GATE 2025 IN Q.37: solution and answer

GATE 2025 · INQ.37 · 2 marks · MCQ

Newton-Raphson method is used to compute the inverse of the number 1.6. Among the following options, the initial guess of the solution that results in non-convergence of the iterative process is

(A)
0.55
(B)
0.75
(C)
1.15
(D)
1.25

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Answer and solution

ANSWER

Option (D): starting from 1.25, the reciprocal iteration $x_{n + 1} = x_{n} (2 - 1.6 x_{n})$ jumps straight to 0 and stays there, so it never converges to $1/1.6 = 0.625$ .

Official answer: DSource: Official answer key, GATE 2025 IN (IIT Roorkee)

Choose a function whose root is $\frac{1}{1.6} = 0.625$ and whose Newton-Raphson update needs no division: $f (x) = \frac{1}{x} - 1.6$ , so $f^{'} (x) = - \frac{1}{x ^{2}}$ . (With $f (x) = 1.6 x - 1$ the update divides by $1.6$ , which is the very thing being computed, and since $f$ is linear the method lands on the root in one step from any guess, so no option could fail. The intended function is therefore $f (x) = \frac{1}{x} - 1.6$ .)
Newton-Raphson update $x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}$ : $x_{n + 1} = x_{n} + x_{n}^{2} (\frac{1}{x _{n}} - 1.6) = x_{n} (2 - 1.6 x_{n})$
Define the relative error $e_{n} = 1 - 1.6 x_{n}$ . Then $1 - 1.6 x_{n + 1} = 1 - 3.2 x_{n} + 2.56 x_{n}^{2} = (1 - 1.6 x_{n})^{2}$ , i.e. $e_{n + 1} = e_{n}^{2} \Rightarrow e_{n} = e_{0}^{2^{n}}$
Convergence condition: $e_{n} \to 0$ (that is, $x_{n} \to 0.625$ ) only if $∣ e_{0} ∣ < 1$ , i.e. $- 1 < 1 - 1.6 x_{0} < 1 ⟺ 0 < x_{0} < \frac{2}{1.6} = 1.25$ .
Check the options: $x_{0} = 0.55 \Rightarrow e_{0} = 0.12$ ; $x_{0} = 0.75 \Rightarrow e_{0} = - 0.2$ ; $x_{0} = 1.15 \Rightarrow e_{0} = - 0.84$ . All have $∣ e_{0} ∣ < 1$ , so all three converge. Even $x_{0} = 1.15$ , the most distant of them, gets there: $x_{1} = 1.15 \times 0.16 = 0.184$ , $x_{2} \approx 0.314$ , $x_{3} \approx 0.470$ , $x_{4} \approx 0.587$ , $x_{5} \approx 0.623$ , heading for $0.625$ .
For $x_{0} = 1.25$ : $e_{0} = 1 - 2 = - 1$ , so $∣ e_{0} ∣ = 1$ and the condition fails. Directly, $x_{1} = 1.25 (2 - 1.6 \times 1.25) = 1.25 \times 0 = 0$ . After that the iteration stays at $x_{2} = x_{3} = \dots = 0$ , and $f (0)$ is not even defined. It never approaches $0.625$ .
So the initial guess that does not converge is $x_{0} = 1.25$ , option (D).

KEY IDEA

To find $1/ a$ with Newton-Raphson, apply it to $f (x) = \frac{1}{x} - a$ . That gives the division-free iteration $x_{n + 1} = x_{n} (2 - a x_{n})$ . Its relative error $e_{n} = 1 - a x_{n}$ squares at every step ( $e_{n + 1} = e_{n}^{2}$ ), so the method converges exactly when $0 < x_{0} < 2/ a$ . Here $2/1.6 = 1.25$ , and that boundary value is one of the options.

WATCH OUT FOR

Using $f (x) = 1.6 x - 1$ instead. That function is linear, so Newton-Raphson reaches the root in one step from any guess and none of the options would fail. A second mistake is treating the boundary $x_{0} = 2/ a = 1.25$ as safe because it 'looks close' to 0.625. At the boundary $∣ e_{0} ∣ = 1$ , the first step sends $x_{1}$ to exactly 0 and the iteration sticks there. A third is picking 0.55 or 1.15 because they look far from the root: both lie inside $(0, 1.25)$ and both converge.

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GATE 2025 IN Q.37: solution and answer

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