Request for different proofs












2














Let $eta=e^{frac{2pi i}n}$, an $n$-th root of unity. For pedagogical reasons and inspiration, I ask to see different proofs (be it elementary, sophisticated, theoretical, etc) for the following product evaluation.




If $T(n)=frac{(3n-2)(n-1)}2$ and $i=sqrt{-1}$ then
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$







gr.group-theory cv.complex-variables soft-question teaching elementary-proofs







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  • How precise would you like the conclusion? Would getting the norm and the parity of $T(n)$ be enough, or do you want $T(n)$ exactly?
    – user44191
    6 hours ago










  • It'd be nice to get it exactly, however even getting to what you described does add insight into the discussion. So, you're welcome to present it.
    – T. Amdeberhan
    6 hours ago








  • 2




    can you change to a more specific title?
    – YCor
    4 hours ago
















2














Let $eta=e^{frac{2pi i}n}$, an $n$-th root of unity. For pedagogical reasons and inspiration, I ask to see different proofs (be it elementary, sophisticated, theoretical, etc) for the following product evaluation.




If $T(n)=frac{(3n-2)(n-1)}2$ and $i=sqrt{-1}$ then
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$







gr.group-theory cv.complex-variables soft-question teaching elementary-proofs







share|cite|improve this question
























  • How precise would you like the conclusion? Would getting the norm and the parity of $T(n)$ be enough, or do you want $T(n)$ exactly?
    – user44191
    6 hours ago










  • It'd be nice to get it exactly, however even getting to what you described does add insight into the discussion. So, you're welcome to present it.
    – T. Amdeberhan
    6 hours ago








  • 2




    can you change to a more specific title?
    – YCor
    4 hours ago














2












2








2







Let $eta=e^{frac{2pi i}n}$, an $n$-th root of unity. For pedagogical reasons and inspiration, I ask to see different proofs (be it elementary, sophisticated, theoretical, etc) for the following product evaluation.




If $T(n)=frac{(3n-2)(n-1)}2$ and $i=sqrt{-1}$ then
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$







gr.group-theory cv.complex-variables soft-question teaching elementary-proofs







share|cite|improve this question















Let $eta=e^{frac{2pi i}n}$, an $n$-th root of unity. For pedagogical reasons and inspiration, I ask to see different proofs (be it elementary, sophisticated, theoretical, etc) for the following product evaluation.




If $T(n)=frac{(3n-2)(n-1)}2$ and $i=sqrt{-1}$ then
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$







gr.group-theory cv.complex-variables soft-question teaching elementary-proofs




gr.group-theory cv.complex-variables soft-question teaching elementary-proofs






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share|cite|improve this question













share|cite|improve this question




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edited 6 hours ago

























asked 6 hours ago









T. Amdeberhan

17k228126




17k228126












  • How precise would you like the conclusion? Would getting the norm and the parity of $T(n)$ be enough, or do you want $T(n)$ exactly?
    – user44191
    6 hours ago










  • It'd be nice to get it exactly, however even getting to what you described does add insight into the discussion. So, you're welcome to present it.
    – T. Amdeberhan
    6 hours ago








  • 2




    can you change to a more specific title?
    – YCor
    4 hours ago


















  • How precise would you like the conclusion? Would getting the norm and the parity of $T(n)$ be enough, or do you want $T(n)$ exactly?
    – user44191
    6 hours ago










  • It'd be nice to get it exactly, however even getting to what you described does add insight into the discussion. So, you're welcome to present it.
    – T. Amdeberhan
    6 hours ago








  • 2




    can you change to a more specific title?
    – YCor
    4 hours ago
















How precise would you like the conclusion? Would getting the norm and the parity of $T(n)$ be enough, or do you want $T(n)$ exactly?
– user44191
6 hours ago




How precise would you like the conclusion? Would getting the norm and the parity of $T(n)$ be enough, or do you want $T(n)$ exactly?
– user44191
6 hours ago












It'd be nice to get it exactly, however even getting to what you described does add insight into the discussion. So, you're welcome to present it.
– T. Amdeberhan
6 hours ago






It'd be nice to get it exactly, however even getting to what you described does add insight into the discussion. So, you're welcome to present it.
– T. Amdeberhan
6 hours ago






2




2




can you change to a more specific title?
– YCor
4 hours ago




can you change to a more specific title?
– YCor
4 hours ago










2 Answers
2






active

oldest

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4














Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.






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    3














    We first find the norm; we then determine the argument.



    Call the product you wrote $A_n$. Then $A_n^2 = prod_{j<k}^{0,n-1} (eta^k - eta^j)^2 = Disc(x^n - 1) = (-1)^{frac{n (n -1)}{2}}Res(x^n - 1, n x^{n - 1})$



    $= (-1)^{frac{n(n-1)}{2}} n^n prod_{0 leq i < n, 0 leq j < n-1} (eta^i - 0)$



    All terms in the expression except $n^n$ have norm $1$, so we have that $|A_n| = n^{frac{n}{2}}$. We therefore only need to figure out the argument of $A_n$.



