Arjun Suresh (talk | contribs) (Created page with "Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is <br> (A) 3.575 &nb...") |
Arjun Suresh (talk | contribs) |
||
| Line 1: | Line 1: | ||
Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is | Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is | ||
| − | + | ||
(A) 3.575 | (A) 3.575 | ||
| − | + | ||
(B) 3.676 | (B) 3.676 | ||
| − | + | ||
(C) 3.667 | (C) 3.667 | ||
| − | + | ||
<b>(D) </b>3.607 | <b>(D) </b>3.607 | ||
==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is
(A) 3.575
(B) 3.676
(C) 3.667
(D) 3.607
We know that, according to Newton-Raphson method,
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
In this question $f(x) = x^2-13$, and so $f'(x) = 2x$.
Here current $x_n = 3.5$, $f(x_n) = f(3.5) = -0.75$, and $f'(x_n) = f'(3.5) = 7$. So
$$x_{n+1} = 3.5 - \frac{-0.75}{7} = 3.607$$
So option (D) is correct.
Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is
(A) 3.575
(B) 3.676
(C) 3.667
(D) 3.607
We know that, according to Newton-Raphson method,
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
In this question $f(x) = x^2-13$, and so $f'(x) = 2x$.
Here current $x_n = 3.5$, $f(x_n) = f(3.5) = -0.75$, and $f'(x_n) = f'(3.5) = 7$. So
$$x_{n+1} = 3.5 - \frac{-0.75}{7} = 3.607$$
So option (D) is correct.
GATECSE