Polynomial Signals
Exercise PLAN-517
primary source nl only. Explicit question version 2 with distinct contract-change review; the predecessor question is not equivalent current gold. Core, family and split are retained. No new Lean verification is claimed for this NL view.
Assumptions and conventions
- T_m(cos beta) = cos(m beta); arcsine real only on from minus one to one
- Outside the interval the first-kind Chebyshev uses the hyperbolic-cosine representation
This question explicitly revises its historical contract. A Lean companion must bind the current contract hash, not the predecessor hash. This is one reserved core, not two exercises.
Question
Let k be a natural number, m=2k+1, and T_m the first-kind Chebyshev polynomial, with T_j(cos beta)=cos(j beta). Grover's normalized-overlap calculation uses T_(2k+1)(a)=(-1)^k sin((2k+1)arcsin(a)) for a in[0,1], where arcsin denotes the real inverse sine.
Determine the full real a-domain on which that real-inverse expression is defined and prove the identity there, including a=0,+1,-1 and k=0. Assess the predecessor's requested 'concrete refutation' at a>1 precisely: is the real expression a false numeric equality, or is it undefined because no real inverse-sine argument exists? At a=2,k=1 compute T_3(2) exactly and explain what the bound of a REAL sine does and does not show. Do not infer any bound for complex sine from the real bound.
Then prove the odd multiple-angle identity T_(2k+1)(sin z)=(-1)^k sin((2k+1)z) for all complex z. For any complex inverse-sine value z satisfying sin z=a, determine whether the same Chebyshev value is recovered; do not assume a globally single-valued holomorphic arcsine exists. Give an explicit complex z with sin z=2, and compute (-1)^1 sin(3z) exactly to settle, rather than presume, the continuation claim.
For arbitrary natural j give the correct real hyperbolic formula on a>=1 and on a<=-1, including the parity sign on the negative side; prove it from the Chebyshev recurrence or equivalent complete algebra. Explain why complex continuation or an unnormalized inner product greater than1 does not make such a a permissible normalized quantum overlap or singular value in a unit-normalized block. Retain the exact(-1)^k phase in the scalar/Grover identity, distinguishing it from probability and from a free controlled-phase convention.
Hint: mathematical route
The odd Chebyshev/sine formula has real-inverse domain[-1,1] but continues correctly through any complex inverse-sine value; signed reflected hyperbolic formulas describe both real exterior branches without creating unphysical normalized overlaps.
- For real a in[-1,1], phi=arcsin(a) exists; endpoints/zero/k0 remain valid rather than requiring an interior overlap.
- The Chebyshev/cosine recurrence holds for complex arguments as well, with beta=pi/2-z.
- Odd order yields the exact(-1)^k sine relation for every complex z; any inverse value of a therefore recovers the same polynomial without a global branch claim.
- T_3(2)=26 is exact, whereas the real-inverse expression at2 is undefined; a bounded real-sine extrapolation cannot produce26.
- The complex choice z=pi/2-i arccosh2 gives sin(z)=2 and (-1)sin(3z)=26, so real-domain failure is not a refutation of complex continuation.
- The real exterior formulas use cosh(j arccosh a) for a>=1 and(-1)^j cosh(j arccosh(-a)) for a<=-1; physical normalization and phase obligations remain separate.
Reference solution
As a REAL inverse-sine expression, the displayed formula is defined for a in[-1,1]. For those a let phi=arcsin(a) in[-pi/2,pi/2] and beta=pi/2-phi. Then cos beta=sin phi=a, and T_(2k+1)(a)=cos((2k+1)(pi/2-phi))=(-1)^k sin((2k+1)phi). The sign follows from cos((2k+1)pi/2)=0 and sin((2k+1)pi/2)=(-1)^k. This includes a0 (both sides0), a1 (both sides1), a-1 (both sides-1) and k0 (T_1(a)=a=sin(arcsin a)). The interval is the domain of the real inverse, not a maximal domain for the polynomial identity after analytic continuation.
For a>1 there is NO real phi with sin phi=a, so real arcsin(a), understood as an inverse, has no value. Consequently there is no defined real numerical equality there to falsify by substitution. For a2,k1, the recurrence gives T_3(x)=4x^3-3x and T_3(2)=26. This refutes any attempted extension that insists its right side remains (-1) times the sine of a REAL argument, hence lies in[-1,1]. It does NOT refute a complex inverse-sine interpretation. Assigning a totalized software function outside its inverse domain would be an extra convention, not the real-inverse identity under discussion. A complex sine is not bounded in modulus by1.
