Value of Vandermonde type determinant












4












$begingroup$


Let $x_1,...,x_n $ are distinct real numbers.



Is it a formula for the Vandermonde type determinant $V(x_1, cdots,x_n)$ whose last column is $x_1^k, cdots, x_n^k$, where $k geq n$, instead of $x_1^{n-1}, cdots, x_n^{n-1}$?



Thanks






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  • $begingroup$
    All the other columns are the same as usual?
    $endgroup$
    – EuYu
    Nov 16 '12 at 15:01










  • $begingroup$
    Yes, the first $n-1$ column are the same.
    $endgroup$
    – Richard
    Nov 16 '12 at 15:01
















4












$begingroup$


Let $x_1,...,x_n $ are distinct real numbers.



Is it a formula for the Vandermonde type determinant $V(x_1, cdots,x_n)$ whose last column is $x_1^k, cdots, x_n^k$, where $k geq n$, instead of $x_1^{n-1}, cdots, x_n^{n-1}$?



Thanks






linear-algebra determinant







share|cite|improve this question











$endgroup$












  • $begingroup$
    All the other columns are the same as usual?
    $endgroup$
    – EuYu
    Nov 16 '12 at 15:01










  • $begingroup$
    Yes, the first $n-1$ column are the same.
    $endgroup$
    – Richard
    Nov 16 '12 at 15:01














4












4








4


2



$begingroup$


Let $x_1,...,x_n $ are distinct real numbers.



Is it a formula for the Vandermonde type determinant $V(x_1, cdots,x_n)$ whose last column is $x_1^k, cdots, x_n^k$, where $k geq n$, instead of $x_1^{n-1}, cdots, x_n^{n-1}$?



Thanks






linear-algebra determinant







share|cite|improve this question











$endgroup$




Let $x_1,...,x_n $ are distinct real numbers.



Is it a formula for the Vandermonde type determinant $V(x_1, cdots,x_n)$ whose last column is $x_1^k, cdots, x_n^k$, where $k geq n$, instead of $x_1^{n-1}, cdots, x_n^{n-1}$?



Thanks






linear-algebra determinant




linear-algebra determinant






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edited Nov 16 '12 at 15:18









uforoboa

4,89811829




4,89811829










asked Nov 16 '12 at 14:58









RichardRichard

1,77712046




1,77712046












  • $begingroup$
    All the other columns are the same as usual?
    $endgroup$
    – EuYu
    Nov 16 '12 at 15:01










  • $begingroup$
    Yes, the first $n-1$ column are the same.
    $endgroup$
    – Richard
    Nov 16 '12 at 15:01


















  • $begingroup$
    All the other columns are the same as usual?
    $endgroup$
    – EuYu
    Nov 16 '12 at 15:01










  • $begingroup$
    Yes, the first $n-1$ column are the same.
    $endgroup$
    – Richard
    Nov 16 '12 at 15:01
















$begingroup$
All the other columns are the same as usual?
$endgroup$
– EuYu
Nov 16 '12 at 15:01




$begingroup$
All the other columns are the same as usual?
$endgroup$
– EuYu
Nov 16 '12 at 15:01












$begingroup$
Yes, the first $n-1$ column are the same.
$endgroup$
– Richard
Nov 16 '12 at 15:01




$begingroup$
Yes, the first $n-1$ column are the same.
$endgroup$
– Richard
Nov 16 '12 at 15:01










4 Answers
4






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8












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Sure, at least you can find such a formula for any fixed $k geqslant n$. Not sure about a general formula for an unknown $k$ though.



Here is a hint: if all the $x_i$ are distinct, vector $(x_1^k,ldots,x_n^k)$ is a linear combination of vectors $(x_1^i,ldots,x_n^i)$, $i=0,,1,ldots,n-1$. If
$$
(x_1^k,ldots,x_n^k) = sum_{i=0}^{n-1} lambda_i (x_1^i,ldots,x_n^i),
$$
then your determinant is simply equal to $lambda_{n-1}$ times the usual Vandermonde determinant. So all you need to do is find $lambda_{n-1}$.






