Doing some computations and plottings I've found out that the function 2 F 1 ( 2 s − 1 , s − 1 2 ; s ; − 1 )behaves for large real s like 4 − s . More precisely: It seems that 2 F 1 ( 2 s − 1 , s − 1 2 ; s ; − 1 ) 4 s grows, but very very slowly. No exponential growth or decay at all. If you modify the 4 only slightly this of course changes and you get exponential groth or decay. So my question is to explain this phenomenon. Is lim s → ∞ 2 F 1 ( 2 s − 1 , s − 1 2 ; s ; − 1 ) 4 s = ∞ ?Is there an easy to detemine for every x&gt;0 the respective b(x)&gt;0 such that you have 2 F 1 ( 2 s − 1 , s − 1 2 ; s ; − x ) ∼ b − s ?What is lim s → ∞ 2 F 1 ( 2 s − 1 , s − 1 2 ; s ; − x ) b s then?

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Doing some computations and plottings I&#039;ve found out that the function          2        F    1    (  2  s  −  1  ,  s  −            1      2        ;  s  ;  −  1  )behaves for large real s like       4          −      s      . More precisely: It seems that          2        F    1    (  2  s  −  1  ,  s  −            1      2        ;  s  ;  −  1  )      4    s  grows, but very very slowly. No exponential growth or decay at all. If you modify the 4 only slightly this of course changes and you get exponential groth or decay. So my question is to explain this phenomenon. Is      lim          s      →      ∞                 2        F    1    (  2  s  −  1  ,  s  −            1      2        ;  s  ;  −  1  )      4    s    =  ∞  ?Is there an easy to detemine for every x&amp;gt;0 the respective b(x)&amp;gt;0 such that you have          2        F    1    (  2  s  −  1  ,  s  −            1      2        ;  s  ;  −  x  )  ∼      b          −      s        ?What is      lim          s      →      ∞                 2        F    1    (  2  s  −  1  ,  s  −            1      2        ;  s  ;  −  x  )      b          s      then?

siriceboynu1 · Accepted Answer

f  (  s  )  =          2        F    1        (    2    s    −    1    ,    s    −          1      2        ;    s    ;    −    1    )    =          2        F    1        (    s    −          1      2        ,    2    s    −    1    ;    s    ;    −    1    )  Even for small values of s, we have a nice logarithmic behaviour  log  ⁡  [  f  (  s  )  ]  ∼  a  −  b    sUsing the data computed for   1  ≤  n  ≤  100, we have, with       R    2    =  0.9999982,                          Estimate                    Standard Error                    Confidence Interval                            a                    1.640373                    0.021344                    {        1.598011        ,        1.682736        }                            b                    1.379308                    0.000367                    {        1.380037        ,        1.378580        }            and, as you noticed, b is quite close to   log  ⁡  (  4  )  =  1.38629Pushing the numerical analysis much further, b is closer and closer to   (  log  ⁡  (  4  )  −  ϵ  ). This is normal since          2        F    1        (    2    s    ,    s    ;    s    ;    −    1    )    =      4          −      s      What is interesting is that, if   log  ⁡  [  f  (  s  )  ] is an increasing function going to infinity while, if   log  ⁡  [  f  (  s  )  ] goes through a maximum value.What is interesting is that      4    s            2        F    1    (  s  −  1  ,  2  s  −  1  ;  s  ;  −  1  )  =  2  +  2      π              Γ      (      s      )              Γ              (        s        −                  1          2                )

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