Question Statement

Find the root of the equation $x-\cos x=0$ in the interval $[0,1]$ using the bisection method, correct to two decimal places.

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Background and Explanation

The bisection method (or interval halving method) is a root-finding algorithm that repeatedly bisects an interval and selects a subinterval in which a root must lie. It requires that the function $f(x)$ is continuous on $[a,b]$ with $f(a)$ and $f(b)$ having opposite signs (Intermediate Value Theorem).

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Solution

Let $f(x)=x-\cos x$. We first verify that a root exists in $[0,1]$ by checking the sign change:

$\begin{array}{rl}f(0)& =0-\cos 0=0-1=-1<0\\ f(1)& =1-\cos 1\approx 0.459>0\end{array}$

Since $f(0)<0$ and $f(1)>0$, by the Intermediate Value Theorem, there exists at least one root in $[0,1]$.

Iteration 0

Bisect the interval $[0,1]$: $x_0=\frac{0+1}{2}=0.5$

Calculate $f(x_0)$: $\begin{array}{rl}f(0.5)& =0.5-\cos (0.5)\\ & \approx -0.377<0\end{array}$

Since $f(0.5)<0$ and $f(1)>0$, the root lies in $[0.5,1]$.

Iteration 1

Bisect $[0.5,1]$: $x_1=\frac{0.5+1}{2}=0.75$

Calculate $f(x_1)$: $\begin{array}{rl}f(0.75)& =0.75-\cos (0.75)\\ & \approx 0.018>0\end{array}$

Since $f(0.5)<0$ and $f(0.75)>0$, the updated interval is $[0.5,0.75]$.

Iteration 2

Bisect $[0.5,0.75]$: $\begin{array}{rl}{x}_2& =\frac{0.5+0.75}{2}\\ & =0.625\end{array}$

Calculate $f(x_2)$: $\begin{array}{rl}f(0.625)& =0.625-\cos (0.625)\\ & \approx -0.185<0\end{array}$

The updated interval is $[0.625,0.75]$.

Iteration 3

Bisect $[0.625,0.75]$: $\begin{array}{rl}{x}_3& =\frac{0.625+0.75}{2}\\ & =0.6875\end{array}$

Calculate $f(x_3)$: $\begin{array}{rl}f(0.6875)& =0.6875-\cos (0.6875)\\ & \approx -0.085<0\end{array}$

The updated interval is $[0.6875,0.75]$.

Iteration 4

Bisect $[0.6875,0.75]$: $\begin{array}{rl}{x}_4& =\frac{0.6875+0.75}{2}\\ & =0.71875\end{array}$

Calculate $f(x_4)$: $\begin{array}{rl}f(0.71875)& =0.71875-\cos (0.71875)\\ & \approx -0.033<0\end{array}$

The updated interval is $[0.71875,0.75]$.

Iteration 5

Bisect $[0.71875,0.75]$: $\begin{array}{rl}{x}_5& =\frac{0.71875+0.75}{2}\\ & =0.734375\end{array}$

Calculate $f(x_5)$: $\begin{array}{rl}f(0.734375)& =0.734375-\cos (0.734375)\\ & \approx -0.007<0\end{array}$

The updated interval is $[0.734375,0.75]$.

Iteration 6

Bisect $[0.734375,0.75]$: $\begin{array}{rl}{x}_6& =\frac{0.734375+0.75}{2}\\ & =0.7421875\end{array}$

Calculate $f(x_6)$: $\begin{array}{rl}f(0.7421875)& =0.7421875-\cos (0.7421875)\\ & \approx 0.0051>0\end{array}$

The updated interval is $[0.734375,0.7421875]$.

Iteration 7

Bisect $[0.734375,0.7421875]$: $\begin{array}{rl}{x}_7& =\frac{0.734375+0.7421875}{2}\\ & =0.73828125\end{array}$

Calculate $f(x_7)$: $\begin{array}{rl}f(0.73828125)& =0.73828125-\cos (0.73828125)\\ & \approx -0.0013<0\end{array}$

The updated interval is $[0.73828125,0.7421875]$.

Iteration 8

Bisect $[0.73828125,0.7421875]$: $\begin{array}{rl}{x}_8& =\frac{0.73828125+0.7421875}{2}\\ & =0.740234375\end{array}$

Calculate $f(x_8)$: $\begin{array}{rl}f(0.740234375)& =0.740234375-\cos (0.740234375)\\ & \approx 0.0019>0\end{array}$

The updated interval is $[0.73828125,0.740234375]$.

Iteration 9

Bisect $[0.73828125,0.740234375]$: $\begin{array}{rl}{x}_9& =\frac{0.73828125+0.740234375}{2}\\ & =0.7392578125\end{array}$

Calculate $f(x_9)$: $\begin{array}{rl}f(0.7392578125)& =0.7392578125-\cos (0.7392578125)\\ & \approx 0.00028>0\end{array}$

The updated interval is $[0.73828125,0.7392578125]$.

Conclusion

At this stage, both endpoints of the interval round to $0.74$ when expressed to two decimal places:

• $0.73828125\approx 0.74$
• $0.7392578125\approx 0.74$

Therefore, the root of the equation $x-\cos x=0$ correct to two decimal places is:

$\boxed{0.74}$

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Key Formulas or Methods Used

• Intermediate Value Theorem: If $f$ is continuous on $[a,b]$ and $f(a)\cdot f(b)<0$, then there exists $c\in (a,b)$ such that $f(c)=0$
• Bisection Formula: $x_n=\frac{a_n+b_n}{2}$ where $[a_n,b_n]$ is the current interval
• Interval Update Rule:
  • If $f(x_n)<0$, new interval is $[x_n,b_n]$
  • If $f(x_n)>0$, new interval is $[a_n,x_n]$

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Summary of Steps

1. Verify sign change: Confirm $f(0)<0$ and $f(1)>0$ to ensure a root exists in $[0,1]$
2. Calculate midpoint: $x_n=\frac{a+b}{2}$ of the current interval $[a,b]$
3. Evaluate function: Compute $f(x_n)$ and determine its sign
4. Update interval: Select the subinterval where the sign change occurs
5. Repeat: Continue bisecting until both endpoints agree to the desired precision (two decimal places)
6. Report result: The midpoint (or either endpoint) of the final interval gives the approximate root