Difference between revisions of "2F1(a,b;a+b+1/2;z)^2=3F2(2a,a+b,2b;a+b+1/2,2a+2b;z)"

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==Theorem==
 
==Theorem==
The following formula holds:
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The following formula (known as Clausen's formula) holds:
 
$${}_2F_1 \left( a,b;a+b +\dfrac{1}{2}; z \right)^2 = {}_3F_2 \left( 2a, a+b, 2b;a+b+\dfrac{1}{2}, 2a+2b;z \right),$$
 
$${}_2F_1 \left( a,b;a+b +\dfrac{1}{2}; z \right)^2 = {}_3F_2 \left( 2a, a+b, 2b;a+b+\dfrac{1}{2}, 2a+2b;z \right),$$
 
where ${}_2F_1$ denotes [[hypergeometric 2F1]] and ${}_3F_2$ denotes [[hypergeometric 3F2]].
 
where ${}_2F_1$ denotes [[hypergeometric 2F1]] and ${}_3F_2$ denotes [[hypergeometric 3F2]].
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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=findme|next=}}: $4.2 (1)$ (formula incorrect in printed edition, but fixed in errata)
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=findme|next=0F1(;r;z)0F1(;s;z)=2F1(r/2+s/2, r/2+s/2-1/2;r,s,r+s-1;4z)}}: $4.2 (1)$ (formula incorrect in printed edition, but fixed in errata)
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 23:25, 3 March 2018

Theorem

The following formula (known as Clausen's formula) holds: $${}_2F_1 \left( a,b;a+b +\dfrac{1}{2}; z \right)^2 = {}_3F_2 \left( 2a, a+b, 2b;a+b+\dfrac{1}{2}, 2a+2b;z \right),$$ where ${}_2F_1$ denotes hypergeometric 2F1 and ${}_3F_2$ denotes hypergeometric 3F2.

Proof

References