Prove $sum_{k=1}^m cot^2 kpi/(2m+1)=m(2m-1)/3$












3












$begingroup$



Prove that $$ sum_{k=1}^m cot^2 frac{kpi}{2m+1}=frac{m(2m-1)}{3} $$




I have tried to use $$sinleft((2m+1)xright)=
left(sin^{2m+1}xright) cdot left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}left(cot^2xright)^{m-j}right)$$
and induction without any success. Thanks for any help!






real-analysis calculus trigonometric-series







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$endgroup$








  • 1




    $begingroup$
    A few good hints here
    $endgroup$
    – rtybase
    Jan 7 at 23:13










  • $begingroup$
    Another hint contained in this question (the polynomial development part, see the question, not the answer).
    $endgroup$
    – rtybase
    Jan 7 at 23:32






  • 2




    $begingroup$
    I guess you can find a copy of Cauchy's Cours d'Analyse - this is the key ingredient of Cauchy's proof of $zeta(2)=frac{pi^2}{6}$. It can be proved in many ways, for instance by applying Vieta's formulas to Chebyshev polynomials of the second kind. You may have a look at my notes too, in the section about the Basel problem.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 23:40


















3












$begingroup$



Prove that $$ sum_{k=1}^m cot^2 frac{kpi}{2m+1}=frac{m(2m-1)}{3} $$




I have tried to use $$sinleft((2m+1)xright)=
left(sin^{2m+1}xright) cdot left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}left(cot^2xright)^{m-j}right)$$
and induction without any success. Thanks for any help!






real-analysis calculus trigonometric-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    A few good hints here
    $endgroup$
    – rtybase
    Jan 7 at 23:13










  • $begingroup$
    Another hint contained in this question (the polynomial development part, see the question, not the answer).
    $endgroup$
    – rtybase
    Jan 7 at 23:32






  • 2




    $begingroup$
    I guess you can find a copy of Cauchy's Cours d'Analyse - this is the key ingredient of Cauchy's proof of $zeta(2)=frac{pi^2}{6}$. It can be proved in many ways, for instance by applying Vieta's formulas to Chebyshev polynomials of the second kind. You may have a look at my notes too, in the section about the Basel problem.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 23:40
















3












3








3


1



$begingroup$



Prove that $$ sum_{k=1}^m cot^2 frac{kpi}{2m+1}=frac{m(2m-1)}{3} $$




I have tried to use $$sinleft((2m+1)xright)=
left(sin^{2m+1}xright) cdot left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}left(cot^2xright)^{m-j}right)$$
and induction without any success. Thanks for any help!






real-analysis calculus trigonometric-series







share|cite|improve this question











$endgroup$





Prove that $$ sum_{k=1}^m cot^2 frac{kpi}{2m+1}=frac{m(2m-1)}{3} $$




I have tried to use $$sinleft((2m+1)xright)=
left(sin^{2m+1}xright) cdot left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}left(cot^2xright)^{m-j}right)$$
and induction without any success. Thanks for any help!






real-analysis calculus trigonometric-series




real-analysis calculus trigonometric-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 0:06









rtybase

10.8k21533




10.8k21533










asked Jan 7 at 23:03









user64066user64066

1,643717




1,643717








  • 1




    $begingroup$
    A few good hints here
    $endgroup$
    – rtybase
    Jan 7 at 23:13










  • $begingroup$
    Another hint contained in this question (the polynomial development part, see the question, not the answer).
    $endgroup$
    – rtybase
    Jan 7 at 23:32






  • 2




    $begingroup$
    I guess you can find a copy of Cauchy's Cours d'Analyse - this is the key ingredient of Cauchy's proof of $zeta(2)=frac{pi^2}{6}$. It can be proved in many ways, for instance by applying Vieta's formulas to Chebyshev polynomials of the second kind. You may have a look at my notes too, in the section about the Basel problem.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 23:40
















  • 1




    $begingroup$
    A few good hints here
    $endgroup$
    – rtybase
    Jan 7 at 23:13










  • $begingroup$
    Another hint contained in this question (the polynomial development part, see the question, not the answer).
    $endgroup$
    – rtybase
    Jan 7 at 23:32






