题目
Problem
Prove by mathematical induction that, for every integer n⩾2,
dxndn(xlnx)=(−1)n(n−2)!x1−n.
(6)
题目中文翻译
用数学归纳法证明:对每一个整数 n⩾2,
dxndn(xlnx)=(−1)n(n−2)!x1−n.
(6)
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
解法一
思路
展开
从n=2验证,不从n=1开始。归纳步骤再求一次导数,指数带出的负因子同时改变符号并补全阶乘。
答题过程
展开
Let f(x)=xlnx, where x>0.
For n=2,
f′(x)=1+lnx,f′′(x)=x−1.
This agrees with (−1)2(2−2)!x1−2=x−1, so the result is true for n=2.
Assume that, for some integer k⩾2,
f(k)(x)=(−1)k(k−2)!x1−k.
Differentiating both sides,
f(k+1)(x)===(−1)k(k−2)!(1−k)x−k(−1)k+1(k−1)!x−k(−1)k+1((k+1)−2)!x1−(k+1).
Thus truth for n=k implies truth for n=k+1. Since the result holds for n=2, by mathematical induction it holds for every integer n⩾2.