The Closed Learning Cycle on Paper

Let us trace the complete, unbroken learning cycle on paper with explicit numbers for our running 2-layer network.

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System Configuration: 'Die Hard in Space'

We evaluate the screenplay for 'Die Hard in Space' across our running $4 \to 2 \to 1$ Multi-Layer Perceptron.

• Input Feature Vector ($x \in \mathbb{R}^{4 \times 1}$):

  $$x = \begin{bmatrix} 1.0 \\ 0.0 \\ 0.5 \\ 1.0 \end{bmatrix} \begin{matrix} \text{(Action Measurement)} \\ \text{(Romance Measurement)} \\ \text{(Comedy Measurement)} \\ \text{(Sci-Fi Measurement)} \end{matrix}$$

• Layer 1 Initial Parameters ($W^{(1)} \in \mathbb{R}^{2 \times 4}, b^{(1)} \in \mathbb{R}^{2 \times 1}$):

  $$W^{(1)} = \begin{bmatrix} 3.0 & -2.0 & 0.0 & 3.0 \\ -2.0 & 3.0 & 2.0 & -1.0 \end{bmatrix}, \qquad b^{(1)} = \begin{bmatrix} -2.0 \\ -1.5 \end{bmatrix}$$

  Hidden activation function: $f(z) = \text{ReLU}(z) = \max(0, z)$.

• Layer 2 Initial Parameters ($W^{(2)} \in \mathbb{R}^{1 \times 2}, b^{(2)} \in \mathbb{R}^{1 \times 1}$):

  $$W^{(2)} = \begin{bmatrix} 1.5 & 1.0 \end{bmatrix}, \qquad b^{(2)} = \begin{bmatrix} -2.0 \end{bmatrix}$$

  Output activation function: $\sigma(z) = \frac{1}{1 + e^{-z}}$.

• Ground-Truth Target: $y = 0.0$ (Box Office Flop).

• Step Size Factor: $\eta = 0.50$.

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Phase 1: Initial Forward Pass ($t = 0$)

Step 1: Layer 1 Affine Transformation

Compute $z^{(1)} = W^{(1)} x + b^{(1)}$:

$$z_1^{(1)} = (3.0)(1.0) + (-2.0)(0.0) + (0.0)(0.5) + (3.0)(1.0) - 2.0 = 3.0 + 0.0 + 0.0 + 3.0 - 2.0 = \mathbf{+4.0}$$ $$z_2^{(1)} = (-2.0)(1.0) + (3.0)(0.0) + (2.0)(0.5) + (-1.0)(1.0) - 1.5 = -2.0 + 0.0 + 1.0 - 1.0 - 1.5 = \mathbf{-3.5}$$ $$z^{(1)} = \begin{bmatrix} +4.0 \\ -3.5 \end{bmatrix}$$

Step 2: Layer 1 Non-Linear Activation

Apply element-wise ReLU activation:

$$a^{(1)} = \text{ReLU}\left(z^{(1)}\right) = \begin{bmatrix} \max(0, 4.0) \\ \max(0, -3.5) \end{bmatrix} = \begin{bmatrix} \mathbf{4.0} \\ \mathbf{0.0} \end{bmatrix} \begin{matrix} \text{(Summer Popcorn Flick Factor)} \\ \text{(Rom-Com Factor)} \end{matrix}$$

Step 3: Layer 2 Affine Transformation

Compute scalar pre-activation $z^{(2)} = W^{(2)} a^{(1)} + b^{(2)}$:

$$z^{(2)} = (1.5)(4.0) + (1.0)(0.0) + (-2.0) = 6.0 + 0.0 - 2.0 = \mathbf{+4.0}$$

Step 4: Layer 2 Sigmoid Probability Squashing

Compute final prediction $\hat{y} = a^{(2)} = \sigma(z^{(2)})$:

$$a^{(2)} = \sigma(4.0) = \frac{1}{1 + e^{-4.0}} = \frac{1}{1 + 0.018316} \approx \mathbf{0.982014} \quad (\mathbf{98.20\%})$$

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Phase 2: Error Quantification & Loss Evaluation

The network predicted a $98.20\%$ Hit probability, but the movie was a box office Flop ($y = 0.0$).

