UPSC 2014 Maths Optional Paper 2 Q5b — Step-by-Step Solution

10 marks · Section B

Newton-Raphson method (convergence, geometric meaning) · Numerical Analysis · asked 7× in 14 yrs · Read the full method →

Question

Apply Newton–Raphson method to determine a root of the equation $\cos x-xe^{x}=0$ correct up to four decimal places.

Standard Newton–Raphson; quadratic convergence near simple root.

Setup. $f(x)=\cos x-xe^{x}$ . $f'(x)=-\sin x-(1+x)e^{x}$ .

Newton iteration: $x_{n+1}=x_n-f(x_n)/f'(x_n)$ .

$f(0)=\cos 0-0=1>0$ . $f(1)=\cos 1-e\approx 0.5403-2.7183=-2.178<0$ .

Root in $(0,1)$ . Refine: $f(0.5)=\cos 0.5-0.5\,e^{0.5}\approx 0.8776-0.8244=0.0532>0$ .

Root in $(0.5,1)$ . Start with $x_0=0.5$ .

Iteration 1. $f(0.5)=0.05322$ , $f'(0.5)=-\sin 0.5-1.5\,e^{0.5}\approx -0.4794-2.4731=-2.9525$ .

x_1=0.5-\dfrac{0.05322}{-2.9525}=0.5+0.01803=0.51803.

Iteration 2. $\cos(0.51803)\approx 0.86880$ , $e^{0.51803}\approx 1.67872$ , $0.51803\cdot 1.67872\approx 0.86961$ .

$f(x_1)\approx 0.86880-0.86961=-0.00081$ .

$\sin(0.51803)\approx 0.49524$ , $f'(x_1)\approx -0.49524-1.51803\cdot 1.67872\approx -3.04358$ .

x_2=0.51803-\dfrac{-0.00081}{-3.04358}=0.51803-0.00027=0.51776.

Iteration 3. Approximate $f(0.51776)\approx 0.00000$ (very small). Quadratic convergence has kicked in.

Successive iterates agree to 4 decimal places after iteration 2.

\boxed{\;\text{Root}\approx 0.5178.\;}