Question Statement

(a) Find the root of the equation $x=\cos x$ in the interval $[0,1]$ correct to two decimal places using the Regula Falsi method.

(b) Using Newton's Raphson method, find a root correct to three decimal places with an initial guess of $x_0=2$.

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Background and Explanation

The Regula Falsi method (False Position method) is a bracketing numerical technique that uses linear interpolation to approximate the root of an equation $f(x)=0$ within an interval $[a,b]$ where the function changes sign. It successively narrows the interval containing the root.

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Solution

Part (a): Regula Falsi Method

We begin by rewriting the equation $x=\cos x$ in standard form: $x-\cos x=0$

Define the function: $f(x)=x-\cos x$

Step 1: Verify the root exists in $[0,1]$

Evaluate the function at the endpoints to check for a sign change: $\begin{array}{rl}& f(x)=x-\cos x\\ & f(0)=0-\cos 0=-1<0\\ & f(1)=1-\cos 1=0.4597>0\end{array}$

Since $f(0)<0$ and $f(1)>0$, the root lies between $0$ and $1$.

Step 2: Apply the Regula Falsi formula

The Regula Falsi formula for finding the next approximation is: $x=\frac{af(b)-bf(a)}{f(b)-f(a)}$

First iteration on the interval $[0,1]$: $\begin{array}{rl}{x}_0& =\frac{0\cdot f(1)-1\cdot f(0)}{f(1)-f(0)}\\ & =\frac{0-(-1)}{0.4597-(-1)}\\ & =0.6851\end{array}$

Evaluate $f(x_0)$: $\begin{array}{rl}f\left(x_0\right)& =f(0.6851)\\ & =0.6851-\cos (0.6851)\\ & =-0.0893<0\end{array}$

Since $f(0.6851)<0$ and $f(1)>0$, the updated interval is $[0.6851,1]$.

Second iteration on the interval $[0.6851,1]$: $\begin{array}{rl}{x}_1& =\frac{0.6851\cdot f(1)-1\cdot f(0.6851)}{f(1)-f(0.6851)}\\ & =\frac{0.6851(0.4597)-(-0.0873)}{0.4597-(-0.0893)}\\ & =0.7363\end{array}$

Evaluate $f(x_1)$: $\begin{array}{rl}f\left(x_1\right)& =f(0.7363)\\ & =0.7363-\cos (0.7363)\\ & =-0.0047<0\end{array}$

The updated interval is $[0.7363,1]$.

Third iteration on the interval $[0.7363,1]$: $\begin{array}{rl}{x}_2& =\frac{0.7363\cdot f(1)-1\cdot f(0.7363)}{f(1)-f(0.7363)}\\ & =\frac{0.7363(0.4597)-(-0.0047)}{0.4597-(-0.0047)}\\ & =0.7390\end{array}$

Conclusion:

Up to two decimal places, the values of $x_1$ and $x_2$ are the same. Hence, correct up to two decimal places, the root of the equation is $\boxed{0.73}$.

Part (b): Newton-Raphson Method

Note: The solution steps for this part were not provided in the source data. The problem requires finding a root of the equation correct to three decimal places using Newton's Raphson method with an initial approximation $x_0=2$.

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Key Formulas or Methods Used

• Regula Falsi (False Position) Formula: $x=\frac{af(b)-bf(a)}{f(b)-f(a)}$
• Intermediate Value Theorem: Used to confirm that a root exists in $[0,1]$ since $f(0)\cdot f(1)<0$
• Newton-Raphson Formula: $x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$ (applicable for Part b, where $f^{\prime }(x)=1+\sin x$ for the given equation)

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Summary of Steps

1. Define $f(x)=x-\cos x$ and verify $f(0)=-1<0$ and $f(1)\approx 0.4597>0$
2. Apply Regula Falsi to $[0,1]$ to obtain $x_0\approx 0.6851$
3. Check sign of $f(0.6851)\approx -0.0893$ and update interval to $[0.6851,1]$
4. Apply Regula Falsi again to get $x_1\approx 0.7363$
5. Check sign of $f(0.7363)\approx -0.0047$ and update interval to $[0.7363,1]$
6. Apply Regula Falsi third time to get $x_2\approx 0.7390$
7. Verify that $x_1\approx 0.74$ and $x_2\approx 0.74$ agree to two decimal places
8. Conclude that the root is $0.73$ (or $0.739$) correct to two decimal places