题目
Problem
The diagram shows the curve with equation y=x−2 for 2≤x≤N together with a set of (N−2) rectangles of unit width.
(a) By considering the sum of the areas of these rectangles, show that
r=1∑Nr21>23−N1+N21.
[5]
(b) Use a similar method to find, in terms of N, an upper bound for
r=1∑Nr21.
[3]
(c) Deduce lower and upper bounds for
r=1∑∞r21.
[2]
题目中文翻译
图中显示曲线 y=x−2,其中 2≤x≤N,以及一组宽度均为 1 的 N−2 个矩形。
(a) 通过考虑这些矩形的面积之和,证明
r=1∑Nr21>23−N1+N21
(5)
(b) 使用类似方法,求关于 N 的下列和式的上界:
r=1∑Nr21
(3)
(c) 推出
r=1∑∞r21
的下界和上界。
(2)
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
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x−2 递减,题图的左端点矩形在曲线上方。面积和从 2 加到 N−1,再补上首尾两项得到题目要求的和。
答题过程
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For N>2, the left-endpoint rectangles of width one lie above f(x)=x−2 on the interior of each interval. The left-endpoint areas are 2−2, 3−2, 4−2, …, (N−2)−2, (N−1)−2. Hence
r=2∑N−1r21>=∫2Nx−2dx[−x−1]2N=21−N1.
Adding 1 and 1/N2 to both sides gives
r=1∑Nr21>23−N1+N21.
(b)
解法一
思路
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换用右端点,矩形面积和小于积分。这个和从 3 加到 N,补上 r=1,2 两项即可。
答题过程
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The right-endpoint areas are 3−2, 4−2, 5−2, …, (N−1)−2, N−2. These rectangles lie below the decreasing curve, so
r=3∑Nr21<=∫2Nx−2dx21−N1.
Adding 1+41 gives
r=1∑Nr21<47−N1.
(c)
解法一
思路
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部分和递增且有上界,因此收敛。让 N 趋于无穷,首尾的 1/N 项消失。若要保留严格不等号,可保留 [2,3] 上固定的矩形误差。
答题过程
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The partial sums increase and are bounded above by 47, so they converge. In (a) the first rectangle exceeds the integral on [2,3] by 41−61=121. In (b) the integral on that same interval exceeds the first rectangle by 61−91=181. These positive gaps remain as N→∞. In particular,
23<r=1∑∞r21<47.