# LSTM complete numerical with supplied gates

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/read/9/tee-c9-s12

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/section/tee-c9-s12/

**Question.** An LSTM cell has input gate iₜ = 0.7, forget gate fₜ = 0.4, output gate oₜ = 0.8, candidate cell content C̃ₜ = 0.9 and previous cell state Cₜ₋₁ = 0.5. Calculate the new cell state Cₜ and hidden state hₜ.

### Given

Activated input gate i = 0.7; forget gate f = 0.4; output gate o = 0.8.

Candidate = 0.9; old cell = 0.5.

### To find

Find the new cell C and hidden state h.

### Formula — which one and why

**LSTM cell and hidden state.** Multiply old memory by the forget gate; multiply candidate content by the input gate; add those two contributions. Apply tanh to the new cell and multiply by the output gate.

f,i,o = activated forget/input/output gates; C\_old = previous cell; C\_tilde = candidate content; C = new cell; h = visible hidden state. ⊙ means elementwise multiplication.



$$
C=f\odot C_{\mathrm{old}}+i\odot\widetilde C
$$





$$
h=o\odot\tanh C
$$



**Yaad rakhne ka cue:** Purana rakho + naya lo; phir tanh aur output gate.

### Solution — step by step

### Equations and supplied-value convention

The question supplies **activated gate outputs** and the candidate content. Do not apply sigmoid or tanh to those supplied values again. First update the cell, then calculate the hidden state:



$$
C_t=f_tC_{t-1}+i_t\widetilde C_t
$$





$$
h_t=o_t\tanh(C_t)
$$



The first equation combines retained old content and admitted candidate content. The second equation is a separate transformation of the updated cell.

![The diagram uses the supplied gate values. The output gate is applied after the cell has been updated and passed through tanh.](https://deep-learning-study-room.vercel.app/assets/figures/v_lstm_given.svg)

Diagram: The diagram uses the supplied gate values. The output gate is applied after the cell has been updated and passed through tanh.. Labels: When the LSTM gate values are already supplied | When the LSTM gate values are already supplied | Example: f = 0.4, old C = 0.5, i = 0.7, candidate = 0.9, o = 0.8. | Old part | 0.4 × 0.5 = | 0.2 | New part | 0.7 × 0.9 = | 0.63 | New cell | 0.2 + 0.63 = | 0.83 | Hidden output | 0.8 × | tanh(0.83) ≈ | 0.544381 | The final hidden output includes tanh; it is not just 0.8 × 0.83.

The diagram uses the supplied gate values. The output gate is applied after the cell has been updated and passed through tanh. Tap the diagram to enlarge.

**Step 1: calculate the retained old-cell contribution.**



$$
f_tC_{t-1}=0.4(0.5)=0.2
$$



**Step 2: calculate the admitted candidate contribution.**



$$
i_t\widetilde C_t=0.7(0.9)=0.63
$$



**Step 3: add the two contributions to obtain the new cell.**



$$
C_t=0.2+0.63=0.83
$$



**Step 4: activate the updated cell.**



$$
\tanh(C_t)=\tanh(0.83)\approx0.680476006
$$



**Step 5: multiply by the output gate.**



$$
h_t=0.8\tanh(0.83)\approx0.544380805
$$



**Final answer:** Cₜ = 0.83 and hₜ ≈ 0.544381. Multiplying the output gate directly by Cₜ would give 0.664 and omit the required tanh operation.

### Related vector calculation: retention alone

**Question.** Given forget gate f = [0.1, 0.9] and previous cell C = [2, 1], calculate the retained old-cell contribution.

### Given

Forget gate = [0.1,0.9]; old cell = [2,1].

Input-gate and candidate values are not supplied.

### To find

Find only the retained old-cell contribution.

### Formula — which one and why

**LSTM retained old memory.** Multiply each old cell value by its corresponding forget-gate value. This gives only the old contribution, not the complete new cell.

f = activated forget gate; C\_old = previous cell state; ⊙ means multiply matching entries.



$$
C_{\mathrm{retained}}=f\odot C_{\mathrm{old}}
$$



**Yaad rakhne ka cue:** Forget gate × old cell = bacha hua purana hissa.

### Solution — step by step

**Step 1: retain the first old-cell component.**



$$
0.1(2)=0.2
$$



**Step 2: retain the second old-cell component.**



$$
0.9(1)=0.9
$$



**Step 3: assemble the two retained components.**

Keep their original order. This gives the following column:



$$
f\odot C=\begin{bmatrix}0.1(2)\\0.9(1)\end{bmatrix}=\begin{bmatrix}0.2\\0.9\end{bmatrix}
$$



**Answer:** [0.2, 0.9]. This is the retained contribution, not necessarily the final new cell. The candidate contribution must also be known to calculate the full cell update.