# 可微函数习题

1. 请证明
a) 椭圆

$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$

在点 (x_0, y_0) 的切线方程

$$\dfrac{xx_0}{a^2} + \dfrac{yy_0}{b^2} = 1$$

b) 由位于半轴为 $a > b > 0$ 的椭圆镜的两个焦点 $F_1 = (- \sqrt{a^2 - b^2}, 0), \ F_2 = (\sqrt{a^2 - b^2}, 0)$ 之一的光源发出的光线汇聚于另一焦点.

a)

$$\begin{split} y^2 - y_{0}^{2} &= \displaystyle \frac{b^2}{a^2}(a^2 - x^2) - \frac{b^2}{a^2}(a^2 - x_0^2) \\ &= \displaystyle - \frac{b^2}{a^2}(x^2 - x_0^2) \end{split}$$

$$\displaystyle \frac{y - y_0}{x - x_0} = - \frac{b^2(x + x_0)}{a^2(y + y_0)}$$

$(x_0, y_0)$ 处的椭圆的切线斜率为 $k =\displaystyle \lim_{x \to x_0}{- \frac{b^2(x + x_0)}{a^2(y + y_0)}}$.

$$\begin{split} &y - y_0 = - \dfrac{b^2x_0}{a^2y_0}(x - x_0) \\ \Rightarrow \ & \dfrac{yy_0}{b^2} + \dfrac{xx_0}{a^2} = \dfrac{y_0^2}{b^2} + \dfrac{x_0^2}{a^2} = 1 \end{split}$$

$\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \ \Rightarrow \ y' = - \dfrac{b^2x}{a^2y}$

b)

$F_1$ 和 $F_2$ 到 $X$ 的向量分别为 ${\bf e}_{f_{1}} = (x_0 + c, y_0), \ {\bf e}_{f_{2}} = (x_0 - c, y_0)$ .

$$\begin{split} \cos{\widehat{{\bf n} \, {\bf e}_{f_{1}}}} &= \dfrac{\langle {\bf n}, {\bf e}_{f_{1}} \rangle}{\mid{\bf n}\mid \, \mid{\bf e}_{f_{1}}\mid} = \dfrac{x_0 + c + \dfrac{a^2y_0^2}{b^2 x_0}}{\mid{\bf n}\mid \, \mid{\bf e}_{f_{1}}\mid} \\ &= \dfrac{x_0 + c + \dfrac{a^2y_0^2}{b^2 x_0}}{\mid{\bf n}\mid \, \sqrt{(x_0 + c)^2 + y_0^2}} \\ &= \dfrac{x_0 + c + \dfrac{a^2y_0^2}{b^2 x_0}}{\mid{\bf n}\mid \, \sqrt{(x_0 + c)^2 + \dfrac{b^2}{a^2}(a^2 - x_0^2)}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{x_0^2 + cx_0 + \dfrac{a^2y_0^2}{b^2}}{\sqrt{a^2(x_0 + c)^2 + {b^2}(a^2 - x_0^2)}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{x_0^2 + cx_0 + a^2 - x_0^2}{\sqrt{a^2x_0^2 + 2a^2x_0c + a^2c^2 + b^2a^2 - b^2x_0^2}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{cx_0 + a^2}{\sqrt{(a^2 - b^2)x_0^2 + 2a^2x_0c + a^2(c^2 + b^2)}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{cx_0 + a^2}{\sqrt{c^2x_0^2 + 2a^2x_0c + a^4}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{cx_0 + a^2}{\sqrt{(cx_0 + a^2)^2}} = \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{cx_0 + a^2}{cx_0 + a^2} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \end{split}$$

