1
$\begingroup$

Assume that $n$ is a positive integer. Is there any simple form of this hypergeometric value $$_2\mathrm{F}_1\left[\frac{1}{2},1,\frac{3+n}{2},-1\right]?$$

$\endgroup$

2 Answers 2

1
$\begingroup$

write $F(n)$ for your formula. Then $$ F \left( 0 \right) =\frac{1}{4}\,\pi \\ F \left( 2 \right) =-\frac{3}{2}+\frac{3}{4}\,\pi \\ F \left( 4 \right) =-5+{\frac {15}{8}}\,\pi \\ F \left( 6 \right) =-{\frac {77}{6}}+{\frac {35}{8}}\,\pi \\ F \left( 8 \right) =-30+{\frac {315}{32}}\,\pi \\ F \left( 10 \right) =-{\frac {671}{10}}+{\frac {693}{32}}\,\pi $$ and $$ F \left( 1 \right) =-2+2\,\sqrt {2}\\ F \left( 3 \right) =-{\frac {20}{3}}+\frac{16}{3}\,\sqrt {2}\\ F \left( 5 \right) =-{\frac {86}{5}}+{\frac {64}{5}}\,\sqrt {2}\\ F \left( 7 \right) =-{\frac {1416}{35}}+{\frac {1024}{35}}\,\sqrt {2}\\ F \left( 9 \right) =-{\frac {5734}{63}}+{\frac {4096}{63}}\,\sqrt {2}\\ F \left( 11 \right) =-{\frac {46124}{231}}+{\frac {32768}{231}}\, \sqrt {2} $$ Now, can you guess the patterns?

$\endgroup$
3
  • 1
    $\begingroup$ @Gerard No I cannot ;) $\endgroup$
    – user66638
    Commented Mar 23, 2015 at 22:17
  • 2
    $\begingroup$ It appears that $F(2n) = a_n + b_n \pi$ where $$a_n ={\frac {- \left( -2\,{n}^{2}+7\,n-6 \right) a _{ n-3} - \left( 9\,{n}^{2}-14\,n+7 \right) a _{ n-2 } - \left( -10\,{n}^{2}+10\,n \right) a_{ n-1} }{3 \,{n}^{2}-3\,n}} $$ while $$b_n = \dfrac{(2n+1)!}{(n!)^2 2^{n+2}}$$ $\endgroup$ Commented Mar 24, 2015 at 5:14
  • 2
    $\begingroup$ $F(2n+1) = c_n + d_n \sqrt{2}$ where $$c_{{m}}={\frac {- \left( 4\,{m}^{2}+4\,m \right) c_{{m-2}}- \left( -6 \,{m}^{2}-5\,m+1 \right) c_{{m-1}}}{2\,{m}^{2}+m}} $$ and $$d_n = \dfrac{n!(n+1)!}{(2n+1)!} 2^{3n+1}$$ $\endgroup$ Commented Mar 24, 2015 at 5:25
1
$\begingroup$

First apply the Pfaff transformation mentioned above: $$ F(2n)=1/2\ _2F_1(1,n+1;\frac{(n+1)+n}{2};1/2). $$ Then use a formula from section 7.3.8 in Prudnikov,Bychkov,Marichev Integral and series, vol.3 to represent $F(2n)$ in terms of $\Gamma$ functions and a finite sum $\sum_1^n$. (In my Russian edition of this vol.3 it is formula 15, section 7.3.8, page 415).

After Pfaff transformation it is also possible to derive rather simple integral form using PBM-3 section 7.3.1 in terms of incomplete Beta functions.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .