How do we simplify a multiple sum that involves determinants?

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Let $d in 2mathbb N$ be an even positive integer, let $pin mathbb N$ be another positive integer and let $T$ be a parameter.
Let $vecq := left( q_xi right)_xi=0^d/2-1 in mathbb N^(d/2)$.
Consider a following multiple sum:
begineqnarray
&&mathcal S_d,p(T):=sumlimits_j=1^d sumlimits_2 le q_0 le cdots le q_d/2-1 le d frac12^ cdot nu(vecq)
left|
beginarrayr
left( (fracT-d-12 + delta_2 xi+1,j p+q_xi)^(eta-q_xi) 1_q_xi le etaright)_eta=1^d\
left(2^-eta+1 (T-d+1)^(p(delta_2xi+1,j+delta_2xi+2,j)+eta+q_xi-3)right)_eta=1^d\
endarray
right|_xi=0^d/2-1
endeqnarray
where $nu(vecq) := prod_eta=2^d p_eta!$
where
beginequation
vecq:= left(
underbrace2,cdots,2_p_2,
underbrace3,cdots,3_p_3,
cdots,
underbraced,cdots,d_p_d
right)
endequation
Note that the term in the multiple sum is a determinant of a $dtimes d$ matrix whose entries are polynomials in $T$.
We stumbled across this sum when solving question Spectral densities of finite dimensional sample covariance matrices .



Now by using Mathematica we have discovered the following pattern:
beginequation
mathcal S_d,p(T)= sumlimits_xi=0^(2d-5) vee 0 (T-d+1)^(p+xi) cdot P_d,xi^(fracd^24-1-xi)(T)
endequation
where $P_d,xi^(fracd^24-1-xi)(T)$ is a polynomial in $T$ of order $(fracd^24-1-xi)$.



Here the polynomials in question read:
begineqnarray
P_2,0^(0)(T) &=& frac12\
hline\
P_4,0^(3)(T) &=& frac1128 (T-3) left(2 p^2 T-2 p^2-p T+3 p+2 T^2-6 T+4right)\
P_4,1^(2)(T) &=& frac1128 left(-4 p^2 T+8 p^2-3 p T+5 p-2 T^2+10 T-12right)\
P_4,2^(1)(T) &=& frac164 left(p^2+2 p-T+3right)\
P_4,3^(0)(T) &=& frac164\
hline\
vdots
endeqnarray



Now my question will be how do we find a closed form expression for the polynomials in question for arbitrary $d$.



Update: Note that we also have:
begineqnarray
mathcal S_d,0(T) &=&
d cdot 2^-1-fracd^24prodlimits_j=0^d/2-1 (T-d+1)^(d-2 j-2) (frac12)^(j) j!\
&=& d cdot 2^-1-fracd^24 left(prodlimits_j=0^d/2-1 (frac12)^(j)j!right) cdot prodlimits_k=2^d-1 (T-k)^lfloor frack2rfloor
endeqnarray



Below is a piece of Mathematica code that "proves" the above relationship.



d = 2 RandomInteger[1, 5]; T =.; p = 0;
mat = Flatten[
 Table[Table[q[xi], xi, 0, d/2 - 1], (-1)^(d/2) 1/2^
 Sum[q[xi], xi, 
 0, (d)/2 - 
 1] 1/(Times @@ (Tally[Table[q[xi], xi, 0, d/2 - 1]][[All, 
 2]]!))
 Sum[Det[
 Riffle[

 Table[Pochhammer[(T - d - 1)/2 + KD[2 xi + 1, j] p + 
 q[xi], (eta - q[xi])] If[q[xi] <= eta, 1, 0], xi, 0, 
 d/2 - 1, eta, 1, d],

 Table[1/2^(eta - 1)
 Pochhammer[T - d + 1, 
 p (KD[2 xi + 1, j] + KD[2 xi + 2, j]) + eta - 1 + q[xi] - 
 2], xi, 0, d/2 - 1, eta, 1, d]]
 ], j, 1, d]
 , Evaluate[
 Sequence @@ 
 Join[, 
 Table[q[xi], If[xi == 0, 2, q[xi - 1] + 1], d, xi, 0, 
 d/2 - 1]]]], d/2 - 1];
(Total[mat[[All, 2]]] - (d Product[
 Pochhammer[T - d + 1, d - 2 j - 2] Pochhammer[1/2, j] (j)!, j, 
 0, d/2 - 1] 2^(-1 - d^2/4))) // Simplify
MatrixForm[mat]


Again, my question is how do we go about proving this relationship rigorously.



