MCA_UNIT-1_Computer Oriented Numerical Statistical Methods

RAI UNIVERSITY, AHMEDABAD 1
Course: MCA
Subject: Computer Oriented Numerical
Statistical Methods
Unit-1
RAI UNIVERSITY, AHMEDABAD

Unit-I- Solution of Non-linear Equations
Sr.
No.
Name of the Topic Page
No.
1 Absolute, Relative and Percentage Error 1
2 Roots ofan equation, Linear and non-Linear equations
(Definition and Difference),
4
3 Iterative Methods for finding roots of non-Linear equations :
Bisection Method
5
Newton-Raphson Method
14
False Position Method
18
secant Method
25
7 Exercise 29
8 References 30

1.1 Errors
In practice, all measurements are subject to errors and it is therefore of importance
to find the total effect of small errors in the observed values of the variables, on a
quantity which depends on these variables.
If 𝛿𝑥 is an error in 𝑋 then
| 𝛿𝑥| = | 𝑇𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒 − 𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑣𝑎𝑙𝑢𝑒| is called an absolute error in 𝑋,
𝛿𝑥
𝑥
is called the relative error in 𝑋 and
𝛿𝑥
𝑥
× 100 is called the percentage error in 𝑋.
Example—
A circle of radius 𝒓 = 𝟓 cm having an error in radius 𝟏𝟎% ,then find
percentage errorin its area.
Solution:
Area of circle 𝐴 = 𝜋𝑟2
∴ 𝐴 = 𝜋𝑟2
⇒ 𝛿𝐴 = 2𝜋𝑟𝛿𝑟
Now
𝛿𝐴
𝐴
=
2𝜋𝑟𝛿𝑟
𝜋 𝑟2
𝛿𝐴
𝐴
=
2𝛿𝑟
𝑟
∴
𝛿𝐴
𝐴
× 100 = 2(
𝛿𝑟
𝑟
× 100)
∴
𝛿𝐴
𝐴
× 100 = 2(10)
∴
𝛿𝐴
𝐴
× 100 = 20
Therefore percentage error in Area of circle is 20.
Example—

A cuboid having error in their sides equal to 𝟐%, 𝟓% 𝒂𝒏𝒅 𝟏%. Then find the
corresponding error in its volume.
Solution:
Volume of cuboid is 𝑉 = 𝑙 × 𝑤 × ℎ
𝛿𝑉 = 𝑤ℎ𝛿𝑙 + 𝑙ℎ𝛿𝑤 + 𝑤𝑙𝛿ℎ
Now we have to given that, 𝑙 = 2%, 𝑤 = 5%,ℎ = 1%
𝛿𝑉
𝑉
=
𝑤ℎ𝛿𝑙+𝑙ℎ𝛿𝑤+𝑤𝑙𝛿ℎ
𝑙×𝑤×ℎ
𝛿𝑉
𝑉
× 100 = (
𝛿𝑙
𝑙
+
𝛿ℎ
ℎ
+
𝛿𝑤
𝑤
) × 100
𝛿𝑉
𝑉
× 100 = (
𝛿𝑙
𝑙
× 100)+ (
𝛿ℎ
ℎ
× 100)+ (
𝛿𝑤
𝑤
× 100)
𝛿𝑉
𝑉
× 100 = 2 + 5 + 1
𝛿𝑉
𝑉
× 100 = 8
Error in Volume of cuboid = 8%
Example—
A right circular cone of radius 𝒓 = 𝟏𝟎 𝒄𝒎 & ℎ = 20 𝑐𝑚 having percentage
error in radius equal to 𝟏% and percentage errorin height equal to 𝟓% then
find
(i) Absolute error
(ii) Relative error
(iii) Percentageerror
in the volume of right circular cone.
Solution:
Volume of right circular cone
𝑉 =
1
3
𝜋𝑟2
ℎ
𝛿𝑉 =
1
3
2𝑟𝛿𝑟ℎ +
1
3
𝜋𝑟2
𝛿ℎ

