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Question Number 97476 by  M±th+et+s last updated on 08/Jun/20

2F1((1/2),(1/2);(1/2);z)=(1−z)^(1/2) ∗∗1  by kummer transformation  2F1((1/2),(1/2);(1/2);z)=2F1((1/2),(1/2);1+(1/2)+(1/2)−(1/2);z)  2F1((1/2),(1/2);(1/2);z)=((sin^(−1) (√(1−z)))/(√(1−z)))∗∗2    why do i get different answer in  ∗∗1 and 2∗∗

2F1(12,12;12;z)=(1z)121bykummertransformation2F1(12,12;12;z)=2F1(12,12;1+12+1212;z)2F1(12,12;12;z)=sin11z1z2whydoigetdifferentanswerin1and2

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