The continuation can be proved without postulating a global branch. The Chebyshev recurrence T_0=1,T_1=x,T_(j+1)=2x T_j-T_(j-1), together with cos((j+1)w)=2cos(w)cos(jw)-cos((j-1)w), proves T_j(cos w)=cos(jw) for every COMPLEX w by induction. Put w=pi/2-z and j=2k+1. The same entire trigonometric relations give T_(2k+1)(sin z)=cos((2k+1)(pi/2-z))=(-1)^k sin((2k+1)z) for all complex z. Therefore any complex number z satisfying sin z=a gives the SAME polynomial value T_(2k+1)(a), independently of which inverse value was chosen. Local inverse branches can be used on appropriate domains, but this does not define a single global holomorphic arcsine through the branch points±1. No branch-independence conclusion relies on applying a real bounded-sine estimate to complex arguments.
For the requested exact value, put eta=arccosh(2)=log(2+sqrt(3))>0 and choose z=pi/2-i*eta. From sin(x+i y)=sin x cosh y+i cos x sinh y, sin z=cosh eta=2. Moreover sin(3z)=sin(3pi/2)cosh(3eta)=-cosh(3eta). The hyperbolic triple-angle identity gives cosh(3eta)=4*cosh(eta)^3-3*cosh(eta)=26, so (-1)^1 sin(3z)=26=T_3(2). This explicitly refutes the predecessor ANSWER'S rejection of complex continuation. It is not a counterexample to the real inverse's domain restriction. In particular k0 also continues correctly to T_1(a)=sin z=a; no exceptional prohibition appears there.
For the real outside-interval representation, if a>=1 write a=cosh eta, eta=arccosh(a)>=0. The functions cosh(j eta) satisfy the same recurrence in j and initial values1,cosh eta as T_j(a), so T_j(a)=cosh(j arccosh a). For a<=-1, write b=-a>=1. The Chebyshev recurrence also proves T_j(-b)=(-1)^j T_j(b). Thus T_j(a)=(-1)^j cosh(j arccosh(-a)). The inverse hyperbolic cosine has a nonnegative REAL argument at least1 on each stated branch. The formula cosh(j arccosh(a)) without reflection is not a real formula for a<-1. At a±1 these formulas give1 and(-1)^j. For the odd order here the negative-side value is -cosh((2k+1)arccosh(-a)); for example T_3(-2)=-26. The original broad hyperbolic convention is retained with this mandatory reflected argument and parity sign, not treated as an unqualified positive-side formula.
Finally, |<u|v>|<=1 for unit vectors by Cauchy-Schwarz, and any clean compression J_out†U J_in of a unitary between isometries has norm<=1. Singular values of an exact unit-normalized encoded block therefore lie in[0,1]. An inner product between unnormalized vectors can exceed1, but calling it an overlap amplitude does not make it a normalized state's amplitude, success probability or valid unit-normalized scalar block. One must normalize or change the block factor and track that change; complex analytic continuation is mathematics, not a physical amplitude-amplification construction outside its normalized domain. In source Example11.4 the scalar a=1/sqrt(N) lies in[0,1] for N>=1, and the qubitization expression carries the exact factor(-1)^k relative to the conventional Grover expression. That factor has no effect on an isolated success probability, but becomes a relative phase under control unless tracked or corrected; it is not silently discarded in the scalar identity.
Assessment rubric
- States[-1,1] as the real-inverse domain, proves the real identity with the correct(-1)^k sign and checks endpoints/zero/k0; does not mistake that domain statement for failure of complex continuation. (1 weight)
- Proves the all-complex odd multiple-angle identity by a complete recurrence or equivalent argument; explains any inverse value recovers the polynomial and distinguishes this from existence of a global holomorphic inverse branch. (1 weight)
- Computes T_3(2)=26 and separates an undefined real arcsine from a false bounded-real-sine extrapolation. Constructs z=pi/2-i arccosh2 and verifies (-1)sin(3z)=26 exactly rather than claiming complex sine bounded by1. (1 weight)
- Derives both signed real hyperbolic branches, including a<-1 and endpoints; keeps normalized overlap/unitary-compression constraints and exact phase bookkeeping separate from analytic continuation and probabilities. (1 weight)
Graph-grounded prerequisites
These are support/source anchors, not an automatically extracted theorem-dependency proof.
- Harmonic analysis (frozen curriculum anchor; see graph registry)
- Approximation theory (frozen curriculum anchor; see graph registry)
- Matrix analysis (frozen curriculum anchor; see graph registry)
Lean view and evidence boundary
No aligned Lean view has been admitted for this native core in this textbook checkpoint. This is not a claim that the underlying mathematics lacks a formal proof. Contributions must preserve this exact target contract and split.
Sources and provenance
- website/research/sources.json
lin-wiebe-2026-qasc
Source SHA-256:9f85f45f20ed5a46931ad0dec3b94ee09748bc56e39fae29b0aefe910440b0ae· MIT metadata only - Primary source
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Item digest: 70cd1d0ed0ffc4439747cbcf81260969b18cff8679d89b37942a46e634fd3b8f
Graph snapshot: qcb-snapshot-ffa644f0da3031a1ce754b9a
Authored support steps: 6. This is not measured graph distance or model difficulty.
Question revision lineage
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