7.2 KiB
Standalone Equation Certificates for Problem 2.8
This sheet replaces every Ore-division or “the remainder simplifies” claim in the proof. Each displayed assertion is an equality in a rational-function field. Multiplying by the listed denominator produces a polynomial whose coefficients are all zero.
The companion program p28_standalone_equations.py recreates these checks
with sparse dictionaries over fractions.Fraction. It implements addition,
multiplication, integer powers, and formal differentiation only. It does not
call a division algorithm, factorizer, simplifier, Gröbner basis, special
function library, root finder, or numerical sampler.
1. Input matrix
Let u and x be indeterminates, r=x^{-1}, and
\omega=u(3u-2)(3u+2). Define
[ \begin{aligned} a_1={}&r(144u^5-288u^4+144u^3) -99u^5+333u^4-229u^3-114u^2+40u+64,\ a_2={}&r(432u^4-864u^3+432u^2) -243u^4+909u^3-868u^2-80u+272,\ a_3={}&r(432u^3-864u^2+432u) -153u^3+648u^2-860u+360,\ a_4={}&144r(u-1)^2,\ b_1={}&-144ru^3+9u^4+63u^3+158u^2+168u+64,\ b_2={}&216ru^2+36u^3-189u^2-316u-168,\ b_3={}&108ru+54u^2-189u-158,\ c_1={}&-288r^2u^3+ r(54u^4+378u^3+948u^2+1008u+384)\ &+18u^5+45u^4-251u^3-1086u^2-1384u-576,\ c_2={}&-432r^2u^2+ r(153u^4-657u^3+1292u^2+2064u+1072)\ &-72u^4+702u^3-1069u^2-2508u-1512,\ c_3={}&-216r^2u+ r(180u^3-891u^2+1450u+1116)\ &-108u^3+864u^2-1385u-1422,\ c_4={}&-4r^2+r(6u^2-33u+\tfrac{536}{9})-4u^2+32u-63. \end{aligned} ]
Then
[ \mathcal M(u,x)= \begin{pmatrix} a_1/\omega&a_2/\omega&a_3/\omega&a_4/\omega\ -u^3&-3u^2&-3u&-1\ xb_1/144&-xb_2/72&-xb_3/36&x(-2r-(2u-7))/2\ x^2c_1/288&x^2c_2/144&x^2c_3/72&x^2c_4/4 \end{pmatrix}. ]
For the recurrence, u=2N+3, so u\ge3.
2. Four Ore identities without division
Put
[ m=\frac{u-1}{2},\qquad P_i(t)=\sum_{s=0}^{3}\mathcal M(u,x)_{i+1,s+1}t^s, ]
[ \mathcal TQ=(1-x)x\partial_xQ+(t+1)Q ]
and
[ L_+(t)=(1-x)t(t+u)^3+ x(t+m+1)(t+m+\tfrac76)(t+m+\tfrac32)(t+m+\tfrac{11}{6}). ]
Define
[ \begin{aligned} \ell_0={}&(u-1)u(3u-2)(3u+2)x/144,\ \ell_1={}&[-576+864u-432u^2+72u^3 +(580-872u+405u^2-36u^3)x]/72,\ \ell_2={}&[432-432u+108u^2 +(-436+405u-54u^2)x]/36,\ \ell_3={}&(-12+6u+11x-2ux)/2,\ q_3={}&-36+(536-297u+54u^2)x +(-567+288u-36u^2)x^2. \end{aligned} ]
The complete equation set is
[ u(3u-2)(3u+2)x(\mathcal TP_0-P_1) -144(u-1)^2L_+=0, ]
[ \mathcal TP_1-P_2+L_+=0, ]
[ 2(\mathcal TP_2-P_3)-(-2+7x-2ux)L_+=0, ]
[ 36(\mathcal TP_3+\ell_3P_3+\ell_2P_2+\ell_1P_1+\ell_0P_0) -q_3L_+=0. ]
These are the quotient-and-zero-remainder claims written as four direct factorizations. No quotient or remainder operation is needed.
3. Lowest and generic tail coefficients
Set
[ A(t)=\frac{144(u-1)^2(t+u)^3}{u(3u-2)(3u+2)},\qquad B(t)=P_0(t)-\frac{A(t)}x, ]
[ \varrho=-\frac{(3u-2)(3u+2)}{144(u-1)^2u^2}, ]
[ \chi_j= \frac{(m+j)(m+\frac16+j)(m+\frac12+j)(m+\frac56+j)} {m(m+\frac16)(m+\frac12)(m+\frac56)} \left(\frac{2m(2m+1)}{(2m+j)(2m+j+1)}\right)^3, ]
[ \psi_j= \frac{(m+j)(m+\frac16+j)(m+\frac12+j)(m+\frac56+j)} {(2m+j)^3(j+1)}. ]
The two identities are
[ \varrho[-A(0)]-1=0 ]
and, for j\ge1,
[ \varrho\left[ -A(j)\chi_j+ \frac{(A(j-1)+B(j-1))\chi_{j-1}}{\psi_{j-1}} \right]-1=0. ]
For u=2N+3\ge3, the integer-dependent denominator factors are nonzero.
Factors of x are cleared before coefficient comparison; no conclusion
is obtained by evaluating a rational expression at x=0.
