# BPTT with every arithmetic step visible

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/read/9/tee-c9-s7

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/section/tee-c9-s7/

**Question.** A scalar RNN uses the linear state equation hₜ = u xₜ + v hₜ₋₁ + b. Its prediction is ŷ = h₂ and its loss is L = ½(ŷ−1)². Given u = 0.2, v = 0.5, b = 0, h₀ = 0, x₁ = 1 and x₂ = 2, calculate the gradients of u, v and b using BPTT, then update them once using learning rate η = 0.1.

### Given

hₜ = uxₜ+vhₜ₋₁+b; prediction = h₂.

u = 0.2; v = 0.5; b = 0; h₀ = 0; x₁ = 1; x₂ = 2.

Target = 1; half squared loss; η = 0.1.

### To find

Find gradients for u,v,b and update each once.

### Formula — which one and why

**Backpropagation through time.** Unroll the recurrence, send the loss derivative backward through its links and add contributions for each shared parameter before updating it once.

For h\_t=ux\_t+vh\_(t−1)+b with loss only at the last state: δ\_t = loss derivative at that state. The linear recurrence has activation slope 1.



$$
\delta_T=h_T-y
$$





$$
\delta_t=v\delta_{t+1}
$$





$$
g_u=\sum_t\delta_tx_t
$$





$$
g_v=\sum_t\delta_th_{t-1}
$$





$$
g_b=\sum_t\delta_t
$$



**Yaad rakhne ka cue:** Time ke har use ka gradient jodo; shared weight ko ek baar update karo.

**Gradient descent.** Evaluate the gradient at the current parameter, multiply by the learning rate and subtract. If the objective is supplied, differentiate it and recompute the gradient after each update.

θ (theta) = trainable parameter; g = derivative of the stated objective J; η = learning rate; t labels the update number.



$$
g_t=J\prime(\theta_{t-1})
$$





$$
\theta_t=\theta_{t-1}-\eta g_t
$$



**Yaad rakhne ka cue:** Naya = purana − learning rate × current gradient.

### Solution — step by step

### Model and dependency structure

u is the input weight, v is the recurrent weight and b is the hidden bias. Each is shared by both time steps. The hidden activation is linear in this question, so its derivative is 1. There is a loss only at time step 2; time step 1 influences that loss through h₂.

![Follow the two shared-parameter state calculations forward, then sum the backward contributions from each use of u, v and b.](https://deep-learning-study-room.vercel.app/assets/figures/v_bptt_worked.svg)

Diagram: Follow the two shared-parameter state calculations forward, then sum the backward contributions from each use of u, v and b.. Labels: The two-step RNN in the worked question | The two-step RNN in the worked question | Shared weights u = 0.2, v = 0.5; bias 0; linear hidden activation. | Start | h₀ = 0 | Time 1: x₁ = 1 | h₁ = 0.2 | Time 2: x₂ = 2 | h₂ = 0.5 | δ₂ = 0.5 − 1 = −0.5 | Backward through v = 0.5 | δ₁ = −0.5 × 0.5 = −0.25 | Input weight gradient: (−0.25 × 1) + (−0.5 × 2) = −1.25 | Recurrent weight gradient: (−0.25 × 0) + (−0.5 × 0.2) = −0.1 | Shared bias gradient: −0.25 + (−0.5) = −0.75 | Each row is a sum over time. Update u, v and b only after calculating these three | gradients.

Follow the two shared-parameter state calculations forward, then sum the backward contributions from each use of u, v and b. Tap the diagram to enlarge.

### Step 1: complete the forward pass



$$
h_1=ux_1+vh_0+b
$$





$$
h_1=0.2(1)+0.5(0)+0=0.2
$$





$$
h_2=ux_2+vh_1+b
$$





$$
h_2=0.2(2)+0.5(0.2)+0=0.5
$$





$$
L=\frac12(0.5-1)^2=0.125
$$



### Step 2: differentiate the loss at the final step

Because the output is h₂ and the state activation is linear:



$$
\delta_2=\frac{\partial L}{\partial h_2}=h_2-1=-0.5
$$



### Step 3: propagate the derivative to the earlier state

The local derivative of h₂ with respect to h₁ is the recurrent weight v = 0.5. Applying the chain rule:



$$
\delta_1=\frac{\partial L}{\partial h_1}=\frac{\partial L}{\partial h_2}\frac{\partial h_2}{\partial h_1}
$$





$$
\delta_1=\delta_2v=(-0.5)(0.5)=-0.25
$$



There is no separate direct loss at time step 1 to add here. Both deltas use the original recurrent weight.

### Step 4: sum the contributions to each shared parameter

u appears in both h₁ and h₂, so its gradient receives one contribution from each use:



$$
\frac{\partial L}{\partial u}=\delta_1x_1+\delta_2x_2
$$





$$
\frac{\partial L}{\partial u}=(-0.25)(1)+(-0.5)(2)=-1.25
$$



For v, the sending activation is the previous hidden state at the corresponding time:



$$
\frac{\partial L}{\partial v}=\delta_1h_0+\delta_2h_1
$$





$$
\frac{\partial L}{\partial v}=(-0.25)(0)+(-0.5)(0.2)=-0.1
$$



The same additive bias is used at both steps. Its two local derivatives are 1:



$$
\frac{\partial L}{\partial b}=\delta_1+\delta_2=-0.25-0.5=-0.75
$$



| Shared parameter | Contribution at time 1 | Contribution at time 2 | Total gradient |
| --- | --- | --- | --- |
| u | −0.25 × 1 = −0.25 | −0.5 × 2 = −1 | −1.25 |
| v | −0.25 × 0 = 0 | −0.5 × 0.2 = −0.1 | −0.1 |
| b | −0.25 | −0.5 | −0.75 |

