Menu Close

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-g




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∗∗
$$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\left(\mathrm{1}−{z}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \ast\ast\mathrm{1} \\ $$$${by}\:{kummer}\:{transformation} \\ $$$$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}};{z}\right) \\ $$$$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\frac{{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−{z}}}{\:\sqrt{\mathrm{1}−{z}}}\ast\ast\mathrm{2} \\ $$$$ \\ $$$${why}\:{do}\:{i}\:{get}\:{different}\:{answer}\:{in} \\ $$$$\ast\ast\mathrm{1}\:{and}\:\mathrm{2}\ast\ast \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *