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CIE 9231 2024 November Paper 13 Q2

A Level / CIE / FP1

CIE 9231 2024 Nov Paper 13 Paper · Question 2

题目

Problem

Prove by mathematical induction that, for all positive integers nn,

dndxn(tan⁡−1x)=Pn(x)(1+x2)−n,\frac{\mathrm{d}^n}{\mathrm{d}x^n}(\tan^{-1}x)=P_n(x)(1+x^2)^{-n},

where Pn(x)P_n(x) is a polynomial of degree n−1n-1.

(6)
题目中文翻译

用数学归纳法证明:对所有正整数 nn,

dndxn(tan⁡−1x)=Pn(x)(1+x2)−n,\frac{\mathrm{d}^n}{\mathrm{d}x^n}(\tan^{-1}x)=P_n(x)(1+x^2)^{-n},

其中 Pn(x)P_n(x) 是一个 n−1n-1 次多项式。

(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

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归纳求导后把分母统一为下一次幂,得到新多项式的递推式。还必须检查最高次项没有抵消,才能证明次数恰好为k,而不只是至多为k。

答题过程

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Let f(x)=tan⁡−1xf(x)=\tan^{-1}x. For n=1n=1,

f′(x)=11+x2=P1(x)(1+x2)−1,f'(x) =\frac1{1+x^2}=P_1(x)(1+x^2)^{-1},

where P1(x)=1P_1(x)=1 has degree 00. Thus the result holds for n=1n=1.

Assume it holds for some integer k⩾1k\geqslant1, with PkP_k of degree k−1k-1. Differentiating,

f(k+1)(x)= Pk′(x)(1+x2)−k −2kxPk(x)(1+x2)−k−1= [(1+x2)Pk′(x)−2kxPk(x)](1+x2)−k−1.\begin{align*} f^{(k+1)}(x) =&\,P_k'(x)(1+x^2)^{-k}\\[4mm] &\,\hspace{2pt}-2kxP_k(x)(1+x^2)^{-k-1}\\[4mm] =&\,\big[(1+x^2)P_k'(x)-2kxP_k(x)\big] (1+x^2)^{-k-1}. \end{align*}

Define

Pk+1(x)=(1+x2)Pk′(x)−2kxPk(x).P_{k+1}(x)=(1+x^2)P_k'(x)-2kxP_k(x).

This is a polynomial of degree at most kk. If the leading coefficient of PkP_k is a≠0a\ne0, the coefficient of xkx^k in Pk+1P_{k+1} is

(k−1)a−2ka=−(k+1)a≠0.(k-1)a-2ka=-(k+1)a\ne0.

This formula also holds for k=1k=1, when P1′=0P_1'=0. Hence Pk+1P_{k+1} has degree exactly kk, and the required formula holds for n=k+1n=k+1.

The statement is true for n=1n=1 and truth for n=kn=k implies truth for n=k+1n=k+1. Therefore, by induction, it holds for every positive integer nn.