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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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

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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.







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                  share|cite|improve this answer








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                      広島県道265号伴広島線

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                      一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
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                      Setup Asymptote in Texstudio

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                      0 Good time of the day! I am trying to compile the following file: documentclass[12pt]{article} usepackage[inline]{asymptote} begin{document} Here is a venn diagram produced with Asymptote, drawn to width 4cm: defA{A} defB{V{B}} begin{figure} begin{center} begin{asy} size(4cm,0); pen colour1=red; pen colour2=green; pair z0=(0,0); pair z1=(-1,0); pair z2=(1,0); real r=1.5; path c1=circle(z1,r); path c2=circle(z2,r); fill(c1,colour1); fill(c2,colour2); picture intersection=new picture; fill(intersection,c1,colour1+colour2); clip(intersection,c2); add(intersection); draw(c1); draw(c2); //draw("$A$",box,z1)
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