### Step 5: update each shared parameter once



$$
u_{\mathrm{new}}=0.2-0.1(-1.25)=0.325
$$





$$
v_{\mathrm{new}}=0.5-0.1(-0.1)=0.51
$$





$$
b_{\mathrm{new}}=0-0.1(-0.75)=0.075
$$



**Final answer:** gradients are −1.25, −0.1 and −0.75; updated parameters are u = 0.325, v = 0.51 and b = 0.075.

### Evaluate the updated model

Restart from the same h₀ = 0 and use all updated parameters together:



$$
h_{1,\mathrm{new}}=0.325(1)+0.51(0)+0.075=0.4
$$





$$
h_{2,\mathrm{new}}=0.325(2)+0.51(0.4)+0.075=0.929
$$





$$
L_{\mathrm{new}}=\frac12(0.929-1)^2=0.0025205
$$



The loss decreases in this particular update. That is consistent with the computed gradient, but a decrease is not guaranteed for every choice of learning rate.

### Extension to a tanh RNN

For tanh, the derivative at each state transition also contains 1−hₜ². A trainable output layer contributes its own local derivatives. If losses occur at several time steps, add the direct output-loss contribution and the future-state contribution before applying the local activation derivative. Truncated BPTT limits the number of time steps traversed backward; it does not create separate copies of the recurrent parameters.

### Second worked example: nonlinear tanh BPTT

**Question.** A scalar RNN uses hₜ = tanh(uxₜ + vhₜ₋₁ + b). Given u = 0.4, v = 0.3, b = 0.1, h₀ = 0 and inputs x₁ = 1, x₂ = −1, its final output is h₂ and target is 0.2. Use only the final-time loss L = ½(h₂ − 0.2)². Calculate both states, all three shared-parameter gradients and one simultaneous update at η = 0.1.

### Given

Shared input weight u = 0.4; shared recurrent weight v = 0.3; shared bias b = 0.1. Initial state h₀ = 0; x₁ = 1, x₂ = −1. Hidden activation = tanh; final target = 0.2; learning rate η = 0.1. There is no separate output layer and no loss at time 1.

### To find

Forward states h₁ and h₂, final loss, gradients with respect to u, v and b, and all three updated parameters.

### Formula — which one and why

Backpropagation must pass through each tanh derivative. Because u, v and b are shared, add their local contributions from both time steps before updating them.



$$
z_t=ux_t+vh_{t-1}+b,\qquad h_t=\tanh(z_t)
$$





$$
\delta_2=(h_2-y)(1-h_2^2),\qquad \delta_1=v\delta_2(1-h_1^2)
$$





$$
\frac{\partial L}{\partial u}=\delta_1x_1+\delta_2x_2
$$





$$
\frac{\partial L}{\partial v}=\delta_1h_0+\delta_2h_1,\qquad \frac{\partial L}{\partial b}=\delta_1+\delta_2
$$



Here δₜ means ∂L/∂zₜ, not ∂L/∂hₜ. The factor v in δ₁ is the **old recurrent weight 0.3**. The derivative of tanh is 1 − hₜ², using that step's activated state.

### Solution — step by step

### Step 1: calculate the first state

z₁ = 0.4(1) + 0.3(0) + 0.1 = 0.5.

h₁ = tanh(0.5) = **0.462117157**.

### Step 2: carry that state into time 2

z₂ = 0.4(−1) + 0.3(0.462117157) + 0.1 = **−0.161364853**.

h₂ = tanh(−0.161364853) = **−0.159978716**.

### Step 3: calculate the final loss and final-time delta

Prediction error h₂ − y = −0.159978716 − 0.2 = **−0.359978716**.

L = ½(−0.359978716)² = **0.064792338**.

δ₂ = (−0.359978716)[1 − (−0.159978716)²] = **−0.350765712**.

### Step 4: backpropagate through the recurrent connection and first tanh

There is no direct loss at time 1. Its gradient arrives entirely through time 2:

δ₁ = 0.3(−0.350765712)[1 − (0.462117157)²] = **−0.082757670**.

### Step 5: sum each shared parameter's contributions over time

| Parameter | Time contributions | Total gradient |
| --- | --- | --- |
| u | (−0.082757670)(1) + (−0.350765712)(−1) | 0.268008043 |
| v | (−0.082757670)(0) + (−0.350765712)(0.462117157) | −0.162094854 |
| b | −0.082757670 + (−0.350765712) | −0.433523382 |

For v, the sending state is hₜ₋₁: h₀ for time 1, h₁ for time 2. The negative second input makes its contribution to the u gradient positive.

### Step 6: update all parameters using the same old forward pass

u\_new = 0.4 − 0.1(0.268008043) = **0.373199196**.

v\_new = 0.3 − 0.1(−0.162094854) = **0.316209485**.

b\_new = 0.1 − 0.1(−0.433523382) = **0.143352338**.

### Step 7: interpret the update with a fresh forward pass

Using the new parameters from h₀ = 0 gives h₁,new = tanh(0.373199196 + 0.143352338) = **0.475034096**.

h₂,new = tanh(−0.373199196 + 0.316209485(0.475034096) + 0.143352338) = **−0.079468645**.

The new loss is ½(−0.079468645 − 0.2)² = **0.039051362**.

**Final answer:** gradients (u,v,b) = **(0.268008043, −0.162094854, −0.433523382)** and updated parameters = **(0.373199196, 0.316209485, 0.143352338)**. This step moves the final prediction toward 0.2 and reduces loss from 0.064792338 to 0.039051362. Both time steps contributed to learning the same parameters.