  • 2




    $begingroup$
    I guess you can find a copy of Cauchy's Cours d'Analyse - this is the key ingredient of Cauchy's proof of $zeta(2)=frac{pi^2}{6}$. It can be proved in many ways, for instance by applying Vieta's formulas to Chebyshev polynomials of the second kind. You may have a look at my notes too, in the section about the Basel problem.
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 23:40










1




1




$begingroup$
A few good hints here
$endgroup$
– rtybase
Jan 7 at 23:13




$begingroup$
A few good hints here
$endgroup$
– rtybase
Jan 7 at 23:13












$begingroup$
Another hint contained in this question (the polynomial development part, see the question, not the answer).
$endgroup$
– rtybase
Jan 7 at 23:32




$begingroup$
Another hint contained in this question (the polynomial development part, see the question, not the answer).
$endgroup$
– rtybase
Jan 7 at 23:32




2




2




$begingroup$
I guess you can find a copy of Cauchy's Cours d'Analyse - this is the key ingredient of Cauchy's proof of $zeta(2)=frac{pi^2}{6}$. It can be proved in many ways, for instance by applying Vieta's formulas to Chebyshev polynomials of the second kind. You may have a look at my notes too, in the section about the Basel problem.
$endgroup$
– Jack D'Aurizio
Jan 7 at 23:40






$begingroup$
I guess you can find a copy of Cauchy's Cours d'Analyse - this is the key ingredient of Cauchy's proof of $zeta(2)=frac{pi^2}{6}$. It can be proved in many ways, for instance by applying Vieta's formulas to Chebyshev polynomials of the second kind. You may have a look at my notes too, in the section about the Basel problem.
$endgroup$
– Jack D'Aurizio
Jan 7 at 23:40












2 Answers
2






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oldest

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$begingroup$

Further to my last comment and given you used
$$sinleft((2m+1)xright)=
left(sin^{2m+1}xright) left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}(cot^2x)^{m-j}right)=
left(sin^{2m+1}xright) cdot Pleft(cot^2{x}right)tag{1}$$
where
$$P(x)=sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}x^{m-j}=binom{2m+1}{1}x^m-binom{2m+1}{3}x^{m-1}+...$$
is a polynomial of degree $m$, with (easy to see from $(1)$ since $sinleft((2m+1)frac{kpi}{2m+1}right)=0$) $cot^2{frac{kpi}{2m+1}}, k=1..m$ as roots. Using Vieta's formulas
$$sum_{k=1}^m cot^2{frac{kpi}{2m+1}}=-frac{-binom{2m+1}{3}}{binom{2m+1}{1}}=frac{(2m+1)(2m)(2m-1)}{(2m+1)cdot 2 cdot 3}=frac{m(2m-1)}{3}$$






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$endgroup$





















    1












    $begingroup$

    Hint:
    The $-icotleft(frac{kpi}{2m+1}right)$, for $1 leq |k| leq m$, are the distinct roots of $(X+1)^{2m+1}-(X-1)^{2m+1}$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






