points$)$ Recall the softmax function for $N$ classes:

$(x_j)=\frac{e^{x_j}}{{\displaystyle \underset{i-1}{\overset{N}{}}}{e}^{x_1}}$

Computing cross entropy loss for this function using the formula

$H(p,q)=-{\displaystyle \underset{\text{x}\text{i}\text{n}\text{x}}{\overset{?}{}}}p(x)logq(x)$

yields terms of the form,

$log(x_j)=log(\frac{e^{x_j}}{{\displaystyle \underset{i-1}{\overset{N}{}}}{e}^{x_i}})$

$=log(e^{x_j})-log({\displaystyle \underset{i=1}{\overset{N}{}}}{e}^{x_i})$

$=x_j-log({\displaystyle \underset{1=1}{\overset{N}{}}}{e}^{x_i})$

Consider calculating $log(x_j)$ for $x=[100001000010000]$ and $x=[-10000-10000-10000].$

$($a$)$ Analytically calculate $log(x_j)$ for these two cases.

$($b$)$ Do you see any problem in numerically computing these values in a computer? $($Hint: You can

try it out yourself in Python.$)$

$($c$)$ If yes, then how can you overcome these problems?