List of top Numerical Analysis Questions asked in GATE MA

The real sequence generated by the iterative scheme
\[ x_n = \frac{x_{n-1}}{2} + \frac{1}{x_{n-1}}, \quad n \geq 1 \]
If \( P(x) \) is a polynomial of degree 5 and \[ \alpha = \sum_{i=0}^{6} P(x_i) \left( \prod_{\substack{j=0
j\neq i}}^{6} (x_i - x_j)^{-1} \right), \] where \( x_0, x_1, \ldots, x_6 \) are distinct points in the interval \([2,3]\), then the value of \( \alpha^2 - \alpha + 1 \) is __________.
The initial value problem \[ \frac{dy}{dx} = f(x, y), \quad y(x_0) = y_0 \] is solved by using the following second-order Runge-Kutta method: \[ K_1 = h f(x_i, y_i) \] \[ K_2 = h f(x_i + \alpha h, y_i + \beta K_1) \] \[ y_{i+1} = y_i + \frac{1}{4} (K_1 + 3 K_2), \quad i \geq 0 \] where \( h \) is the uniform step length between the points \( x_0, x_1, \dots, x_n \) and \( y_i = y(x_i) \). The value of the product \( \alpha \beta \) is __________ (round off to TWO decimal places).
The first derivative of a function \( f \in C^\infty(-3, 3) \) is approximated by an interpolating polynomial of degree 2, using the data \[ (-1, f(-1)), (0, f(0)) \text{ and } (2, f(2)). \] It is found that \[ f'(0) \approx -\frac{2}{3} f(-1) + \alpha f(0) + \beta f(2). \] Then, the value of \( \frac{1}{\alpha \beta} \) is
The first derivative of a function \( f \in C^\infty(-3, 3) \) is approximated by an interpolating polynomial of degree 2, using the data \[ (-1, f(-1)), (0, f(0)) \text{ and } (2, f(2)). \] It is found that \[ f'(0) \approx -\frac{2}{3} f(-1) + \alpha f(0) + \beta f(2). \] Then, the value of \( \frac{1}{\alpha \beta} \) is

If the polynomial \[ p(x) = \alpha + \beta (x+2) + \gamma (x+2)(x+1) + \delta (x+2)(x+1)x \] interpolates the data \[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -2 & 2 \\ -1 & -1 \\ 0 & 8 \\ 1 & 5 \\ 2 & -34 \\ \hline \end{array} \] then \( \alpha + \beta + \gamma + \delta = \underline{\hspace{1cm}}\) .

The quadrature formula \[ \int_0^2 x f(x) \, dx \approx \alpha f(0) + \beta f(1) + \gamma f(2) \] is exact for all polynomials of degree \( \leq 2 \). Then \( 2 \beta - \gamma = \underline{\hspace{1cm}} \).

The initial value problem \[ \frac{dy}{dt} = f(t, y), t > 0, y(0) = 1, \] where \( f(t, y) = -10 y \), is solved by the following Euler method: \[ y_{n+1} = y_n + h f(t_n, y_n), n \geq 0, \text{with step-size} \, h. \] Then \( y_n \to 0 \) as \( n \to \infty \), provided
Let \( f(x) = x^4 + 2x^3 - 11x^2 - 12x + 36 \) for \( x \in \mathbb{R}. \) The order of convergence of the Newton-Raphson method \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, n \geq 0, \] with \( x_0 = 2.1 \), for finding the root \( \alpha = 2 \) of the equation \( f(x) = 0 \) is \(\underline{\hspace{1cm}}\) .
Consider the fixed-point iteration \[ x_{n+1} = \varphi(x_n), n \geq 0, \] with \[ \varphi(x) = 3 + (x - 3)^3, x \in (2.5, 3.5), \] and the initial approximation \( x_0 = 3.25 \). Then, the order of convergence of the fixed-point iteration method is