已知:
Zb×m[l]=Wb×a[l]⋅Aa×m[l−1]+v⃗b×1Ab×m[l]=g(Z[l])
Z_{b\times m}^{[l]} = W_{b\times a}^{[l]} \cdot A_{a\times m}^{[l-1]} + \vec{v}_{b \times 1}\\
A_{b\times m}^{[l]} = g(Z^{[l]})
Zb×m[l]=Wb×a[l]⋅Aa×m[l−1]+vb×1Ab×m[l]=g(Z[l])
g(.)g(.)g(.) 是 activation function, dWdWdW 表示 dJdW\frac{dJ}{dW}dWdJ,其他的类似。
已知dA[l]dA^{[l]}dA[l],则:
dZb×m[l]=dA[l]∗g′(Z[l])dWb×a[l]=∂L∂W[l]=1mdZ[l]A[l−1]Tdbb×1[l]=∂L∂b[l]=1m∑i=1mdZ[l](i)(横向求和)dAa×m[l−1]=∂L∂A[l−1]=W[l]TdZ[l] dZ^{[l]}_{b\times m} = dA^{[l]} * g'(Z^{[l]})\\ dW^{[l]}_{b\times a} = \frac{\partial \mathcal{L} }{\partial W^{[l]}} = \frac{1}{m} dZ^{[l]} A^{[l-1] T}\\ db^{[l]}_{b\times 1} = \frac{\partial \mathcal{L} }{\partial b^{[l]}} = \frac{1}{m} \sum_{i = 1}^{m} dZ^{[l](i)} (\text{横向求和})\\ dA^{[l-1]}_{a\times m} = \frac{\partial \mathcal{L} }{\partial A^{[l-1]}} = W^{[l] T} dZ^{[l]} dZb×m[l]=dA[l]∗g′(Z[l])dWb×a[l]=∂W[l]∂L=m1dZ[l]A[l−1]Tdbb×1[l]=∂b[l]∂L=m1i=1∑mdZ[l](i)(横向求和)dAa×m[l−1]=∂A[l−1]∂L=W[l]TdZ[l]