Now we have to given that 𝑟 = 10 𝑐𝑚,ℎ = 20 𝑐𝑚 ,
𝛿𝑟
𝑟
× 100 = 1,
𝛿ℎ
ℎ
× 100 = 5
𝛿𝑟 =
1×𝑟
100
& 𝛿ℎ =
5×ℎ
100
𝛿𝑉 =
1
3
𝜋 [2 × 𝑟 × ℎ × (
1×𝑟
100
) + 𝑟2
(
5×ℎ
100
)]
𝛿𝑉 =
1
3
𝜋 [2 × 10 × 20 ×
1×10
100
+
10×10×5×20
100
]
𝛿𝑉 =
1
3
𝜋[40+ 100]
𝛿𝑉 =
1
3
𝜋[140]
𝛿𝑉 = 146.53
∴Absolute error 𝛿𝑉 = 146.53
⇒
𝛿𝑉
𝑉
=
146.53
1
3
𝜋𝑟2ℎ
=
3×146.53
𝜋×100×20
∴
𝛿𝑉
𝑉
=
439.59
6283
= 0.06996 ≅ 0.07
∴ Relative error
𝛿𝑉
𝑉
= 0.07
𝛿𝑉
𝑉
× 100 = 0.07× 100 = 7
∴ Percentage error = 7%
2.1 Linear equation— General form of a linear equation is—
𝑎𝑥 + 𝑏𝑦 = 𝑐, where 𝑎, 𝑏 and 𝑐 are constants.
2.2 Graph of a linear equation— An equation 𝑎𝑥 + 𝑏𝑦 = 𝑐 represents a straight
line.
2.3 Solution of two linear equations— Solution of two linear equations 𝑎1 𝑥 +
𝑏1 𝑦 = 𝑐1 and 𝑎2 𝑥 + 𝑏2 𝑦 = 𝑐2 is given by—
𝑥 =
𝑐1 𝑏2−𝑐2 𝑏1
𝑎1 𝑏2−𝑎2 𝑏1
and 𝑦 =
𝑐2 𝑎1−𝑐1 𝑎2
𝑎1 𝑏2−𝑎2 𝑏1
.
Graphically this solution gives the point of intersection of these straight lines.

Example—
Find the solution of the equations 𝒙 + 𝟐𝒚 = 𝟑 and 𝟑𝒙 + 𝒚 = 𝟒.
Solution— given straight lines are 𝑥 + 2𝑦 = 3 and 3𝑥 + 𝑦 = 4.
Here, 𝑎1 = 1, 𝑏1 = 2, 𝑐1 = 3, 𝑎2 = 3, 𝑏2 = 1, 𝑐2 = 4
Hence, 𝑥 =
𝑐1 𝑏2−𝑐2 𝑏1
𝑎1 𝑏2−𝑎2 𝑏1
=
3.1−4.2
1.1−3.2
=
3−8
1−6
= 1 and 𝑦 =
𝑐2 𝑎1−𝑐1 𝑎2
𝑎1 𝑏2−𝑎2 𝑏1
=
4.1−3.3
1.1−3.2
=
4−9
1−6
= 1
2.4 Definition: Non-linear equation:
A polynomial of degree at least 2 is called nonlinear equation.
Graph of a non-linear equation is not a straight line.
e.g.𝑥3
− 3𝑥 + 1 = 0, 𝑐𝑜𝑠𝑥 − 𝑥𝑒 𝑥
= 0.
2.5 Here we use Numerical Methods for solving Non-linear equations like
Bisection Method
Regula-Falsi Method
Newton-Raphson Method
Secant Method
3.1 Introduction—
In the Applied mathematics and engineering the problem of determining the roots
of an equation 𝑓(𝑥) = 0 has a great importance.
 If 𝑓(𝑥) is a polynomial of degree two or more , we have formulae to find
solution. But, if f(x) is a transcendental function, we do not have formulae to
obtain solutions. When such type of equations are there, we have some
methods like Bisection method, Newton-Raphson Method and The method

of false position. Thosemethods are solved by using a theorem in theory of
equations, i.e., “If f(x) is continuous in the interval (𝑎, 𝑏) and if 𝑓(𝑎) and
𝑓(𝑏) are of opposite signs, then the equation 𝑓(𝑥) = 0 will have at least
one real root between 𝑎and 𝑏”.
3.2 Definition—
A number ξ is a solution of f(x) = 0 if f(ξ) = 0 . Such a solution ξ is a rootof
f(x) = 0.
3.3 Iterative Method—
This method is based on the idea of successiveapproximation.i.e. Starting with one
or more initial approximations to the root, we obtain a sequence of approximations
or iterates { 𝑥 𝑘 }, which in the limit converges to the root.
Definition—
A sequence of iterates { 𝑥 𝑘 } is said to converge to the root 𝜉, if lim
𝑘⟶∞
| 𝑥 𝑘 − ξ| =
0 𝑜𝑟 lim
𝑘⟶∞
𝑥 𝑘 = ξ.
3.4 BisectionMethod—
Let us supposewe have an equation of the form 𝑓(𝑥) = 0 in which solution lies
between in the range (𝑎, 𝑏).
Also 𝑓(𝑥) is continuous and it can be algebraic or transcendental. If 𝑓(𝑎) and
𝑓(𝑏)are oppositesigns, then there exist at least one real root between 𝑎and 𝑏.

