NUMERICAL METHODS
(Meng2063)
CHAPTER TWO
ROOT OF EQUATION

2.1. INTRODUCTION
• Students should be familiar with from ordinary algebra is finding the root
of an equation 𝑓(𝑥) = 0, (the value of the argument that makes f zero).
• More precisely, if the function is defined as y = f(x), we seek the value α
such that.
𝑓 𝛼 = 0 (2.1.1)
• A polynomial of degree n has exactly n roots, real or complex, simple or
multiple, whereas a transcendental equation may have one root, infinite
root, or no root.
• Given one or two initial approximation to the root we require suitable
iteration for 𝑓 𝑥 , such that the sequence of iterates 𝑥𝑘 converges to
exact root.

Cont . . .
Criteria’s to terminate iteration procedures:
• Since we cannot perform infinite number of iterations, we need a criterion to stop the iteration.
• Use either one or two of the following criterion:
i. The equation 𝑓 𝑥 = 0 is satisfied to a given accuracy or error bound 𝜀.
𝑓(𝑥𝑖) ≤ 𝜀
ii. The magnitude of the difference between two successive iterates is smaller than a given accuracy
𝑥𝑖+1 − 𝑥𝑖 ≤ 𝜀
• Generally we will discuss the basic methods for finding the point 𝛼, bisection method,
Newton’s method, and Secant methods.

2.2. Bisection Method
• Bisection is a root finding method with the approach of bisecting the intervals to
approximate the real root.
• Suppose we have intervals a and b, know that f(a)f(b) < 0. This means that f is
negative at one point and positive at the other.
• If we assume that f is continuous, then from Intermediate Value Theorem it follows
that there must be some value between a and b at which f is zero.
• Now let's try to use these ideas to find 𝜶. Let c be the midpoint of the interval [a, b]
𝒄 =
𝟏
𝟐
𝒂 + 𝒃 𝟐. 𝟐. 𝟏

Procedure for Bisection method:
1. Choose 𝒂 as the lower and 𝒃 as the upper guesses for the root. Such that the
function changes sign over the interval. This can be done by evaluating the
𝑓(𝑥) at a and b or by plotting the graph of the function.
2. Estimate the root c from equation 2.2.1, and find 𝒇(𝒄)
3. Use the following evaluations to determine the next interval where the root lies
i. If f(a) f(c) > 0; hence the root is between c and b, i.e., 𝜶 ∈ [c, b].
ii. If f(a) f(c) < 0; hence the real root lies between a and c, i.e., 𝜶 ∈ [a, c].
iii. If f(c) = 0; if we assume that we already know 𝜶 ≠ 𝟎, this means that
𝑓(𝑐) = 0, if this is the condition, we have found a root and terminate the
solution

Example 2.2.1: If 𝑓 𝑥 = 2 − 𝑒𝑥, and we take the original interval to be [a, b] = [0, 1]. Find
the most approximate solution with absolute percent error less than 0.1%.
y
f(x)
x
a b
f(b)
f(a)
c1
f(c1)
c2
f(c2)
c3
c4 c5
f(c3)
f(c4)
𝐶1 =
1
2
𝑎 + 𝑏
𝐶2 =
1
2
(𝑎 + 𝐶1)
𝐶3 =
1
2
(𝐶2 + 𝐶1)
𝐶4 =
1
2
(𝐶2 + 𝐶3)
𝐶5 =
1
2
(𝐶4 + 𝐶3)
𝑪 =
𝟏
𝟐
𝒂 + 𝒃

2.3. Newton’s Method
• It is the classic algorithm for finding roots of functions. It is often introduced in
the calculus sequence as an application of the derivative of a function.
• There are two good derivations of Newton's method, geometric and analytic.
• Here the geometric derivation is discussed.
• Consider the figure, We wish to find a
root 𝛼 of 𝑦 = 𝑓(𝑥), given an "initial guess"
of 𝑥𝑛.
• The fundamental idea in Newton's method
is to use the tangent line approximation to
the function f at point (𝑥𝑛, 𝑓(𝑥𝑛).

