Question

一道大学数学求极限题目

Answer 1

cos(x/2)cos(x/4)...cos(x/2^(n-1))cos(x/2^n)
= cos(x/2)cos(x/4)...cos(x/2^(n-1))cos(x/2^n)sin(x/2^n)/sin(x/2^n)
= cos(x/2)cos(x/4)...cos(x/2^(n-1))sin(x/2^(n-1))/(2sin(x/2^n))
= cos(x/2)cos(x/4)...sin(x/2^(n-2))/(4sin(x/2^n))
...
= cos(x/2)sin(x/2)/(2^(n-1)·sin(x/2^n))
= sin(x)/(2^n·sin(x/2^n))
= sin(x)/(x·sin(x/2^n)/(x/2^n)).
n → ∞时, x/2^n → 0, 从而sin(x/2^n)/(x/2^n) → 1.
因此所求极限为sin(x)/x.

追问

能帮我看看我提问的一道行列式的问题吗，万分感谢！！

追答

请给下问题链接或者完整标题, 我搜一下(用户名没法搜).

追问

http://zhidao.baidu.com/question/361470934543368932.html?fr=index_nav&ssid=646999c579a1d5c55a99b3f193be2dc5.3.1413940608.1.XhaRFZqlRWaE&uid=wapp_1413509567843_768&step=2

追答

推荐答案是对的, 我再解释一下:
设该行列式为D[n], 将其按第一行展开得:
D[n] = (a+b)·M[1,1]-ab·M[1,2],
其中M[i,j]表示去掉i行j列后的余子式.
注意到M[1,1]恰好就是D[n-1],
而M[1,2]再按其第一列展开得M[1,2] = D[n-2].
因此D[n] = (a+b)·D[n-1]-ab·D[n-2].

其实做到这一步用归纳法就行了,
验证D[n] = ∑{0 ≤ i ≤ n} a^i·b^(n-i)对n = 1, 2成立.
假设对n < k成立, 则对n = k, 有:
D[k] = (a+b)·D[k-1]-ab·D[k-2]
= (a+b)·∑{0 ≤ i ≤ k-1} a^i·b^(k-1-i) - ab·∑{0 ≤ i ≤ k-2} a^i·b^(k-2-i)
= (a+b)·∑{0 ≤ i ≤ k-1} a^i·b^(k-1-i) - a·∑{0 ≤ i ≤ k-2} a^i·b^(k-1-i)
= a^k+ ∑{0 ≤ i ≤ k-1} a^i·b^(k-i)
= ∑{0 ≤ i ≤ k} a^i·b^(k-i),
即n = k时也成立, 有数学归纳法知对任意正整数n成立.

那位回答者用的是直接求解的方法:
由D[n] = (a+b)·D[n-1]-ab·D[n-2],
可知D[n]-a·D[n-1] = b·(D[n-1]-a·D[n-2]),
即数列D[n]-a·D[n-1]是以b为公比的等比数列.
代入D[1], D[2]得首项D[2]-a·D[1] = b^2,
从而D[n]-a·D[n-1] = (D[2]-a·D[1])·b^(n-2) = b^n ①.
同理, D[n]-b·D[n-1]是以a为公比的等比数列,
类似可得D[n]-b·D[n-1] = a^n ②.
联立①②解得:
D[n] = (a^(n+1)-b^(n+1))/(a-b) = ∑{0 ≤ i ≤ n} a^i·b^(n-i).
(严格来说这个解法需要a ≠ b, 对a = b要单独讨论).

追问

太棒了，谢谢你！