19/05/04 19:50:30.09 QKBukbpD.net
>>758
√(1-xx) = y とおく。
2/{(1-y)/x +1}
= 2x/(1+x-y)
= 2x(1+x+y)/{(1+x)^2 - yy}
= 2x(1+x+y)/{2x(1+x)}
= 1 + y/(1+x),
2/{(1-y)/x -1}
= 2x/(1-x-y)
= 2x(1-x+y)/{(1-x)^2 - yy}
= 2x(1-x+y)/{-2x(1-x)}
= -1 - y/(1-x),
(与式) = -1 - y/(1+x) + (1/2){-1 -y/(1-x)}^2 - (1/6){-1 -y/(1-x)}^3
= -1 - y/(1+x) + (1/2){1 + y/(1-x)}^2 - (1/6){1 + y/(1-x)}^3
= -1 - y/(1+x) - 1/3 + 2/(1-x) - (1/3)y(5-4x)/(1-x)^2
= -4/3 + 2/(1-x) - y/(1+x) - (1/3)y(5-4x)/(1-x)^2,
綺麗にならねぇ・・・・