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: | + | 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]]. | ||
Revision as of 19:33, 17 June 2017
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
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous): $4.2 (1)$ (formula incorrect in printed edition, but fixed in errata)
