Hypergeometric function 3F2 with unit argument












1












$begingroup$


Recently I obtained the following expression
$${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1), $$
with $b>a>0$ and $ninmathbb{N}$.



My question is: If someone knows a closed form solution to the above expression (either in terms of the gamma function or the rising factorials).



I'm aware of the Saalschütz's theorem which states that
$${}_3F_2(-n,a,b; c, 1+a+b-n-c; 1) = frac{(c-a)_{n}(c-b)_{n}}{(c)_{n}(c-a-b)_{n}}, $$
or the Dixon's identity, however the derived equation (while only based on 3 parameters) does not have the necessary forms. I also tried a lot of identities listed here: http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/17/02/06/ but no success so far.



Alternatively, I can also obtain the equation
$${}_3F_2(-n,-a-1 ,1-b-n; b + 1, 1-a-n; -1), $$
by using few operations in the early stage of the problem, but the list of formulas is much larger for $z=1$.



I'm not really an expert on the hypergeometric functions, so a help from a trained eye in this problematic would be appreciated and very helpfull. From what I have read, the ${}_3F_2(-n,...;...; 1)$ form of the hypergeometric function ${}_3F_2$ frequently appears in many problems.






sequences-and-series special-functions hypergeometric-function sums-of-squares







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    1












    $begingroup$


    Recently I obtained the following expression
    $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1), $$
    with $b>a>0$ and $ninmathbb{N}$.



    My question is: If someone knows a closed form solution to the above expression (either in terms of the gamma function or the rising factorials).



    I'm aware of the Saalschütz's theorem which states that
    $${}_3F_2(-n,a,b; c, 1+a+b-n-c; 1) = frac{(c-a)_{n}(c-b)_{n}}{(c)_{n}(c-a-b)_{n}}, $$
    or the Dixon's identity, however the derived equation (while only based on 3 parameters) does not have the necessary forms. I also tried a lot of identities listed here: http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/17/02/06/ but no success so far.



    Alternatively, I can also obtain the equation
    $${}_3F_2(-n,-a-1 ,1-b-n; b + 1, 1-a-n; -1), $$
    by using few operations in the early stage of the problem, but the list of formulas is much larger for $z=1$.



    I'm not really an expert on the hypergeometric functions, so a help from a trained eye in this problematic would be appreciated and very helpfull. From what I have read, the ${}_3F_2(-n,...;...; 1)$ form of the hypergeometric function ${}_3F_2$ frequently appears in many problems.






    sequences-and-series special-functions hypergeometric-function sums-of-squares







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Recently I obtained the following expression
      $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1), $$
      with $b>a>0$ and $ninmathbb{N}$.



      My question is: If someone knows a closed form solution to the above expression (either in terms of the gamma function or the rising factorials).



      I'm aware of the Saalschütz's theorem which states that
      $${}_3F_2(-n,a,b; c, 1+a+b-n-c; 1) = frac{(c-a)_{n}(c-b)_{n}}{(c)_{n}(c-a-b)_{n}}, $$
      or the Dixon's identity, however the derived equation (while only based on 3 parameters) does not have the necessary forms. I also tried a lot of identities listed here: http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/17/02/06/ but no success so far.



      Alternatively, I can also obtain the equation
      $${}_3F_2(-n,-a-1 ,1-b-n; b + 1, 1-a-n; -1), $$
      by using few operations in the early stage of the problem, but the list of formulas is much larger for $z=1$.



      I'm not really an expert on the hypergeometric functions, so a help from a trained eye in this problematic would be appreciated and very helpfull. From what I have read, the ${}_3F_2(-n,...;...; 1)$ form of the hypergeometric function ${}_3F_2$ frequently appears in many problems.






      sequences-and-series special-functions hypergeometric-function sums-of-squares







      share|cite|improve this question









      $endgroup$




      Recently I obtained the following expression
      $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1), $$
      with $b>a>0$ and $ninmathbb{N}$.



      My question is: If someone knows a closed form solution to the above expression (either in terms of the gamma function or the rising factorials).