share|cite|improve this answer









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    5












    $begingroup$

    To continue Dan's answer, we want
    $$ x_i^k = sum_{j=0}^{n-1} lambda_j x_i^j, qquad 1 le i le n $$
    That is the polynomial $p(x) := sum_{j=0}^{n-1} lambda_j x^j$ interpolates $x^k$ at $x_0, ldots, x_{n-1}$. Lagrange interpolation gives
    $$ p(x) = sum_{j=0}^{n-1} x_j^k cdot prod_{ell ne j} frac{x-x_ell}{x_j - x_ell} $$
    $lambda_{n-1}$ is the coefficient of $x^{n-1}$, so it is
    $$ lambda_{n-1} = sum_{j=0}^{n-1} x_j^k prod_{ellne j} frac 1{x_j - x_ell}. $$






    share|cite|improve this answer









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      3












      $begingroup$

      Perform Laplace expansion along the last column. As the deletion of the $ell$-th row and the last column gives a $(n-1)times(n-1)$ Vandermonde matrix (in the original flavour), we get
      $$
      sum_{ell=1}^n (-1)^{ell+n} x_ell^kprod_{i<j,textrm{ and },i,jnot=ell}(x_j-x_i).
      $$






      share|cite|improve this answer









      $endgroup$





















        2












        $begingroup$

        Actually there is a general formula but not an easy way to describe it.



        Let me first "define" the polynomials $f_m(x_1,x_2,ldots,x_k)$ through some examples.




        • $f_0(x_1,x_2,x_3)=1.$

        • $f_1(x_1,x_2,x_3)=x_1+x_2+x_3.$

        • $f_2(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$

        • $f_3(x_1,x_2,x_3)=x_1^3+x_2^3+x_3^3+x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2+x_1x_2x_3.$

        • $f_3(x_1,x_2,x_3,x_4)=x_1^3+x_1^2x_2+x_1^2x_3+x_1^2x_4+x_1x_2^2+x_1x_2x_3+x_1x_2x_4+x_1x_3^2+x_1x_3x_4+x_1x_4^2+x_2^3+x_2^2x_3+x_2^2x_4+x_2x_3^2+x_2x_3x_4+x_2x_4^2+x_3^3+x_3^2x_4+x_3x_4^2+x_4^3
          .$

        • $ldots$


        Let $V(x_1,x_2,ldots,x_n,k)$ denote the Vandermonde type determinant whose last column is $x_1^k,x_2^k,ldots,x_n^k$ where $kgeq n-1$. So $V(x_1,x_2,ldots,x_n,n-1)$ is the usual Vandermonde. Then $$V(x_1,x_2,ldots,x_n,k)=V(x_1,x_2,ldots,x_n,n-1)cdot f_{k-n+1}(x_1,x_2,ldots,x_n).$$






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          4 Answers
          4






          active

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          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8












          $begingroup$

          Sure, at least you can find such a formula for any fixed $k geqslant n$. Not sure about a general formula for an unknown $k$ though.



          Here is a hint: if all the $x_i$ are distinct, vector $(x_1^k,ldots,x_n^k)$ is a linear combination of vectors $(x_1^i,ldots,x_n^i)$, $i=0,,1,ldots,n-1$. If
          $$
          (x_1^k,ldots,x_n^k) = sum_{i=0}^{n-1} lambda_i (x_1^i,ldots,x_n^i),
          $$
          then your determinant is simply equal to $lambda_{n-1}$ times the usual Vandermonde determinant. So all you need to do is find $lambda_{n-1}$.






          share|cite|improve this answer









          $endgroup$


















            8












            $begingroup$

            Sure, at least you can find such a formula for any fixed $k geqslant n$. Not sure about a general formula for an unknown $k$ though.



            Here is a hint: if all the $x_i$ are distinct, vector $(x_1^k,ldots,x_n^k)$ is a linear combination of vectors $(x_1^i,ldots,x_n^i)$, $i=0,,1,ldots,n-1$. If
            $$
            (x_1^k,ldots,x_n^k) = sum_{i=0}^{n-1} lambda_i (x_1^i,ldots,x_n^i),
            $$
            then your determinant is simply equal to $lambda_{n-1}$ times the usual Vandermonde determinant. So all you need to do is find $lambda_{n-1}$.






            share|cite|improve this answer









            $endgroup$
















              8












              8








              8





              $begingroup$

              Sure, at least you can find such a formula for any fixed $k geqslant n$. Not sure about a general formula for an unknown $k$ though.



