In function “def gradient_check(x, theta, epsilon=1e-7, print_msg=False):”
I’m not solving the “grad” equation. I just put grad=gradapprox and all the tests are passed 
. Kindly assist me how to tackle this problem
Hi, @hark99.
That function returns the difference between the gradient computed by backward_propagation and the approximated gradient. If you set them to the same value, the difference will be 0. That’s why the tests passed.
The comments tell you how to compute grad:
# Check if gradapprox is close enough to the output of backward_propagation()
#(approx. 1 line)
# grad =
I have to calculate the derivative of ‘gradapprox’ or something else after setting
J_pLus = (theta + epsilon) * x
I am still facing issues in calculating the gradient of gradapprox. Give me a hint if possible.
@nramon
I am not sure how to write a code for “Check if gradapprox is close enough to the output of backward_propagation()”. Can you please guide me?
Hi @sima_ranjbari,
You have already computed gradapprox, right? Now call backward_propagation() (you are given x and theta) and follow these steps (from the notebook) to compute the difference:
- Compute the numerator using
np.linalg.norm(numerator = ...). - Compute the denominator. You will need to call
np.linalg.normtwice (denominator = ...). - Divide them (
difference = ...).
Let me know if you need help with a specific step. Good luck 
Hi there, i have this same issue. Though i dont know what ‘close enoughh’ mean, i think i have this correct (as i just use the backward_prop function with the argument as discussed above). Function compiles, but test fails with: AssertionError: You are not using np.linalg.norm for numerator or denominator.. . However i am using these functions. Any thoughts on what might be wrong? Many thanks! 
Whoops fixed it. typo earlier with mistake in gradapprox formula. all good now thanks
This thread is four years old, I’ll close it.
New reports should be posted in their own new thread.