Question:
Newton-Raphson iterative formula for finding
\[
\frac{1}{\sqrt{N}}, (N>0)
\]
is
Newton-Raphson iterative formula for finding
\[
\frac{1}{\sqrt{N}}, (N>0)
\]
is
Show Hint
In Newton-Raphson, choose \( f(x) \) so that its root corresponds to your target expression.
Updated On: May 23, 2025
- \( x_{n+1} = x_n (2 - N x_n) \)
- \( x_{n+1} = \frac{1}{2} \left( x_n + \frac{N}{x_n} \right) \)
- \( x_{n+1} = \frac{1}{2} \left( x_n + \frac{1}{N x_n} \right) \)
\( x_{n+1} = \frac{1}{2} \left( x_n + \frac{N}{x} \right) \)
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The Correct Option is C
Solution and Explanation
Using Newton-Raphson’s method for solving \( f(x) = 0 \), the iterative formula is:
\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]
Let
\[
f(x) = \frac{1}{x^2} - N
\]
Then,
\[
f'(x) = -\frac{2}{x^3}
\]
Substituting in the formula:
\[
x_{n+1} = x_n - \frac{\frac{1}{x_n^2} - N}{-2/x_n^3}
\]
Simplify:
\[
= x_n + \frac{x_n (1 - N x_n^2)}{2}
\]
Simplify further:
\[
= \frac{1}{2} \left( x_n + \frac{1}{N x_n} \right)
\]
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