题目
The constant is such that .
(a) Show that .
(b) Verify by calculation that lies between and .
(c) Use an iterative formula based on the equation in part (a) to determine correct to decimal places. Give the result of each iteration to decimal places.
题目中文翻译
常数 满足 。
(a) 证明 。
(b) 通过计算验证 位于 与 之间。
(c) 使用基于 (a) 问中方程的迭代公式,求 的值,精确到小数点后 位。每次迭代的结果保留小数点后 位。
原卷通用作答要求
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。
解答
(a)
解法一
思路
展开
分部积分先求原函数,再代入上下限。注意零下限处指数项的值不为零。
答题过程
展开
Using and , integration by parts gives
Thus the given condition becomes
Hence
giving
(b)
解法一
思路
展开
把重排方程两边之差设成连续函数,计算两端并检查变号。
答题过程
展开
Let . Then
Since is continuous and changes sign, a root lies in . Moreover,
so it is the unique root. Hence .
(c)
解法一
思路
展开
按(a)的重排建立迭代,取区间中的值作为初值。输出四位小数,内部计算保留完整精度。
答题过程
展开
Use
with . The iterations are
| to 4 decimal places | |
|---|---|
| 0 | 2.3000 |
| 1 | 2.3166 |
| 2 | 2.3140 |
| 3 | 2.3144 |
| 4 | 2.3144 |
| 5 | 2.3144 |
Therefore to 2 decimal places.