How do we simplify a multiple sum that involves determinants?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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












    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




      3,916726

























          active

          oldest

          votes











          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );








           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863562%2fhow-do-we-simplify-a-multiple-sum-that-involves-determinants%23new-answer', 'question_page');

          );

          Post as a guest






















          active

          oldest

          votes













          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes










           

          draft saved


          draft discarded


























           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863562%2fhow-do-we-simplify-a-multiple-sum-that-involves-determinants%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          Color the edges and diagonals of a regular polygon

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne
          Read more

          Relationship between determinant of matrix and determinant of adjoint?

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/… – Hector Blandin Nov 17 &
          Read more

          What is the equation of a 3D cone with generalised tilt?

          Image
          Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
          Read more