Unexpected result in the numerical calculation – (invalid value encountered in double_scalar )

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from numpy import log as ln
z = 3
k = 2

x = 1 - ln(1 + z) / ln(1 + k)
y = 1/5
print("The x=", x)

Q = x**y

print(Q)

The result is

The x= -0.26185950714291484
c:Users-Desktop... RuntimeWarning: invalid value 
encountered in double_scalars
  Q = x**y
nan

I am having troubling to understand the problem here. It’s clear that I can print the x, but I cannot take the power of it.

Meanwhile when I write the code as

x = -0.26185950714291484
y = 1/5
print("The x=", x)
Q = x**y

print(Q)

I get

The x= -0.26185950714291484
(0.6188299348758349+0.44960626529008196j)

I am literally shocked. I cannot understand what is going on. If I just put the numbers manually I can calculate the result. But if I do them computationally I get an error ???

Answer

Well they look the same by they are not the same, with the code:

from numpy import log as ln
z = 3
k = 2

x = 1 - ln(1 + z) / ln(1 + k)
y = 1/5
print("The x=", x)

Q = x**y

print(Q)

you have a variable ‘x’ of type <class 'numpy.float64'>, whereas with:

x = -0.26185950714291484
y = 1/5
print("The x=", x)
Q = x**y

print(Q)

the variable ‘x’ is of type <class 'float'>.

Then you applied the exponentiation with a value of different types, namely <type one of type ‘numpy.float64’> (i.e., ‘x’) with another of type <class ‘float’> (i.e., ‘y’). You want to avoid the problems is to convert ‘x’ to float was well:

from numpy import log as ln
z = 3
k = 2

x = 1 - ln(1 + z) / ln(1 + k)
y = 1/5

Q = float(x)**y

print(Q)

Output:

(0.6188299348758349+0.44960626529008196j)


Source: stackoverflow