回覆 1# ben1006123 的帖子
第 3 題
令\(\displaystyle f(x)=\frac{1}{2}+cosx+cos(2x)+\ldots+cos(10x)\)。
(a)試證明\(f(x)=\frac{\displaystyle sin\left(\frac{21x}{2}\right)}{\displaystyle 2sin\left(\frac{x}{2}\right)}\)。
提示：用\(z=cosx+isinx\)將\(cosx\)表示成\(\displaystyle cosx=\frac{z+\overline{z}}{2}\)。
(b)在滿足\(\displaystyle \frac{\pi}{2}\le x\le \pi\)的條件下，試求出方程式\(f(x)=0\)解的個數。
[解答]
(a) z = cosx + isinx = e^(ix)
令
s = 1 + cosx + cos(2x) + ... + cos(10x)
t = 1 + sinx + sin(2x) + ... + sin(10x)
s + it = 1 + e^(ix) + e^(i2x) + ... + e^(i10x)
= [1 - e^(i11x)]/[1 - e^(ix)]
= {[1 - e^(i11x)][1 - e^(-ix)]}/{[1 - e^(ix)][1 - e^(-ix)]}
= [1 - e^(-ix) - e^(i11x) + e^(i10x)]/(2 - 2cosx)
s = 1 + cosx + cos(2x) + ... + cos(10x)
= [1 - cosx - cos(11x) + cos(10x)]/(2 - 2cosx)
= (1/2) - [cos(11x) - cos(10x)]/(2 - 2cosx)
= (1/2) + [2sin(21x/2)sin(x/2)]/4[sin(x/2)]^2
= (1/2) + sin(21x/2)/[2sin(x/2)]
f(x) = s - 1/2 = sin(21x/2)/[2sin(x/2)]
(b) π/2 ≦ x ≦ π
5.25π ≦ (21/2)x ≦ 10.5π
sin(21x/2) = 0，有 5 個解
