LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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.

The radix-two even/odd factorisation of the DFT

Statement

Let $M \geq 1$ , $N = 2 M$ and $f \in C^{Z / N}$ . Define the even and odd parts $e, o \in C^{Z / M}$ by

$e (r) : = f (2 r), o (r) : = f (2 r + 1), r \in Z / M Z,$

where $2 r$ and $2 r + 1$ denote the classes in $Z / N Z$ of the corresponding integers; the maps $r \mapsto 2 r$ and $r \mapsto 2 r + 1$ from $Z / M Z$ to $Z / N Z$ are well defined and injective with disjoint images, which together exhaust $Z / N Z$ . Let $X (f)$ , $X (e)$ , $X (o)$ be the unnormalised transforms of The unnormalised engineering DFT and its conversion to the unitary transform, the latter two extended periodically to $Z$ . Then for every $k \in Z$

$X_{k} (f) = X_{k} (e) + e^{- 2 π i k / N} X_{k} (o), X_{k + M} (f) = X_{k} (e) - e^{- 2 π i k / N} X_{k} (o) .$

Thus an $N$ -point transform is computed from the two $M$ -point transforms of its even and odd parts together with the twiddle factors $e^{- 2 π i k / N}$ ; this is the radix-two (decimation-in-time) step of the fast Fourier transform, and it holds for every input class list, with no hypothesis beyond $N = 2 M$ .

Facts & Assumptions

Given: Natural numbers $M \geq 1$ and $N = 2 M$ , a function $f \in C^{Z / N}$ , and integers $k, r, r^{'}$ ; the classes are those of The congruence class $[a]_{n}$ and the quotient set $Z / n$ .

[F1]

$X_{k} (u) = \sum_{x = 0}^{L - 1} u ([x]_{L}) e^{- 2 π i k x / L}$ for a function $u$ on $Z / L Z$ , and $X_{k} (u)$ depends on $k$ only modulo $L$ (The unnormalised engineering DFT and its conversion to the unitary transform).

[F2]

$[u]_{n} = [v]_{n}$ exactly when $n ∣ (u - v)$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ); if $r - r^{'} = M t$ then $2 r - 2 r^{'} = N t$ and $2 r + 1 - (2 r^{'} + 1) = N t$ , and conversely $2 M ∣ 2 s$ forces $M ∣ s$ by cancellation in $Z$ (The integers have no zero divisors; multiplicative cancellation). The divisors of $1$ in $Z$ are exactly $1$ and $- 1$ , and $0 < 1 < 2$ , $- 1 < 0$ in the ordered ring $Z$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ , The integers form a totally ordered ring).

[F3]

Division with remainder: for every integer $x$ and the positive divisor $2$ there are unique integers $q, r$ with $x = 2 q + r$ and $0 \leq r < 2$ , hence $r \in {0, 1}$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ); and the classes $[0]_{L}, \dots, [L - 1]_{L}$ enumerate $Z / L Z$ without repetition (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[F4]

Finite sums over $Z / L Z$ : computed from any enumeration, invariant under reindexing along a bijection, and additive over disjoint splittings (A finite sum in a commutative monoid indexed by an arbitrary finite set, Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[L1]

$\exp (u + v) = \exp u \exp v$ , $\exp w = 1$ exactly when $w \in 2 π i Z$ , and $e^{- π i} = - 1$ , so $e^{- 2 π i (k + M) / N} = e^{- 2 π i k / N} e^{- π i} = - e^{- 2 π i k / N}$ because $2 π i M / N = π i$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The complex exponential by its power series).

[L2]

Since $N = 2 M$ with $M > 0$ , division in the embedded real field gives $2 r / N = r / M$ and $(2 r + 1) / N = r / M + 1 / N$ . The frequency $k + M$ differs from $k$ by the period $M$ of each shorter transform in [F1].

Proof

technique · direct

1.1F2F3L2

The index maps are well defined with the stated images. If $r \equiv r^{'} (m o d M)$ , then $r - r^{'} = M t$ and [F2] gives $2 r \equiv 2 r^{'} (m o d N)$ and $2 r + 1 \equiv 2 r^{'} + 1 (m o d N)$ , so $e$ and $o$ are well-defined functions on $Z / M Z$ . If $2 r \equiv 2 r^{'} (m o d N)$ , then $N ∣ 2 (r - r^{'})$ , that is $2 M ∣ 2 (r - r^{'})$ , so $M ∣ (r - r^{'})$ by cancellation [F2], hence $r = r^{'}$ in $Z / M Z$ : the doubling map is injective, and likewise $r \mapsto 2 r + 1$ . If $2 r \equiv 2 r^{'} + 1 (m o d N)$ then $N ∣ (2 r - 2 r^{'} - 1)$ , say $2 (r - r^{'}) - 1 = 2 M t$ , so $1 = 2 (r - r^{'} - M t)$ and $2 ∣ 1$ ; by [F2] this forces $2 \in {1, - 1}$ , contradicting $0 < 1 < 2$ and $- 1 < 0$ . Hence the two images are disjoint. Finally, every class of $Z / N Z$ is $[x]_{N}$ for a unique $0 \leq x < N = 2 M$ [F3]; writing $x = 2 q + r$ with $r \in {0, 1}$ [F3] gives $x = 2 q$ or $x = 2 q + 1$ , and $x < 2 M$ forces $q < M$ (if $q \geq M$ then $x \geq 2 q \geq 2 M$ ), so the class is in one of the two images; the images therefore exhaust $Z / N Z$ .

