1. Roots Of The EquationsKarakotiBy- Digvijay Singh

2. Roots Of Equations A root or solutions of equation f(x)=0 are the values of x for which the equation holds true. Sometimes roots of equations are called as zeros of the equation. Numerical methods for finding roots of equations can often be easily programmed. 3. Methods Under Roots Of The Equation Bisection Method The False-position Method The Newton Raphson Method 4. The Bisection Method:The bisection method is one type of incremental search method where the interval is always divided in the half. The root is determined as lying in the midpoint of the subinterval within which the sign change occurs. The process is iterative. 5. Bisection Method Example: Consider finding the root of f(x) = x2 - 3. Let step = 0.01, abs = 0.01 and start with the interval [1, 2]. Table 1. Bisection method applied to f(x)=x2 - 3.Thus, with the seventh iteration, we note that the final interval, [1.7266, 1.7344], has a width less than 0.01 and |f(1.7344)| < 0.01, and therefore we chose b = 1.7344 to be our approximation of the root. 6. The False-position Method:An alternative method that exploits its graphical insight is to join f(xi) and f(xu) by a straight line. The intersection of this line with the x-axis represents an improved estimate of the root. The fact that the replacement of the curve by a straight line gives a false position section of the line with the x-axis can be estimated as 7. False Position Method Example: Consider finding the root of f(x) = x2 - 3. Let step = 0.01, abs = 0.01 and start with the interval [1, 2]. Table 1. False-position method applied to f(x)=x2 - 3. 8. False Position Method Example Continues.. Thus, with the third iteration, we note that the last step 1.7273 1.7317 is less than 0.01 and |f(1.7317)| < 0.01, and therefore we chose b = 1.7317 to be our approximation of the root. Note: After three iterations of the false-position method, we have an acceptable answer (1.7317 where f(1.7317) = -0.0044) whereas with the bisection method, it took seven iterations to find a (notable less accurate) acceptable answer (1.71344 where f(1.73144) = 0.0082) 9. The Newton Raphson Method:If the initial guess at the root is xi, a tangent can be extended from the point {xi,x(xi)}. The point where this tangent crosses the x-axis usually represents an improved estimate of the root. The Newton Raphson formula is: 10. Newton Raphson Example: As an example of Newton's method, suppose we wish to find a root of the function f(x)=cos(x)+2sin(x)+x2. A closed form solution for x does not exist so we must use a numerical technique. We will use x0=0 as our initial approximation. We will let the two values step=0.001 and abs=0.001 and we will halt after a maximum of N=100 iterations. From calculus, we know that the derivative of the given function is f(1)(x)=sin(x)+2cos(x)+2x. 11. Newton Raphson Example continues.. We will use four decimal digit arithmetic to find a solution and the resulting iteration is shown in Table 1. Table 1. Newton's method applied to f(x)=cos(x)+2sin(x)+x2.Thus, with the last step, both halting conditions are met, and therefore, after four iterations, our approximation to the root is -0.6598.