The following equation needs to be numerically solved using the Newton $$-$$ Raphson method $${x^3} + 4x - 9 = 0.\,\,$$ The iterative equation for this purpose is ($$k$$ indicates the iteration level)

Given $$a>0,$$ we wish to calculate it reciprocal value $${1 \over a}$$ by using Newton - Raphson method for $$f(x)=0.$$ The Newton - Raphson algorithm for the function will be

Let $$\,\,f\left( x \right) = x - \cos \,x.\,\,\,$$ Using Newton-Raphson method at the $$\,{\left( {n + 1} \right)^{th}}$$ iteration, the point $$\,{x_{n + 1}}$$ is computed from $${x_n}$$ as

Given the differential equation $${y^1} = x - y$$ with initial condition $$y(0)=0.$$ The value of $$y(0.1)$$ calculated numerically upto the third place of decimal by the $${2^{nd}}$$ order Runge-Kutta method with step size $$h=0.1$$ is