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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'?

Jean Marie
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Dana
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  • I haven't an answer. But have a look at the interesting document \url{http://www.fuw.edu.pl/~derezins/hyper-published.pdf} around paragraph (4.9). Maybe Kummer's relationship is a step... – Jean Marie Nov 16 '18 at 08:07
  • @ Jean Marie:Unfortunately, I did not see anything in this field. – Dana Nov 16 '18 at 08:40
  • Have you had a look at {https://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Kummer.27s_equation} ? – Jean Marie Nov 16 '18 at 08:51
  • Another source again {http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/17/02/01/} – Jean Marie Nov 16 '18 at 09:01
  • I did not find anything related. – Dana Nov 16 '18 at 14:21
  • Who else is there to help me? – Dana Nov 17 '18 at 11:38
  • How about an example reduction? For instance, https://dlmf.nist.gov/13.3.E20 can be applied and then the derivative computed. Of course, repeated application of contiguous relations can force something. Both lead to series expansions. – rrogers Nov 17 '18 at 13:46
  • My major question is in link:https://math.stackexchange.com/questions/3002323/inverse-laplace-of-hypergeometric-confluent-function – Dana Nov 17 '18 at 13:52
  • I do not want to derive from this function after this change. I want to get Laplace's reverse. Or 2a converted to a. Or a converted to 2 a. Both should have a coefficient in the Laplace variable. – Dana Nov 17 '18 at 14:00
  • Why haven't you said in your initial question plainly the origin of your problem and its link with Laplace Transform ? It would certainly help people to help you. If you mean Laplace transform of a 1F1, it can be expressed using other hypergeometric functions {http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/22/01/} but I stop there because I am not at all an expert on hypergeometric functions... – Jean Marie Nov 17 '18 at 21:52
  • @G Cab: to use the inverse Lpalace transform from hypergeometric confluent function, I need to make Pochhamer symble identical. If you can look at this link:https://math.stackexchange.com/questions/3002323/inverse-laplace-of-hypergeometric-confluent-function – Dana Nov 18 '18 at 05:37

1 Answers1

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

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

Now, the denominator equals $$ \eqalign{ & \left( {2a} \right)^{\,\overline {\,k\,} } = \prod\limits_{0\, \le \,j\, \le \,k - 1} {\left( {2a + j} \right)} = \cr & = \prod\limits_{0\, \le \,2j\, \le \,\left\lfloor {{{k - 1} \over 2}} \right\rfloor = \left\lceil {{k \over 2}} \right\rceil - 1} {\left( {2a + 2j} \right)} \prod\limits_{0\, \le \,2j\, \le \,\left\lfloor {{k \over 2}} \right\rfloor - 1} {\left( {2a + 1 + 2j} \right)} = \cr & = 2^{\,k} \prod\limits_{0\, \le \,2j\, \le \,\left\lfloor {{{k - 1} \over 2}} \right\rfloor = \left\lceil {{k \over 2}} \right\rceil - 1} {\left( {a + j} \right)} \prod\limits_{0\, \le \,2j\, \le \,\left\lfloor {{k \over 2}} \right\rfloor - 1} {\left( {a + 1/2 + j} \right)} = \cr & = 2^{\,k} a^{\,\overline {\,\left\lceil {{k \over 2}} \right\rceil \,} } \left( {a + 1/2} \right)^{\,\overline {\,\left\lfloor {{k \over 2}} \right\rfloor \,} } \cr} $$

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 $a^{\,\overline {\,k \,} }$.

G Cab
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