The Modern Mathematics of Deep Learning

Circular frame reconstruction

Penguin learning project

The project uses 342 complete Palmer penguin records: 240 training-pool and 102 held-out rows. The selected training rows determine feature scaling and model parameters. Source CSV

This project trains weights with softmax, unlike the homepage lab, which uses nearest species averages. Same penguin measurements; different classification methods. Predictions can differ.

Frames and sparse signals · Encode or reconstruct signals with frames and dictionaries.

The project scales bill and flipper measurements using training-row means and scales; its feature vector has two entries.

Consider this concept alongside the project; its source assumptions may differ. These observations do not prove a theorem.

Input: 12 × 2; Held-out evaluation: 102 · #58 · 40.6 mm, 193 mm. Known answer: Adélie; Prediction: Adélie · [≈0.7383, ≈0.2444, ≈0.01737].

Used in these source sections: 6.3 Taking apart and rebuilding

The decoder adds weighted synthesis vectors and divides by the window length.

These operations define structured network layers and their analysis.

Source formula example

x=\frac{1}{n}\sum_{i=1}^{p}\sum_{j=1}^{q}\langle x,u_{i}\ast_{\circ}v_{j}\rangle\tilde{u}_{i}\ast_{\circ}\tilde{v}_{j}.

illustration · The invented finite inputs illustrate this expression. They do not prove a general identity, bound, or existence claim.

Symbols

$x$: Input indexed on a finite circular group.
$uᵢ,vⱼ$: Analysis frame vectors.
$\tilde{u}ᵢ,\tilde{v}ⱼ$: Dual synthesis frame vectors.
$n$: Positive window length.
$∗\circ$: Circular convolution.

Source formula example

This source example uses its own inputs and assumptions. Editing it does not change the project.

Edit a value or the code, then run the example. This is a finite demonstration, not a proof of the paper’s theorem.

First coordinate

This example’s data flow

You choose

input.first = 1

Initial values in code

basisWindow = [ [1, 0], [0, 0], [0, 1], ]

Calculated here

Result expression: { coefficients, trace, traceCoordinate: trace / 2, reconstructed }

x = [input.first, 2, 3] Uses: input.first
coefficients = [ [x[0], x[1]], [x[1], x[2]], [x[2], x[0]], ] Uses: x
trace = basisWindow.reduce( (sum, row, i) => sum + row.reduce((v, z, j) => v + z * coefficients[i][j], 0), 0, ) Uses: basisWindow, coefficients
reconstructed = [ (coefficients[0][0] + coefficients[2][1]) / 2, (coefficients[0][1] + coefficients[1][0]) / 2, (coefficients[1][1] + coefficients[2][0]) / 2, ] Uses: coefficients

Returns

Example result

coefficients: array (3 items): [array (2 items): [1, 2], array (2 items): [2, 3], array (2 items): [3, 1]]
trace: 2
traceCoordinate: 1
reconstructed: array (3 items): [1, 2, 3]

JavaScript

Read numbers from input and return the result. JavaScript basics.

Result

Enable JavaScript to run this example.

Technical statement

The cyclic group has d = 3 positions and the window has n = 2, so n < d. Standard basis analysis and synthesis frames produce the six displayed coefficients. The trace and coefficient sum for the first coordinate both equal 2x₁. Each coordinate appears twice; dividing by n reconstructs x. The general identity requires dual-frame assumptions.

Source: Source expression