    Let $eta' = e^frac{2 pi i}{2n}$ be the square root of $eta$. We can rewrite $A_n = prod_{0leq j<k<n} eta'^{k + j} (eta'^{k - j} - eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates with positive imaginary part, and therefore will always have argument $frac{pi}{2}$. So let us concentrate on the argument of the first term, $prod_{0 leq j < k < n} eta'^{k +j}$. We can do this by finding $sum_{0 leq j < k < n} j + k$.



    $sum_{0 leq j < k < n} j + k = left(sum_{0 leq j < k < n} jright) + left(sum_{0 leq j < k < n} kright)$



    $= left(sum_{0 leq j <n} (n - j - 1)jright) + left(sum_{0leq k<n} k*kright)$



    $= sum_{0 leq j < n} (n - j - 1)j + j*j = sum_{0 leq j < n} (n - 1)j$



    $= (n - 1) frac{n (n - 1)}{2}$



    We therefore end up with an argument of $frac{n(n - 1)}{2} frac{pi}{2} + frac{n (n - 1)^2}{2} frac{2 pi}{2n} = frac{(3n^2 - 5n + 2)pi}{4}$. We finally have that:



    The norm of $A_n$ is $n^frac{n}{2}$, and the argument is $frac{(3n^2 - 5n + 2)pi}{4}$. Correspondingly, we have that $A_n = n^{frac{n}{2}} i^{T(n)}$, as desired.






    share|cite|improve this answer























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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      4














      Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.






      share|cite|improve this answer


























        4














        Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.






        share|cite|improve this answer
























          4












          4








          4






          Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.






          share|cite|improve this answer












          Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          Alexey Ustinov

          6,68745778




          6,68745778























              3














              We first find the norm; we then determine the argument.



              Call the product you wrote $A_n$. Then $A_n^2 = prod_{j<k}^{0,n-1} (eta^k - eta^j)^2 = Disc(x^n - 1) = (-1)^{frac{n (n -1)}{2}}Res(x^n - 1, n x^{n - 1})$



              $= (-1)^{frac{n(n-1)}{2}} n^n prod_{0 leq i < n, 0 leq j < n-1} (eta^i - 0)$



              All terms in the expression except $n^n$ have norm $1$, so we have that $|A_n| = n^{frac{n}{2}}$. We therefore only need to figure out the argument of $A_n$.



              Let $eta' = e^frac{2 pi i}{2n}$ be the square root of $eta$. We can rewrite $A_n = prod_{0leq j<k<n} eta'^{k + j} (eta'^{k - j} - eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates with positive imaginary part, and therefore will always have argument $frac{pi}{2}$. So let us concentrate on the argument of the first term, $prod_{0 leq j < k < n} eta'^{k +j}$. We can do this by finding $sum_{0 leq j < k < n} j + k$.



              $sum_{0 leq j < k < n} j + k = left(sum_{0 leq j < k < n} jright) + left(sum_{0 leq j < k < n} kright)$



              $= left(sum_{0 leq j <n} (n - j - 1)jright) + left(sum_{0leq k<n} k*kright)$



              $= sum_{0 leq j < n} (n - j - 1)j + j*j = sum_{0 leq j < n} (n - 1)j$



              $= (n - 1) frac{n (n - 1)}{2}$



              We therefore end up with an argument of $frac{n(n - 1)}{2} frac{pi}{2} + frac{n (n - 1)^2}{2} frac{2 pi}{2n} = frac{(3n^2 - 5n + 2)pi}{4}$. We finally have that:



              The norm of $A_n$ is $n^frac{n}{2}$, and the argument is $frac{(3n^2 - 5n + 2)pi}{4}$. Correspondingly, we have that $A_n = n^{frac{n}{2}} i^{T(n)}$, as desired.






              share|cite|improve this answer




























                3














                We first find the norm; we then determine the argument.



                Call the product you wrote $A_n$. Then $A_n^2 = prod_{j<k}^{0,n-1} (eta^k - eta^j)^2 = Disc(x^n - 1) = (-1)^{frac{n (n -1)}{2}}Res(x^n - 1, n x^{n - 1})$



                $= (-1)^{frac{n(n-1)}{2}} n^n prod_{0 leq i < n, 0 leq j < n-1} (eta^i - 0)$



                All terms in the expression except $n^n$ have norm $1$, so we have that $|A_n| = n^{frac{n}{2}}$. We therefore only need to figure out the argument of $A_n$.



                Let $eta' = e^frac{2 pi i}{2n}$ be the square root of $eta$. We can rewrite $A_n = prod_{0leq j<k<n} eta'^{k + j} (eta'^{k - j} - eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates with positive imaginary part, and therefore will always have argument $frac{pi}{2}$. So let us concentrate on the argument of the first term, $prod_{0 leq j < k < n} eta'^{k +j}$. We can do this by finding $sum_{0 leq j < k < n} j + k$.