$$\begin{split} \cos{\widehat{{\bf n} \, {\bf e}_{f_{2}}}} &= \dfrac{\langle {\bf n}, {\bf e}_{f_{2}} \rangle}{\mid{\bf n}\mid \, \mid{\bf e}_{f_{2}}\mid} = \dfrac{x_0 - c + \dfrac{a^2y_0^2}{b^2 x_0}}{\mid{\bf n}\mid \, \mid{\bf e}_{f_{1}}\mid} \\ &= \dfrac{x_0 - c + \dfrac{a^2y_0^2}{b^2 x_0}}{\mid{\bf n}\mid \, \sqrt{(x_0 - c)^2 + y_0^2}} \\ &= \dfrac{x_0 - c + \dfrac{a^2y_0^2}{b^2 x_0}}{\mid{\bf n}\mid \, \sqrt{(x_0 - c)^2 + \dfrac{b^2}{a^2}(a^2 - x_0^2)}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{x_0^2 - cx_0 + \dfrac{a^2y_0^2}{b^2}}{\sqrt{a^2(x_0 - c)^2 + {b^2}(a^2 - x_0^2)}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{x_0^2 - cx_0 + a^2 - x_0^2}{\sqrt{a^2x_0^2 - 2a^2x_0c + a^2c^2 + b^2a^2 - b^2x_0^2}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{a^2 - cx_0}{\sqrt{(a^2 - b^2)x_0^2 + 2a^2x_0c + a^2(c^2 + b^2)}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{a^2 - cx_0}{\sqrt{c^2x_0^2 - 2a^2x_0c + a^4}} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{a^2 - cx_0}{\sqrt{(cx_0 - a^2)^2}} = \dfrac{a}{\mid{\bf n}\mid x_0} \cdot \dfrac{a^2 - cx_0}{a^2 - cx_0} \\ &= \dfrac{a}{\mid{\bf n}\mid x_0} \end{split}$$

1. 请写出近似计算公式
a) $\sin(\dfrac{\pi}{6} + \alpha)$ , 其中 $\alpha$ 的值接近零;
b) $\sin(30^{\circ} + \alpha^{\circ})$ , 其中 $\alpha$ 的值接近零;
c) $\cos(\dfrac{\pi}{4} + \alpha)$ , 其中 $\alpha$ 的值接近零;
d) $\sin(45^{\circ} + \alpha^{\circ})$ , 其中 $\alpha$ 的值接近零;

a)

$$\displaystyle \lim_{\alpha \to 0} \dfrac{\sin(\dfrac{\pi}{6} + \alpha) - \sin(\dfrac{\pi}{6})}{\alpha} = \displaystyle \lim_{\alpha \to 0} \dfrac{\cos(\dfrac{\dfrac{\pi}{3} + \alpha}{2}) \sin(\dfrac{\alpha}{2})}{\dfrac{\alpha}{2}} = \cos(\dfrac{\pi}{6}) = \dfrac{\sqrt{3}}{2}$$

b)

$$\begin{split} \displaystyle \lim_{\alpha \to 0} \dfrac{\sin(30^{\circ} + \alpha^{\circ}) - \sin(30^{\circ}) }{\alpha^{\circ}} &= \displaystyle \lim_{\alpha \to 0} \dfrac{\cos(\dfrac{\dfrac{\pi}{3} + \dfrac{\pi \cdot \alpha}{180}}{2}) \sin(\dfrac{\pi \cdot \alpha}{360})}{\dfrac{\alpha}{2}} \\ &= \displaystyle \dfrac{\pi}{180} \lim_{\alpha \to 0} \dfrac{\cos(\dfrac{\dfrac{\pi}{3} + \dfrac{\pi \cdot \alpha}{180}}{2}) \sin(\dfrac{\pi \cdot \alpha}{360})}{\dfrac{\pi \alpha}{360}} \\ &= \dfrac{\pi}{180} \cos(\dfrac{\pi}{6}) \\ &= \dfrac{\sqrt{3} \pi}{360} \end{split}$$

c)
$\cos(\dfrac{\pi}{4} + \alpha) \thickapprox \dfrac{\sqrt{2} \alpha + \sqrt{2}}{2}$

d) $\cos(\dfrac{\pi}{4} + \alpha) \thickapprox \dfrac{\sqrt{2} \pi}{360} \alpha + \dfrac{\sqrt{2}}{2}$

1. 一个可以当做质点物体在重力作用下从一个光滑的山坡上滑下, 我们认为山坡是可微函数 $y = f(x)$ 的图像.
a) 请求出物体在点 $(x_0, y_0)$ 的加速度向量的水平分量和竖直分量.
b) 当 $f(x) = x^2$ 且物体从高处滑下时, 请在抛物线 $y = x^2$ 上求出使加速度水平分量最大的点.