Update 2:
By simplifying the relationship in Update 1 we end up with a following different relationship:
begineqnarray
&&sumlimits_2 le q_0 < cdots < q_d/2-1 le d (-1)^d^2/2
prodlimits_j=-d+1^-2 left[
frac(T+j)^sumlimits_xi=0^d/2-1 1_-2 xi-1 le j le q_xi-d-2 cdot 1_q_xige d-2 xi
(T+j)^sumlimits_xi=0^d/2-1 1_q_xi-d-1 le j le -2xi-2 cdot 1_q_xile d-2 xi
right] cdot left|
beginarrayr
left(prodlimits_j=q_xi^q_xi+eta-2(T-d-1+1 cdot j)right)_eta=1^d\
left(prodlimits_j=q_xi^eta-1 (T-d-1+2 cdot j)1_q_xi le etaright)_eta=1^d
endarray
right|_xi=0^d/2-1
=\
&&2^fracd^24-1 cdot prodlimits_xi=0^d/2-1 (frac12)^(xi) xi!
endeqnarray



For example if we take $d=4$ the left hand side reads:



enter image description here



which is independent of $T$ and reads $-4$.



Likewise for $d=6$ the left hand side reads:



enter image description here



which again is independent of $T$ and reads $192$.



Now, again how do we prove that relationship rigorously?




summation determinant




share|cite|improve this question

























    up vote
    2
    down vote

    favorite












    Let $d in 2mathbb N$ be an even positive integer, let $pin mathbb N$ be another positive integer and let $T$ be a parameter.
    Let $vecq := left( q_xi right)_xi=0^d/2-1 in mathbb N^(d/2)$.
    Consider a following multiple sum:
    begineqnarray
    &&mathcal S_d,p(T):=sumlimits_j=1^d sumlimits_2 le q_0 le cdots le q_d/2-1 le d frac12^ cdot nu(vecq)
    left|
    beginarrayr
    left( (fracT-d-12 + delta_2 xi+1,j p+q_xi)^(eta-q_xi) 1_q_xi le etaright)_eta=1^d\
    left(2^-eta+1 (T-d+1)^(p(delta_2xi+1,j+delta_2xi+2,j)+eta+q_xi-3)right)_eta=1^d\
    endarray
    right|_xi=0^d/2-1
    endeqnarray
    where $nu(vecq) := prod_eta=2^d p_eta!$
    where
    beginequation
    vecq:= left(
    underbrace2,cdots,2_p_2,
    underbrace3,cdots,3_p_3,
    cdots,
    underbraced,cdots,d_p_d
    right)
    endequation
    Note that the term in the multiple sum is a determinant of a $dtimes d$ matrix whose entries are polynomials in $T$.
    We stumbled across this sum when solving question Spectral densities of finite dimensional sample covariance matrices .



    Now by using Mathematica we have discovered the following pattern:
    beginequation
    mathcal S_d,p(T)= sumlimits_xi=0^(2d-5) vee 0 (T-d+1)^(p+xi) cdot P_d,xi^(fracd^24-1-xi)(T)
    endequation
    where $P_d,xi^(fracd^24-1-xi)(T)$ is a polynomial in $T$ of order $(fracd^24-1-xi)$.



