# Why does a $\sum_k$ appear when using the chain rule to derive $\delta^L_j?$

I'm following along this book on machine learning.

At the moment, the author is proving that

$$\begin{eqnarray} \delta^L_j = \frac{\partial C}{\partial a^L_j} \sigma'(z^L_j) \nonumber\end{eqnarray}$$

• $$\delta$$ is the output error of the $$j^{\rm th}$$ sigmoid neuron in the $$l^{\rm th}$$ layer ($$L$$ is the last layer in the neural network)
• $$C$$ is the cost for a single training example
• $$a$$ is the output from the $$j^{\rm th}$$ neuron in the $$l^{\rm th}$$ layer
• $$z$$ is the unweighted input for the $$j^{\rm th}$$ neuron in the $$l^{\rm th}$$ layer
• $$\sigma'$$ is the derivative of the sigmoid function

I'm concerned with $$(2)$$ in the author's proof:

$$\begin{eqnarray} \delta^L_j = \frac{\partial C}{\partial z^L_j}. \tag{1}\end{eqnarray}$$

$$\begin{eqnarray} \delta^L_j = \sum_k \frac{\partial C}{\partial a^L_k} \frac{\partial a^L_k}{\partial z^L_j}, \tag{2}\end{eqnarray}$$

$$\begin{eqnarray} \delta^L_j = \frac{\partial C}{\partial a^L_j} \frac{\partial a^L_j}{\partial z^L_j}. \tag{3}\end{eqnarray}$$

$$\begin{eqnarray} \delta^L_j = \frac{\partial C}{\partial a^L_j} \sigma'(z^L_j), \tag{4}\end{eqnarray}$$

The author's explanation for the simplification from step 2 to 3 is:

the output activation $$a^L_k$$ of the $$k^{\rm th}$$ neuron depends only on the weighted input $$z^L_j$$ for the $$j^{\rm th}$$ neuron when $$k=j$$. And so $$\partial a^L_k / \partial z^L_j$$ vanishes when $$k≠j$$.

Where does the $$\sum_k$$ come from in $$(2)$$? Why can't we just skip straight from $$(1)$$ to $$(3)$$ and ignore the $$\sum_k$$ used in $$(2)$$?

The author is using the chain rule. We want the partial of $$C$$ with respect to the $$z$$'s, but we only know the partial of $$C$$ with respect to the $$a$$'s. It is an important claim about the model that each $$a$$ only depends on the corresponding $$z$$. The bullets you quote support that claim because the output of a given neuron should only depend on the inputs to that neuron but if you represent $$C$$ as a function of $$a$$'s you need this step, then to argue that the second partial is zero unless $$j=k$$.
• Thanks for the explanation. "If you represent C as a function of $a$'s you need this step"— this is clearly an aspect of multivariable calculus I don't understand. Could you please point me to some resources that dive into this? (Just knowing what to google would be enough)
• It is mainstream multivariable calculus. If you have a function $C(x,y)$ and ask for $\frac {\partial C(x,y)}{\partial z}$ it is $\frac {\partial C(x,y)}{\partial x}\frac {\partial x}{\partial z}$ plus the obvious other term by the chain rule. Commented Feb 22, 2019 at 3:41