The Modern Mathematics of Deep Learning

A signed-measure functional and Fourier test

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.

Approximation and depth · Study approximation error and network size under stated conditions.

The project supplies finite observations and fitted parameters. A held-out sample does not establish population bounds or prove a theorem.

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: 1.2.2 Can the network express the answer?

A signed measure defines a linear functional by integrating a function against it. The source proof studies when that functional vanishes on all activated-affine functions and then on Fourier exponentials.

The toy uses three points with signed weights 1, −2, 1.

The calculation illustrates how a Fourier test can detect a nonzero signed measure.

Source formula example

\int_{K}e^{-2\pi i(\langle w,x\rangle+b)}\,\mathrm{d}\mu(x)=0

illustration · This evaluates a finite signed measure at selected functions and one frequency. It does not establish the source’s vanishing condition for all frequencies or prove the uniqueness theorem.

Symbols

$F$: A continuous linear functional; in the toy it is a finite signed weighted sum.
$μ$: A signed measure.
$w,b$: Frequency and offset; b is zero in the toy.

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.

Frequency w

This example’s data flow

You choose

input.w = 0.25

Initial values in code

points = [0, 1, 2]
weights = [1, -2, 1]

Calculated here

Result expression: { points, weights, quadraticFunctional, real, imaginary }

quadraticFunctional = points.reduce((sum, x, i) => sum + weights[i] * x ** 2, 0) Uses: points, weights
real = points.reduce( (sum, x, i) => sum + weights[i] * Math.cos(-2 * Math.PI * input.w * x), 0, ) Uses: points, weights, input.w
imaginary = points.reduce( (sum, x, i) => sum + weights[i] * Math.sin(-2 * Math.PI * input.w * x), 0, ) Uses: points, weights, input.w

Returns

Example result

points: array (3 items): [0, 1, 2]
weights: array (3 items): [1, -2, 1]
quadraticFunctional: 2
real: 0
imaginary: 2

JavaScript

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

Result

Enable JavaScript to run this example.

Technical statement

The toy functional is nonzero because its value on x² is 2. The computed Fourier integral uses w=1/4 by default and is nonzero, showing that this measure does not satisfy the all-frequency vanishing condition. The source proof uses continuous duality and Fourier uniqueness in a general compact-domain setting.

Source: Source expression