Algebra

Hypergeometrical Series

1+𝛼⋅𝛽1⋅𝛾𝑥+𝛼(𝛼+1)𝛽(𝛽+1)1⋅2⋅𝛾(𝛾+1)𝑥²+𝛼(𝛼+1)(𝛼+2)𝛽(𝛽+1)(𝛽+2)1⋅2⋅3⋅𝛾(𝛾+1)(𝛾+2)𝑥³+⋯ is convergent if 𝑥 is <1, and divergent if 𝑥>1; by (239 ii.) and if 𝑥=1, the series is convergent if 𝛾−𝛼−𝛽 is positive, divergent if 𝛾−𝛼−𝛽 is negative, (239 iv) and divergent if 𝛾−𝛼−𝛽 is zero (239 v) 291 Let the hypergeometrical series (291) be denoted by 𝐹(𝛼,𝛽,𝛾); then, the series being convergent, it is shewn by induction that 𝐹(𝛼,𝛽+1,𝛾+1)𝐹(𝛼,𝛽,𝛾)=11−𝑘₁1−𝑘₂1−⋯ concluding with 1−𝑘_2𝑟−11−𝑘_2𝑟𝑧_2𝑟 where 𝑘₁, 𝑘₂, 𝑘₃, ⋯ with 𝑧_2𝑟, are given by the formula 𝑘_2𝑟−1=(𝛼+𝑟−1)(𝛾+𝑟−1−𝛽)𝑥(𝛾+2𝑟−2)(𝛾+2𝑟−1) 𝑘_2𝑟=(𝛽+𝑟)(𝛾+𝑟−𝛼)𝑥(𝛾+2𝑟−1)(𝛾+2𝑟) 𝑧_2𝑟=𝐹(𝛼+𝑟,𝛽+𝑟+1,𝛾+2𝑟+1)𝐹(𝛼+𝑟,𝛽+𝑟,𝛾+2𝑟) The continued fraction may be concluded at any point with 𝑘_2𝑟𝑧_2𝑟. When 𝑟 is infinite, 𝑧_2𝑟=1 and the continued fraction is infinite. 292 Let 𝑓(𝛾)≡1+𝑥²1⋅𝛾+𝑥⁴1⋅2⋅𝛾(𝛾+1)+𝑥⁶1⋅2⋅3⋅𝛾(𝛾+1)(𝛾+2)+⋯ the result of substituting 𝑥²𝛼𝛽 for 𝑥 in (291), and making 𝛽=𝛼=∞. Then, by last, or independently by induction, 𝑓(𝛾+1)𝑓(𝛾)=11+𝑝₁1+𝑝₂1+⋯ +𝑝_𝑚1+⋯ with 𝑝_𝑚=𝑥²(𝛾+𝑚−1)(𝛾+𝑚) 293 In this result put 𝛾=12 and 𝑦2 for 𝑥, and we obtain by Exp. Th. (150), 𝑒^𝑦−^−𝑦𝑒^𝑦+^−𝑦=𝑦1+𝑦²3+𝑦²5+⋯ the 𝑟^th component being 𝑦²2𝑟−1. Or the continued fraction may be formed by ordinary division of one series by the other. 294 𝑒^𝑚𝑛 is incommensurable, 𝑚 and 𝑛 being integers. From the last and (174), by putting 𝑥𝑚𝑛. 295

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive