Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
||
| (4 intermediate revisions by the same user not shown) | |||
| Line 17: | Line 17: | ||
So option <b>(D)</b> is correct. | So option <b>(D)</b> is correct. | ||
{{Template:FBD}} | {{Template:FBD}} | ||
| − | [[Category:Newton-Raphson method]] | + | [[Category:Newton-Raphson method ]] |
[[Category: GATE2010]] | [[Category: GATE2010]] | ||
| − | [[Category: | + | [[Category: Calculus questions from GATE]] |
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