Let 𝑓(𝑎) be positive and 𝑓(𝑏) negative. Which implies at least one root exits
between a and b. We assume that root to be
𝑥0 =
(𝑎+𝑏)
2
.
Check the sign of 𝑓(𝑥0). If 𝑓(𝑥0)is negative, the root lies between a and xo. If
𝑓( 𝑥0)is positive, the root lies between 𝑥0and b. Subsequently any one of this case
occur.
𝑥1 =
𝑎+𝑥0
2
or𝑥1 =
𝑥0+𝑏
2
When 𝑓(𝑥1)is negative, the root lies between 𝑥0and 𝑥1 and let the root be 𝑥2 =
𝑥0+𝑥1
2
. Again𝑓(𝑥2)negative then the root lies between 𝑥0and 𝑥2, let 𝑥3 =
𝑥1+𝑥2
2
and
so on.
Repeat the process𝑥0,𝑥1, 𝑥2,…Whoselimit of convergence
is the exact root.
3.5 Steps—

1. Find a and b in which f(a) and f(b) are oppositesigns for the given equation
using trial and error method.
2. Assume initial root as
𝑥0 =
(𝑎 + 𝑏)
2
3. If 𝑓(𝑥0) is negative, the root lies between a and 𝑥0 . And takes the root as
𝑥1 =
(𝑎 + 𝑥0)
2
4. If 𝑓(𝑥0)is positive, then the root lies between 𝑥0 and b.And takes the root as
𝑥1 =
(𝑥0 + 𝑏)
2
5. If 𝑓(𝑥1) is negative, the root lies between 𝑥0and 𝑥1 . And takes the root as
𝑥2 =
(𝑥0 + 𝑥1)
2
6. If 𝑓(𝑥1) is positive, the root lies between 𝑥0and 𝑥2 . And takes the root as
𝑥2 =
(𝑥0 + 𝑥2)
2
7. Repeat the process until any two consecutive values are equal and hence the
root.

This method is simple to use and the sequence of approximations always
converges to the root for any 𝑓(𝑥) which is continuous in the interval that contains
the root.
If the permissible error is𝜀, then the approximation number of iteratives required
may be determined from the relation
𝑏0−𝑎0
2 𝑛
≤ 𝜀.
The minimum number of iterations required for converging to a root in the interval
(0,1) for a given 𝜀 are listed as
Number of iterations
𝜺 𝟏𝟎−𝟐
𝟏𝟎−𝟑
𝟏𝟎−𝟒
𝟏𝟎−𝟓
𝟏𝟎−𝟔
𝟏𝟎−𝟕
n 7 10 14 17 20 24
The bisection method requires a large number of iterations to achieve a reasonable
degree of accuracy of the root. It requires one function at each iteration.
Example—
Find the positive root of 𝒙 𝟑
− 𝒙 = 𝟏correctto four decimal places by bisection
method.
Solution—
Let 𝑓( 𝑥) = 𝑥3
− 𝑥 − 1
𝑓(0) = 03
− 0 − 1 = −1 = −𝑣𝑒
𝑓(1) = 13
− 1 − 1 = −1 = −𝑣𝑒
𝑓(2) = 23
− 2 − 1 = 5 = +𝑣𝑒
So the root lies between the 1 and 2.
We can take 𝑥0 =
1+2
2
=
3
2
= 1.5 as initial root and proceed.
i.e., 𝑓(1.5) = 0.8750 = +𝑣𝑒
and 𝑓(1) = −1 = −𝑣𝑒

So the root lies between 1 and 1.5.
𝑥1 =
1 + 1.5
2
=
2.5
2
= 1.25
as initial root and proceed.
𝑓(1.25) = −0.2969
So the root 𝑥2lies between 1.25 and 1.5.
Now
𝑥2 =
1.25+ 1.5
2
=
2.75
2
= 1.3750
𝑓(1.375) = 0.2246 = +𝑣𝑒
So the root 𝑥3lies between 1.25 and 1.375.
Now
𝑥3 =
1.25+ 1.3750
2
=
2.6250
2
= 1.3125
𝑓(1.3125) = −0.051514 = −𝑣𝑒
Therefore, root lies between 1.375and 1.3125
Now
𝑥4 = ( 1.375 + 1.3125) / 2 = 1.3438
𝑓(1.3438) = 0.082832 = +𝑣𝑒
So root lies between 1.3125 and 1.3438
Now
𝑥5 = ( 1.3125 + 1.3438) / 2 = 1.3282