4. Terminating denominator induction
For n\ge1, define
[ \mathcal L_n(t)=t(t+2n-1)^3 -z(t+n)(t+n+\tfrac16)(t+n+\tfrac12)(t+n+\tfrac56) =\sum_{j=0}^4c_jt^j ]
and, for operator polynomials with coefficients on the left,
[ \Theta Q=z\partial_zQ+tQ. ]
The horizontal row is reconstructed by
[ \pi_0=1,\qquad \pi_3=\frac{c_4}{c_0}t, ]
[ \pi_2=\frac{c_3}{c_4}\pi_3-\Theta\pi_3,\qquad \pi_1=\frac{c_2}{c_4}\pi_3-\Theta\pi_2. ]
Its final equation is
[ \Theta\pi_1+1-\frac{c_1}{c_4}\pi_3 +\frac{72}{n(2n+1)(6n+1)(6n+5)z} \left[ t(t-2n)^3 -z(t-n)(t-n-\tfrac16)(t-n-\tfrac12)(t-n-\tfrac56) \right]=0. ]
Let
[ \mathcal C_n(z)= \begin{pmatrix} 0&1&0&0\ 0&0&1&0\ 0&0&0&1\ -c_0/c_4&-c_1/c_4&-c_2/c_4&-c_3/c_4 \end{pmatrix}. ]
The complete differential-gauge equation is
[ \mathcal C_n[-z\mathcal M(2n+1,-z/(1-z))] -\theta[-z\mathcal M(2n+1,-z/(1-z))] -[-z\mathcal M(2n+1,-z/(1-z))]\mathcal C_{n+1}=0. ]
The checker cross-multiplies all sixteen entries independently.
The exact matrix-to-scalar bridge is
[ \sum_{r=0}^3\pi_r(t) \left[-z\mathcal M(2n+1,-z/(1-z))\right]_{r+1,1} -d_0(t)-zd_1(t)=0. ]
The checker expands all five reconstruction/bridge equations before it
checks any hypergeometric coefficient identity. Thus d_0+zd_1 is tied
to the displayed challenge matrix and is not a guessed surrogate.
Now define
[ \nu_n=\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)}, ]
[ d_0(t)= \frac{72(2n+1)^2(2n-t)^3}{n(6n+1)(6n+5)} ]
and
[ d_1(t)=-\frac{P(n,t)}{n(2n+1)(6n+1)(6n+5)}, ]
where
[ \begin{aligned} P(n,t)={}&-5n-76n^2+1404n^3+4360n^4+4320n^5+1440n^6\ &+(5+127n-1760n^2-6536n^3-7632n^4-3024n^5)t\ &+(-51+659n+3086n^2+4500n^3+2232n^4)t^2\ &+(-72-432n-864n^2-576n^3)t^3. \end{aligned} ]
Let
[ h_{n,k}= \frac{(-n)_k(-n-\frac16)_k(-n-\frac12)_k(-n-\frac56)_k} {(1-2n)_k^3k!}, \qquad0\le k\le n. ]
The induction is exactly:
[ d_0(0)-\nu_n=0, ]
[ d_0(k)+d_1(k-1)\frac{h_{n,k-1}}{h_{n,k}} -\nu_n\frac{h_{n+1,k}}{h_{n,k}}=0 \qquad(1\le k\le n), ]
[ d_1(n)+ \nu_n\frac{(n+\frac76)(n+\frac32)(n+\frac{11}{6})} {8(2n+1)^3}=0. ]
Together with
[ q_0(x)=18/x+159/4,\qquad Q_0(x)=xq_0(x), ]
[ p_1(z)=(1-z)Q_0(-z/(1-z)) =18(1-\tfrac{77}{24}z),\qquad p_{n+1}=(d_0(\theta)+zd_1(\theta))p_n, ]
[ -zC(-z/(1-z)) =\left.(\pi_0(\theta)p_1,\pi_1(\theta)p_1, \pi_2(\theta)p_1,\pi_3(\theta)p_1)\right|_{n=1}, ]
[ \left[ \theta(\theta-2)^3 -z(\theta-1)(\theta-\tfrac76) (\theta-\tfrac32)(\theta-\tfrac{11}{6}) \right]p_1=0, ]
[ \theta[-zC(-z/(1-z))] +[-zC(-z/(1-z))]\mathcal C_1=0, ]
and
[ \deg((d_0(\theta)+zd_1(\theta))p_n)\le n+1, ]
these equations prove every coefficient, including the top boundary. The
free z^{2n} branch of the fourth-order differential equation is never
invoked.
For n\ge1 and 1\le k\le n, all integer- and
Pochhammer-dependent factors in the coefficient identities are nonzero.
The z^{-1} factors in the horizontal reconstruction are handled in
\mathbb Q(n,z)(t) and cross-multiplied first; the checked base functions
have removable limits at z=0.
5. Short hypergeometric identities
For
[ y(z)=\sum_{k\ge0} \frac{(\frac16)_k(\frac12)_k(\frac56)_k}{(k!)^3}z^k, ]
the coefficient ratio is
[ \frac{[z^{k+1}]y}{[z^k]y} =\frac{(k+\frac16)(k+\frac12)(k+\frac56)}{(k+1)^3}. ]
Therefore
[ y-1=\frac5{72}z, {}_4F_3\left( \begin{matrix}1,\frac76,\frac32,\frac{11}{6}\2,2,2\end{matrix};z \right) ]
and coefficient comparison gives
[ \left[\theta^3- z(\theta+\tfrac16)(\theta+\tfrac12)(\theta+\tfrac56)\right]y=0. ]
With z=-x/(1-x), this expands to
[ 72\theta^3y+108x\theta^2y+46x\theta y+5xy=0. ]
6. Replay
From the submission directory:
python3 certificates/p28_standalone_equations.py
Success means every denominator-cleared numerator has an empty coefficient dictionary. A stored transcript is not consulted.