      I'm aware of the Saalschütz's theorem which states that
      $${}_3F_2(-n,a,b; c, 1+a+b-n-c; 1) = frac{(c-a)_{n}(c-b)_{n}}{(c)_{n}(c-a-b)_{n}}, $$
      or the Dixon's identity, however the derived equation (while only based on 3 parameters) does not have the necessary forms. I also tried a lot of identities listed here: http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/17/02/06/ but no success so far.



      Alternatively, I can also obtain the equation
      $${}_3F_2(-n,-a-1 ,1-b-n; b + 1, 1-a-n; -1), $$
      by using few operations in the early stage of the problem, but the list of formulas is much larger for $z=1$.



      I'm not really an expert on the hypergeometric functions, so a help from a trained eye in this problematic would be appreciated and very helpfull. From what I have read, the ${}_3F_2(-n,...;...; 1)$ form of the hypergeometric function ${}_3F_2$ frequently appears in many problems.






      sequences-and-series special-functions hypergeometric-function sums-of-squares




      sequences-and-series special-functions hypergeometric-function sums-of-squares






      share|cite|improve this question













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      asked Jan 25 at 10:01









      K. KeeperK. Keeper

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          $begingroup$

          I am not an expert on hypergeometric functions, but just incrementing by $n$ in Mathematica and looking for a pattern I find conjecturally that up to $n=5$ (and further if the pattern continues):



          $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1)= sum _{k=0}^n frac{(-1)^{k} binom{n}{k} left(prod _{j=1}^k (-b+j-n)right)left(prod _{j=1}^k (a-b+j-1)right) }{ left(prod _{j=1}^k (-a+j-n)right)left(prod _{j=1}^k (b+j)right) }$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't this also the case just by the definition of ${}_3F_2$? I was hoping more for a solution not involving a sum, if there is any. Something in the lines of mathworld.wolfram.com/GausssHypergeometricTheorem.html, which is derived for more general case of a simpler hypergeometric function. But thanks a lot anyway.
            $endgroup$
            – K. Keeper
            Jan 25 at 13:39










          • $begingroup$
            Thanks for clarifying. For $n=3$ you get $$frac{(b+2) left(-6 a^3-6 a^2 b^2+6 a^2 b+15 a b^3+18 a b^2+15 a b+6 a+b^5-2 b^4-b^3+2 b^2right)}{(-a-2) (-a-1) a (b+1) (b+2) (b+3)}$$ where the $(b+2)$ cancels the numerator can't be factorised further; it doesn't look promising.
            $endgroup$
            – James Arathoon
            Jan 25 at 14:48










          • $begingroup$
            Hmm, interesting. Actually (using also now Mathematica to evaluate it) the second function (the one with -1) seems to have a nicer form as it returns much more stable fractions but so far I could not find the pattern.
            $endgroup$
            – K. Keeper
            Jan 25 at 23:00











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          0












          $begingroup$

          I am not an expert on hypergeometric functions, but just incrementing by $n$ in Mathematica and looking for a pattern I find conjecturally that up to $n=5$ (and further if the pattern continues):



          $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1)= sum _{k=0}^n frac{(-1)^{k} binom{n}{k} left(prod _{j=1}^k (-b+j-n)right)left(prod _{j=1}^k (a-b+j-1)right) }{ left(prod _{j=1}^k (-a+j-n)right)left(prod _{j=1}^k (b+j)right) }$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't this also the case just by the definition of ${}_3F_2$? I was hoping more for a solution not involving a sum, if there is any. Something in the lines of mathworld.wolfram.com/GausssHypergeometricTheorem.html, which is derived for more general case of a simpler hypergeometric function. But thanks a lot anyway.
            $endgroup$
            – K. Keeper
            Jan 25 at 13:39










          • $begingroup$
            Thanks for clarifying. For $n=3$ you get $$frac{(b+2) left(-6 a^3-6 a^2 b^2+6 a^2 b+15 a b^3+18 a b^2+15 a b+6 a+b^5-2 b^4-b^3+2 b^2right)}{(-a-2) (-a-1) a (b+1) (b+2) (b+3)}$$ where the $(b+2)$ cancels the numerator can't be factorised further; it doesn't look promising.
            $endgroup$
            – James Arathoon
            Jan 25 at 14:48










