UPSC 2020 Maths Optional Paper 2 Q5b — Step-by-Step Solution

10 marks · Section B

Newton-Raphson method (convergence, geometric meaning) · Numerical Analysis · asked 6× in 13 yrs · Read the full method →

Question

Show that the equation $f(x)=\cos\frac{\pi(x+1)}{8}+0.148x-0.9062=0$ has one root in the interval $(-1,0)$ and one in $(0,1)$ . Calculate the negative root correct to four decimal places using Newton–Raphson method.

Technique

IVT for existence (sign changes at $-1,0,1$ ); Newton–Raphson $x_{n+1}=x_n-f/f'$ for the negative root.

Solution

Step 1 — Existence of the two roots (sign changes)

f(x)=\cos\frac{\pi(x+1)}{8}+0.148x-0.9062.

Evaluate at the endpoints:

$f(-1)=\cos 0+0.148(-1)-0.9062=1-0.148-0.9062=-0.0542<0.$
$f(0)=\cos\frac{\pi}{8}+0-0.9062=0.923880-0.9062=+0.017680>0.$
$f(1)=\cos\frac{2\pi}{8}+0.148-0.9062=\cos\frac{\pi}{4}+0.148-0.9062=0.707107-0.7582=-0.051093<0.$

Since $f$ is continuous,

f(-1)<0,\ f(0)>0\ \Rightarrow\ \text{a root in }(-1,0);\qquad f(0)>0,\ f(1)<0\ \Rightarrow\ \text{a root in }(0,1).

By the Intermediate Value Theorem there is at least one root in each interval. $\blacksquare$

Step 2 — Newton–Raphson setup for the negative root

f'(x)=-\frac{\pi}{8}\sin\frac{\pi(x+1)}{8}+0.148,\qquad x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.

Take $x_0=-0.5$ (interior of $(-1,0)$ ).

Step 3 — Iterations

Working to 6 decimals (angle $\theta_n=\tfrac{\pi(x_n+1)}{8}$ ):

$n$	$x_n$	$f(x_n)$	$f'(x_n)$	$x_{n+1}$
0	$-0.500000$	$+0.000585$	$0.071388$	$-0.508199$
1	$-0.508199$	$-0.000005$	$0.072629$	$-0.508129$
2	$-0.508129$	$\approx 0$	$0.072618$	$-0.508129$

The iterate is stable from $n=2$ :

x\to -0.508129\ (\text{to 6 d.p.}).

(Convergence is quadratic: starting at $x_0=-0.5$ the value is correct to 4 d.p. after a single step.)

Answer

\boxed{\;x_{\text{neg}}\approx -0.5081\;}