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State Preparation / Structured Preparation

Structured Preparation

Exercise SP-103

advanced · proof · train

local source supported. Current content-bound mathematical second pass accepted. No new Lean verification is claimed for this NL view.

Assumptions and conventions

      Question

      Consider the generating power series obtained as the product of the exponential series exp(X) and the series 1/(1-X)^{k+1}. Let its degree-r coefficient be a_{k,r}. In the coordinate t, define A_k(t) as the truncation of this series to degrees 0..k, i.e. A_k(t) = sum_{r=0}^{k} a_{k,r} t^r. The source polynomial, in the shifted coordinate t = p+1, is a positive combination S(p) = exp(-1) * (1-t)^{k+1} * A_k(t) + t^{k+1} * A_k(1-t). Prove that A_k(t) > 0 for every t >= 0, and deduce that S(p) > 0 for every p in the closed interval [-1, 0].

      Hint: mathematical route

      The source Hermite bridge polynomial is strictly positive on [-1,0] because its generating-series coefficients are nonnegative with constant term one.

      1. The series coefficient a_{k,r} is a Cauchy product of 1/m! and binomials, hence nonnegative with a_{k,0}=1.
      2. A finite nonnegative sum whose constant term is 1 is at least 1, so A_k(t)>0 for t>=0.
      3. With t=p+1 in [0,1], both A_k(t) and A_k(1-t) are positive and the prefactors nonnegative.
      4. A case split on t=0 vs t>0 keeps one strictly positive term, giving S(p)>0 on [-1,0].
      Reference solution

      Step 1 (coefficients are nonnegative): the coefficient a_{k,r} equals the Cauchy product sum_{m=0}^{r} (1/m!) * C(k+r-m, k), where C denotes a binomial coefficient coming from 1/(1-X)^{k+1} = sum C(k+i, k) X^i and 1/m! from exp. Each summand is a quotient of nonnegative reals, so a_{k,r} >= 0, and a_{k,0} = C(k,k)/0! = 1 > 0. Step 2 (positivity of A_k on t >= 0): A_k(t) = sum_{r=0}^{k} a_{k,r} t^r is a finite sum of nonnegative terms a_{k,r} t^r (products of nonnegative a_{k,r} and nonnegative t^r). The r=0 term is a_{k,0} = 1, which is strictly positive, so the whole sum is at least 1 > 0. Step 3 (positivity of S on [-1,0]): put t = p+1. For p in [-1,0] we have t in [0,1], so both t >= 0 and 1-t >= 0. Then A_k(t) > 0 and A_k(1-t) > 0 by Step 2. The prefactors exp(-1) > 0, (1-t)^{k+1} >= 0, t^{k+1} >= 0 are nonnegative. Step 4 (strictness): if t = 0 the second term vanishes but the first is exp(-1) * 1 * A_k(0) > 0; otherwise t > 0 makes t^{k+1} * A_k(1-t) > 0, so one term is always strictly positive while the other is nonnegative. Hence S(p) > 0 throughout [-1,0].

      Assessment rubric
      1. Computes a_{k,r} as the Cauchy product sum of 1/m! and binomial C(k+r-m,k), establishing nonnegativity and a_{k,0}=1. (1 weight)
      2. Concludes A_k(t) >= 1 > 0 for all t >= 0 because the degree-0 term is 1 and all other terms are nonnegative. (1 weight)
      3. Substitutes t=p+1, noting p in [-1,0] gives t and 1-t both in [0,1] so both A_k factors are positive. (1 weight)
      4. Handles strictness by a case split on t=0 vs t>0 so that at least one of the two nonnegative terms is strictly positive. (1 weight)

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      Sources and provenance
      • QuantumBlockEncoding/HermitePolynomial.lean

        lines 168-205: def coefficientSeries (k) := PowerSeries.exp R * (invOneSubPow R (k+1)).val; theorem coefficientSeries_coeff : coeff r = sum_{m in range (r+1)} C(k+r-m,k)/m!; coefficientSeries_coeff_nonneg; coefficientSeries_coeff_zero = 1. lines 254-286: coefficientPolynomial_pos (k) (t) (ht: 0<=t) : 0 < eval t; sourceInterpolant_pos (k) (p) (hp : p in Icc (-1) 0) : 0 < (sourceInterpolant k).eval p

        Source SHA-256: f8a3e1d5f29aed1228b6ede3577a9af1e76301559172ee67a266fcb0d56fee1e · MIT (repository LICENSE)

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      Graph snapshot: qcb-snapshot-ffa644f0da3031a1ce754b9a

      Authored support steps: 4. This is not measured graph distance or model difficulty.