a)

$${\bf s}(t) = (s_x(t), s_y(t)) = (x(t) - x(0), y(t) - y(0))$$

$${\bf v}(t) = (v_x(t), v_y(t)) = (\dot{s}_x(t), \dot{s}_y(t))$$

$$\dfrac{1}{2}m{\bf v}^2(t) - \dfrac{1}{2}m{\bf v}^2(t_0)= m{\bf g}(y(t) - y(t_0))$$

$${\bf v}^2(t) - {\bf v}^2(t_0) = 2{\bf g}(s_y(t) - s_y(t_0)) \tag{1}$$

${\bf v}^2(t) = v_x^2(t) + v_y^2(t)$ ,

$$v_x^2(t) + v_y^2(t) - v_x^2(t_0) - v_y^2(t_0) = 2{\bf g}(s_y(t) - s_y(t_0)) \tag{2}$$

$$\displaystyle \lim_{t \to t_0}{\dfrac{{\bf v}^2(t) - {\bf v}^2(t_0)}{t - t_0}} = 2{\bf g}\lim_{t \to t_0}{\dfrac{( s_y(t) - s_y(t_0) )}{t - t_0}} \tag{3}$$

$$\displaystyle \lim_{t \to t_0}{\dfrac{{v_x^2(t) - v_x^2(t_0) + v_y^2(t) - v_y^2(t_0)}}{t - t_0}} = 2{\bf g} \lim_{t \to t_0}{\dfrac{( s_y(t) - s_y(t_0) )}{t - t_0}} \tag{4}$$

$${\bf a}(t_0) = \displaystyle \lim_{t \to 0}{\dfrac{{\bf v}(t) - {\bf v}(t_0)}{t - t_0}} \tag{5}$$
$$a_x(t_0) = \displaystyle \lim_{t \to 0}{\dfrac{v_x(t) - v_x(t_0)}{t - t_0}} \tag{6}$$
$$a_y(t_0) = \displaystyle \lim_{t \to 0}{\dfrac{v_y(t) - v_y(t_0)}{t - t_0}} \tag{7}$$

$$v_y(t_0) = \displaystyle \lim_{t \to t_0}{\dfrac{s_y(t) - s_y(t_0)}{t - t_0}} \tag{8}$$

$(5), (8)$ 代入 $(3)$, 得:
$${\bf a}(t_0) \, {\bf v}(t_0) = {\bf g} \, v_y(t_0)$$

$${\bf a}^2(t_0) \, {\bf v}^2(t_0) = {\bf g}^2 \, v_y^2(t_0) \Rightarrow a_x^2(t_0) + a_y^2(t_0) = {\bf g}^2 \, \dfrac{v_y^2(t_0)}{v_x^2(t_0) + v_y^2(t_0)} \tag{9}$$

$$v_y(t) = f^{'}(x_t) \cdot v_x(t)$$

$$v_y(t_0) = f^{'}(x_0) \cdot v_x(t_0) \tag{10}$$

$(10)$ 代入 $(9)$ 得:

$$a_x^2(t_0) + a_y^2(t_0) = {\bf g}^2 \, \dfrac{(f^{'}(x_0))^2}{1 + (f^{'}(x_0))^2} \tag{11}$$

$(6), (7), (8)$ 代入 $(4)$ 式, 得:

$$a_x(t_0) \cdot v_x(t_0) + a_y(t_0) \cdot v_y(t_0) = {\bf g} v_y(t_0) \tag{12}$$

$(10)$ 代入 $(12)$ 得:
$$a_x(t_0) + a_y(t_0) \ f^{'}(x_0) = {\bf g} \ f^{'}(x_0) \tag{13}$$

$$\begin{split} \begin{cases} a_x^2(t_0) + a_y^2(t_0) = {\bf g}^2 \, \dfrac{(f^{'}(x_0))^2}{1 + (f^{'}(x_0))^2} \\ a_x(t_0) + a_y(t_0) \ f^{'}(x_0) = {\bf g} \ f^{'}(x_0) \end{cases} \end{split} \tag{14}$$

$$\begin{cases} a_x(t_0) = \dfrac{{\bf g} \, f^{'}(x_0)}{1 + (f^{'}(x_0))^2} \\ a_y(t_0) = {\bf g} - \dfrac{{\bf g}}{1 + (f^{'}(x_0))^2} \end{cases}$$

b)

$$a_x(t) = \dfrac{{\bf g} \, f^{'}(x_t)}{1 + (f^{'}(x_t))^2} = \dfrac{{\bf g}}{\left( \dfrac{1}{\sqrt{\mid f^{'}(x_t) \mid}} - \sqrt{\mid f^{'}(x_t) \mid} \right)^2 + 2} \le \dfrac{{\bf g}}{2}$$

[总结]:

$(13)$ 式也可以通过力学分析得出:

\begin{aligned} \begin{cases} a_x(t) &= \dfrac{F_x(t)}{m} \\ a_y(t) &= \dfrac{F_y(t) + m{\bf g}}{m} \\ F_y(t) &= - \dfrac{1}{f^{'}(x_t)} F_x(t) , \quad f^{'}(x_t) \ne 0 \\ F_y(t) &= - m{\bf g} , \quad f^{'}(t) = 0 \end{cases} \quad \Rightarrow a_y(t) = \begin{cases} - \dfrac{a_x(t)}{f^{'}(t)} + {\bf g}, \quad &f^{'}(t) \ne 0 \\ 0, &f^{'}(t) = 0 \end{cases} \end{aligned}

$$a_x(t) m + a_y(t) \ f^{'}(x_t) m = m {\bf g} \ f^{'}(x_t)$$

$F_x(t) = a_x(t) m$. <\br>

$F_y + {\bf G} = a_y(t) m$ .

$F_x(t) + (F_y(t) + {\bf G}) \, f^{'}(x_t) = {\bf G} \ f^{'}(x_t) \Rightarrow \dfrac{F_y(t)}{F_x(t)} = - \dfrac{1}{ f^{'}(x_t)}$ .
${\bf F}(t)$ 垂直于坡面.

1. \begin{aligned} \psi_0(x) = \begin{cases} x , \, &0 \leq x \leq \dfrac{1}{2}, \\ • x , \, &\dfrac{1}{2} \leq x \leq 1, \end{cases} \end{aligned} 以 $1$ 为周期延拓该函数至全部数轴, 并记之为 $\varphi_0$. 再设 $$\varphi_n(x) = \dfrac{1}{4^n}\varphi_0(4^n x) .$$ 函数 $\varphi_n$ 的周期为 $4^{-n}$ , 并且在点 $x = \dfrac{k}{2^{2n+1}} \, (k \in \Bbb{Z})$ 意外处处有导数, 导数等于 $+1$ 或 $-1$ . 设 $$f(x) = \sum_{n = 1}^{\infty}\varphi_n(x) .$$ 请证明: 函数 $f$ 在 $\Bbb{R}$ 上定义且连续, 但处处没有导数.

$\varphi_0(x)$ 以 周期为$1$ 延拓至整个 $\Bbb{R}$ 有: $\varphi_0(x) = \varphi_0(x + 1)$

$$\varphi_n(x) = \dfrac{1}{4^n}\varphi_0(4^n x) = \dfrac{1}{4^n}\varphi_0(4^n(x + \dfrac{1}{4^n})) = \varphi_n(x + \dfrac{1}{4^n}) \tag{1}$$

$\varphi_n(\Bbb{R}) = [0, \dfrac{1}{2 \times 4^n}]$ .

$$\displaystyle p(a;x) = \sum_{n = 1}^{a}\varphi_n(x), \, (a \in \Bbb{Z}^+, x \in \Bbb{R}) \tag{2}$$

$$\displaystyle g(a;x) = \sum_{n = a}^{\infty}\varphi_n(x), \, (a \in \Bbb{Z}^+, x \in \Bbb{R}) \tag{3}$$

$$\displaystyle 0 \le g(a;x) \le \dfrac{1}{2}\sum_{n = a}^{\infty}\dfrac{1}{4^n} = \dfrac{2}{3 \times 4^a} < \dfrac{1}{4^a} \tag{4}$$

$|p(a_0; x) - p(a_0; x_0)| < \dfrac{\varepsilon}{2}$ 成立.

\begin{aligned} |f(x) - f(x_0)| &= |p(a_0; x) + g(a_0 + 1; x) - p(a_0; x_0) - g(a_0 + 1; x_0)|\\ & < |p(a_0; x) - p(a_0; x_0)| + g(a_0 + 1; x) + g(a_0 + 1; x_0) \\ & < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{4} + \dfrac{\varepsilon}{4} = \varepsilon \end{aligned}

