GATE 2022 IN Q.34: solution and answer

GATE 2022 · INQ.34 · 1 mark · NAT

The Newton-Raphson method is applied to determine the solution of $f (x) = 0$ where $f (x) = x - cos (x)$ . If the initial guess of the solution is $x_{0} = 0$ , the value of the next approximation $x_{1}$ is ________ (round off to two decimal places)

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Answer and solution

ANSWER

$x_{1} = 1.00$ .

Official answer: 0.99 to 1.01Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Newton-Raphson iteration: $x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}$ .
Differentiate: $f (x) = x - cos x \Rightarrow f^{'} (x) = 1 + sin x$ (since $\frac{d}{d x} (- cos x) = + sin x$ ).
At the initial guess $x_{0} = 0$ (radians): $f (0) = 0 - cos 0 = - 1$ and $f^{'} (0) = 1 + sin 0 = 1$ .
One iteration: $x_{1} = x_{0} - \frac{f ( x _{0} )}{f ^{'} ( x _{0} )} = 0 - \frac{- 1}{1} = 1$
Rounded to two decimal places, $x_{1} = 1.00$ . (Check: the root of $x = cos x$ is near $0.739$ , and $x_{1} = 1$ moves from $x_{0} = 0$ toward it, as expected for one step.)

KEY IDEA

One Newton-Raphson step replaces $f$ by its tangent line at $x_{0}$ and takes that line's zero: $x_{1} = x_{0} - f (x_{0}) / f^{'} (x_{0})$ . At $x_{0} = 0$ we have $f (0) = - 1$ and $f^{'} (0) = 1$ , so $x_{1} = 1$ .

WATCH OUT FOR

Answering with the converged root ( $\approx 0.74$ ) when the question asks for only the first iterate $x_{1}$ . Another slip is getting the sign of the update wrong ( $x_{0} + f / f^{'}$ ), which gives $- 1$ . Note that the sign of $\frac{d}{d x} cos x$ makes no difference here because $sin 0 = 0$ , but it does matter at every later step.

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GATE 2022 IN Q.34: solution and answer

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