1. Roots Of The Equations
Karakoti
By- 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 εstep = 0.01, εabs = 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 εstep = 0.01, εabs = 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) + 2 sin(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 εstep = 0.001 and εabs = 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) + 2 cos(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) + 2 sin(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.