ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Linear and polynomial kernels as finite feature inner products

Example

For $x, z \in R^{p}$ with $p \geq 1$ , the linear kernel $k_{l i n} (x, z) = x^{T} z$ and, for each integer $d \geq 0$ , the inhomogeneous polynomial kernel $k_{d} (x, z) = (1 + x^{T} z)^{d}$ are positive semidefinite. When $p = 2$ and $d = 2$ , the polynomial kernel has the explicit feature map $h (x) = (1, \sqrt{2} x_{1}, \sqrt{2} x_{2}, x_{1}^{2}, x_{2}^{2}, \sqrt{2} x_{1} x_{2})$ with $k_{2} (x, z) = h (x)^{T} h (z)$ .

Example references

Hastie, Tibshirani, and Friedman, The Elements of Statistical Learning, §5.8.2

Positive semidefinite kernels and reproducing kernel Hilbert spaces

Facts & Assumptions

Given: Integers $p \geq 1$ and $d \geq 0$ , vectors in $R^{p}$ , and the two kernels in the Example.

[F1]

Positive semidefiniteness means symmetry and nonnegative finite Gram quadratic forms (Positive semidefinite kernels and reproducing kernel Hilbert spaces).

Verification

1.1givenF1algebra

For any finite sample $x_{1}, \dots, x_{m} \in R^{p}$ and coefficients $c_{1}, \dots, c_{m} \in R$ , the linear Gram quadratic form is $\sum_{i, j} c_{i} c_{j} x_{i}^{T} x_{j} = ∥ \sum_{i} c_{i} x_{i} ∥_{2}^{2} \geq 0$ . Thus $k_{l i n}$ satisfies the nonnegative-form condition in [F1], and it is symmetric because $x^{T} z = z^{T} x$ .

1.2givenconstructalgebra

For $α = (α_{0}, \dots, α_{p}) \in N_{0}^{p + 1}$ with $∣ α ∣ = d$ , define $ψ_{α} (x) = \sqrt{d! / α_{0}! \dots α_{p}!} \prod_{r = 1}^{p} x_{r}^{α_{r}}$ . The multinomial theorem gives $(1 + x^{T} z)^{d} = \sum_{∣ α ∣ = d} ψ_{α} (x) ψ_{α} (z)$ .

2.1step 1.2F1algebra

Therefore, for every finite sample and coefficients, $\sum_{i, j} c_{i} c_{j} k_{d} (x_{i}, x_{j}) = \sum_{∣ α ∣ = d} (\sum_{i} c_{i} ψ_{α} (x_{i}))^{2} \geq 0$ . The polynomial kernel is symmetric and positive semidefinite by [F1].

3.1step 1.2algebra∎

When $p = 2$ and $d = 2$ , the six multi-indices in [step 1.2] give $h (x) = (1, \sqrt{2} x_{1}, \sqrt{2} x_{2}, x_{1}^{2}, x_{2}^{2}, \sqrt{2} x_{1} x_{2})$ , so the stated feature identity holds.

Depends on

Positive semidefinite kernels and reproducing kernel Hilbert spaces

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Hastie, Tibshirani, and Friedman, The Elements of Statistical Learning, 2nd ed., corrected 12th printing (standard reference, not scraped)