Every use of a weight contributes

STATEMENT

Derive backpropagation through time and compute every parameter gradient for the forward pass in I.9.B01.

GIVEN

Use (I.9.1) and a sum of per-step losses for the derivation. For the numerical case use the linear scalar cell, inputs $(1,2,0)$, $h_0=0$, $(u,w,b,v,c)=(1,1/2,0,2,0)$, and final target one.

FIND

The recurrent backward rule, the shared-weight accumulation rule, all five scalar parameter gradients, input derivatives and initial-state derivative.

STRATEGY

First add the two incoming backward signals at each state. Differentiate the activation, then accumulate the local parameter contributions. Specialise to the identity activation only after deriving the tanh rule.

SOLUTION

The readout at time $t$ contributes $\mat V^{\mathsf T}\vec e_t$ to its state. The next cell contributes $\mat W^{\mathsf T}\vec\delta_{t+1}$. Multiplying their sum by the coordinatewise tanh derivative gives (I.9.3).

For a shared recurrent-weight entry $W_{ij}$, the local differential is $d a_{t,i}=h_{t-1,j}\,dW_{ij}$. Summing all uses gives

This is the entrywise derivation of (I.9.4), not an average over time. If the objective is an average, its factor is already in $\vec e_t$.

In the linear example, $z_3-y=3/2$ and the activation derivative is one. Only the final readout has a loss. Starting from the end,

The local contributions to $w$ are therefore $0$, $3/2$ and $15/2$:

Input derivatives are $u\delta_t$, giving $(3/4,3/2,3)$. The initial-state derivative is $w\delta_1=3/8$. We compute the latter even though this example does not optimise $h_0$.

For an independent algebraic check, with $b=c=h_0=0$ the final state is $h_3=u(w^2+2w)$. Thus $\partial_w z_3=vu(2w+2)=6$, and $\partial_w\mathcal L=(3/2)6=9$.

The reproduction script perturbs each parameter by $10^{-6}$, recomputes the full forward loss and compares central differences with the analytic derivative. It additionally checks tanh cells with losses at multiple steps and a two-coordinate state with non-symmetric recurrent weights. The latter catches transpose errors that a scalar example cannot.

The script and run command are in the numeric check on I.9.B01.

Check — sanity

Since $h_0=0$, the first local $w$ contribution is zero, but the first input still receives a nonzero derivative. “This use contributes zero” does not mean “this time step is disconnected.”