              Here is a hint: if all the $x_i$ are distinct, vector $(x_1^k,ldots,x_n^k)$ is a linear combination of vectors $(x_1^i,ldots,x_n^i)$, $i=0,,1,ldots,n-1$. If
              $$
              (x_1^k,ldots,x_n^k) = sum_{i=0}^{n-1} lambda_i (x_1^i,ldots,x_n^i),
              $$
              then your determinant is simply equal to $lambda_{n-1}$ times the usual Vandermonde determinant. So all you need to do is find $lambda_{n-1}$.






              share|cite|improve this answer









              $endgroup$



              Sure, at least you can find such a formula for any fixed $k geqslant n$. Not sure about a general formula for an unknown $k$ though.



              Here is a hint: if all the $x_i$ are distinct, vector $(x_1^k,ldots,x_n^k)$ is a linear combination of vectors $(x_1^i,ldots,x_n^i)$, $i=0,,1,ldots,n-1$. If
              $$
              (x_1^k,ldots,x_n^k) = sum_{i=0}^{n-1} lambda_i (x_1^i,ldots,x_n^i),
              $$
              then your determinant is simply equal to $lambda_{n-1}$ times the usual Vandermonde determinant. So all you need to do is find $lambda_{n-1}$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 16 '12 at 15:17









              Dan ShvedDan Shved

              12.7k2045




              12.7k2045























                  5












                  $begingroup$

                  To continue Dan's answer, we want
                  $$ x_i^k = sum_{j=0}^{n-1} lambda_j x_i^j, qquad 1 le i le n $$
                  That is the polynomial $p(x) := sum_{j=0}^{n-1} lambda_j x^j$ interpolates $x^k$ at $x_0, ldots, x_{n-1}$. Lagrange interpolation gives
                  $$ p(x) = sum_{j=0}^{n-1} x_j^k cdot prod_{ell ne j} frac{x-x_ell}{x_j - x_ell} $$
                  $lambda_{n-1}$ is the coefficient of $x^{n-1}$, so it is
                  $$ lambda_{n-1} = sum_{j=0}^{n-1} x_j^k prod_{ellne j} frac 1{x_j - x_ell}. $$






                  share|cite|improve this answer









                  $endgroup$


















                    5












                    $begingroup$

                    To continue Dan's answer, we want
                    $$ x_i^k = sum_{j=0}^{n-1} lambda_j x_i^j, qquad 1 le i le n $$
                    That is the polynomial $p(x) := sum_{j=0}^{n-1} lambda_j x^j$ interpolates $x^k$ at $x_0, ldots, x_{n-1}$. Lagrange interpolation gives
                    $$ p(x) = sum_{j=0}^{n-1} x_j^k cdot prod_{ell ne j} frac{x-x_ell}{x_j - x_ell} $$
                    $lambda_{n-1}$ is the coefficient of $x^{n-1}$, so it is
                    $$ lambda_{n-1} = sum_{j=0}^{n-1} x_j^k prod_{ellne j} frac 1{x_j - x_ell}. $$






                    share|cite|improve this answer









                    $endgroup$
















                      5












                      5








                      5





                      $begingroup$

                      To continue Dan's answer, we want
                      $$ x_i^k = sum_{j=0}^{n-1} lambda_j x_i^j, qquad 1 le i le n $$
                      That is the polynomial $p(x) := sum_{j=0}^{n-1} lambda_j x^j$ interpolates $x^k$ at $x_0, ldots, x_{n-1}$. Lagrange interpolation gives
                      $$ p(x) = sum_{j=0}^{n-1} x_j^k cdot prod_{ell ne j} frac{x-x_ell}{x_j - x_ell} $$
                      $lambda_{n-1}$ is the coefficient of $x^{n-1}$, so it is
                      $$ lambda_{n-1} = sum_{j=0}^{n-1} x_j^k prod_{ellne j} frac 1{x_j - x_ell}. $$






                      share|cite|improve this answer









                      $endgroup$



                      To continue Dan's answer, we want
                      $$ x_i^k = sum_{j=0}^{n-1} lambda_j x_i^j, qquad 1 le i le n $$
                      That is the polynomial $p(x) := sum_{j=0}^{n-1} lambda_j x^j$ interpolates $x^k$ at $x_0, ldots, x_{n-1}$. Lagrange interpolation gives
                      $$ p(x) = sum_{j=0}^{n-1} x_j^k cdot prod_{ell ne j} frac{x-x_ell}{x_j - x_ell} $$
                      $lambda_{n-1}$ is the coefficient of $x^{n-1}$, so it is
                      $$ lambda_{n-1} = sum_{j=0}^{n-1} x_j^k prod_{ellne j} frac 1{x_j - x_ell}. $$







