(a) 通过画出一对合适的图像的草图，证明方程 $\operatorname{cosec}\frac12x=\mathrm{e}^x-3$ 在区间 $0<x<\pi$ 内恰有一个根，记为 $\alpha$。

(b) 通过计算验证 $\alpha$ 位于 $1$ 与 $2$ 之间。

(c) 证明：如果由迭代公式

给出的、取值在区间 $0<x<\pi$ 内的数列收敛，那么它收敛于 $\alpha$。

(d) 使用此迭代公式，以 $1.4$ 为初始值，求 $\alpha$，结果保留 $2$ 位小数。每次迭代的结果保留 $4$ 位小数。

(e) 写出使用此初始值求得精确到 $2$ 位小数的 $\alpha$ 所需计算的最少迭代次数。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

在整个给定区间，余割曲线递减而指数曲线递增。两端的相对高低交换，故有且只有一个交点。

Sketch $y=\operatorname{cosec}(x/2)$ and $y=e^x-3$ for $0<x<\pi$.

The cosecant curve decreases from $+\infty$ towards $(\pi,1)$. The exponential curve increases, passing through $(0,-2)$ and tending towards $(\pi,e^\pi-3)$.

Near $x=0$ the cosecant curve is above the exponential curve. Near $x=\pi$, it is below it since $e^\pi-3>1$.

Continuity gives an intersection, and opposite strict monotonicity gives exactly one. Mark that intersection on the sketch. Hence there is $\boxed{\text{exactly one root }\alpha}$.

用两边之差计算端点值，弧度制下的变号确定区间。

Let $f(x)=\operatorname{cosec}(x/2)-(e^x-3)$. Using radians,

Since $f$ is continuous on $[1,2]$, the sign change proves that $\boxed{1<\alpha<2}$.

收敛的极限满足不动点方程，指数化后即为原方程。

Suppose $x_n\to L$. Since $0<x_n<\pi$, the recurrence gives $x_{n+1}\geqslant\ln4$, so $\ln4\leqslant L\leqslant\pi$. The iteration function is continuous throughout this closed interval. Taking limits gives

Exponentiating and rearranging,

If $L=\pi$, the fixed-point equation would give $\pi=\ln4$, which is false. Thus $0<L<\pi$, and by the uniqueness in (a), $\boxed{L=\alpha}$.

使用指定初值，逐步计算四位小数。为回答(e)，保留从第一次迭代开始的编号。

Starting with $x_0=1.4$, apply

in radians, keeping full internal precision:

The last two values agree to 2 decimal places. Hence $\boxed{\alpha=1.50}$.

第三次与第四次迭代第一次连续给出相同的两位小数，因此需要算到第四次。

The third and fourth calculated iterates are the first consecutive pair which both round to $1.50$.

Therefore the minimum number of calculated iterations is $\boxed{4}$.