1-n次根号下cosmx的等价无穷小是多少？

Answer 1

x->0
cost = 1-(1/2)t^2 +o(t^2)
代入 t=mx
cosmx = 1-(1/2)(mx)^2 +o(x^2)
[cosmx]^(1/n)
代入上面等价
=[1-(1/2)(mx)^2 +o(x^2)]^(1/n)
=1 - [m^2/(2n)]x^2 +o(x^2)
1-[cosmx]^(1/n)
代入上面等价
=1 -[1 - [m^2/(2n)]x^2 +o(x^2)]
=[m^2/(2n)]x^2 +o(x^2)
所以
1-[cosmx]^(1/n) 等价于[m^2/(2n)]x^2