LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.

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Finite geometric sum for phase-estimation amplitudes

Statement

For $n \in N$ and $z \in C$ , define $S_{n} (z) : = \sum_{j = 0}^{n - 1} z^{j},$ with the empty sum $S_{0} (z) = 0$ . Then $S_{n} (z) = {\begin{cases} ι_{C} (n), & z = 1, \\ (1 - z^{n}) / (1 - z), & z \neq 1, \end{cases}$ where $ι_{C}$ is the canonical natural embedding of $N$ into $C$ .

For $Q = 2^{m}$ , $θ \in [0, 1)$ , and integer $y$ , setting $z = \exp (2 π i (θ - y / Q))$ gives $z^{Q} = \exp (2 π i Q θ)$ ; moreover $z = 1$ exactly when $Q θ - y \in Q Z$ .

Facts & Assumptions

Given: $n \in N$ , $z \in C$ , and, for the specialization, $m \geq 1$ , $Q = 2^{m}$ , $θ \in [0, 1)$ , and $y \in Z$ .

[F1]

A finite sum is independent of its enumeration and splits over a disjoint union; for the initial segment ${0, \dots, n}$ , this gives $S_{n + 1} (z) = S_{n} (z) + z^{n}$ (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).

[F2]

Complex powers satisfy $z^{0} = 1$ and $z^{n + 1} = z^{n} z$ (Integer powers in the complex field).

[F3]

The complex numbers form a field, so multiplication distributes over addition and every nonzero element has an inverse ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[F4]

The canonical natural is defined recursively by $ι_{C} (0) = 0$ , $ι_{C} (n + 1) = ι_{C} (n) + 1$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[F5]

Induction holds on $N$ (The principle of mathematical induction).

[F6]

The complex exponential satisfies $\exp (a + b) = \exp (a) \exp (b)$ , and $\exp (w) = 1$ exactly when $w \in 2 π i Z$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , The complex exponential by its power series).

Proof

1.1F1F2basealgebra

At $n = 0$ , the sum is empty, so $(1 - z) S_{0} (z) = 0 = 1 - z^{0}$ by [F1] and [F2].

1.2givenih

Fix $n \in N$ and assume $(1 - z) S_{n} (z) = 1 - z^{n}$ ; this is the induction hypothesis.

2.1F1F2F3F5step 1.2discharge-induction

By [F1] and [F2], $(1 - z) S_{n + 1} = (1 - z) S_{n} + (1 - z) z^{n} = (1 - z^{n}) + (z^{n} - z^{n + 1}) = 1 - z^{n + 1}$ , using distributivity in [F3] and the induction hypothesis [step 1.2]. Thus [F5] proves $(1 - z) S_{n} (z) = 1 - z^{n}$ for every $n$ .

3.1F1F3F4F5step 2.1

If $z = 1$ , every summand is $1$ , so [F1] gives $S_{0} (1) = 0$ and $S_{n + 1} (1) = S_{n} (1) + 1$ . At $n = 0$ , this equals $ι_{C} (0) = 0$ by [F4]; if $S_{n} (1) = ι_{C} (n)$ , the two successor recursions give $S_{n + 1} (1) = S_{n} (1) + 1 = ι_{C} (n) + 1 = ι_{C} (n + 1)$ . Induction [F5] proves $S_{n} (1) = ι_{C} (n)$ for every $n$ . If $z \neq 1$ , then $1 - z \neq 0$ , so divide the identity from step 2.1 by $1 - z$ in the field [F3] to obtain $S_{n} (z) = (1 - z^{n}) / (1 - z)$ .

4.1F2F6step 2.1discharge-induction∎

In the QPE specialization, repeated use of [F6] and the power recursion [F2] gives $z^{Q} = \exp (2 π i Q (θ - y / Q)) = \exp (2 π i (Q θ - y))$ . Since $y \in Z$ , [F6] gives $\exp (- 2 π i y) = 1$ , hence $z^{Q} = \exp (2 π i Q θ)$ . The same kernel statement gives $z = 1$ exactly when $2 π i (θ - y / Q) \in 2 π i Z$ , equivalently $θ - y / Q \in Z$ , or $Q θ - y \in Q Z$ .

Depends on

Used by

Dependency tree · two levels

41 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

Andrew M. Childs, Lecture Notes on Quantum Algorithms, §4.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, version 6, §4.6 (standard reference, not scraped)