解 设 $\mathcal{L}x_n$ 表示第 $(n + 1)$ 年年初账户中的金额。则

第 1 年年初的金额      $$x_0 = 1000$$

第 2 年年初的金额 $$x_1 = 1000(1 + \frac{8.5}{100}) + 1000 = 2085$$

第 3 年年初的金额 $$x_2 = 2085(1 + 0.085) + 1000 = 3262.22$$

第 4 年年初的金额 $$x_3 = 3262.22(1.085) + 1000 = 4539.51$$

我们可以看出，一般地

$$x_n = 1.085x_{n-1} + 1000$$

这是一个递推关系，它用序列中前一个元素的值来给出每个元素的值。

例 7.2 再次考虑将若干根相同直径的电缆穿入导管的问题 $d$ （例 7.2）。圆形横截面的最小导管直径 $D_n$ 取决于数量 $n$ 需封闭的电缆数量，如图 7.1 所示：

$$\begin{aligned} D_0 &= 0, \quad D_1 = d, \quad D_2 = 2d, \quad D_3 = (1 + 2\sqrt{3})d, \quad D_4 = (1 + \sqrt{2})d \ D_ $$D_0 = 0, \quad D_1 = d, \quad D_2 = 2d, \quad D_3 = (1 + 2\sqrt{3})d, \quad D_4 = (1 + \sqrt{2})d$$

$$D_5 = \frac{1}{4}\sqrt{2(5 - \sqrt{5})}d, \quad D_6 = 3d, \quad D_7 = 3d, \dots$$

Thus then duct diameters form a sequence of values ${D_1, D_2, D_3, \dots} = {D_n}_{n=1}^\infty$ .

Figure 7.1
Enclosing a
number of cables
in a circular duct.

Example 7.3 A computer simulation of the crank and connecting rod mechanism considered in Example 2.44 evaluates the position of the end $Q$ of the connecting rod at equal intervals of the angle $x^\circ$ . Given that the displacement $y$ of $Q$ satisfies

$$y = r \cos x^\circ + \sqrt{(l^2 - r^2 \sin^2 x^\circ)}$$ find the sequence of values of $y$ wh $$y = r \cos x^\circ + \sqrt{(l^2 - r^2 \sin^2 x^\circ)}$$

find the sequence of values of $y$ where $r = 5$ , $l = 10$ and the interval between successive values of $x^\circ$ is $1^\circ$ .$$