# Cost tends to infinity when relu activation is used

I have implemented a neural network with 1 hidden layer using sigmoid activation unit but after watching a video on how relu activation function can be much faster a tried implementing it but the cost function is either Nan or inf in python. I have found that many people have the same problem but I couldn't find any solution on the internet. I have added '#changed' comment to the lines that I have changed from the previous version in which I used sigmoid activation function. Related python code snippets:

def relu(arg):  #I have tried both relu and leaky relu
return 1*(arg<0)*0.0001*arg + (arg>=0)*arg

for i in range(arg.shape):
for j in range(arg.shape):
if arg[i][j]>0:
arg[i][j]=1
else:
arg[i][j]=0
return arg
def softmax(x):
x = x.transpose()
e_x = np.exp(x - np.max(x))
return (e_x / e_x.sum(axis=0)).transpose()

#forward prop:
a1 = np.insert(data,0,np.ones(len(data)),1).astype(np.float64)
z2 = a1.dot(theta1)
a2 = relu(z2) #changed
a2 = np.insert(a2,0,np.ones(len(a2)),1)
z3 = a2.dot(theta2)
a3 = softmax(z3) #changed
#compute the cost:
cost = -(output*(np.log(a3))+(1-output)*(np.log(1-a3))).sum()
cost = (1/len(data))*cost + (lamb/(2*len(data)))*((np.delete(theta1,0,0)**2).sum() + (np.delete(theta2,0,0)**2).sum())

backProp:
sigma3 = a3-output
sigma2 = np.delete(sigma2,0,1)
delta2 = (np.transpose(a2)).dot(sigma3)
delta1 = (np.transpose(a1)).dot(sigma2)

#update theta

If a3 is 0 or 1, np.log(a3) or np.log(1-a3) is going to give an error as $\log(0)$ is undefined, and $\lim_{x -> 0} \log(x) = -\infty$
• @AyushChaurasia I'm not even sure that is the problem, can you print out the log values to verify? Also see here: datascience.stackexchange.com/questions/9302/… for the suggestion to add a small value $10^{-15}$ in the log function so that it never reaches zero. – geometrikal Jul 4 '17 at 6:32