𝑓(1.3282) = 0.014898 = +𝑣𝑒
Now
𝑥6 = ( 1.3125 + 1.3282) / 2 = 1.3204
𝑓(1.3204) = −0.018340 = −𝑣𝑒
Now
𝑥7 = ( 1.3204 + 1.3282) / 2 = 1.3243
𝑓(1.3243) = −𝑣𝑒
So root liesbetween 1.3243and 1.3282
Now
𝑥8 = ( 1.3243 + 1.3282) / 2 = 1.3263
𝑓(1.3263) = +𝑣𝑒
Now
𝑥9 = ( 1.3243 + 1.3263) / 2 = 1.3253
𝑓(1.3253) = +𝑣𝑒
Now
𝑥10 = ( 1.3243 + 1.3253) / 2 = 1.3248
𝑓(1.3248) = +𝑣𝑒
Now

𝑥11 = ( 1.3243 + 1.3248) / 2 = 1.3246
𝑓(1.3246) = −𝑣𝑒
Now
𝑥12 = ( 1.3248 + 1.3246) / 2 = 1.3247
𝑓(1.3247) = −𝑣𝑒
Now
𝑥13 = ( 1.3247 + 1.3247) / 2 = 1.32475
Hence the approximate root is 1.32475 .
Example—
Find the positive root of 𝒙– 𝒄𝒐𝒔 𝒙 = 𝟎 by bisection method.
Solution—
Let
𝑓( 𝑥) = 𝑥– 𝑐𝑜𝑠 𝑥 , 𝑓(0) = 0 − 𝑐𝑜𝑠 (0) = 0 − 1 = −1 = −𝑣𝑒
𝑓(0.5) = 0.5 –𝑐𝑜𝑠 (0.5) = −0.37758 = −𝑣𝑒
𝑓(1) = 1 – 𝑐𝑜𝑠 (1) = 0.42970 = +𝑣𝑒
So root lies between 0.5 and 1
Let
𝑥0 =
(0.5 + 1)
2
= 0.75
as initial root and proceed.
𝑓(0.75) = 0.75 – 𝑐𝑜𝑠 (0.75) = 0.018311 = +𝑣𝑒

𝑥1 = (0.5 + 0.75)/2 = 0.62
𝑓(0.625) = 0.625 – 𝑐𝑜𝑠 (0.625) = −0.18596
𝑥2 = (0.625 + 0.750) /2 = 0.6875
𝑓(0.6875) = −0.085335
𝑥3 = (0.6875 + 0.750) /2 = 0.71875
𝑓(0.71875) = 0.71875 − 𝑐𝑜𝑠(0.71875) = −0.033879
𝑥4 = (0.71875 + 0.750) /2 = 0.73438
𝑓(0.73438) = −0.0078664 = −𝑣𝑒
𝑥5 = 0.742190
𝑓(0.742190) = 0.0051999 = + 𝑣𝑒
𝑥6 = (0.73438 + 0.742190) /2 = 0.73829
𝑓(0.73829) = −0.0013305
𝑥7 = (0.73829+ 0.74219) = 0.7402
𝑓(0.7402) = 0.7402
𝑐𝑜𝑠(0.7402) = 0.0018663
𝑥8 = 0.73925
𝑓(0.73925) = 0.00027593

𝑥9 = 0.7388
The rootis 0.7388.
3.6 Advantages of bisection method—
a) The bisection method is always convergent. Since the method brackets the
root, the method is guaranteed to converge.
b) As iterations are conducted, the interval gets halved. So one can guarantee
the error in the solution of the equation.
3.7 Drawbacksofbisectionmethod—
a) The convergence of the bisection method is slow as it is simply based on
halving the interval.
b) If one of the initial guesses is closer to the root, it will take larger number of
iterations to reach the root.
4.1 Newton-Raphsonmethod (or Newton’s method)—
Supposewe have an equation of the form 𝑓(𝑥) = 0, whose solution lies between
the range( 𝑎, 𝑏).Also 𝑓(𝑥)is continuous and it can be algebraic or transcendental. If
𝑓(𝑎)and 𝑓(𝑏)are oppositesigns, then there exist at least one real root between
𝑎and 𝑏.

