Question:
The iterative formula for finding \(2^{\frac{1}{5}}\) using Newton-Raphson method is
The iterative formula for finding \(2^{\frac{1}{5}}\) using Newton-Raphson method is
Show Hint
For Newton-Raphson method,
\[
\boxed{
x_{n+1}
=
x_n-\frac{f(x_n)}{f'(x_n)}
}
\]
Choose \(f(x)\) such that its root is the required quantity.
Updated On: Jul 14, 2026
- \(\displaystyle x_{n+1}=\frac45x_n+\frac{2}{5x_n^4}\)
- \(\displaystyle x_{n+1}=\frac{5x_n^4}{x_n^5-2}\)
- \(\displaystyle x_{n+1}=x_n+\frac{x_n^5-2}{5x_n^4}\)
- \(\displaystyle x_{n+1}=x_n+\frac{5x_n^4}{x_n^5}\)
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The Correct Option is A
Solution and Explanation
Step 1: Form the equation whose root is to be obtained.
To find
\[
2^{\frac15},
\]
consider
\[
f(x)=x^5-2.
\]
Then,
\[
f'(x)=5x^4.
\]
Step 2: Apply the Newton-Raphson formula. The Newton-Raphson iterative formula is \[ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}. \] Substituting, \[ x_{n+1} = x_n-\frac{x_n^5-2}{5x_n^4}. \]
Step 3: Simplify the expression. \[ x_{n+1} = x_n-\frac{x_n}{5}+\frac{2}{5x_n^4} \] \[ = \frac45x_n+\frac{2}{5x_n^4}. \] Therefore, \[ \boxed{ x_{n+1} = \frac45x_n+\frac{2}{5x_n^4} } \] is the required iterative formula. Thus, \[ \boxed{(A)} \] is the correct answer.
Step 2: Apply the Newton-Raphson formula. The Newton-Raphson iterative formula is \[ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}. \] Substituting, \[ x_{n+1} = x_n-\frac{x_n^5-2}{5x_n^4}. \]
Step 3: Simplify the expression. \[ x_{n+1} = x_n-\frac{x_n}{5}+\frac{2}{5x_n^4} \] \[ = \frac45x_n+\frac{2}{5x_n^4}. \] Therefore, \[ \boxed{ x_{n+1} = \frac45x_n+\frac{2}{5x_n^4} } \] is the required iterative formula. Thus, \[ \boxed{(A)} \] is the correct answer.
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