Question

cosx的平方,展开成(x-1)的幂级数展开式？

Answer 1

(cosx)^2 = (1/2)(1+cos2x) = (1/2)[1+cos(2x-2+2)]
= (1/2)[1+cos(2x-2)cos2-sin(2x-2)sin2]
= 1/2 + (cos2/2)cos2(x-1) - (sin2/2)sin2(x-1)
= 1/2 + (cos2/2)∑<n=0, ∞>(-1)^n[2(x-1)]^(2n)/(2n)!
- (sin2/2)∑<n=0, ∞>(-1)^n[2(x-1)]^(2n+1)/(2n+1)!
= 1/2 + (cos2)∑<n=0, ∞>(-1)^n 2^(2n-1) (x-1)^(2n)/(2n)!
- (sin2)∑<n=0, ∞>(-1)^n 2^(2n) (x-1)^(2n+1)/(2n+1)!
x 为全体实数。