1.2F1F4

Splitting the defining sum: with $x$ running over $0, \dots, 2 M - 1$ , the list is the disjoint union of the even numbers $2 r$ and the odd numbers $2 r + 1$ , $0 \leq r < M$ ; hence by the splitting and reindexing rules [F4] $X_{k} (f) = \sum_{x = 0}^{2 M - 1} f ([x]_{N}) e^{- 2 π i k x / N} = \sum_{r = 0}^{M - 1} f ([2 r]_{N}) e^{- 2 π i k (2 r) / N} + \sum_{r = 0}^{M - 1} f ([2 r + 1]_{N}) e^{- 2 π i k (2 r + 1) / N}$ .

1.3F1L1L2

Evaluating the two pieces: by $2 r / N = r / M$ and the addition law [L1], $\sum_{r = 0}^{M - 1} f ([2 r]_{N}) e^{- 2 π i k (2 r) / N} = \sum_{r = 0}^{M - 1} e ([r]_{M}) e^{- 2 π i k r / M} = X_{k} (e)$ ; and $\sum_{r = 0}^{M - 1} f ([2 r + 1]_{N}) e^{- 2 π i k (2 r + 1) / N} = e^{- 2 π i k / N} \sum_{r = 0}^{M - 1} o ([r]_{M}) e^{- 2 π i k r / M} = e^{- 2 π i k / N} X_{k} (o)$ , where the exponents combine by [L1] using $(2 r + 1) / N = r / M + 1 / N$ .

2.1step 1.2step 1.3

Adding the two evaluations of step 1.3 gives $X_{k} (f) = X_{k} (e) + e^{- 2 π i k / N} X_{k} (o)$ for every $k \in Z$ , which is the first displayed identity.

3.1F1L1step 2.1∎

For the second identity, replace $k$ by $k + M$ in step 2.1: $X_{k + M} (e) = X_{k} (e)$ and $X_{k + M} (o) = X_{k} (o)$ because the transforms of the $M$ -point functions depend on $k$ only modulo $M$ [F1], while $e^{- 2 π i (k + M) / N} = - e^{- 2 π i k / N}$ by [L1]; hence $X_{k + M} (f) = X_{k} (e) - e^{- 2 π i k / N} X_{k} (o)$ .

Remarks

This is the declared decimation-in-time form; Taylor's Proposition 12.1 and the MIT lecture's heading 4 are its mirror. Taylor splits the input into the halves $f (ω^{j}) \pm f (ω^{j + n / 2})$ and reads off the output parities, and the MIT lecture's heading 4 likewise splits the coefficient list into its two halves and combines them with a twiddle; both are the decimation-in-frequency description of the same pair of identities, the transposed statement of the even/odd-coefficient form proved here. No second independent result is being recorded.
Sign caveat in Taylor. Formula (12.1) uses $ω^{- j ℓ}$ , but (12.9) prints $ω^{j}$ in the odd-frequency branch. With $ω = e^{2 π i / N}$ , the negative-sign odd-frequency sum instead factors as $\sum_{j} ω^{- j} (f (ω^{j}) - f (ω^{j + N / 2})) (ω^{2})^{- j k}$ . Thus that branch needs $ω^{- j}$ , consistent with Taylor's four-point factor $- i$ in (12.6). The local proof above derives its signs directly rather than importing the inconsistent printed general twiddle.
The twiddle factor is the price of the odd subproblem. The even half reuses the $M$ -point transform unchanged; the odd half carries the factor $e^{- 2 π i k / N}$ , and at $k + M$ that factor changes sign, which is exactly what produces the second displayed identity. Both identities are used in the recursion definition later on this page.

Depends on

Used by

The radix-two split fails for odd N Counterexample
The recursive radix-two fast Fourier transform Definition
The four-point radix-two FFT executed in full Example
Correctness of the recursive radix-two FFT Theorem

Dependency tree · two levels

84 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

MIT 18.310 lecture 23, The Finite Fourier Transform and the Fast Fourier Transform Algorithm (course page) (standard reference, not scraped)
Michael E. Taylor, Fourier Analysis, Distributions, and Constant-Coefficient Linear PDE (author PDF) (standard reference, not scraped)