Step 1: Raw Prediction Error

$$e = y - \hat{y} = 0.0 - 0.982014 = \mathbf{-0.982014}$$

Step 2: Mean Squared Error Loss

$$L_1 = (a^{(2)} - y)^2 = (0.982014 - 0.0)^2 \approx \mathbf{0.964351}$$

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Phase 3: Backward Pass & Gradient Attribution

Step 1: Output Layer Error Attribution

1. Loss Sensitivity:

   $$\frac{\partial L}{\partial a^{(2)}} = 2(a^{(2)} - y) = 2(0.982014 - 0.0) = \mathbf{+1.964028}$$

2. Activation Function Slope:

   $$\sigma'(z^{(2)}) = a^{(2)}(1 - a^{(2)}) = 0.982014 \times (1 - 0.982014) = 0.982014 \times 0.017986 = \mathbf{0.017663}$$

3. Output Error Delta ($\delta^{(2)}$):

   $$\delta^{(2)} = \frac{\partial L}{\partial a^{(2)}} \cdot \sigma'(z^{(2)}) = (+1.964028) \times (0.017663) = \mathbf{+0.034690}$$

4. Output Parameter Gradients:

   $$\frac{\partial L}{\partial W^{(2)}} = \delta^{(2)} (a^{(1)})^T = 0.034690 \times \begin{bmatrix} 4.0 & 0.0 \end{bmatrix} = \begin{bmatrix} \mathbf{0.138760} & \mathbf{0.000000} \end{bmatrix}$$ $$\frac{\partial L}{\partial b^{(2)}} = \delta^{(2)} = \mathbf{+0.034690}$$

Step 2: Hidden Layer Error Backpropagation

1. Transposed Weight Projection:

   $$(W^{(2)})^T \delta^{(2)} = \begin{bmatrix} 1.5 \\ 1.0 \end{bmatrix} (0.034690) = \begin{bmatrix} 1.5 \times 0.034690 \\ 1.0 \times 0.034690 \end{bmatrix} = \begin{bmatrix} \mathbf{0.052035} \\ \mathbf{0.034690} \end{bmatrix}$$

2. Hidden Activation Slopes:

   $$f'(z^{(1)}) = \begin{bmatrix} \text{ReLU}'(+4.0) \\ \text{ReLU}'(-3.5) \end{bmatrix} = \begin{bmatrix} \mathbf{1.0} \\ \mathbf{0.0} \end{bmatrix}$$

3. Hidden Error Delta ($\delta^{(1)}$):

   $$\delta^{(1)} = \left( (W^{(2)})^T \delta^{(2)} \right) \odot f'(z^{(1)}) = \begin{bmatrix} 0.052035 \\ 0.034690 \end{bmatrix} \odot \begin{bmatrix} 1.0 \\ 0.0 \end{bmatrix} = \begin{bmatrix} \mathbf{0.052035} \\ \mathbf{0.000000} \end{bmatrix}$$

4. Hidden Layer Parameter Gradients:

   $$\frac{\partial L}{\partial W^{(1)}} = \delta^{(1)} x^T = \begin{bmatrix} 0.052035 \\ 0.000000 \end{bmatrix} \begin{bmatrix} 1.0 & 0.0 & 0.5 & 1.0 \end{bmatrix} = \begin{bmatrix} \mathbf{0.052035} & \mathbf{0.000000} & \mathbf{0.026018} & \mathbf{0.052035} \\ \mathbf{0.000000} & \mathbf{0.000000} & \mathbf{0.000000} & \mathbf{0.000000} \end{bmatrix}$$ $$\frac{\partial L}{\partial b^{(1)}} = \delta^{(1)} = \begin{bmatrix} \mathbf{0.052035} \\ \mathbf{0.000000} \end{bmatrix}$$

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Phase 4: Single Parameter Update Step ($\eta = 0.50$)

We update all parameters along the direction of steepest descent ($-\nabla L$):

1. Layer 2 Updates

$$W^{(2)}_{\text{new}} = W^{(2)}_{\text{old}} - \eta \frac{\partial L}{\partial W^{(2)}} = \begin{bmatrix} 1.5 & 1.0 \end{bmatrix} - 0.5 \begin{bmatrix} 0.138760 & 0.000000 \end{bmatrix} = \begin{bmatrix} \mathbf{1.430620} & \mathbf{1.000000} \end{bmatrix}$$ $$b^{(2)}_{\text{new}} = b^{(2)}_{\text{old}} - \eta \frac{\partial L}{\partial b^{(2)}} = -2.0 - 0.5(0.034690) = \mathbf{-2.017345}$$

