答案说明

• 未来技术学院未公开发布试卷内容，回忆版试卷内容请浏览往届老生整理的词条：未来技术学院 2023 年选拔考试（数学） ；
• 未来技术学院未公开发布试卷标准答案，本词条内容为往届老生整理的参考答案，无法保证答案的完全准确性与最优性，仅供学习参考；
• 欢迎指出本套答案的不足点，或提供更好的解题思路；

参考答案

一.

1.

证明：

$S_{\triangle OUV} = \dfrac{1}{2}||\mathbf{U}||||\mathbf{V}||\sin(t)$

$= \dfrac{1}{2}\sqrt{(||\mathbf{U}||||\mathbf{V}||\sin(t))^2}$

$= \dfrac{1}{2}\sqrt{(||\mathbf{U}||||\mathbf{V}||)^2(1 - (\cos(t))^2)}$

$= \dfrac{1}{2}\sqrt{(||\mathbf{U}||||\mathbf{V}||)^2 - (\mathbf{U}\mathbf{V})^2}$

图片：问题 1 示意图

（图片来源：彭宇阳）

2.

证明：

$S_{\triangle OUV} = \dfrac{1}{2}\sqrt{(||\mathbf{U}||||\mathbf{V}||)^2 - (\mathbf{U}\mathbf{V})^2}$

$= \dfrac{1}{2}\sqrt{(u_1^2 + u_2^2)(v_1^2 + v_2^2)-(u_1v_1 + u_2v_2)^2}$

$= \dfrac{1}{2}\sqrt{(u_1v_2 - u_2v_1)^2}$

$= \dfrac{1}{2}|u_1v_2 - u_2v_1|$

二.

1.

证明：

采用数学归纳法：

① n = 2 时：

LHS = $\displaystyle\sum^2_{k=1}a_kb_k$ = $a_1b_1 + a_2b_2$ = RHS

② n ≥ 3 时：

LHS = $\displaystyle\sum^{n-1}_{k=1}a_kb_k + a_nb_n$

= $S_{n-1}b_{n-1} + \displaystyle\sum^{n-2}_{k=1}S_k(b_k-b_{k+1}) + a_nb_n$

=$S_{n-1}b_{n-1} + \displaystyle\sum^{n-1}_{k=1}S_k(b_k-b_{k+1}) - S_{n-1}(b_{n-1}-b_n) + a_nb_n$

= $(S_{n-1}b_n + a_nb_n) + \displaystyle\sum^{n-1}_{k=1}S_k(b_k-b_{k+1})$

=$S_nb_n+\displaystyle\sum^{n-1}_{k=1}S_k(b_k-b_{k+1})$

2.

证明：

设 $a_k = \cos(kx)$ ,$b_k = \dfrac{1}{k}$

由欧拉公式可得：

$\displaystyle\sum^{n}_{k=1}\cos(kx) + i\displaystyle\sum^{n}_{k=1}\sin(kx)$

= $\displaystyle\sum^{n}_{k=1}(\cos(kx) + i\sin(kx))$

= $\displaystyle\sum^{n}_{k=1}e^{ikx}$

= $\dfrac{(e^{inx} - 1)e^{ix}}{e^{ix} - 1}$

= $\dfrac{2ie^{\tfrac{inx}{2}}e^{ix}\sin(\dfrac{nx}{2})}{2ie^{\tfrac{ix}{2}}\sin(\dfrac{x}{2})}$

= $\dfrac{\sin(\dfrac{nx}{2})}{\sin(\dfrac{x}{2})}e^{\tfrac{i(n+1)x}{2}}$

= $\dfrac{\sin(\dfrac{nx}{2})}{\sin(\dfrac{x}{2})}(\cos(\dfrac{(n+1)x}{2}) + i\sin(\dfrac{(n+1)x}{2}))$

∴ $S_k = a_1 + a_2 +\cdots + a_k = \dfrac{\sin(\dfrac{kx}{2})\cos(\dfrac{(k+1)x}{2})}{\sin(\dfrac{x}{2})}$

∴ LHS = $\displaystyle\sum^n_{k=1}\dfrac{\cos({kx})}{k}$

= $\dfrac{\sin(\dfrac{nx}{2})\cos(\dfrac{(n+1)x}{2})}{n\sin(\dfrac{x}{2})} + \displaystyle\sum^{n-1}_{k=1}(\dfrac{1}{n} -\dfrac{1}{n+1})\dfrac{\sin(\dfrac{kx}{2})\cos(\dfrac{(k+1)x}{2})}{\sin(\dfrac{x}{2})}$

≤ $\dfrac{1}{n\sin(\dfrac{x}{2})} + \displaystyle\sum^{n-1}_{k=1}(\dfrac{1}{k} -\dfrac{1}{k+1})\dfrac{1}{\sin(\dfrac{x}{2})} = \dfrac{1}{\sin(\dfrac{x}{2})}$

3.

证明：

$\displaystyle\sum^N_{k=1}a_kb_k = S_Nb_N - \displaystyle\sum^{N-1}_{k=1}S^{(1)}_k \Delta ^{(1)} b_k$

由于 $k>N$ 时，$b_k=0$

∴ 原式 = $(S_Nb_N - S_Nb_{N+1}) - \displaystyle\sum^{N-1}_{k=1}S^{(1)}_k \Delta ^{(1)} b_k$

= $-\displaystyle\sum^N_{k=1}S^{(1)}_k \Delta ^{(1)} b_k$

= $-((\displaystyle\sum^N_{k=1}S^{(1)}_k)\Delta ^{(1)} b_k + \displaystyle\sum^{N-1}_{k=1}(\displaystyle\sum^N_{k=1}S^{(1)}_k)(\Delta ^{(1)} b_k - \Delta ^{(1)} b_{k+1}))$

= $-S^{(2)}_k\Delta ^{(1)} b_k + \displaystyle\sum^{N-1}_{k=1}S^{(2)}_k \Delta ^{(2)} b_k$

= $(-1)^2\displaystyle\sum^N_{k=1}S^{(2)}_k \Delta ^{(2)} b_k$

以此类推可得证。

三.

答：

两周期函数之和不一定为周期函数。考虑以下反例（答案不唯一）：

$f_1(x) = \sin x$ , $f_2(x) = \sin(\alpha x)$, 其中 $\alpha$ 为无理数

$g(x) = f_1(x) + f_2(x) = \sin x + \sin(\alpha x)$

若 $g(x)$ 周期为 $T$，则有 $g(x) = g(x+T)$，即

$\sin x + \sin(\alpha x) = \sin(x+T) + \sin(\alpha(x+T))$

LHS = $-2\cos(\alpha x+\dfrac{\alpha T}{2})\sin(\dfrac{\alpha T}{2}) = 2\cos(x+\dfrac{T}{2})\sin(\dfrac{T}{2})$ = RHS

取 $x = \dfrac{π}{2} - \dfrac{T}{2}$，则 $\sin(\dfrac{\alpha T}{2}) = 0$

故 $\alpha T = 2pπ$

取 $\alpha x = \dfrac{π}{2} - \dfrac{\alpha T}{2}$，则 $\sin(\dfrac{T}{2}) = 0$

故 $T = 2qπ$，其中 $p$、$q$ 为非零整数

故 $\alpha T = 2pπ = 2\alpha qπ$，即 $\alpha = \dfrac{p}{q}$，这与$\alpha$为无理数相矛盾。

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