281828459045
Just to add, numpy also has np.e
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.