Proving identity generalizing $frac1{(a-b)(a-c)}+frac1{(b-a)(b-c)} + frac1{(c-a)(c-b)} = 0$












4












$begingroup$


It is simple to see $frac{1}{a-b} + frac{1}{b-a}=0$ and $frac1{(a-b)(a-c)}+frac1{(b-a)(b-c)} + frac1{(c-a)(c-b)} = 0$, but the generalization



$$
sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}=0$$



where the hat denotes omission, isn't as simple for me. I was unable to prove by induction. I also tried to show the derivative w.r.t. each $a_i$ vanishes, but this too is difficult. I've also tried to relate this to the determinant of a matrix with entries of the form $1/(a_i-a_j)$ but this didn't get me anywhere.



Any help would be much appreciated.






combinatorics functions







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    4












    $begingroup$


    It is simple to see $frac{1}{a-b} + frac{1}{b-a}=0$ and $frac1{(a-b)(a-c)}+frac1{(b-a)(b-c)} + frac1{(c-a)(c-b)} = 0$, but the generalization



    $$
    sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}=0$$



    where the hat denotes omission, isn't as simple for me. I was unable to prove by induction. I also tried to show the derivative w.r.t. each $a_i$ vanishes, but this too is difficult. I've also tried to relate this to the determinant of a matrix with entries of the form $1/(a_i-a_j)$ but this didn't get me anywhere.



    Any help would be much appreciated.






    combinatorics functions







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      It is simple to see $frac{1}{a-b} + frac{1}{b-a}=0$ and $frac1{(a-b)(a-c)}+frac1{(b-a)(b-c)} + frac1{(c-a)(c-b)} = 0$, but the generalization



      $$
      sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}=0$$



      where the hat denotes omission, isn't as simple for me. I was unable to prove by induction. I also tried to show the derivative w.r.t. each $a_i$ vanishes, but this too is difficult. I've also tried to relate this to the determinant of a matrix with entries of the form $1/(a_i-a_j)$ but this didn't get me anywhere.



      Any help would be much appreciated.






      combinatorics functions







      share|cite|improve this question









      $endgroup$




      It is simple to see $frac{1}{a-b} + frac{1}{b-a}=0$ and $frac1{(a-b)(a-c)}+frac1{(b-a)(b-c)} + frac1{(c-a)(c-b)} = 0$, but the generalization



      $$
      sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}=0$$



      where the hat denotes omission, isn't as simple for me. I was unable to prove by induction. I also tried to show the derivative w.r.t. each $a_i$ vanishes, but this too is difficult. I've also tried to relate this to the determinant of a matrix with entries of the form $1/(a_i-a_j)$ but this didn't get me anywhere.



      Any help would be much appreciated.






      combinatorics functions




      combinatorics functions






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      asked Dec 18 '18 at 2:07









      DwaggDwagg

      307111




      307111






















          2 Answers
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          8












          $begingroup$

          Consider the function
          $$ f(x) := sum_{j=1}^n frac{(x-a_1) cdots widehat{(x-a_j)} cdots (x-a_n)}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}. $$



          This is a polynomial function of degree at most $n-1$. Also, $f(a_1) = f(a_2) = cdots = f(a_n) = 1$. Therefore, $f(x) equiv 1$. If $n ge 2$, then that implies that the coefficient of $x^{n-1}$ in this polynomial must be 0. However, the coefficient of $x^{n-1}$ is exactly
          $$sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very elegant.${}$
            $endgroup$
            – Cheerful Parsnip
            Dec 18 '18 at 2:24



















          2












          $begingroup$

          Another proof is the following. Let



          $$p(x) = (z-a_1)cdots (z-a_n)$$



          with $a_1,ldots, a_n$ distinct. By the residue theorem we have



          $$int_{z=|R|} frac{1}{p(z)} dz = 2pi i sum_{|a_j|<R} text{Res}left(frac{1}{p(z)}, a_jright).$$



          If $ngeq 2$ then the absolute value of the integral is bounded by $sim 2pi R/R^n to 0$ as $Rto infty$. Since the RHS is independent of $R$ for large $R$, this implies



          $$sum_{j} text{Res}left(frac{1}{p(z)}, a_jright) = sum_j frac{1}{p'(a_j)}=0$$



          which is equivalent to the identity.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

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            active

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            active

            oldest

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            8












            $begingroup$

            Consider the function
            $$ f(x) := sum_{j=1}^n frac{(x-a_1) cdots widehat{(x-a_j)} cdots (x-a_n)}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}. $$



            This is a polynomial function of degree at most $n-1$. Also, $f(a_1) = f(a_2) = cdots = f(a_n) = 1$. Therefore, $f(x) equiv 1$. If $n ge 2$, then that implies that the coefficient of $x^{n-1}$ in this polynomial must be 0. However, the coefficient of $x^{n-1}$ is exactly
            $$sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Very elegant.${}$
              $endgroup$
              – Cheerful Parsnip
              Dec 18 '18 at 2:24
















            8












            $begingroup$

            Consider the function
            $$ f(x) := sum_{j=1}^n frac{(x-a_1) cdots widehat{(x-a_j)} cdots (x-a_n)}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}. $$



