The function $f (x) = e^{x} - 1$ is tto be solved using Newton-Raphson method. If the initial value of $x_{0}$ is taken as 1.0, then the absolute error observed at 2nd iteration is
0.06

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Solution

The correct option is A 0.06
$f (x) = e^{x} - 1, x_{0} = 1$
$f^{'} (x) = e^{x}$
By Newton-Raphson method
$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})}$
$n = 0, 1, 2, 3, . . . .$
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$= 1 - \frac{(e - 1)}{e} = \frac{1}{e}$
$x_{2} = x_{1} - \frac{f (x_{1})}{f^{'} (x_{1})} = \frac{1}{e} - \frac{(e^{1 / e} - 1)}{e^{1 / e}}$
$= \frac{1}{e} + \frac{1}{e^{1 / e}} - 1$
$= 0.37 + 0.69 - 1$
$= 0.06$
Absolute error after 2nd iteration
= |0-0.06| = 0.06.

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The function f(x)=ex−1 is tto be solved using Newton-Raphson method. If the initial value of x0 is taken as 1.0, then the absolute error observed at 2nd iteration is0.06

The correct option is A 0.06 f(x)=ex−1,x0=1 f′(x)=ex By Newton-Raphson method xn+1=xn−f(xn)f′(xn) n=0,1,2,3,.... x1=x0−f(x0)f′(x0) =1−(e−1)e=1e x2=x1−f(x1)f′(x1)=1e−(e1/e−1)e1/e =1e+1e1/e−1 =0.37+0.69−1 =0.06 Absolute error after 2nd iteration = |0-0.06| = 0.06.

The function $f (x) = e^{x} - 1$ is tto be solved using Newton-Raphson method. If the initial value of $x_{0}$ is taken as 1.0, then the absolute error observed at 2nd iteration is
0.06