    Here the polynomials in question read:
    begineqnarray
    P_2,0^(0)(T) &=& frac12\
    hline\
    P_4,0^(3)(T) &=& frac1128 (T-3) left(2 p^2 T-2 p^2-p T+3 p+2 T^2-6 T+4right)\
    P_4,1^(2)(T) &=& frac1128 left(-4 p^2 T+8 p^2-3 p T+5 p-2 T^2+10 T-12right)\
    P_4,2^(1)(T) &=& frac164 left(p^2+2 p-T+3right)\
    P_4,3^(0)(T) &=& frac164\
    hline\
    vdots
    endeqnarray



    Now my question will be how do we find a closed form expression for the polynomials in question for arbitrary $d$.



    Update: Note that we also have:
    begineqnarray
    mathcal S_d,0(T) &=&
    d cdot 2^-1-fracd^24prodlimits_j=0^d/2-1 (T-d+1)^(d-2 j-2) (frac12)^(j) j!\
    &=& d cdot 2^-1-fracd^24 left(prodlimits_j=0^d/2-1 (frac12)^(j)j!right) cdot prodlimits_k=2^d-1 (T-k)^lfloor frack2rfloor
    endeqnarray



    Below is a piece of Mathematica code that "proves" the above relationship.



    d = 2 RandomInteger[1, 5]; T =.; p = 0;
    mat = Flatten[
     Table[Table[q[xi], xi, 0, d/2 - 1], (-1)^(d/2) 1/2^
     Sum[q[xi], xi, 
     0, (d)/2 - 
     1] 1/(Times @@ (Tally[Table[q[xi], xi, 0, d/2 - 1]][[All, 
     2]]!))
     Sum[Det[
     Riffle[

     Table[Pochhammer[(T - d - 1)/2 + KD[2 xi + 1, j] p + 
     q[xi], (eta - q[xi])] If[q[xi] <= eta, 1, 0], xi, 0, 
     d/2 - 1, eta, 1, d],

     Table[1/2^(eta - 1)
     Pochhammer[T - d + 1, 
     p (KD[2 xi + 1, j] + KD[2 xi + 2, j]) + eta - 1 + q[xi] - 
     2], xi, 0, d/2 - 1, eta, 1, d]]
     ], j, 1, d]
     , Evaluate[
     Sequence @@ 
     Join[, 
     Table[q[xi], If[xi == 0, 2, q[xi - 1] + 1], d, xi, 0, 
     d/2 - 1]]]], d/2 - 1];
    (Total[mat[[All, 2]]] - (d Product[
     Pochhammer[T - d + 1, d - 2 j - 2] Pochhammer[1/2, j] (j)!, j, 
     0, d/2 - 1] 2^(-1 - d^2/4))) // Simplify
    MatrixForm[mat]


    Again, my question is how do we go about proving this relationship rigorously.



    Update 2:
    By simplifying the relationship in Update 1 we end up with a following different relationship:
    begineqnarray
    &&sumlimits_2 le q_0 < cdots < q_d/2-1 le d (-1)^d^2/2
    prodlimits_j=-d+1^-2 left[
    frac(T+j)^sumlimits_xi=0^d/2-1 1_-2 xi-1 le j le q_xi-d-2 cdot 1_q_xige d-2 xi
    (T+j)^sumlimits_xi=0^d/2-1 1_q_xi-d-1 le j le -2xi-2 cdot 1_q_xile d-2 xi
    right] cdot left|
    beginarrayr
    left(prodlimits_j=q_xi^q_xi+eta-2(T-d-1+1 cdot j)right)_eta=1^d\
    left(prodlimits_j=q_xi^eta-1 (T-d-1+2 cdot j)1_q_xi le etaright)_eta=1^d
    endarray
    right|_xi=0^d/2-1
    =\
    &&2^fracd^24-1 cdot prodlimits_xi=0^d/2-1 (frac12)^(xi) xi!
    endeqnarray



    For example if we take $d=4$ the left hand side reads:



    enter image description here



    which is independent of $T$ and reads $-4$.



    Likewise for $d=6$ the left hand side reads:



    enter image description here



    which again is independent of $T$ and reads $192$.



