Engineering Mat...

The following equation needs to be numerically solved by using Newton – Raphson method.

x log₁₀ x – 1.2 = 0

what is the type of convergence of Secant method

In case of Newton-Raphson algorithm, the type of convergence is

Newton-Raphson method is to be used to find the root of equation 4x – e^2x + sin x = 0. If the initial trial value for the root is taken as 0.5, the next approximation for the root would be _______ (corrected up to three decimal places)

If the differential equation \(\frac{{dy}}{{dx}} = x + 2y,\;y\left( 0 \right) = 0\) is solved using the Euler’s method with step size h = 0.25, then y₂ is equal to ______ (up to three decimal places).

The Newton-Raphson iteration \({X_{n + 1}} = \frac{1}{3}\left( {2{x_n} + \frac{N}{{X_n^2}}} \right)\) can be used to compute

The second approximation to a real root of the equation x³ – 2x – 5 = 0 by method of false position between 2 and 3 is ______

Only one of the real roots of f(x) lies in the interval 2 ≤ x ≤ 4 and bisection method is used to find its value. The minimum number of iterations required to achieve an accuracy of 0.2% is ______

An equation \(\sin x = \frac{1}{x}\) is required to be solved by the Bisection method where x lies between 1 and 1.5 (x is in radian). The approximate root of the fourth iteration will be: (Correct upto four decimal places)

Consider the first order initial value problem y’ + y = 0, y(0) = 1 for x = 0.1, the solution obtained using a single iteration  of the third order Runge Kutta method with step-size h = 0.1 is _________.