对任意自然数n,和实数θ,∏{0 ≤ i ≤ n} sin(2^i·θ) ≤ (√3/2)^n.
设f(x) = 4x²(1-x).关于这个函数有如下两个结论:
①f(x)在[0,1]上的最大值是16/27;
②当x ≥ 3/4,有f(x) ≤ f(3/4) = 9/16.
最容易想到的方法是求导证明.
如果没学导数,可以用均值不等式证明①,用作差证明②.具体过程略.
用数学归纳法.
首先n = 0时,显然|sin(θ)| ≤ 1,结论成立.
∵|sin(θ)sin(2θ)|² = (2sin²(θ)cos(θ))² = 4·sin²(θ)·sin²(θ)·(1-sin²(θ)) = f(sin²(θ)),
∴|sin(θ)sin(2θ)| ≤ √(16/27) = 4√3/9 < √3/2 (由①),
假设n = k-2,k-1时结论成立.分两种情况:
1) 若|sin(θ)| ≤ √3/2.
在n = k-1的归纳假设中以2θ替换θ,得:
∏{1 ≤ i ≤ k} |sin(2^i·θ)| = ∏{0 ≤ i ≤ k-1} |sin(2^i·2θ)| ≤ (√3/2)^(k-1).
于是∏{0 ≤ i ≤ k} |sin(2^i·θ)| ≤ (√3/2)^(k-1)·|sin(θ)| ≤ (√3/2)^k.
2) 若|sin(θ)| ≥ √3/2.
则sin²(θ) ≥ 3/4,由②得|sin(θ)sin(2θ)|² = f(sin²(θ)) ≤ 9/16.
故|sin(θ)sin(2θ)| ≤ 3/4 = (√3/2)².
在n = k-2的归纳假设中以4θ替换θ,得:
∏{2 ≤ i ≤ k} |sin(2^i·θ)| = ∏{0 ≤ i ≤ k-2} |sin(2^i·4θ)| ≤ (√3/2)^(k-2).
于是∏{0 ≤ i ≤ k} |sin(2^i·θ)| ≤ (√3/2)^(k-2)·|sin(θ)sin(2θ)| ≤ (√3/2)^k.
综合两种情况,无论θ取值如何,都能证明n = k时结论成立.
由数学归纳法,结论对任意自然数n成立,证毕.