Showing $binom{2n}{n} = (-4)^n binom{-1/2}{n}$












4












$begingroup$


Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?



Let $n$ be a non-negative integer.
$$binom{2n}{n} = (-4)^n binom{-frac{1}{2}}{n}$$






binomial-coefficients factorial







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
    $endgroup$
    – KCd
    Jan 13 '15 at 2:17
















4












$begingroup$


Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?



Let $n$ be a non-negative integer.
$$binom{2n}{n} = (-4)^n binom{-frac{1}{2}}{n}$$






binomial-coefficients factorial







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
    $endgroup$
    – KCd
    Jan 13 '15 at 2:17














4












4








4


1



$begingroup$


Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?



Let $n$ be a non-negative integer.
$$binom{2n}{n} = (-4)^n binom{-frac{1}{2}}{n}$$






binomial-coefficients factorial







share|cite|improve this question









$endgroup$




Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?



Let $n$ be a non-negative integer.
$$binom{2n}{n} = (-4)^n binom{-frac{1}{2}}{n}$$






binomial-coefficients factorial




binomial-coefficients factorial






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 13 '15 at 1:47









MarkusMarkus

234




234








  • 1




    $begingroup$
    Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
    $endgroup$
    – KCd
    Jan 13 '15 at 2:17














  • 1




    $begingroup$
    Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
    $endgroup$
    – KCd
    Jan 13 '15 at 2:17








1




1




$begingroup$
Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
$endgroup$
– KCd
Jan 13 '15 at 2:17




$begingroup$
Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
$endgroup$
– KCd
Jan 13 '15 at 2:17










4 Answers
4






active

oldest

votes


















3












$begingroup$

You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.



The term $binom{-1/2}{n}$ is equal to
$$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
$$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
The rest of the calculation is straightforward.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    $$begin{align*}
    (-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
    &=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
    &=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
    &=frac{2^nn!(2n-1)!!}{(n!)^2}\\
    &=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
    &=frac{(2n)!}{(n!)^2}\\
    &=binom{2n}n
    end{align*}$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      $$begin{align}
      binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
      /n! \
      &= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
      &= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
      &= (-1)^n (2n)!/ (2^n n!)^2 \
      &= (-1/4)^n binom{2n}{n}.
      end{align}$$



      Hence:
      $$ (1-4X)^{-1/2} $$
      generates (link)
      $$ binom{2n}{n}.$$






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.



        begin{align}
        (-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
        end{align}






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.



          The term $binom{-1/2}{n}$ is equal to
          $$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
          Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
          $$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
          The rest of the calculation is straightforward.






          share|cite|improve this answer











          $endgroup$


















            3












            $begingroup$

            You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.



            The term $binom{-1/2}{n}$ is equal to
            $$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
            Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
            $$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
            The rest of the calculation is straightforward.






            share|cite|improve this answer











            $endgroup$
















              3












              3








              3





              $begingroup$

              You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.



              The term $binom{-1/2}{n}$ is equal to
              $$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
              Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
              $$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
              The rest of the calculation is straightforward.






              share|cite|improve this answer











              $endgroup$



              You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.



              The term $binom{-1/2}{n}$ is equal to
              $$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
              Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
              $$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
              The rest of the calculation is straightforward.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 13 '15 at 2:11

























              answered Jan 13 '15 at 2:03









              André NicolasAndré Nicolas

              453k36427813




              453k36427813























                  2












                  $begingroup$

                  $$begin{align*}
                  (-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
                  &=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
                  &=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
                  &=frac{2^nn!(2n-1)!!}{(n!)^2}\\
                  &=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
                  &=frac{(2n)!}{(n!)^2}\\
                  &=binom{2n}n
                  end{align*}$$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    $$begin{align*}
                    (-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
                    &=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
                    &=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
                    &=frac{2^nn!(2n-1)!!}{(n!)^2}\\
                    &=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
                    &=frac{(2n)!}{(n!)^2}\\
                    &=binom{2n}n
                    end{align*}$$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      $$begin{align*}
                      (-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
                      &=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
                      &=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
                      &=frac{2^nn!(2n-1)!!}{(n!)^2}\\
                      &=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
                      &=frac{(2n)!}{(n!)^2}\\
                      &=binom{2n}n
                      end{align*}$$






                      share|cite|improve this answer









                      $endgroup$



                      $$begin{align*}
                      (-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
                      &=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
                      &=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
                      &=frac{2^nn!(2n-1)!!}{(n!)^2}\\
                      &=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
                      &=frac{(2n)!}{(n!)^2}\\
                      &=binom{2n}n
                      end{align*}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 13 '15 at 2:09









                      Brian M. ScottBrian M. Scott

                      458k38511913




                      458k38511913























                          2












                          $begingroup$

                          $$begin{align}
                          binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
                          /n! \
                          &= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
                          &= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
                          &= (-1)^n (2n)!/ (2^n n!)^2 \
                          &= (-1/4)^n binom{2n}{n}.
                          end{align}$$



                          Hence:
                          $$ (1-4X)^{-1/2} $$
                          generates (link)
                          $$ binom{2n}{n}.$$






                          share|cite|improve this answer











                          $endgroup$


















                            2












                            $begingroup$

                            $$begin{align}
                            binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
                            /n! \
                            &= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
                            &= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
                            &= (-1)^n (2n)!/ (2^n n!)^2 \
                            &= (-1/4)^n binom{2n}{n}.
                            end{align}$$



                            Hence:
                            $$ (1-4X)^{-1/2} $$
                            generates (link)
                            $$ binom{2n}{n}.$$






                            share|cite|improve this answer











                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              $$begin{align}
                              binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
                              /n! \
                              &= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
                              &= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
                              &= (-1)^n (2n)!/ (2^n n!)^2 \
                              &= (-1/4)^n binom{2n}{n}.
                              end{align}$$



                              Hence:
                              $$ (1-4X)^{-1/2} $$
                              generates (link)
                              $$ binom{2n}{n}.$$






                              share|cite|improve this answer











                              $endgroup$



                              $$begin{align}
                              binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
                              /n! \
                              &= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
                              &= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
                              &= (-1)^n (2n)!/ (2^n n!)^2 \
                              &= (-1/4)^n binom{2n}{n}.
                              end{align}$$



                              Hence:
                              $$ (1-4X)^{-1/2} $$
                              generates (link)
                              $$ binom{2n}{n}.$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 22 '18 at 3:03









                              Shaun

                              9,241113684




                              9,241113684










                              answered Jan 13 '15 at 2:08









                              ir7ir7

                              4,14311015




                              4,14311015























                                  1












                                  $begingroup$

                                  I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.



                                  begin{align}
                                  (-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
                                  end{align}






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.



                                    begin{align}
                                    (-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
                                    end{align}






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.



                                      begin{align}
                                      (-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
                                      end{align}






                                      share|cite|improve this answer









                                      $endgroup$



                                      I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.



                                      begin{align}
                                      (-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
                                      end{align}







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 13 '15 at 2:56









                                      Benjamin RoycraftBenjamin Roycraft

                                      40628




                                      40628






























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