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      active

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      active

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      2












      $begingroup$

      Further to my last comment and given you used
      $$sinleft((2m+1)xright)=
      left(sin^{2m+1}xright) left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}(cot^2x)^{m-j}right)=
      left(sin^{2m+1}xright) cdot Pleft(cot^2{x}right)tag{1}$$
      where
      $$P(x)=sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}x^{m-j}=binom{2m+1}{1}x^m-binom{2m+1}{3}x^{m-1}+...$$
      is a polynomial of degree $m$, with (easy to see from $(1)$ since $sinleft((2m+1)frac{kpi}{2m+1}right)=0$) $cot^2{frac{kpi}{2m+1}}, k=1..m$ as roots. Using Vieta's formulas
      $$sum_{k=1}^m cot^2{frac{kpi}{2m+1}}=-frac{-binom{2m+1}{3}}{binom{2m+1}{1}}=frac{(2m+1)(2m)(2m-1)}{(2m+1)cdot 2 cdot 3}=frac{m(2m-1)}{3}$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Further to my last comment and given you used
        $$sinleft((2m+1)xright)=
        left(sin^{2m+1}xright) left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}(cot^2x)^{m-j}right)=
        left(sin^{2m+1}xright) cdot Pleft(cot^2{x}right)tag{1}$$
        where
        $$P(x)=sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}x^{m-j}=binom{2m+1}{1}x^m-binom{2m+1}{3}x^{m-1}+...$$
        is a polynomial of degree $m$, with (easy to see from $(1)$ since $sinleft((2m+1)frac{kpi}{2m+1}right)=0$) $cot^2{frac{kpi}{2m+1}}, k=1..m$ as roots. Using Vieta's formulas
        $$sum_{k=1}^m cot^2{frac{kpi}{2m+1}}=-frac{-binom{2m+1}{3}}{binom{2m+1}{1}}=frac{(2m+1)(2m)(2m-1)}{(2m+1)cdot 2 cdot 3}=frac{m(2m-1)}{3}$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Further to my last comment and given you used
          $$sinleft((2m+1)xright)=
          left(sin^{2m+1}xright) left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}(cot^2x)^{m-j}right)=
          left(sin^{2m+1}xright) cdot Pleft(cot^2{x}right)tag{1}$$
          where
          $$P(x)=sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}x^{m-j}=binom{2m+1}{1}x^m-binom{2m+1}{3}x^{m-1}+...$$
          is a polynomial of degree $m$, with (easy to see from $(1)$ since $sinleft((2m+1)frac{kpi}{2m+1}right)=0$) $cot^2{frac{kpi}{2m+1}}, k=1..m$ as roots. Using Vieta's formulas
          $$sum_{k=1}^m cot^2{frac{kpi}{2m+1}}=-frac{-binom{2m+1}{3}}{binom{2m+1}{1}}=frac{(2m+1)(2m)(2m-1)}{(2m+1)cdot 2 cdot 3}=frac{m(2m-1)}{3}$$






          share|cite|improve this answer











          $endgroup$



          Further to my last comment and given you used
          $$sinleft((2m+1)xright)=
          left(sin^{2m+1}xright) left(sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}(cot^2x)^{m-j}right)=
          left(sin^{2m+1}xright) cdot Pleft(cot^2{x}right)tag{1}$$
          where
          $$P(x)=sum_{j=0}^m (-1)^j binom{2m+1}{2j+1}x^{m-j}=binom{2m+1}{1}x^m-binom{2m+1}{3}x^{m-1}+...$$
          is a polynomial of degree $m$, with (easy to see from $(1)$ since $sinleft((2m+1)frac{kpi}{2m+1}right)=0$) $cot^2{frac{kpi}{2m+1}}, k=1..m$ as roots. Using Vieta's formulas
          $$sum_{k=1}^m cot^2{frac{kpi}{2m+1}}=-frac{-binom{2m+1}{3}}{binom{2m+1}{1}}=frac{(2m+1)(2m)(2m-1)}{(2m+1)cdot 2 cdot 3}=frac{m(2m-1)}{3}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 0:11

























          answered Jan 8 at 0:03









          rtybasertybase

          10.8k21533




          10.8k21533























              1












              $begingroup$

              Hint:
              The $-icotleft(frac{kpi}{2m+1}right)$, for $1 leq |k| leq m$, are the distinct roots of $(X+1)^{2m+1}-(X-1)^{2m+1}$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint:
                The $-icotleft(frac{kpi}{2m+1}right)$, for $1 leq |k| leq m$, are the distinct roots of $(X+1)^{2m+1}-(X-1)^{2m+1}$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint:
                  The $-icotleft(frac{kpi}{2m+1}right)$, for $1 leq |k| leq m$, are the distinct roots of $(X+1)^{2m+1}-(X-1)^{2m+1}$.






                  share|cite|improve this answer









                  $endgroup$



                  Hint:
                  The $-icotleft(frac{kpi}{2m+1}right)$, for $1 leq |k| leq m$, are the distinct roots of $(X+1)^{2m+1}-(X-1)^{2m+1}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 23:12









                  MindlackMindlack

                  3,39217




                  3,39217






























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