x x squared F(n,x) = 1 + --------------- + -------------------------------- + ... ([gamma] + n)1! ([gamma] + n)([gamma] + n + 1)2! We have x F(n + 1,x) - F(n,x) = - ------------------------------ F(n + 2,x), ([gamma] + n)([gamma] + n + 1) whence we obtain F(1,x) 1 x/[gamma]([gamma] + 1) x/([gamma] + 1)([gamma] + 2) ------ = -- ---------------------- ---------------------------- F(0,x) 1 + 1 + 1 + ..., which may also be written [gamma] x x ------- ----------- ----------- [gamma] + [gamma] + 1 + [gamma] + 2 + ... By putting +- x squared/4 for x in F(0,x) and F(1,x), and putting at the same time [gamma] = 1/2, we obtain x x squared x squared x squared x x squared x squared x squared tan x = -- -- -- -- tanh x = -- -- -- -- 1 - 3 - 5 - 7 - ... 1 + 3 + 5 + 7 + ... These results were given by Lambert, and used by him to prove that [pi] and [pi] squared incommensurable, and also any commensurable power of e. Gauss in his famous memoir on the hypergeometric series F([alpha], [beta], [gamma], x) = [alpha].[beta] [alpha]([alpha] + 1)[beta]([beta] + 1) --------------x + -------------------------------------- x squared + ... 1.[gamma] 1.2.[gamma].([gamma] + 1) gave the expression for F([alpha], [beta] + 1, [gamma] + 1, x) / F([alpha], [beta], [gamma], x) as a continued fraction, from which if we put [beta] = 0 and write [gamma] - 1 for [gamma], we get the transformation [alpha] [alpha]([alpha] + 1) 1 + -------x + --------------------x squared + [gamma] [gamma]([gamma] + 1) [alpha]([alpha] + 1)([alpha] + 2) ---------------------------------x cubed + ... = [gamma]([gamma] + 1)([gamma] + 2) 1 [beta]1 x [beta]2 x -- --------- --------- where 1 - 1 - 1 - ... [alpha] ([alpha] + 1)[gamma] [beta]1 = -------, [beta]3 = --------------------------, ..., [gamma] ([gamma] + 1)([gamma] + 2) ([alpha] + n - 1)([gamma] + n - 2) [beta]_{2n-1} = ------------------------------------, ([gamma] + 2n - 3)([gamma] + 2n - 2) [gamma] - [alpha] 2([gamma] + 1 - [alpha]) [beta]2 = --------------------, [beta]4 = --------------------------, [gamma]([gamma] + 1) ([gamma] + 2)([gamma] + 3) n([gamma] + n - 1 - [alpha]) ..., [beta]_{2n} = ------------------------------------. ([gamma] + 2n - 2)([gamma] + 2n - 1)