Question

将函数cos(x-π/4)展开成(x-π/4)的幂级数？

Answer 1

f(x) = cos(x-π/4) =>f(π/4)= 1
f'(x) = -sin(x-π/4) =>f'(π/4)/1!= 0
f''(x)= -cos(x-π/4) =>f''(π/4)/2!= -1/2
...
f(x)
=f(π/4) + [f'(π/4)/1!](x-π/4)+....
= 1 -(1/2)(x-π/4)^2 +....+ [(-1)^n /(2n)!](x-π/4)^(2n) +......

Answer 2

cosu = 1 - u^2/2! + u^4/4! - u^6/6! + u^8/8! - ......
所以，
cos(x-π/4) = 1 - (x-π/4)^2/2! + (x-π/4)^4/4! - (x-π/4)^6/6! + (x-π/4)^8/8! - ......