          • $begingroup$
            Hmm, interesting. Actually (using also now Mathematica to evaluate it) the second function (the one with -1) seems to have a nicer form as it returns much more stable fractions but so far I could not find the pattern.
            $endgroup$
            – K. Keeper
            Jan 25 at 23:00
















          0












          $begingroup$

          I am not an expert on hypergeometric functions, but just incrementing by $n$ in Mathematica and looking for a pattern I find conjecturally that up to $n=5$ (and further if the pattern continues):



          $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1)= sum _{k=0}^n frac{(-1)^{k} binom{n}{k} left(prod _{j=1}^k (-b+j-n)right)left(prod _{j=1}^k (a-b+j-1)right) }{ left(prod _{j=1}^k (-a+j-n)right)left(prod _{j=1}^k (b+j)right) }$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Isn't this also the case just by the definition of ${}_3F_2$? I was hoping more for a solution not involving a sum, if there is any. Something in the lines of mathworld.wolfram.com/GausssHypergeometricTheorem.html, which is derived for more general case of a simpler hypergeometric function. But thanks a lot anyway.
            $endgroup$
            – K. Keeper
            Jan 25 at 13:39










          • $begingroup$
            Thanks for clarifying. For $n=3$ you get $$frac{(b+2) left(-6 a^3-6 a^2 b^2+6 a^2 b+15 a b^3+18 a b^2+15 a b+6 a+b^5-2 b^4-b^3+2 b^2right)}{(-a-2) (-a-1) a (b+1) (b+2) (b+3)}$$ where the $(b+2)$ cancels the numerator can't be factorised further; it doesn't look promising.
            $endgroup$
            – James Arathoon
            Jan 25 at 14:48










          • $begingroup$
            Hmm, interesting. Actually (using also now Mathematica to evaluate it) the second function (the one with -1) seems to have a nicer form as it returns much more stable fractions but so far I could not find the pattern.
            $endgroup$
            – K. Keeper
            Jan 25 at 23:00














          0












          0








          0





          $begingroup$

          I am not an expert on hypergeometric functions, but just incrementing by $n$ in Mathematica and looking for a pattern I find conjecturally that up to $n=5$ (and further if the pattern continues):



          $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1)= sum _{k=0}^n frac{(-1)^{k} binom{n}{k} left(prod _{j=1}^k (-b+j-n)right)left(prod _{j=1}^k (a-b+j-1)right) }{ left(prod _{j=1}^k (-a+j-n)right)left(prod _{j=1}^k (b+j)right) }$$






          share|cite|improve this answer











          $endgroup$



          I am not an expert on hypergeometric functions, but just incrementing by $n$ in Mathematica and looking for a pattern I find conjecturally that up to $n=5$ (and further if the pattern continues):



          $${}_3F_2(-n,a - b ,1-b-n; b + 1, 1-a-n; 1)= sum _{k=0}^n frac{(-1)^{k} binom{n}{k} left(prod _{j=1}^k (-b+j-n)right)left(prod _{j=1}^k (a-b+j-1)right) }{ left(prod _{j=1}^k (-a+j-n)right)left(prod _{j=1}^k (b+j)right) }$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 25 at 13:06

























          answered Jan 25 at 12:52









          James ArathoonJames Arathoon

          1,546423




          1,546423












          • $begingroup$
            Isn't this also the case just by the definition of ${}_3F_2$? I was hoping more for a solution not involving a sum, if there is any. Something in the lines of mathworld.wolfram.com/GausssHypergeometricTheorem.html, which is derived for more general case of a simpler hypergeometric function. But thanks a lot anyway.
            $endgroup$
            – K. Keeper
            Jan 25 at 13:39










          • $begingroup$
            Thanks for clarifying. For $n=3$ you get $$frac{(b+2) left(-6 a^3-6 a^2 b^2+6 a^2 b+15 a b^3+18 a b^2+15 a b+6 a+b^5-2 b^4-b^3+2 b^2right)}{(-a-2) (-a-1) a (b+1) (b+2) (b+3)}$$ where the $(b+2)$ cancels the numerator can't be factorised further; it doesn't look promising.
            $endgroup$
            – James Arathoon
            Jan 25 at 14:48










