One page per topic area. Each formula has a plain-English translation, because knowing what it means is what the quizzes test.

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Losses

The one insight: cross-entropy = negative log-likelihood. Minimising it is maximum likelihood estimation.

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Activations

The 0.25 is why sigmoid causes vanishing gradients: 0.25¹⁰ ≈ 10⁻⁶.

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Optimisation

Gradient descent      w := w − α·∂J/∂w
Momentum              v := ρv + ∂J/∂w  ;  w := w − αv
L2 / Ridge            J + α·Σθ²                shrinks toward 0
L1 / Lasso            J + α·Σ|θ|               drives to exactly 0 (feature selection)
Weight decay          same as L2, set on the optimiser in PyTorch

C = 1/α in sklearn's LogisticRegression and SVC. Small C = strong regularisation.

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Classification metrics

              Predicted
             Neg    Pos
Actual Neg |  TN  |  FP  |
       Pos |  FN  |  TP  |

Accuracy    = (TP+TN)/total
Precision   = TP/(TP+FP)        when I say yes, how often right
Recall/TPR  = TP/(TP+FN)        of real yeses, how many caught
Specificity = TN/(TN+FP)
FPR         = FP/(FP+TN)
F1          = 2PR/(P+R)         harmonic mean

ROC = TPR vs FPR across thresholds. AUC: 1.0 perfect, 0.5 random. R² = 1 − SS_res/SS_tot. Silhouette = (b−a)/max(a,b).

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Trees

Gini             1 − Σp²
Entropy          −Σ p·log₂p
Information gain entropy(parent) − weighted-avg entropy(children)

Both are 0 at purity, maximal at an even mix.

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Convolution

Output size    out = floor((in − k + 2p)/s) + 1
Transpose conv out = (in − 1)·s − 2p + k
Conv params    k·k·C_in·C_out + C_out
Dense params   in·out + out

Worked: input 28, k=5, p=2, s=1 → (28−5+4)/1+1 = 28 (same padding). Conv2d(3,16,5) → 5·5·3·16 + 16 = 1,216.

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Batch norm

x̂ = (x − μ_B)/√(σ²_B + ε)
y  = γ·x̂ + β                γ, β learned

Train uses batch statistics; eval uses running averages. That's what model.eval() switches.

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Initialisation

Xavier/Glorot   Var(W) = 2/(n_in + n_out)     with tanh, sigmoid
He/Kaiming      Var(W) = 2/n_in               with ReLU

He uses twice the variance because ReLU zeroes half the activations.

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Attention

Attention(Q,K,V) = softmax(Q·Kᵀ/√d_k)·V

head_i    = Attention(QW_i^Q, KW_i^K, VW_i^V)
MultiHead = Concat(head_1…head_h)·W^O

√d_k prevents softmax saturation. Cost is O(n²·d) in sequence length.

Positional encoding:

PE(pos,2i)   = sin(pos/10000^(2i/d))
PE(pos,2i+1) = cos(pos/10000^(2i/d))

Causal mask: set the upper triangle to −inf before softmax.

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Language modelling

Perplexity = exp(cross-entropy loss)          lower is better
N-gram     P(wₙ|wₙ₋₂,wₙ₋₁) = count(w₋₂,w₋₁,wₙ)/count(w₋₂,w₋₁)
Cosine sim = (a·b)/(‖a‖·‖b‖)

Perplexity ≈ the effective number of choices the model is deciding between per token.

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LoRA

W' = W₀ + (α/r)·B·A

W₀ : d×k frozen
A  : r×k trainable, random init
B  : d×r trainable, ZERO init          ← so BA=0 at step 0

Parameters: r·(d+k) instead of d·k. d=k=4096, r=8 → 65k vs 16.7M.

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Reinforcement learning

Return       G_t = R_{t+1} + γR_{t+2} + γ²R_{t+3} + …
Bellman/Q    Q(s,a) ← Q(s,a) + α[r + γ·max_a' Q(s',a') − Q(s,a)]

The bracket is the temporal-difference error.

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Alignment

Bradley-Terry     P(y_c ≻ y_r) = σ(r(x,y_c) − r(x,y_r))
Reward loss       −log σ(r(x,y_c) − r(x,y_r))

RLHF objective    max E[r(x,y)] − β·KL(π ‖ π_ref)

PPO clipped       L = min(ratio·Â, clip(ratio, 1−ε, 1+ε)·Â)
                  ratio = π_new(a|s)/π_old(a|s),  ε ≈ 0.2

Optimal policy    π*(y|x) = (1/Z(x))·π_ref(y|x)·exp(r(x,y)/β)

DPO loss          −log σ( β·log[πθ(y_c|x)/π_ref(y_c|x)]
                         − β·log[πθ(y_r|x)/π_ref(y_r|x)] )

Log-derivative    ∇θ E[f(x)] = E[f(x)·∇θ log πθ(x)]

The DPO trick: invert the optimal-policy equation to express reward in terms of the policy, substitute into Bradley-Terry, and β·log Z(x) cancels because it's identical for both responses to the same prompt. No reward model, no RL.

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Numbers to have memorised