题目
Problem
The function f is such that f′′(x)=f(x).
Prove by mathematical induction that, for every positive integer n,
dx2n−1d2n−1(xf(x))=xf′(x)+(2n−1)f(x).
(7)
题目中文翻译
函数 f 满足 f′′(x)=f(x)。
用数学归纳法证明:对于每一个正整数 n,
dx2n−1d2n−1(xf(x))=xf′(x)+(2n−1)f(x).
(7)
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
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一次归纳要跨越两个导数阶数,因此从假设式连续求导两次,每次都用f的二阶导数等于自身。
答题过程
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For n=1, the product rule gives
dxd(xf(x))=xf′(x)+f(x),
which is the required expression, so the base case holds.
Assume that, for some positive integer k,
dx2k−1d2k−1(xf(x))=xf′(x)+(2k−1)f(x).
Differentiating once and using f′′=f,
==dx2kd2k(xf(x))xf′′(x)+f′(x)+(2k−1)f′(x)xf(x)+2kf′(x).
Differentiating again,
==dx2k+1d2k+1(xf(x))xf′(x)+f(x)+2kf′′(x)xf′(x)+(2k+1)f(x).
Since 2k+1=2(k+1)−1, this is the required formula for n=k+1. The formula holds for n=1 and truth for n=k implies truth for n=k+1, so it follows by induction for every positive integer n.