#?GRADED?FUNCTION:?gradient_check def?gradient_check(x,?theta,?epsilon?=?1e-7): ????""" ????Implement?the?backward?propagation?presented?in?Figure?1. ???? ????Arguments: ????x?--?a?real-valued?input ????theta?--?our?parameter,?a?real?number?as?well ????epsilon?--?tiny?shift?to?the?input?to?compute?approximated?gradient?with?formula(1) ???? ????Returns: ????difference?--?difference?(2)?between?the?approximated?gradient?and?the?backward?propagation?gradient ????""" ???? ????#?Compute?gradapprox?using?left?side?of?formula?(1).?epsilon?is?small?enough,?you?don't?need?to?worry?about?the?limit. ????###?START?CODE?HERE?###?(approx.?5?lines) ????thetaplus?=?theta+0.01???????????????????????????????#?Step?1 ????thetaminus?=?theta-0.01??????????????????????????????#?Step?2 ????J_plus?=?forward_propagation(x,thetaplus)??????????????????????????????????#?Step?3 ????J_minus?=?forward_propagation(x,thetaminus)?????????????????????????????????#?Step?4 ????gradapprox?=?(J_plus-J_minus)/(2*0.01)??????????????????????????????#?Step?5 ????###?END?CODE?HERE?### ???? ????#?Check?if?gradapprox?is?close?enough?to?the?output?of?backward_propagation() ????###?START?CODE?HERE?###?(approx.?1?line) ????grad?=?backward_propagation(x,theta) ????###?END?CODE?HERE?### ???? ????###?START?CODE?HERE?###?(approx.?1?line) ????numerator?=?np.linalg.norm(gradapprox-grad)???????????????????????????????#?Step?1' ????denominator?=?np.linalg.norm(grad)+np.linalg.norm(gradapprox)?????????????????????????????#?Step?2' ????difference?=?numerator/denominator??????????????????????????????#?Step?3' ????###?END?CODE?HERE?### ???? ????if?difference?<?1e-7: ????????print?("The?gradient?is?correct!") ????else: ????????print?("The?gradient?is?wrong!") ???? ????return?difference我懷疑是下面圖的地方出錯(cuò)了，但不知道怎么修改呀？解決：epsilon?=?1e-7應(yīng)該被使用，而不是0.01