                      share|cite|improve this answer












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                      share|cite|improve this answer










                      answered Nov 16 '12 at 15:29









                      martinimartini

                      70.6k45991




                      70.6k45991























                          3












                          $begingroup$

                          Perform Laplace expansion along the last column. As the deletion of the $ell$-th row and the last column gives a $(n-1)times(n-1)$ Vandermonde matrix (in the original flavour), we get
                          $$
                          sum_{ell=1}^n (-1)^{ell+n} x_ell^kprod_{i<j,textrm{ and },i,jnot=ell}(x_j-x_i).
                          $$






                          share|cite|improve this answer









                          $endgroup$


















                            3












                            $begingroup$

                            Perform Laplace expansion along the last column. As the deletion of the $ell$-th row and the last column gives a $(n-1)times(n-1)$ Vandermonde matrix (in the original flavour), we get
                            $$
                            sum_{ell=1}^n (-1)^{ell+n} x_ell^kprod_{i<j,textrm{ and },i,jnot=ell}(x_j-x_i).
                            $$






                            share|cite|improve this answer









                            $endgroup$
















                              3












                              3








                              3





                              $begingroup$

                              Perform Laplace expansion along the last column. As the deletion of the $ell$-th row and the last column gives a $(n-1)times(n-1)$ Vandermonde matrix (in the original flavour), we get
                              $$
                              sum_{ell=1}^n (-1)^{ell+n} x_ell^kprod_{i<j,textrm{ and },i,jnot=ell}(x_j-x_i).
                              $$






                              share|cite|improve this answer









                              $endgroup$



                              Perform Laplace expansion along the last column. As the deletion of the $ell$-th row and the last column gives a $(n-1)times(n-1)$ Vandermonde matrix (in the original flavour), we get
                              $$
                              sum_{ell=1}^n (-1)^{ell+n} x_ell^kprod_{i<j,textrm{ and },i,jnot=ell}(x_j-x_i).
                              $$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 16 '12 at 15:39









                              user1551user1551

                              73.5k566129




                              73.5k566129























                                  2












                                  $begingroup$

                                  Actually there is a general formula but not an easy way to describe it.



                                  Let me first "define" the polynomials $f_m(x_1,x_2,ldots,x_k)$ through some examples.




                                  • $f_0(x_1,x_2,x_3)=1.$

                                  • $f_1(x_1,x_2,x_3)=x_1+x_2+x_3.$

                                  • $f_2(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$

                                  • $f_3(x_1,x_2,x_3)=x_1^3+x_2^3+x_3^3+x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2+x_1x_2x_3.$

                                  • $f_3(x_1,x_2,x_3,x_4)=x_1^3+x_1^2x_2+x_1^2x_3+x_1^2x_4+x_1x_2^2+x_1x_2x_3+x_1x_2x_4+x_1x_3^2+x_1x_3x_4+x_1x_4^2+x_2^3+x_2^2x_3+x_2^2x_4+x_2x_3^2+x_2x_3x_4+x_2x_4^2+x_3^3+x_3^2x_4+x_3x_4^2+x_4^3
                                    .$

                                  • $ldots$


                                  Let $V(x_1,x_2,ldots,x_n,k)$ denote the Vandermonde type determinant whose last column is $x_1^k,x_2^k,ldots,x_n^k$ where $kgeq n-1$. So $V(x_1,x_2,ldots,x_n,n-1)$ is the usual Vandermonde. Then $$V(x_1,x_2,ldots,x_n,k)=V(x_1,x_2,ldots,x_n,n-1)cdot f_{k-n+1}(x_1,x_2,ldots,x_n).$$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    2












                                    $begingroup$

                                    Actually there is a general formula but not an easy way to describe it.



                                    Let me first "define" the polynomials $f_m(x_1,x_2,ldots,x_k)$ through some examples.