            This is a polynomial function of degree at most $n-1$. Also, $f(a_1) = f(a_2) = cdots = f(a_n) = 1$. Therefore, $f(x) equiv 1$. If $n ge 2$, then that implies that the coefficient of $x^{n-1}$ in this polynomial must be 0. However, the coefficient of $x^{n-1}$ is exactly
            $$sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Very elegant.${}$
              $endgroup$
              – Cheerful Parsnip
              Dec 18 '18 at 2:24














            8












            8








            8





            $begingroup$

            Consider the function
            $$ f(x) := sum_{j=1}^n frac{(x-a_1) cdots widehat{(x-a_j)} cdots (x-a_n)}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}. $$



            This is a polynomial function of degree at most $n-1$. Also, $f(a_1) = f(a_2) = cdots = f(a_n) = 1$. Therefore, $f(x) equiv 1$. If $n ge 2$, then that implies that the coefficient of $x^{n-1}$ in this polynomial must be 0. However, the coefficient of $x^{n-1}$ is exactly
            $$sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}.$$






            share|cite|improve this answer









            $endgroup$



            Consider the function
            $$ f(x) := sum_{j=1}^n frac{(x-a_1) cdots widehat{(x-a_j)} cdots (x-a_n)}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}. $$



            This is a polynomial function of degree at most $n-1$. Also, $f(a_1) = f(a_2) = cdots = f(a_n) = 1$. Therefore, $f(x) equiv 1$. If $n ge 2$, then that implies that the coefficient of $x^{n-1}$ in this polynomial must be 0. However, the coefficient of $x^{n-1}$ is exactly
            $$sum_{j=1}^n frac{1}{(a_j-a_1) cdots widehat{(a_j-a_j)} cdots (a_j-a_n)}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 2:20









            Daniel ScheplerDaniel Schepler

            9,1191721




            9,1191721












            • $begingroup$
              Very elegant.${}$
              $endgroup$
              – Cheerful Parsnip
              Dec 18 '18 at 2:24


















            • $begingroup$
              Very elegant.${}$
              $endgroup$
              – Cheerful Parsnip
              Dec 18 '18 at 2:24
















            $begingroup$
            Very elegant.${}$
            $endgroup$
            – Cheerful Parsnip
            Dec 18 '18 at 2:24




            $begingroup$
            Very elegant.${}$
            $endgroup$
            – Cheerful Parsnip
            Dec 18 '18 at 2:24











            2












            $begingroup$

            Another proof is the following. Let



            $$p(x) = (z-a_1)cdots (z-a_n)$$



            with $a_1,ldots, a_n$ distinct. By the residue theorem we have



            $$int_{z=|R|} frac{1}{p(z)} dz = 2pi i sum_{|a_j|<R} text{Res}left(frac{1}{p(z)}, a_jright).$$



            If $ngeq 2$ then the absolute value of the integral is bounded by $sim 2pi R/R^n to 0$ as $Rto infty$. Since the RHS is independent of $R$ for large $R$, this implies



            $$sum_{j} text{Res}left(frac{1}{p(z)}, a_jright) = sum_j frac{1}{p'(a_j)}=0$$



            which is equivalent to the identity.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Another proof is the following. Let



              $$p(x) = (z-a_1)cdots (z-a_n)$$



              with $a_1,ldots, a_n$ distinct. By the residue theorem we have



              $$int_{z=|R|} frac{1}{p(z)} dz = 2pi i sum_{|a_j|<R} text{Res}left(frac{1}{p(z)}, a_jright).$$



              If $ngeq 2$ then the absolute value of the integral is bounded by $sim 2pi R/R^n to 0$ as $Rto infty$. Since the RHS is independent of $R$ for large $R$, this implies



              $$sum_{j} text{Res}left(frac{1}{p(z)}, a_jright) = sum_j frac{1}{p'(a_j)}=0$$



              which is equivalent to the identity.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Another proof is the following. Let



                $$p(x) = (z-a_1)cdots (z-a_n)$$



                with $a_1,ldots, a_n$ distinct. By the residue theorem we have



                $$int_{z=|R|} frac{1}{p(z)} dz = 2pi i sum_{|a_j|<R} text{Res}left(frac{1}{p(z)}, a_jright).$$



                If $ngeq 2$ then the absolute value of the integral is bounded by $sim 2pi R/R^n to 0$ as $Rto infty$. Since the RHS is independent of $R$ for large $R$, this implies



                $$sum_{j} text{Res}left(frac{1}{p(z)}, a_jright) = sum_j frac{1}{p'(a_j)}=0$$



                which is equivalent to the identity.






                share|cite|improve this answer









                $endgroup$



                Another proof is the following. Let



                $$p(x) = (z-a_1)cdots (z-a_n)$$



                with $a_1,ldots, a_n$ distinct. By the residue theorem we have



                $$int_{z=|R|} frac{1}{p(z)} dz = 2pi i sum_{|a_j|<R} text{Res}left(frac{1}{p(z)}, a_jright).$$



                If $ngeq 2$ then the absolute value of the integral is bounded by $sim 2pi R/R^n to 0$ as $Rto infty$. Since the RHS is independent of $R$ for large $R$, this implies



                $$sum_{j} text{Res}left(frac{1}{p(z)}, a_jright) = sum_j frac{1}{p'(a_j)}=0$$



                which is equivalent to the identity.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 16:41









                DwaggDwagg

                307111




                307111






























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