    Now, again how do we prove that relationship rigorously?




    summation determinant




    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $d in 2mathbb N$ be an even positive integer, let $pin mathbb N$ be another positive integer and let $T$ be a parameter.
      Let $vecq := left( q_xi right)_xi=0^d/2-1 in mathbb N^(d/2)$.
      Consider a following multiple sum:
      begineqnarray
      &&mathcal S_d,p(T):=sumlimits_j=1^d sumlimits_2 le q_0 le cdots le q_d/2-1 le d frac12^ cdot nu(vecq)
      left|
      beginarrayr
      left( (fracT-d-12 + delta_2 xi+1,j p+q_xi)^(eta-q_xi) 1_q_xi le etaright)_eta=1^d\
      left(2^-eta+1 (T-d+1)^(p(delta_2xi+1,j+delta_2xi+2,j)+eta+q_xi-3)right)_eta=1^d\
      endarray
      right|_xi=0^d/2-1
      endeqnarray
      where $nu(vecq) := prod_eta=2^d p_eta!$
      where
      beginequation
      vecq:= left(
      underbrace2,cdots,2_p_2,
      underbrace3,cdots,3_p_3,
      cdots,
      underbraced,cdots,d_p_d
      right)
      endequation
      Note that the term in the multiple sum is a determinant of a $dtimes d$ matrix whose entries are polynomials in $T$.
      We stumbled across this sum when solving question Spectral densities of finite dimensional sample covariance matrices .



      Now by using Mathematica we have discovered the following pattern:
      beginequation
      mathcal S_d,p(T)= sumlimits_xi=0^(2d-5) vee 0 (T-d+1)^(p+xi) cdot P_d,xi^(fracd^24-1-xi)(T)
      endequation
      where $P_d,xi^(fracd^24-1-xi)(T)$ is a polynomial in $T$ of order $(fracd^24-1-xi)$.



      Here the polynomials in question read:
      begineqnarray
      P_2,0^(0)(T) &=& frac12\
      hline\
      P_4,0^(3)(T) &=& frac1128 (T-3) left(2 p^2 T-2 p^2-p T+3 p+2 T^2-6 T+4right)\
      P_4,1^(2)(T) &=& frac1128 left(-4 p^2 T+8 p^2-3 p T+5 p-2 T^2+10 T-12right)\
      P_4,2^(1)(T) &=& frac164 left(p^2+2 p-T+3right)\
      P_4,3^(0)(T) &=& frac164\
      hline\
      vdots
      endeqnarray



      Now my question will be how do we find a closed form expression for the polynomials in question for arbitrary $d$.



      Update: Note that we also have:
      begineqnarray
      mathcal S_d,0(T) &=&
      d cdot 2^-1-fracd^24prodlimits_j=0^d/2-1 (T-d+1)^(d-2 j-2) (frac12)^(j) j!\
      &=& d cdot 2^-1-fracd^24 left(prodlimits_j=0^d/2-1 (frac12)^(j)j!right) cdot prodlimits_k=2^d-1 (T-k)^lfloor frack2rfloor
      endeqnarray



      Below is a piece of Mathematica code that "proves" the above relationship.



      d = 2 RandomInteger[1, 5]; T =.; p = 0;
      mat = Flatten[
       Table[Table[q[xi], xi, 0, d/2 - 1], (-1)^(d/2) 1/2^
       Sum[q[xi], xi, 
       0, (d)/2 - 
       1] 1/(Times @@ (Tally[Table[q[xi], xi, 0, d/2 - 1]][[All, 
       2]]!))
       Sum[Det[
       Riffle[

       Table[Pochhammer[(T - d - 1)/2 + KD[2 xi + 1, j] p + 
       q[xi], (eta - q[xi])] If[q[xi] <= eta, 1, 0], xi, 0, 
       d/2 - 1, eta, 1, d],

       Table[1/2^(eta - 1)
       Pochhammer[T - d + 1, 
       p (KD[2 xi + 1, j] + KD[2 xi + 2, j]) + eta - 1 + q[xi] - 
       2], xi, 0, d/2 - 1, eta, 1, d]]
       ], j, 1, d]
       , Evaluate[
       Sequence @@ 
       Join[, 
       Table[q[xi], If[xi == 0, 2, q[xi - 1] + 1], d, xi, 0, 
       d/2 - 1]]]], d/2 - 1];
      (Total[mat[[All, 2]]] - (d Product[
       Pochhammer[T - d + 1, d - 2 j - 2] Pochhammer[1/2, j] (j)!, j, 
       0, d/2 - 1] 2^(-1 - d^2/4))) // Simplify
      MatrixForm[mat]


      Again, my question is how do we go about proving this relationship rigorously.