Let 𝑓(𝑎)be positive and 𝑓(𝑏)negative. Which implies at least one rootexits in
between 𝑎and 𝑏. We assume that root to be either a or b, in which the value of
𝑓(𝑎)or𝑓(𝑏) is very close to zero. That number is assumed to be initial root. Then
we iterate the process byusing the following formula until the value is converges.
𝑥 𝑛+1 = 𝑥 𝑛 −
𝑓(𝑥 𝑛)
𝑓′(𝑥 𝑛)
4.2 Steps—
1. Find 𝑎and 𝑏 in which 𝑓(𝑎)and 𝑓(𝑏) are opposite signs for the given
equation using trial and error method.
2. Assume initial root as 𝑥0 = 𝑎 i.e., if 𝑓(𝑎)is very close to zero or 𝑥0 = 𝑏
if 𝑓(𝑎)is very close to zero.
3. Find 𝑥1 by using the formula
𝑥1 = 𝑥0 −
𝑓(𝑥0)
𝑓′(𝑥0)
4. Find x2 by using the following formula
𝑥2 = 𝑥1 −
𝑓( 𝑥1)
𝑓′( 𝑥1)
5. Find x3, x4, .. . xn until any two successivevalues are equal .
Example 1—
Find the positive root of 𝒇(𝒙) = 𝟐𝒙 𝟑
− 𝟑𝒙 − 𝟔 = 𝟎by Newton– Raphson
method correctto five decimal places.
Solution—
Let 𝑓(𝑥) = 2𝑥3
− 3𝑥 – 6 ; 𝑓 ′(𝑥) = 6𝑥2
– 3

𝑥 𝑛+1 = 𝑥 𝑛 −
𝑓( 𝑥 𝑛)
𝑓′( 𝑥 𝑛)
= 𝑥 𝑛 −
2𝑥 𝑛
3
− 3𝑥 𝑛 – 6
6𝑥 𝑛
2 – 3
=
(6𝑥 𝑛
3
− 3𝑥 𝑛)–(2𝑥 𝑛
3
− 3𝑥 𝑛 – 6)
6𝑥 𝑛
2 – 3
=
6𝑥 𝑛
3
− 3𝑥 𝑛–2𝑥 𝑛
3
+ 3𝑥 𝑛 + 6
6𝑥 𝑛
2 – 3
=
4𝑥 𝑛
3
+ 6
6𝑥 𝑛
2 – 3
𝑓(1) = 2 − 3 − 6 = −7 = −𝑣𝑒
𝑓(2) = 16 – 6 − 6 = 4 = +𝑣𝑒
So, a root between 1 and 2 .
In which 4 is closer to 0. Hence we assume initial root as 2.
Consider 𝑥0 = 2
So
𝑥1 =
4𝑥0
3
+ 6
6𝑥0
2 – 3
=
4 × 23
+ 6
6 × 22 − 3
=
38
21
= 1.809524
𝑥2 = (4(1.809524)3
+ 6)/(6(1.809524)2
− 3) = 29.700256/16.646263
= 1.784200
𝑥3 = (4(1.784200)3
+ 6)/(6(1.784200)2
− 3) = 28.719072/16.100218
= 1.783769
𝑥4 = (4(1.783769)3
+ 6)/(6(1.783769)2
− 3) = 28.702612/16.090991
= 1.783769
Since 𝑥3 and 𝑥4 are equal, hence rootis 1.783769 .

Example 2—
Using Newton’s method, find the rootbetween0 and 1 of 𝒙 𝟑
= 𝟔𝒙 – 𝟒 correct
to 5 decimal places.
Solution—
Let 𝑓( 𝑥) = 𝑥3
− 6𝑥 + 4
𝑥 𝑛+1 = 𝑥 𝑛 −
𝑓( 𝑥 𝑛)
𝑓′( 𝑥 𝑛)
= 𝑥 𝑛 −
𝑥 𝑛
3
− 6𝑥 𝑛 + 4
3𝑥 𝑛
2 − 6
=
(3𝑥 𝑛
3
− 6𝑥 𝑛) − ( 𝑥 𝑛
3
− 6𝑥 𝑛 + 4)
3𝑥 𝑛
2 − 6
=
3𝑥 𝑛
3
− 6𝑥 𝑛 − 𝑥 𝑛
3
+ 6𝑥 𝑛 − 4
3𝑥 𝑛
2 − 6
=
2𝑥 𝑛
3
− 4
3𝑥 𝑛
2 − 6
𝑓(0) = 4 = +𝑣𝑒; 𝑓(1) = −1 = −𝑣𝑒
So a root lies between 0 and 1
𝑓(1)is nearer to 0. Therefore we take initial root as 𝑥0 = 1
𝑥1 =
2𝑥 𝑛
3
− 4
3𝑥 𝑛
2 − 6
=
2 × 13
− 4
3 × 12 − 6
=
2 − 4
3 − 6
=
−2
−3
= 0.66666
𝑥2 = (2(0.66666)3
– 4 )/(3(0.66666)2
− 6) = 0.73016
𝑥3 = (2(0.73015873)3
– 4 )/(3(0.73015873)2
− 6)
= (3.22145837/ 4.40060469) = 0.73205
𝑥4 = (2(0.73204903)3
– 4 )/(3(0.73204903)2
− 6)
= (3.21539602/ 4.439231265) = 0.73205
The root is 0.73205 correct to 5 decimal places.
4.3 Advantages of NewtonRaphsonmethod—

