# python – numpy division with RuntimeWarning: invalid value encountered in double_scalars

## python – numpy division with RuntimeWarning: invalid value encountered in double_scalars

You cant solve it. Simply `answer1.sum()==0`, and you cant perform a division by zero.

This happens because `answer1` is the exponential of 2 very large, negative numbers, so that the result is rounded to zero.

`nan` is returned in this case because of the division by zero.

Now to solve your problem you could:

• go for a library for high-precision mathematics, like mpmath. But thats less fun.
• as an alternative to a bigger weapon, do some math manipulation, as detailed below.
• go for a tailored `scipy/numpy` function that does exactly what you want! Check out @Warren Weckesser answer.

Here I explain how to do some math manipulation that helps on this problem. We have that for the numerator:

``````exp(-x)+exp(-y) = exp(log(exp(-x)+exp(-y)))
= exp(log(exp(-x)*[1+exp(-y+x)]))
= exp(log(exp(-x) + log(1+exp(-y+x)))
= exp(-x + log(1+exp(-y+x)))
``````

where above `x=3* 1089` and `y=3* 1093`. Now, the argument of this exponential is

`-x + log(1+exp(-y+x)) = -x + 6.1441934777474324e-06`

For the denominator you could proceed similarly but obtain that `log(1+exp(-z+k))` is already rounded to `0`, so that the argument of the exponential function at the denominator is simply rounded to `-z=-3000`. You then have that your result is

``````exp(-x + log(1+exp(-y+x)))/exp(-z) = exp(-x+z+log(1+exp(-y+x))
= exp(-266.99999385580668)
``````

which is already extremely close to the result that you would get if you were to keep only the 2 leading terms (i.e. the first number `1089` in the numerator and the first number `1000` at the denominator):

``````exp(3*(1089-1000))=exp(-267)
``````

For the sake of it, lets see how close we are from the solution of Wolfram alpha (link):

``````Log[(exp[-3*1089]+exp[-3*1093])/([exp[-3*1000]+exp[-3*4443])] -> -266.999993855806522267194565420933791813296828742310997510523
``````

The difference between this number and the exponent above is `+1.7053025658242404e-13`, so the approximation we made at the denominator was fine.

The final result is

``````exp(-266.99999385580668) = 1.1050349147204485e-116
``````

``````1.105034914720621496.. × 10^-116 # Wolfram alpha.
``````

and again, it is safe to use numpy here too.

You can use `np.logaddexp` (which implements the idea in @gg349s answer):

``````In [33]: d = np.array([[1089, 1093]])

In [34]: e = np.array([[1000, 4443]])

In [36]: log_res
Out[36]: -266.99999385580668

In [37]: res = exp(log_res)

In [38]: res
Out[38]: 1.1050349147204485e-116
``````

Or you can use `scipy.special.logsumexp`:

``````In [52]: from scipy.special import logsumexp

In [53]: res = np.exp(logsumexp(-3*d) - logsumexp(-3*e))

In [54]: res
Out[54]: 1.1050349147204485e-116
``````
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