      Update 2:
      By simplifying the relationship in Update 1 we end up with a following different relationship:
      begineqnarray
      &&sumlimits_2 le q_0 < cdots < q_d/2-1 le d (-1)^d^2/2
      prodlimits_j=-d+1^-2 left[
      frac(T+j)^sumlimits_xi=0^d/2-1 1_-2 xi-1 le j le q_xi-d-2 cdot 1_q_xige d-2 xi
      (T+j)^sumlimits_xi=0^d/2-1 1_q_xi-d-1 le j le -2xi-2 cdot 1_q_xile d-2 xi
      right] cdot left|
      beginarrayr
      left(prodlimits_j=q_xi^q_xi+eta-2(T-d-1+1 cdot j)right)_eta=1^d\
      left(prodlimits_j=q_xi^eta-1 (T-d-1+2 cdot j)1_q_xi le etaright)_eta=1^d
      endarray
      right|_xi=0^d/2-1
      =\
      &&2^fracd^24-1 cdot prodlimits_xi=0^d/2-1 (frac12)^(xi) xi!
      endeqnarray



      For example if we take $d=4$ the left hand side reads:



      enter image description here



      which is independent of $T$ and reads $-4$.



      Likewise for $d=6$ the left hand side reads:



      enter image description here



      which again is independent of $T$ and reads $192$.



      Now, again how do we prove that relationship rigorously?




      summation determinant




      share|cite|improve this question













      Let $d in 2mathbb N$ be an even positive integer, let $pin mathbb N$ be another positive integer and let $T$ be a parameter.
      Let $vecq := left( q_xi right)_xi=0^d/2-1 in mathbb N^(d/2)$.
      Consider a following multiple sum:
      begineqnarray
      &&mathcal S_d,p(T):=sumlimits_j=1^d sumlimits_2 le q_0 le cdots le q_d/2-1 le d frac12^ cdot nu(vecq)
      left|
      beginarrayr
      left( (fracT-d-12 + delta_2 xi+1,j p+q_xi)^(eta-q_xi) 1_q_xi le etaright)_eta=1^d\
      left(2^-eta+1 (T-d+1)^(p(delta_2xi+1,j+delta_2xi+2,j)+eta+q_xi-3)right)_eta=1^d\
      endarray
      right|_xi=0^d/2-1
      endeqnarray
      where $nu(vecq) := prod_eta=2^d p_eta!$
      where
      beginequation
      vecq:= left(
      underbrace2,cdots,2_p_2,
      underbrace3,cdots,3_p_3,
      cdots,
      underbraced,cdots,d_p_d
      right)
      endequation
      Note that the term in the multiple sum is a determinant of a $dtimes d$ matrix whose entries are polynomials in $T$.
      We stumbled across this sum when solving question Spectral densities of finite dimensional sample covariance matrices .



      Now by using Mathematica we have discovered the following pattern:
      beginequation
      mathcal S_d,p(T)= sumlimits_xi=0^(2d-5) vee 0 (T-d+1)^(p+xi) cdot P_d,xi^(fracd^24-1-xi)(T)
      endequation
      where $P_d,xi^(fracd^24-1-xi)(T)$ is a polynomial in $T$ of order $(fracd^24-1-xi)$.