Fast convergence!
4.4 DisadvantagesofNewtonRaphsonmethod—
a) May not producea root unless the starting value 𝑥0 is close to the actual root
of the function.
b) May not producea root if for instance the iterations get to a point 𝑥 𝑛−1such
that 𝑓′( 𝑥 𝑛−1) = 0 . then the method is false.
5.1 Method of False Position ( or RegulaFalsi Method )—
Let us consider the equation f(x) = 0and f(a) and f(b) are of oppositesigns. Also
let a < 𝑏.
The graph y = f(x) will meet the x-axis at some point between A(a, f(a)) and
B (b,f(b)).The equation of the chord joining the two points A(a,f(a)) and
B (b,f(b)) is
𝑦 – 𝑓(𝑎)
𝑥 − 𝑎
=
𝑓(𝑎) − 𝑓(𝑏)
𝑎 − 𝑏
The x- Coordinate of the point of intersection of this chord with the x-axis gives an
approximate value for the of 𝑓(𝑥) = 0. Taking y = 0 in the chord equation, we get
– 𝑓(𝑎)
𝑥 − 𝑎
=
𝑓(𝑎) − 𝑓(𝑏)
𝑎 − 𝑏
⟹ 𝑥[𝑓(𝑎) − 𝑓(𝑏)] − 𝑎 𝑓(𝑎) + 𝑎 𝑓(𝑏) = −𝑎 𝑓(𝑎) + 𝑏 𝑓(𝑎)
⟹ 𝑥[ 𝑓( 𝑎) − 𝑓( 𝑏)] = 𝑏 𝑓( 𝑎) − 𝑎 𝑓( 𝑏)
⟹ 𝑥 =
𝑏 𝑓( 𝑎) − 𝑎 𝑓( 𝑏);
𝑓( 𝑎) − 𝑓( 𝑏)
This 𝑥1gives an approximate value of the root (𝑥) = 0.(𝑎 < 𝑥1 < 𝑏) .

Now 𝑓(𝑥1) and 𝑓(𝑎) are of opposite signs or 𝑓(𝑥1) and 𝑓(𝑏) are oppositesigns.
If𝑓( 𝑥1).𝑓(𝑎) < 0 . Then 𝑥2 lies between 𝑥1 and a.
𝑥2 =
𝑎 𝑓( 𝑥1)– 𝑥1 𝑓( 𝑏)
𝑓( 𝑥1) − 𝑓( 𝑎)
This process ofcalculation of 𝑥3, 𝑥4, 𝑥5 … is continued till any two successive
values are equal and subsequently we get the solution of the given equation.
5.2 Steps—
1. Find a and b in which f(a) and f(b) are oppositesigns for the given equation
using trial and error method.
2. Therefore root lies between 𝑎and 𝑏 if f(a)is very close to zero select and
compute 𝑥1 by using the following formula:
𝑥1 =
𝑎𝑓( 𝑏) − 𝑏𝑓(𝑎)
𝑓( 𝑏) − 𝑓(𝑎)

3. If (𝑥1 ),𝑓(𝑎) < 0 . then root lies between 𝑥1 and 𝑎 .Compute 𝑥2 by using
the
following formula:
𝑥2 =
𝑎𝑓( 𝑥1) − 𝑥1 𝑓(𝑏)
𝑓( 𝑥1) − 𝑓(𝑎)
4. Calculate the values of 𝑥3
, 𝑥4
, 𝑥5
, … by using the above formula until any
two successivevalues are equal and subsequently we get the solution of the
given equation.
Example —
Solve for a positive rootof 𝒙 𝟑
− 𝟒𝒙 + 𝟏 = 𝟎 by and Regula Falsimethod
Solution—
Let 𝑓(𝑥) = 𝑥3
− 4𝑥 + 1 = 0
𝑓(0) = 03
− 4 (0) + 1 = 1 = +𝑣𝑒
𝑓(1) = 13
− 4(1) + 1 = −2 = −𝑣𝑒
So a root lies between 0 and 1
We shall find the root that lies between 0 and 1.
Here 𝑎 = 0, 𝑏 = 1
𝑥1 =
0 × 𝑓(1) − 1 × 𝑓(0)
𝑓(1) − 𝑓(0)
=
0 × (−2) − 1 × 1
(−2) − 1
=
0 − 1
−3
=
1
3
= 0.333333
𝑓(𝑥1) = 𝑓(1/3) = (1/27) − (4/3) + 1 = −0.2963
Now 𝑓(0) and 𝑓(1/3) are oppositein sign.
Hence the root lies between 0 and 1/3.

