Deep Learning · MT2 · section

Drill 10 Sigmoid neuron gradient

MT2 numerical practice from direct to mixed

Question

Question. x = 2, w = 0.5, b = 0, y = 1, half squared loss, η = 0.1. Find one weight and bias update.

Answer — step by step

Given

x = 2; w = 0.5; b = 0; target y = 1.

Activation: sigmoid; loss: half squared error.

Learning rate η = 0.1.

To find

Find the output, delta, weight/bias gradients and one update.

Formula — which one and why

Weighted sum and activation. Each input is multiplied by its connection weight. Add these contributions, add the bias once, then apply the stated activation.

xᵢ = input i; wᵢ = its weight; b = bias; z = sum before activation; a = output after activation; f = the specified activation function.

$z=\sum_iw_ix_i+b$ LaTeX: z=\sum_iw_ix_i+b

$a=f(z)$ LaTeX: a=f(z)

Yaad rakhne ka cue: Input × weight, sabka sum, bias once, phir activation.

Sigmoid and its slope. Use sigmoid when the neuron is specified to squash its preactivation into (0,1). Once its output is known, its derivative needs only output × (1−output).

z = preactivation; p = sigmoid output; e is the exponential base, not a prediction error in this formula.

$p=\frac{1}{1+e^{-z}}$ LaTeX: p=\frac{1}{1+e^{-z}}

$\frac{dp}{dz}=p(1-p)$ LaTeX: \frac{dp}{dz}=p(1-p)

Yaad rakhne ka cue: Sigmoid output p; uska slope p × (1−p).

Half summed squared loss. This loss convention takes one-half of the sum of squared prediction errors. Use it when the question specifies it, or explicitly state it as an assumption when the paper omits the loss. Differentiating cancels the factor one-half against the square derivative.

oⱼ = output j; yⱼ = target j; L = half summed squared loss under the stated or explicitly assumed convention.

$L=\frac12\sum_j(o_j-y_j)^2$ LaTeX: L=\frac12\sum_j(o_j-y_j)^2

$\frac{\partial L}{\partial o_j}=o_j-y_j$ LaTeX: \frac{\partial L}{\partial o_j}=o_j-y_j

Yaad rakhne ka cue: Half aur square ka 2 cancel: output derivative = prediction − target.

Backpropagation. The loss tells us how each output must change. Multiply by that neuron’s activation slope to obtain its delta. A weight gradient is receiving delta × sending value. Hidden deltas must include every outgoing path.

δ (delta) = derivative of loss with respect to a neuron’s preactivation; f′ = activation derivative; aᵢ = sending activation; Wⱼᵢ = connection weight. Use the exact loss scaling in the question.

$\delta_{o_j}=\frac{\partial L}{\partial o_j}f_o\prime(z_{o_j})$ LaTeX: \delta_{o_j}=\frac{\partial L}{\partial o_j}f_o\prime(z_{o_j})

$\delta_{h_i}=f_h\prime(z_{h_i})\sum_jW_{ji}^{(2)}\delta_{o_j}$ LaTeX: \delta_{h_i}=f_h\prime(z_{h_i})\sum_jW_{ji}^{(2)}\delta_{o_j}

$\frac{\partial L}{\partial W_{ji}}=\delta_ja_i$ LaTeX: \frac{\partial L}{\partial W_{ji}}=\delta_ja_i

$\frac{\partial L}{\partial b_j}=\delta_j$ LaTeX: \frac{\partial L}{\partial b_j}=\delta_j

$W_{ji,\mathrm{new}}=W_{ji}-\eta\frac{\partial L}{\partial W_{ji}}$ LaTeX: W_{ji,\mathrm{new}}=W_{ji}-\eta\frac{\partial L}{\partial W_{ji}}

Yaad rakhne ka cue: Receiving delta × sending value = us connection ka gradient.

Solution — step by step

Meaning and formulas

Here the loss is half squared error. It is different from binary cross-entropy, so we include the sigmoid derivative explicitly.

The symbol δ (delta) is a short name for the loss gradient with respect to the neuron's raw score z.

$\delta=(p-y)p(1-p)$ LaTeX: \delta=(p-y)p(1-p)

$\frac{\partial L}{\partial w}=\delta x,\quad \frac{\partial L}{\partial b}=\delta$ LaTeX: \frac{\partial L}{\partial w}=\delta x,\quad \frac{\partial L}{\partial b}=\delta