$\varphi_n(x)$ 是以 $4^{-n}$ 为周期的函数, 因此其周期可以表示为

$$\left[\dfrac{k}{4^n}, \dfrac{k + 1}{4^n}\right), k, n \in \Bbb{Z}, n \ge 0 \tag{5}$$

$$A_n(x) = \left[\dfrac{[4^n x]}{4^n}, \dfrac{[4^n x] + 1}{4^n}\right) \tag{6}$$

$$B_n(x) = \left[\dfrac{[4^n x]}{4^n}, \dfrac{[4^n x] + \dfrac{1}{2}}{4^n}\right) \tag{7}$$
$$C_n(x) = \left[\dfrac{[4^n x] + \dfrac{1}{2}}{4^n}, \dfrac{[4^n x] + 1}{4^n}\right) \tag{8}$$

\begin{aligned} \dfrac{\varphi_n(x_1) - \varphi_n(x_2)}{x_1 - x_2} = \begin{cases} 1 , \, & x_1, x_2 \in B_n(x_1), x_1 \ne x_2,\\ -1, \, & x_1, x_2 \in C_n(x_1), x_1 \ne x_2, \end{cases} \end{aligned} \tag{9}

$$\dfrac{[4^n x] + \dfrac{1}{2}}{4^n} - \dfrac{[4^n x]}{4^n} = \dfrac{[4^n x] + 1}{4^n} - \dfrac{[4^n x] + \dfrac{1}{2}}{4^n} = \dfrac{1}{2 \times 4^n}$$

$$\dfrac{\varphi_n(x + h_n) - \varphi_n(x)}{h_n} = \pm1 \tag{10}$$ .

$$\dfrac{[4^n x_0]}{4^n} \le x < \dfrac{[4^n x_0] + 1}{4^n} \tag{11}$$

$$0 \le pt - p[t] < p \Rightarrow 0 \le [pt - p[t]] < p$$

$$0 \le [pt] - p[t] \le p - 1, t \in \Bbb{R}, p \in \Bbb{Z}^+ \tag{12}$$

$$0 \le [4^n x_0] - 4[4^{n - 1} x_0] \le 3 \tag{13}$$

$$0 \le \dfrac{[4^n x_0]}{4^n} - \dfrac{4[4^{n - 1} x_0]}{4^n} \le \dfrac{3}{4^n} \Rightarrow \dfrac{[4^n x_0] + 1}{4^n} \le \dfrac{[4^{n -1} x_0] + 1}{4^{n - 1}} \tag{14}$$

$$\dfrac{[4^{n - 1} x_0]}{4^{n - 1}} \le \dfrac{[4^n x_0]}{4^n} \le x_0 < \dfrac{[4^n x_0] + 1}{4^n} \le \dfrac{[4^{n -1} x_0] + 1}{4^{n - 1}} \tag{15}$$

$$\dfrac{[4^n x_0]}{4^n} < \dfrac{[4^{n - 1} x_0] +\dfrac{1}{2}}{4^{n - 1}} \Rightarrow \dfrac{4[4^{n - 1} x_0] + 2}{4^n} - \dfrac{[4^n x_0]}{4^n} > 0 \Rightarrow 4[4^{n - 1} x_0] - [4^n x_0] > -2$$

$4[4^{n - 1} x_0] - [4^n x_0] \ge -1$ , 即
$$4[4^{n - 1} x_0] - [4^n x_0] + 1 \ge 0 \tag{16}$$
$B_{n - 1}(x_0)$ 的末端位置减去 $A_n(x_0)$ 的末端位置:
$$\dfrac{[4^{n - 1} x_0] + \dfrac{1}{2}}{4^{n - 1}} - \dfrac{[4^n x_0] +1}{4^n} = \dfrac{4[4^{n - 1} x_0] - [4^n x_0] + 1}{4^n}$$

$$\dfrac{[4^n x_0]}{4^n} \in B_{n - 1}(x_0) \Rightarrow A_{n} \subset B_{n - 1}(x_0) \tag{17}$$

$$\dfrac{[4^n x_0]}{4^n} \in C_{n - 1}(x_0) \Rightarrow A_{n} \subset C_{n - 1}(x_0) \tag{18}$$

$$\dfrac{\varphi_m(x_0 + h_n) - \varphi_m(x_0)}{h_n} = \pm 1, (m, n \in \Bbb{N}, m \le n) \tag{19}$$

$$\varphi_t(x + h_n) = \varphi_t(x) , \ t, n \in \Bbb{Z}, t > n \tag{20}$$

$$\displaystyle g_k = \dfrac{\sum_{n = 1}^{\infty}\varphi_n(x_0 + h_k) - \sum_{n = 1}^{\infty}\varphi_n(x_0)}{h_k} = \sum_{n = 1}^{\infty}\dfrac{\varphi_n(x_0 + h_k) - \varphi_n(x_0)}{h_k} \tag{21}$$

$$g_k = \sum_{n = 1}^{k}\dfrac{\varphi_n(x_0 + h_k) - \varphi_n(x_0)}{h_k} \tag{22}$$

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