lated from the user.Get the desired precision.While more numbers remain, calculate firs guess, x0.RepeatXn = 0.5 * (X (n-1) + Number/ X (n-1))Until abs (Xn - X (n 1)) *= Desired precisionGet a positive number whose square root is to be calculated from the user.Get the desired precision.End whileNewton-Raphson Method1.Set number of iterations num_iter to zero.2.Set previous estimate of the root (x_prev) to initial guess (x_init).3.Set current estimate of the root (x_curr) to initial guess (x_init).4.While num_iter * max_iter do the followinga.Compute the value of the derivative at x_prev from derx = df (x_curr);b.If derx is less than epsilon (a value close to zero) return (1)c.Compute new estimate of the root from x_curr = x_prev f (x_prev)/ df (x_prev)d. Increment number of iterationse. Set new value of root equal to x_currf.If estimate of root is within desired tolerance then return (0)5.Return (2)Gauss-Seidel MethodX1 (1) = b2/a11 = 10/10 = 1.0X2 (1) = b2/a22 = 20/5 = 4.0X3 (1) = b3/a33 = -12/6 = -2.0The second approximation to the solution isX1 (2) = (b1 a12x2 (1) a13x3 (1))/a11 = [10 2(4) 3(-2.0)]/10X1 (2) = 8/10 = .8X2 (2) = (b2 a21x1 (2) a23x3 (1))/a22 = [20 1(.8) 2(-2.0]/5 = 4.64X3 (2) = b3 a31x1 (2)-a32x2 (2)/a33 =-12 3(0.8)-2(4.64)/6= -3.947Substituting the above values, we obtain the third approximation fromX1 (3) = b1-a12x2 (2) a13x3 (2)/a11 = 10 2(4.64) 3(-3.947)/10 = 1.256X2 (3) = b2-a21x1 (3) a23x3 (2)/a22 = 20 1(1.256) 2(-3.947)/ 5 = 5.328X3 (3) = b3 a31x1 (3) a32x2 (3)/a33 = -12 3(1.256) 2(5.328)/6 = -4.404...
