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CIE 9231 2022 November Paper 12 Q4

A Level / CIE / FP1

CIE 9231 2022 Nov Paper 12 Paper · Question 4

题目

Problem

The function f\mathrm f is such that f′′(x)=f(x)\mathrm f''(x)=\mathrm f(x).

Prove by mathematical induction that, for every positive integer nn,

d2n−1dx2n−1(xf(x))=xf′(x)+(2n−1)f(x).\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\big(x\mathrm f(x)\big)=x\mathrm f'(x)+(2n-1)\mathrm f(x).
(7)
题目中文翻译

函数 f\mathrm f 满足 f′′(x)=f(x)\mathrm f''(x)=\mathrm f(x)。

用数学归纳法证明:对于每一个正整数 nn,

d2n−1dx2n−1(xf(x))=xf′(x)+(2n−1)f(x).\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}\big(x\mathrm f(x)\big)=x\mathrm f'(x)+(2n-1)\mathrm 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=1n=1, the product rule gives

ddx(xf(x))=xf′(x)+f(x),\frac{\mathrm{d}}{\mathrm{d}x}\big(xf(x)\big) =xf'(x)+f(x),

which is the required expression, so the base case holds.

Assume that, for some positive integer kk,

d2k−1dx2k−1(xf(x))=xf′(x)+(2k−1)f(x).\frac{\mathrm{d}^{2k-1}}{\mathrm{d}x^{2k-1}}\big(xf(x)\big) =xf'(x)+(2k-1)f(x).

Differentiating once and using f′′=ff''=f,

 d2kdx2k(xf(x))= xf′′(x)+f′(x)+(2k−1)f′(x)= xf(x)+2kf′(x).\begin{align*} &\,\frac{\mathrm{d}^{2k}}{\mathrm{d}x^{2k}}\big(xf(x)\big)\\[4mm] =&\,xf''(x)+f'(x)+(2k-1)f'(x)\\[4mm] =&\,xf(x)+2kf'(x). \end{align*}

Differentiating again,

 d2k+1dx2k+1(xf(x))= xf′(x)+f(x)+2kf′′(x)= xf′(x)+(2k+1)f(x).\begin{align*} &\,\frac{\mathrm{d}^{2k+1}}{\mathrm{d}x^{2k+1}}\big(xf(x)\big)\\[4mm] =&\,xf'(x)+f(x)+2kf''(x)\\[4mm] =&\,xf'(x)+(2k+1)f(x). \end{align*}

Since 2k+1=2(k+1)−12k+1=2(k+1)-1, this is the required formula for n=k+1n=k+1. The formula holds for n=1n=1 and truth for n=kn=k implies truth for n=k+1n=k+1, so it follows by induction for every positive integer nn.