@@ -98,7 +98,7 @@ First of all, let's focus on the backpropagation of the gradient with an example
The weight matrix of the layer $`L`$ is denoted $`W^{(L)}`$. The bias vector of the layer $`L`$ is denoted $`B^{(L)}`$. We choose the sigmoid function, denoted $`\sigma`$, as the activation function. The output vector of the layer $`L`$ before activation is denoted $`Z^{(L)}`$. The output vector of the layer $`L`$ after activation is denoted $`A^{(L)}`$. By convention, we note $`A^{(0)}`$ the network input vector. Thus $`Z^{(L+1)} = W^{(L+1)}A^{(L)} + B^{(L+1)}`$ and $`A^{(L+1)} = \sigma\left(Z^{(L+1)}\right)`$. Let's consider a network with one hidden layer. Thus, the output is $`\hat{Y} = A^{(2)}`$.
Let $`Y`$ be the target (desired output). We use mean squared error (MSE) as the cost function. Thus, the cost is $`C = \frac{1}{N_{out}}\sum_{i=1}^{N_{out}} (\hat{y_i} - y_i)^2`$.
1.Prove that $`\sigma' = \sigma \times (1-\sigma)`$
1.We admit that for the sigmoide function $`\sigma`$, the derivative is $`\sigma' = \sigma \times (1-\sigma)`$
2. Express $`\frac{\partial C}{\partial A^{(2)}}`$, i.e. the vector of $`\frac{\partial C}{\partial a^{(2)}_i}`$ as a function of $`A^{(2)}`$ and $`Y`$.
3. Using the chaining rule, express $`\frac{\partial C}{\partial Z^{(2)}}`$, i.e. the vector of $`\frac{\partial C}{\partial z^{(2)}_i}`$ as a function of $`\frac{\partial C}{\partial A^{(2)}}`$ and $`A^{(2)}`$.
4. Similarly, express $`\frac{\partial C}{\partial W^{(2)}}`$, i.e. the matrix of $`\frac{\partial C}{\partial w^{(2)}_{i,j}}`$ as a function of $`\frac{\partial C}{\partial Z^{(2)}}`$ and $`A^{(1)}`$.
