# numpy division with RuntimeWarning: invalid value encountered in double_scalars

You can’t solve it. Simply `answer1.sum()==0`, and you can’t 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 that’s 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, let’s 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
```

From wolfram alpha is

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

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