𝑥2 =
0 × 𝑓 (
1
3
) −
1
3
𝑓(0)
𝑓 (
1
3
) − 𝑓(0)
=
−0.3333
−0.2963 − 1
=
−0.3333
−1.2963
= 0.25714
Now 𝑓(𝑥2) = 𝑓(0.25714) = − 0.011558 = −𝑣𝑒
So the root lies between 0 and 0.25714
𝑥3 =
0 × 𝑓(0.25714) − 0.25714 × 𝑓(0)
𝑓(0.25714) – 𝑓(0)
= −
0.25714
−1.011558
= 0.25420
𝑓(𝑥3) = 𝑓(0.25420) = −0.0003742
So the root lies between 0 and 0.25420
𝑥4 =
0 × 𝑓(0.25420) − 0.25420 × 𝑓(0)
𝑓(0.25420) – 𝑓(0)
= −0.25420 / −1.0003742
= 0.25410
𝑓(𝑥4) = 𝑓(0.25410) = − 0.000012936
The rootlies between 0 and 0.25410
𝑥5 =
0 × 𝑓(0.25410) − 0.25410 × 𝑓(0)
𝑓(0.25410) – 𝑓(0)
= −0.25410 / −1.000012936
= 0.25410
Hence the root is 0.25410.
Example 2—
Find an approximate root of x log 10𝑥 – 1.2 = 0 by False position method.

Solution—
Let 𝑓(𝑥) = 𝑥 𝑙𝑜𝑔10𝑥 – 1.2
𝑓(1) = −1.2 = −𝑣𝑒; 𝑓(2) = 2 × 0.30103 − 1.2 = − 0.597940
𝑓(3) = 3 × 0.47712 – 1.2 = 0.231364 = +𝑣𝑒
So, the rootlies between 2 and 3.
𝑥1 =
2𝑓(3) – 3𝑓(2)
𝑓(3) – 𝑓(2)
=
2 × 0.23136 – 3 × (−0.59794)
0.23136 + 0.597
= 2.721014
𝑓( 𝑥1) = 𝑓(2.7210) = − 0.017104 = −𝑣𝑒
The rootlies between 𝑥1 and 3.
𝑥2 =
𝑥1 × f(3)– 3 × f( 𝑥1)
f(3)– f( 𝑥1)
=
2.721014 × 0.231364 – 3 × (−0.017104)
0.23136 + 0.017104
=
0.629544+ 0.051312
0.24846
=
0.680856
0.248464
= 2.740260
𝑓( 𝑥2) = 𝑓(2.7402) = 2.7402 × 𝑙𝑜𝑔(2.7402)– 1.2 = − 0.00038905 = −𝑣𝑒
So the root lies between 2.740260 and 3.
𝑥3 =
2.7402 × f(3) – 3 × f(2.7402)
f(3) – f(2.7402)
=
2.7402 x 0.231336 + 3 x (0.00038905)
0.23136 + 0.00038905
=
0.63514
0.23175
= 2.740627

𝑓( 𝑥3) = 𝑓(2.7406) = 0.00011998 = +𝑣𝑒
So the root lies between 2.740260 and 2.740627
𝑥4 =
2.7402 x f(2.7406) – 2.7406 x f(2.7402)
f(2.7406) – f(2.7402)
=
2.7402 x 0.00011998 + 2.7406 x 0.00038905
0.00011998 + 0.00038905
=
0.0013950
0.00050903
= 2.7405
Hence the root is 2.7405.
5.3 Advantages of RegulaFalsiMethod—
No need to calculate a complicated derivative (as in Newton's method).
5.4 Disadvantages of RegulaFalsi Method—
a) May converge slowly for functions with big curvatures.
b) Newton-Raphson may be still faster if we can apply it.
5.5 Rate Of Convergence —
Here we discuss the rate at which the iteration method converges if the initial
approximation to the root is sufficiently close to the desired root.
5.5.1Definition—
An iterative method is said to be of order 𝑝 or has the rate of 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑝, if 𝑝
is the largest positive real number for which there exists a finite constant 𝐶 ≠ 0
such that