                $sum_{0 leq j < k < n} j + k = left(sum_{0 leq j < k < n} jright) + left(sum_{0 leq j < k < n} kright)$



                $= left(sum_{0 leq j <n} (n - j - 1)jright) + left(sum_{0leq k<n} k*kright)$



                $= sum_{0 leq j < n} (n - j - 1)j + j*j = sum_{0 leq j < n} (n - 1)j$



                $= (n - 1) frac{n (n - 1)}{2}$



                We therefore end up with an argument of $frac{n(n - 1)}{2} frac{pi}{2} + frac{n (n - 1)^2}{2} frac{2 pi}{2n} = frac{(3n^2 - 5n + 2)pi}{4}$. We finally have that:



                The norm of $A_n$ is $n^frac{n}{2}$, and the argument is $frac{(3n^2 - 5n + 2)pi}{4}$. Correspondingly, we have that $A_n = n^{frac{n}{2}} i^{T(n)}$, as desired.






                share|cite|improve this answer


























                  3












                  3








                  3






                  We first find the norm; we then determine the argument.



                  Call the product you wrote $A_n$. Then $A_n^2 = prod_{j<k}^{0,n-1} (eta^k - eta^j)^2 = Disc(x^n - 1) = (-1)^{frac{n (n -1)}{2}}Res(x^n - 1, n x^{n - 1})$



                  $= (-1)^{frac{n(n-1)}{2}} n^n prod_{0 leq i < n, 0 leq j < n-1} (eta^i - 0)$



                  All terms in the expression except $n^n$ have norm $1$, so we have that $|A_n| = n^{frac{n}{2}}$. We therefore only need to figure out the argument of $A_n$.



                  Let $eta' = e^frac{2 pi i}{2n}$ be the square root of $eta$. We can rewrite $A_n = prod_{0leq j<k<n} eta'^{k + j} (eta'^{k - j} - eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates with positive imaginary part, and therefore will always have argument $frac{pi}{2}$. So let us concentrate on the argument of the first term, $prod_{0 leq j < k < n} eta'^{k +j}$. We can do this by finding $sum_{0 leq j < k < n} j + k$.



                  $sum_{0 leq j < k < n} j + k = left(sum_{0 leq j < k < n} jright) + left(sum_{0 leq j < k < n} kright)$



                  $= left(sum_{0 leq j <n} (n - j - 1)jright) + left(sum_{0leq k<n} k*kright)$



                  $= sum_{0 leq j < n} (n - j - 1)j + j*j = sum_{0 leq j < n} (n - 1)j$



                  $= (n - 1) frac{n (n - 1)}{2}$



                  We therefore end up with an argument of $frac{n(n - 1)}{2} frac{pi}{2} + frac{n (n - 1)^2}{2} frac{2 pi}{2n} = frac{(3n^2 - 5n + 2)pi}{4}$. We finally have that:



                  The norm of $A_n$ is $n^frac{n}{2}$, and the argument is $frac{(3n^2 - 5n + 2)pi}{4}$. Correspondingly, we have that $A_n = n^{frac{n}{2}} i^{T(n)}$, as desired.






                  share|cite|improve this answer














                  We first find the norm; we then determine the argument.



                  Call the product you wrote $A_n$. Then $A_n^2 = prod_{j<k}^{0,n-1} (eta^k - eta^j)^2 = Disc(x^n - 1) = (-1)^{frac{n (n -1)}{2}}Res(x^n - 1, n x^{n - 1})$



                  $= (-1)^{frac{n(n-1)}{2}} n^n prod_{0 leq i < n, 0 leq j < n-1} (eta^i - 0)$



                  All terms in the expression except $n^n$ have norm $1$, so we have that $|A_n| = n^{frac{n}{2}}$. We therefore only need to figure out the argument of $A_n$.



                  Let $eta' = e^frac{2 pi i}{2n}$ be the square root of $eta$. We can rewrite $A_n = prod_{0leq j<k<n} eta'^{k + j} (eta'^{k - j} - eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates with positive imaginary part, and therefore will always have argument $frac{pi}{2}$. So let us concentrate on the argument of the first term, $prod_{0 leq j < k < n} eta'^{k +j}$. We can do this by finding $sum_{0 leq j < k < n} j + k$.



                  $sum_{0 leq j < k < n} j + k = left(sum_{0 leq j < k < n} jright) + left(sum_{0 leq j < k < n} kright)$



                  $= left(sum_{0 leq j <n} (n - j - 1)jright) + left(sum_{0leq k<n} k*kright)$



                  $= sum_{0 leq j < n} (n - j - 1)j + j*j = sum_{0 leq j < n} (n - 1)j$



                  $= (n - 1) frac{n (n - 1)}{2}$



                  We therefore end up with an argument of $frac{n(n - 1)}{2} frac{pi}{2} + frac{n (n - 1)^2}{2} frac{2 pi}{2n} = frac{(3n^2 - 5n + 2)pi}{4}$. We finally have that:



                  The norm of $A_n$ is $n^frac{n}{2}$, and the argument is $frac{(3n^2 - 5n + 2)pi}{4}$. Correspondingly, we have that $A_n = n^{frac{n}{2}} i^{T(n)}$, as desired.







                  share|cite|improve this answer














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                  share|cite|improve this answer








                  answered 5 hours ago


























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