2. Layer 1 Updates

$$W^{(1)}_{\text{new}} = \begin{bmatrix} 3.0 & -2.0 & 0.0 & 3.0 \\ -2.0 & 3.0 & 2.0 & -1.0 \end{bmatrix} - 0.5 \begin{bmatrix} 0.052035 & 0.000000 & 0.026018 & 0.052035 \\ 0.000000 & 0.000000 & 0.000000 & 0.000000 \end{bmatrix}$$ $$W^{(1)}_{\text{new}} = \begin{bmatrix} \mathbf{2.973982} & \mathbf{-2.000000} & \mathbf{-0.013009} & \mathbf{2.973982} \\ \mathbf{-2.000000} & \mathbf{3.000000} & \mathbf{2.000000} & \mathbf{-1.000000} \end{bmatrix}$$ $$b^{(1)}_{\text{new}} = \begin{bmatrix} -2.0 \\ -1.5 \end{bmatrix} - 0.5 \begin{bmatrix} 0.052035 \\ 0.000000 \end{bmatrix} = \begin{bmatrix} \mathbf{-2.026018} \\ \mathbf{-1.500000} \end{bmatrix}$$

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Phase 5: Second Forward Pass ($t = 1$) & Proof of Loss Reduction

To prove that the closed cycle reduced error, we re-evaluate the forward pass using the updated parameter matrices:

Step 1: Updated Hidden Layer Forward Pass

Compute $z^{(1)}_{\text{new}} = W^{(1)}_{\text{new}} x + b^{(1)}_{\text{new}}$:

$$z_{1, \text{new}}^{(1)} = (2.973982)(1.0) + (-2.0)(0.0) + (-0.013009)(0.5) + (2.973982)(1.0) - 2.026018$$ $$z_{1, \text{new}}^{(1)} = 2.973982 + 0.0 - 0.006505 + 2.973982 - 2.026018 = \mathbf{+3.915441}$$ $$z_{2, \text{new}}^{(1)} = (-2.0)(1.0) + (3.0)(0.0) + (2.0)(0.5) + (-1.0)(1.0) - 1.500000 = \mathbf{-3.500000}$$

Apply ReLU activation:

$$a^{(1)}_{\text{new}} = \begin{bmatrix} \max(0, 3.915441) \\ \max(0, -3.5) \end{bmatrix} = \begin{bmatrix} \mathbf{3.915441} \\ \mathbf{0.000000} \end{bmatrix}$$

Step 2: Updated Output Layer Forward Pass

Compute $z^{(2)}_{\text{new}} = W^{(2)}_{\text{new}} a^{(1)}_{\text{new}} + b^{(2)}_{\text{new}}$:

$$z^{(2)}_{\text{new}} = (1.430620)(3.915441) + (1.0)(0.0) - 2.017345 = 5.601510 - 2.017345 = \mathbf{+3.584165}$$

Pass through Sigmoid:

$$a^{(2)}_{\text{new}} = \sigma(3.584165) = \frac{1}{1 + e^{-3.584165}} = \frac{1}{1 + 0.027760} \approx \mathbf{0.972990} \quad (\mathbf{97.30\%})$$

Step 3: Updated Loss Evaluation

Compute the new Mean Squared Error loss:

$$L_2 = (a^{(2)}_{\text{new}} - y)^2 = (0.972990 - 0.0)^2 \approx \mathbf{0.946710}$$

Algebraic Proof of Strict Loss Reduction

$$\Delta L = L_2 - L_1 = 0.946710 - 0.964351 = \mathbf{-0.017641} < 0 \quad (\text{Strict Loss Reduction})$$ $$\text{Relative Error Reduction} = \frac{0.017641}{0.964351} \times 100\% = \mathbf{1.83\%}$$

The closed learning cycle successfully nudged the predicted probability from $98.20\%$ down to $97.30\%$, strictly reducing Mean Squared Error loss by $1.83\%$ on paper in a single update step.

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