Cont . . .
• The tangent line to the graph of the function 𝑓(𝑥) at (𝑥𝑛, 𝑓(𝑥𝑛)), is defined as straight line
determined by the limit position of the secant line as the point (𝒙𝒏, 𝒇(𝒙𝒏)), tends
to (𝒙𝒏+𝟏, 𝒇(𝒙𝒏+𝟏)) along the graph
• If α is the angle between the x-axis and the tangent line, then 𝒇’(𝒙𝒏) = 𝒕𝒂𝒏𝜶 is the slope
of the tangent line to the curve at point 𝒙𝒏, 𝒚𝒏 .
• Then the point-slope formula for the equation of the straight line gives us
𝑓′ 𝑥𝑛 =
𝑦−𝑦𝑛
(𝑥−𝑥𝑛)
(2.3.1)
• From equation (2.3.1) we have;
𝑦 − 𝑦𝑛 = 𝑓′ 𝑥𝑛 𝑥 − 𝑥𝑛 (2.3.2)
• Recall 𝑦𝑛 = 𝑓 𝑥𝑛 ; therefore we have a straight line with equation
𝑦 = 𝑓 𝑥𝑛 + 𝑓′ 𝑥𝑛 𝑥 − 𝑥𝑛
• To find where this crosses the axis, we set y = 0 and solve for x
𝒙 = 𝒙𝒏 −
𝒇 𝒙𝒏
𝒇′ 𝒙𝒏
, 𝑓′ 𝑥𝑛 ≠ 0 (2.3.3)
Example 2.3.1. Use Newton
Raphson method to find approximate
root of 𝑓 𝑥 = 𝑥6
− 𝑥 − 1.Use initial
guess of 𝑥0 =1. Conduct the iteration
until the approximate percent error
drops below 0.1%.

2.4. Secant Method
• The general formula to find estimate of the root of a function in secant method is derived by
using equation of straight line. Equation of straight line is given by;
𝑦 = 𝑚𝑥 + 𝑧 (2.4.1)
• Consider the figure; and the slope m is given by;
𝑚 =
𝑓(𝑥𝑖)−𝑓(𝑥𝑖−1)
𝑥𝑖−𝑥𝑖−1
(2.4.2)
• Y – intercept Z can be obtained as
𝑧 = 𝑓 𝑥𝑖−1 − 𝑚(𝑥𝑖−1)
𝑜𝑟 𝑧 = 𝑓(𝑥𝑖) − 𝑚 𝑥𝑖
2.4.3
• Substituting eq (2.4.2) and (2.4.3) into (2.4.1)
𝑦 =
𝑓 𝑥𝑖 −𝑓 𝑥𝑖−1
𝑥𝑖−𝑥𝑖−1
𝑥 + 𝑓 𝑥𝑖−1 −
𝑓 𝑥𝑖 −𝑓 𝑥𝑖−1
𝑥𝑖−𝑥𝑖−1
𝑥𝑖−1
𝑦 =
𝑓 𝑥𝑖 −𝑓 𝑥𝑖−1
𝑥𝑖−𝑥𝑖−1
𝑥 − 𝑥𝑖−1 + 𝑓 𝑥𝑖−1 (2.4.4)

Cont . . .
• Since we are looking for the value which makes the function equals to zero,
set 𝑦 = 0.
0 = 𝑓 𝑥𝑖 − 𝑓 𝑥𝑖−1 𝑥 − 𝑥𝑖 + 𝑓 𝑥𝑖 𝑥𝑖 − 𝑥𝑖−1 (2.4.5)
0 = 𝑥 − 𝑥𝑖 +
𝑓 𝑥𝑖 𝑥𝑖−𝑥𝑖−1
𝑓 𝑥𝑖 −𝑓 𝑥𝑖−1
(2.4.6)
𝑥 = 𝑥𝑖 − 𝑓 𝑥𝑖
𝑥𝑖−𝑥𝑖−1
𝑓 𝑥𝑖 −𝑓 𝑥𝑖−1
(2.4.7)
• Thus equation (2.4.7) is the used to find the next approximate root in secant
method;
𝒙𝒊+𝟏 = 𝒙𝒊 − 𝒇 𝒙𝒊
𝒙𝒊−𝒙𝒊−𝟏
𝒇 𝒙𝒊 −𝒇 𝒙𝒊−𝟏
(2.4.8)
Example 2.6.1:
Determine the real root of 𝑓 𝑥 = 4𝑥3 − 2, using secant method to locate the root. Employ initial
guesses of 𝑥0 = 0, and 𝑥1= 1. Iterate until the absolute percent error fall below 0.5%

THANK YOU!