题目

Problem

(a) Show that the equation

4x-2\sin x-1=0,

where $x$ is in radians, has a root $\alpha$ in the interval $[0.2, 0.6]$

(2)

(b) Starting with the interval $[0.2, 0.6]$ , use interval bisection twice to find an interval of width $0.1$ in which $\alpha$ lies.

(3)

题目中文翻译

(a) 证明方程

4x-2\sin x-1=0

（其中 $x$ 以弧度为单位）在区间 $[0.2, 0.6]$ 内有一个根 $\alpha$ 。

(2)

(b) 从区间 $[0.2, 0.6]$ 开始，使用区间二分法两次，求宽度为 $0.1$ 且包含 $\alpha$ 的区间。

(3)

解答

(a)

解法一

思路

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令

f(x)=4x-2\sin x-1.

计算区间两端的函数值；由于 $f$ 连续且两端函数值异号，根据介值定理，区间内至少有一个根。

答题过程

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Let

f(x)=4x-2\sin x-1.

Then, using radians,

\begin{align*} &\,f(0.2)\\ =&\,4(0.2)-2\sin(0.2)-1\\ =&\,-0.597338\ldots<0. \end{align*}

\begin{align*} &\,f(0.6)\\ =&\,4(0.6)-2\sin(0.6)-1\\ =&\,0.270715\ldots>0. \end{align*}

The function $f$ is continuous, and $f(0.2)$ and $f(0.6)$ have opposite signs. Therefore, by the intermediate value theorem, there is a root $\alpha$ in

\boxed{[0.2,0.6]}.

(b)

解法一

思路

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从 $[0.2,0.6]$ 开始，第一次取中点 $0.4$ ，根据函数值符号把区间缩为 $[0.4,0.6]$ ；第二次取中点 $0.5$ ，再次利用异号端点，把区间缩为宽度 $0.1$ 的区间。

答题过程

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For the first bisection, the midpoint is

\frac{0.2+0.6}{2}=0.4.

Then

\begin{align*} &\,f(0.4)\\ =&\,4(0.4)-2\sin(0.4)-1\\ =&\,-0.178836\ldots<0. \end{align*}

Since $f(0.4)<0$ and $f(0.6)>0$ , the root lies in $[0.4,0.6]$ .

For the second bisection, the midpoint is

\frac{0.4+0.6}{2}=0.5.

Then

\begin{align*} &\,f(0.5)\\ =&\,4(0.5)-2\sin(0.5)-1\\ =&\,0.041148\ldots>0. \end{align*}

Since $f(0.4)<0$ and $f(0.5)>0$ ,

\boxed{0.4<\alpha<0.5}.

This interval has width $0.5-0.4=0.1$ .

FP1 2021 Jan Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程