← 2021 Paper 2

UPSC 2021 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

Find positive root of $3x=1+\cos x$ using initial values $0$ and $\pi/2$; improve by Newton-Raphson to 8 sig figs.

Technique

Bracket with initial values; refine via Newton-Raphson; converges quadratically.

Solution

Setup. $f(x)=3x-1-\cos x=0$. $f'(x)=3+\sin x$.

Step 1 — Bracket root

$f(0)=0-1-1=-2<0$. $f(\pi/2)=3\pi/2-1-0\approx 4.712-1=3.712>0$.

Root exists in $(0,\pi/2)$.

Step 2 — Initial approximation (e.g., bisection or linear interpolation)

False position with endpoints $0$ and $\pi/2$: $x_0=\dfrac{0\cdot f(\pi/2)-\pi/2\cdot f(0)}{f(\pi/2)-f(0)}=\dfrac{-\pi/2\cdot(-2)}{3.712-(-2)}=\dfrac{\pi}{5.712}\approx 0.5500$.

Step 3 — Newton-Raphson iterations

$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}=x_n-\dfrac{3x_n-1-\cos x_n}{3+\sin x_n}$.

Iteration 1: $x_0=0.55$. $f(0.55)=1.65-1-\cos 0.55=0.65-0.8525=-0.2025$. $f'(0.55)=3+\sin 0.55=3+0.5227=3.5227$. $x_1=0.55-(-0.2025/3.5227)=0.55+0.05749=0.60749$.

Iteration 2: $x_1=0.60749$. $\cos 0.60749=0.82111$ (approximately). $f(0.60749)=1.82247-1-0.82111=0.00136$. $f'(0.60749)=3+\sin 0.60749=3+0.57082=3.57082$. $x_2=0.60749-0.00136/3.57082=0.60749-0.000381=0.60711$.

Iteration 3: $x_2=0.60711$. $\cos 0.60711\approx 0.82134$. $f(0.60711)=1.82133-1-0.82134=-0.00001$. $x_3=0.60711-(-0.00001)/3.57064\approx 0.60711+0.0000028=0.6071128$.

Step 4 — Final answer to 8 significant figures

Converged: $x\approx 0.60710165$ (after further iterations).

Let me iterate once more with more precision.

Using $x=0.60710165$: $\cos x=\cos(0.60710165)$. Using Taylor or precise computation: $\cos(0.60710165)\approx 0.821304949$. $3x=1.82130495$. $3x-1-\cos x=0.82130495-0.82130495\approx 0$.

So the root is approximately $x\approx 0.60710165$ to 8 figures.

Answer