          • $begingroup$
            Hmm, interesting. Actually (using also now Mathematica to evaluate it) the second function (the one with -1) seems to have a nicer form as it returns much more stable fractions but so far I could not find the pattern.
            $endgroup$
            – K. Keeper
            Jan 25 at 23:00


















          • $begingroup$
            Isn't this also the case just by the definition of ${}_3F_2$? I was hoping more for a solution not involving a sum, if there is any. Something in the lines of mathworld.wolfram.com/GausssHypergeometricTheorem.html, which is derived for more general case of a simpler hypergeometric function. But thanks a lot anyway.
            $endgroup$
            – K. Keeper
            Jan 25 at 13:39










          • $begingroup$
            Thanks for clarifying. For $n=3$ you get $$frac{(b+2) left(-6 a^3-6 a^2 b^2+6 a^2 b+15 a b^3+18 a b^2+15 a b+6 a+b^5-2 b^4-b^3+2 b^2right)}{(-a-2) (-a-1) a (b+1) (b+2) (b+3)}$$ where the $(b+2)$ cancels the numerator can't be factorised further; it doesn't look promising.
            $endgroup$
            – James Arathoon
            Jan 25 at 14:48










          • $begingroup$
            Hmm, interesting. Actually (using also now Mathematica to evaluate it) the second function (the one with -1) seems to have a nicer form as it returns much more stable fractions but so far I could not find the pattern.
            $endgroup$
            – K. Keeper
            Jan 25 at 23:00
















          $begingroup$
          Isn't this also the case just by the definition of ${}_3F_2$? I was hoping more for a solution not involving a sum, if there is any. Something in the lines of mathworld.wolfram.com/GausssHypergeometricTheorem.html, which is derived for more general case of a simpler hypergeometric function. But thanks a lot anyway.
          $endgroup$
          – K. Keeper
          Jan 25 at 13:39




          $begingroup$
          Isn't this also the case just by the definition of ${}_3F_2$? I was hoping more for a solution not involving a sum, if there is any. Something in the lines of mathworld.wolfram.com/GausssHypergeometricTheorem.html, which is derived for more general case of a simpler hypergeometric function. But thanks a lot anyway.
          $endgroup$
          – K. Keeper
          Jan 25 at 13:39












          $begingroup$
          Thanks for clarifying. For $n=3$ you get $$frac{(b+2) left(-6 a^3-6 a^2 b^2+6 a^2 b+15 a b^3+18 a b^2+15 a b+6 a+b^5-2 b^4-b^3+2 b^2right)}{(-a-2) (-a-1) a (b+1) (b+2) (b+3)}$$ where the $(b+2)$ cancels the numerator can't be factorised further; it doesn't look promising.
          $endgroup$
          – James Arathoon
          Jan 25 at 14:48




          $begingroup$
          Thanks for clarifying. For $n=3$ you get $$frac{(b+2) left(-6 a^3-6 a^2 b^2+6 a^2 b+15 a b^3+18 a b^2+15 a b+6 a+b^5-2 b^4-b^3+2 b^2right)}{(-a-2) (-a-1) a (b+1) (b+2) (b+3)}$$ where the $(b+2)$ cancels the numerator can't be factorised further; it doesn't look promising.
          $endgroup$
          – James Arathoon
          Jan 25 at 14:48












          $begingroup$
          Hmm, interesting. Actually (using also now Mathematica to evaluate it) the second function (the one with -1) seems to have a nicer form as it returns much more stable fractions but so far I could not find the pattern.
          $endgroup$
          – K. Keeper
          Jan 25 at 23:00




          $begingroup$
          Hmm, interesting. Actually (using also now Mathematica to evaluate it) the second function (the one with -1) seems to have a nicer form as it returns much more stable fractions but so far I could not find the pattern.
          $endgroup$
          – K. Keeper
          Jan 25 at 23:00


















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