      Here the polynomials in question read:
      begineqnarray
      P_2,0^(0)(T) &=& frac12\
      hline\
      P_4,0^(3)(T) &=& frac1128 (T-3) left(2 p^2 T-2 p^2-p T+3 p+2 T^2-6 T+4right)\
      P_4,1^(2)(T) &=& frac1128 left(-4 p^2 T+8 p^2-3 p T+5 p-2 T^2+10 T-12right)\
      P_4,2^(1)(T) &=& frac164 left(p^2+2 p-T+3right)\
      P_4,3^(0)(T) &=& frac164\
      hline\
      vdots
      endeqnarray



      Now my question will be how do we find a closed form expression for the polynomials in question for arbitrary $d$.



      Update: Note that we also have:
      begineqnarray
      mathcal S_d,0(T) &=&
      d cdot 2^-1-fracd^24prodlimits_j=0^d/2-1 (T-d+1)^(d-2 j-2) (frac12)^(j) j!\
      &=& d cdot 2^-1-fracd^24 left(prodlimits_j=0^d/2-1 (frac12)^(j)j!right) cdot prodlimits_k=2^d-1 (T-k)^lfloor frack2rfloor
      endeqnarray



      Below is a piece of Mathematica code that "proves" the above relationship.



      d = 2 RandomInteger[1, 5]; T =.; p = 0;
      mat = Flatten[
       Table[Table[q[xi], xi, 0, d/2 - 1], (-1)^(d/2) 1/2^
       Sum[q[xi], xi, 
       0, (d)/2 - 
       1] 1/(Times @@ (Tally[Table[q[xi], xi, 0, d/2 - 1]][[All, 
       2]]!))
       Sum[Det[
       Riffle[

       Table[Pochhammer[(T - d - 1)/2 + KD[2 xi + 1, j] p + 
       q[xi], (eta - q[xi])] If[q[xi] <= eta, 1, 0], xi, 0, 
       d/2 - 1, eta, 1, d],

       Table[1/2^(eta - 1)
       Pochhammer[T - d + 1, 
       p (KD[2 xi + 1, j] + KD[2 xi + 2, j]) + eta - 1 + q[xi] - 
       2], xi, 0, d/2 - 1, eta, 1, d]]
       ], j, 1, d]
       , Evaluate[
       Sequence @@ 
       Join[, 
       Table[q[xi], If[xi == 0, 2, q[xi - 1] + 1], d, xi, 0, 
       d/2 - 1]]]], d/2 - 1];
      (Total[mat[[All, 2]]] - (d Product[
       Pochhammer[T - d + 1, d - 2 j - 2] Pochhammer[1/2, j] (j)!, j, 
       0, d/2 - 1] 2^(-1 - d^2/4))) // Simplify
      MatrixForm[mat]


      Again, my question is how do we go about proving this relationship rigorously.



      Update 2:
      By simplifying the relationship in Update 1 we end up with a following different relationship:
      begineqnarray
      &&sumlimits_2 le q_0 < cdots < q_d/2-1 le d (-1)^d^2/2
      prodlimits_j=-d+1^-2 left[
      frac(T+j)^sumlimits_xi=0^d/2-1 1_-2 xi-1 le j le q_xi-d-2 cdot 1_q_xige d-2 xi
      (T+j)^sumlimits_xi=0^d/2-1 1_q_xi-d-1 le j le -2xi-2 cdot 1_q_xile d-2 xi
      right] cdot left|
      beginarrayr
      left(prodlimits_j=q_xi^q_xi+eta-2(T-d-1+1 cdot j)right)_eta=1^d\
      left(prodlimits_j=q_xi^eta-1 (T-d-1+2 cdot j)1_q_xi le etaright)_eta=1^d
      endarray
      right|_xi=0^d/2-1
      =\
      &&2^fracd^24-1 cdot prodlimits_xi=0^d/2-1 (frac12)^(xi) xi!
      endeqnarray



      For example if we take $d=4$ the left hand side reads:



      enter image description here



      which is independent of $T$ and reads $-4$.



      Likewise for $d=6$ the left hand side reads:



      enter image description here



      which again is independent of $T$ and reads $192$.



      Now, again how do we prove that relationship rigorously?





      summation determinant





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago
























      asked Jul 26 at 16:17









      Przemo

      3,916726




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