An iterative formula to find √y (where Y is a positive number) by the Newton-Raphson technique is given by the expression

Question

TuteeHUB · Accepted Answer

${X_{i + 1}} = \frac{1}{4}\left( {{X_i} - \frac{Y}{{{X_i}}}} \right)$

TuteeHUB · Answer

${X_{i + 1}} = \frac{1}{4}\left( {{X_i} + \frac{Y}{{{X_i}}}} \right)$

TuteeHUB · Answer

${X_{i + 1}} = \frac{1}{3}\left( {{X_i} + \frac{Y}{{{X_i}}}} \right)$

TuteeHUB · Answer

${X_{i + 1}} = \frac{1}{2}\left( {{X_i} + \frac{Y}{{{X_i}}}} \right)$

1.	An iterative formula to find √y (where Y is a positive number) by the Newton-Raphson technique is given by the expression
A.	\({X_{i + 1}} = \frac{1}{4}\left( {{X_i} + \frac{Y}{{{X_i}}}} \right)\)
B.	\({X_{i + 1}} = \frac{1}{3}\left( {{X_i} + \frac{Y}{{{X_i}}}} \right)\)
C.	\({X_{i + 1}} = \frac{1}{2}\left( {{X_i} + \frac{Y}{{{X_i}}}} \right)\)
D.	\({X_{i + 1}} = \frac{1}{4}\left( {{X_i} - \frac{Y}{{{X_i}}}} \right)\)
Answer» D. \({X_{i + 1}} = \frac{1}{4}\left( {{X_i} - \frac{Y}{{{X_i}}}} \right)\)

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