+1 vote
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The Newton-Raphson method is used to find the root of the equation $X^2-2=0$. If the iterations are started from -1, the iterations will

1. converge to -1

2. converge to $\sqrt{2}$

3. converge to $\sqrt{-2}$

4. not converge

+1 vote

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{(x_n^2 - 2)}{2x_n} = \frac{2x_n^2 - x_n^2 + 2}{2x_n} = \frac{x_n^2 + 2}{2x_n}$

First iteration:

$x_{n+1} = \frac{x_n^2 + 2}{2x_n} = \frac{(-1)^2+2}{2(-1)} = \frac{-3}{2} = -1.5$

Second iteration:

$x_{n+1} = \frac{x_n^2 + 2}{2x_n} = \frac{(-1.5)^2+2}{2(-1.5)} = \frac{-4.25}{3} = -1.417$

Third iteration:

$x_{n+1} = \frac{x_n^2 + 2}{2x_n} = \frac{(-1.417)^2+2}{2(-1.417)} = -1.414$

–1 vote
Answer is B. General application of the formula will do.

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