hypergeometric function with special Pochhammer symbol











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What is the relationship between these two hypergeometric functions? Can the following function be written as another function of some hypergeometric functions ?
$$1F1(a+b,2a,x)$$ and $$1F1(a+b,a,x)$$ Can I convert 'a' to '2a'?










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  • I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
    – Jean Marie
    2 days ago












  • @ Jean Marie:Unfortunately, I did not see anything in this field.
    – Dana
    2 days ago










  • Have you had a look at {en.wikipedia.org/wiki/…} ?
    – Jean Marie
    2 days ago










  • Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
    – Jean Marie
    2 days ago












  • I did not find anything related.
    – Dana
    2 days ago















up vote
0
down vote

favorite












What is the relationship between these two hypergeometric functions? Can the following function be written as another function of some hypergeometric functions ?
$$1F1(a+b,2a,x)$$ and $$1F1(a+b,a,x)$$ Can I convert 'a' to '2a'?










share|cite|improve this question
























  • I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
    – Jean Marie
    2 days ago












  • @ Jean Marie:Unfortunately, I did not see anything in this field.
    – Dana
    2 days ago










  • Have you had a look at {en.wikipedia.org/wiki/…} ?
    – Jean Marie
    2 days ago










  • Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
    – Jean Marie
    2 days ago












  • I did not find anything related.
    – Dana
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











What is the relationship between these two hypergeometric functions? Can the following function be written as another function of some hypergeometric functions ?
$$1F1(a+b,2a,x)$$ and $$1F1(a+b,a,x)$$ Can I convert 'a' to '2a'?










share|cite|improve this question















What is the relationship between these two hypergeometric functions? Can the following function be written as another function of some hypergeometric functions ?
$$1F1(a+b,2a,x)$$ and $$1F1(a+b,a,x)$$ Can I convert 'a' to '2a'?







hypergeometric-function






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share|cite|improve this question








edited 18 hours ago









Jean Marie

28k41848




28k41848










asked 2 days ago









Dana

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84












  • I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
    – Jean Marie
    2 days ago












  • @ Jean Marie:Unfortunately, I did not see anything in this field.
    – Dana
    2 days ago










  • Have you had a look at {en.wikipedia.org/wiki/…} ?
    – Jean Marie
    2 days ago










  • Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
    – Jean Marie
    2 days ago












  • I did not find anything related.
    – Dana
    2 days ago


















  • I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
    – Jean Marie
    2 days ago












  • @ Jean Marie:Unfortunately, I did not see anything in this field.
    – Dana
    2 days ago










  • Have you had a look at {en.wikipedia.org/wiki/…} ?
    – Jean Marie
    2 days ago










  • Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
    – Jean Marie
    2 days ago












  • I did not find anything related.
    – Dana
    2 days ago
















I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
– Jean Marie
2 days ago






I haven't an answer. But have a look at the interesting document url{fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step...
– Jean Marie
2 days ago














@ Jean Marie:Unfortunately, I did not see anything in this field.
– Dana
2 days ago




@ Jean Marie:Unfortunately, I did not see anything in this field.
– Dana
2 days ago












Have you had a look at {en.wikipedia.org/wiki/…} ?
– Jean Marie
2 days ago




Have you had a look at {en.wikipedia.org/wiki/…} ?
– Jean Marie
2 days ago












Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
– Jean Marie
2 days ago






Another source again {functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/…}
– Jean Marie
2 days ago














I did not find anything related.
– Dana
2 days ago




I did not find anything related.
– Dana
2 days ago










1 Answer
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We have that
$$
{}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
$$



where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.



Now, the denominator equals
$$
eqalign{
& left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
& = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
& = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
& = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
$$



or, alternatively
$$
left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
$$



In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.






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    1 Answer
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    1 Answer
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    up vote
    0
    down vote













    We have that
    $$
    {}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
    $$



    where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.



    Now, the denominator equals
    $$
    eqalign{
    & left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
    & = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
    prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
    & = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
    prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
    & = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
    $$



    or, alternatively
    $$
    left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
    $$



    In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      We have that
      $$
      {}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
      $$



      where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.



      Now, the denominator equals
      $$
      eqalign{
      & left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
      & = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
      prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
      & = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
      prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
      & = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
      $$



      or, alternatively
      $$
      left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
      $$



      In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        We have that
        $$
        {}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
        $$



        where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.



        Now, the denominator equals
        $$
        eqalign{
        & left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
        & = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
        prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
        & = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
        prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
        & = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
        $$



        or, alternatively
        $$
        left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
        $$



        In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.






        share|cite|improve this answer












        We have that
        $$
        {}_1F_1 (a + b,2a,x) = sumlimits_{0, le ,k} {{{left( {a + b} right)^{,overline {,k,} } } over {left( {2a} right)^{,overline {,k,} } }}{{x^{,k} } over {k!}}}
        $$



        where the exponent overlined is a way to indicate the Rising Factorial or Pochammer symbol.



        Now, the denominator equals
        $$
        eqalign{
        & left( {2a} right)^{,overline {,k,} } = prodlimits_{0, le ,j, le ,k - 1} {left( {2a + j} right)} = cr
        & = prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {2a + 2j} right)}
        prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {2a + 1 + 2j} right)} = cr
        & = 2^{,k} prodlimits_{0, le ,2j, le ,leftlfloor {{{k - 1} over 2}} rightrfloor = leftlceil {{k over 2}} rightrceil - 1} {left( {a + j} right)}
        prodlimits_{0, le ,2j, le ,leftlfloor {{k over 2}} rightrfloor - 1} {left( {a + 1/2 + j} right)} = cr
        & = 2^{,k} a^{,overline {,leftlceil {{k over 2}} rightrceil ,} } left( {a + 1/2} right)^{,overline {,leftlfloor {{k over 2}} rightrfloor ,} } cr}
        $$



        or, alternatively
        $$
        left( {2a} right)^{,overline {,k,} } = left( {2a} right)^{,overline {,k + a - a,} } = left( {2a} right)^{,overline {, - a,} } left( a right)^{,overline {,k + a,} }
        $$



        In both versions it is not possible to simply relate the above results with $a^{,overline {,k ,} }$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 17 hours ago









        G Cab

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