                                    • $f_0(x_1,x_2,x_3)=1.$

                                    • $f_1(x_1,x_2,x_3)=x_1+x_2+x_3.$

                                    • $f_2(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$

                                    • $f_3(x_1,x_2,x_3)=x_1^3+x_2^3+x_3^3+x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2+x_1x_2x_3.$

                                    • $f_3(x_1,x_2,x_3,x_4)=x_1^3+x_1^2x_2+x_1^2x_3+x_1^2x_4+x_1x_2^2+x_1x_2x_3+x_1x_2x_4+x_1x_3^2+x_1x_3x_4+x_1x_4^2+x_2^3+x_2^2x_3+x_2^2x_4+x_2x_3^2+x_2x_3x_4+x_2x_4^2+x_3^3+x_3^2x_4+x_3x_4^2+x_4^3
                                      .$

                                    • $ldots$


                                    Let $V(x_1,x_2,ldots,x_n,k)$ denote the Vandermonde type determinant whose last column is $x_1^k,x_2^k,ldots,x_n^k$ where $kgeq n-1$. So $V(x_1,x_2,ldots,x_n,n-1)$ is the usual Vandermonde. Then $$V(x_1,x_2,ldots,x_n,k)=V(x_1,x_2,ldots,x_n,n-1)cdot f_{k-n+1}(x_1,x_2,ldots,x_n).$$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      Actually there is a general formula but not an easy way to describe it.



                                      Let me first "define" the polynomials $f_m(x_1,x_2,ldots,x_k)$ through some examples.




                                      • $f_0(x_1,x_2,x_3)=1.$

                                      • $f_1(x_1,x_2,x_3)=x_1+x_2+x_3.$

                                      • $f_2(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$

                                      • $f_3(x_1,x_2,x_3)=x_1^3+x_2^3+x_3^3+x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2+x_1x_2x_3.$

                                      • $f_3(x_1,x_2,x_3,x_4)=x_1^3+x_1^2x_2+x_1^2x_3+x_1^2x_4+x_1x_2^2+x_1x_2x_3+x_1x_2x_4+x_1x_3^2+x_1x_3x_4+x_1x_4^2+x_2^3+x_2^2x_3+x_2^2x_4+x_2x_3^2+x_2x_3x_4+x_2x_4^2+x_3^3+x_3^2x_4+x_3x_4^2+x_4^3
                                        .$

                                      • $ldots$


                                      Let $V(x_1,x_2,ldots,x_n,k)$ denote the Vandermonde type determinant whose last column is $x_1^k,x_2^k,ldots,x_n^k$ where $kgeq n-1$. So $V(x_1,x_2,ldots,x_n,n-1)$ is the usual Vandermonde. Then $$V(x_1,x_2,ldots,x_n,k)=V(x_1,x_2,ldots,x_n,n-1)cdot f_{k-n+1}(x_1,x_2,ldots,x_n).$$






                                      share|cite|improve this answer











                                      $endgroup$



                                      Actually there is a general formula but not an easy way to describe it.



                                      Let me first "define" the polynomials $f_m(x_1,x_2,ldots,x_k)$ through some examples.




                                      • $f_0(x_1,x_2,x_3)=1.$

                                      • $f_1(x_1,x_2,x_3)=x_1+x_2+x_3.$

                                      • $f_2(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$

                                      • $f_3(x_1,x_2,x_3)=x_1^3+x_2^3+x_3^3+x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2+x_1x_2x_3.$

                                      • $f_3(x_1,x_2,x_3,x_4)=x_1^3+x_1^2x_2+x_1^2x_3+x_1^2x_4+x_1x_2^2+x_1x_2x_3+x_1x_2x_4+x_1x_3^2+x_1x_3x_4+x_1x_4^2+x_2^3+x_2^2x_3+x_2^2x_4+x_2x_3^2+x_2x_3x_4+x_2x_4^2+x_3^3+x_3^2x_4+x_3x_4^2+x_4^3
                                        .$

                                      • $ldots$


                                      Let $V(x_1,x_2,ldots,x_n,k)$ denote the Vandermonde type determinant whose last column is $x_1^k,x_2^k,ldots,x_n^k$ where $kgeq n-1$. So $V(x_1,x_2,ldots,x_n,n-1)$ is the usual Vandermonde. Then $$V(x_1,x_2,ldots,x_n,k)=V(x_1,x_2,ldots,x_n,n-1)cdot f_{k-n+1}(x_1,x_2,ldots,x_n).$$







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                                      edited Jan 21 at 18:52

























                                      answered Nov 16 '12 at 16:41









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