| 𝜀 𝑘+1| ≤ 𝐶| 𝜀 𝑘 | 𝑝
Where 𝜀 𝑘 = 𝑥 𝑘 − 𝜉 is the error in the kth iteratation. The constant 𝐶 is called the
asymptotic error constantand usually depends on the derivative of 𝑓(𝑥) at 𝑥 = 𝜉.
5.5.2 Newton-RaphsonsMethod—
We know
𝑥 𝑘+1 = 𝑥 𝑘 −
𝑓(𝑥 𝑘)
𝑓′(𝑥 𝑘)
, 𝑘 = 1,2,…
On substituting 𝑥 𝑘 = 𝜉 + 𝜀 𝑘, expanding 𝑓( 𝜉 + 𝜀 𝑘), 𝑓′
(𝜉 + 𝜀 𝑘 ) in Taylor series
about the point 𝜉, we get
𝜀 𝑘+1 = 𝜀 𝑘 −
𝜀 𝑘 𝑓′( 𝜉) +
1
2
𝜀 𝑘
2
𝑓′′( 𝜉) + ⋯
𝑓′( 𝜉) + 𝜀 𝑘 𝑓′′( 𝜉) + ⋯
= 𝜀 𝑘 − [
1
2
𝑓′′( 𝜉)
𝑓′( 𝜉)
𝜀 𝑘
2
+ ⋯] [1 +
𝑓′′( 𝜉)
𝑓′( 𝜉)
𝜀 𝑘 + ⋯]
−1
𝜀 𝑘+1 =
1
2
𝑓′′( 𝜉)
𝑓′( 𝜉)
𝜀 𝑘
2
+ 𝑂( 𝜀 𝑘
3).
On neglecting the 𝜀 𝑘
3
and higher power of 𝜀 𝑘 , we get
𝜀 𝑘+1 = 𝐶𝜀 𝑘
2
Where 𝐶 =
1
2
𝑓′′( 𝜉)
𝑓′( 𝜉)
.
Thus, the Newton-Raphsons Method has second order convergence.
5.5.3 Regula FalsiMethod—
If the function 𝑓(𝑥) in the equation 𝑓( 𝑥) = 0 is convex in the interval (𝑥0, 𝑥1)that
contains the root, then one of the points 𝑥0or𝑥1 is always fixed and the other point
varies with 𝑘 . If the point 𝑥0is fixed , then the function 𝑓(𝑥) is approximated by
the line passing through the points (𝑥0, 𝑓(𝑥0)) and (𝑥 𝑘 ,𝑓(𝑥 𝑘 )) , 𝑘 = 1,2,3,… the
error of the equation becomes

𝜀 𝑘+1 = 𝐶𝜀0 𝜀 𝑘
Where 𝐶 =
1
2
𝑓′′( 𝜉)
𝑓′( 𝜉)
and 𝜀0 = 𝑥0 − ξ is independent of 𝑘. Therefore we can write
𝜀 𝑘+1 = 𝐶∗
𝜀 𝑘
Where 𝐶∗
= 𝐶𝜀0 is the asymptotic error constant. Hence the Regula-Falsi method
has linear rate of convergence.
6.1 SecantMethod
Although the Newton-Raphson method is very power full to solve non-linear
equations, evaluating of the function derivative is the major difficulty of this
method. To overcome this deficiency, the secant method starts the iteration by
employing two starting points and approximates the function derivative by
evaluating of the slope of the line passing through these points. The secant method
has been shown in Fig. 1. As it is illustrated in Fig. 1, the new guess of the root of
the function f(x) can be found as follows:

In the first glance, the secant method may be seemed similar to linear interpolation
method, but there is a major difference between these two methods. In the secant
method, it is not necessary that two starting points to be in opposite sign.
Therefore, the secant method is not a kind of bracketing method but an open
method. This major difference causes the secant method to be possibly divergent in
some cases, but when this method is convergent, the convergent speed of this
method is better than linear interpolation method in most of the problems.
The algorithm of the secant method can be written as follows:
6.2 Steps—
Step 1: Choose two starting points x0 and x1.
Step 2: Let

Step 3: if |x2-x1|<e then let root = x2, else x0 = x1 ; x1 = x2 ; go to step 2.
Step 4: End.
e: Acceptable approximated error.
Example—
A Real root of the equation 𝒇( 𝒙) = 𝒙 𝟑
− 𝟓𝒙 + 𝟏 = 𝟎 lies in the interval ( 𝟎, 𝟏).
Perform four iterations of the secant method.
Solution:
We have 𝑥0 = 0, 𝑥1 = 1, 𝑓0 = 𝑓( 𝑥0) = 1, 𝑓1 = 𝑓( 𝑥1) = −3
𝑥2 = 𝑥1 − [
𝑥0−𝑥1
𝑓0−𝑓1
] 𝑓1 = 1 − [
0−1
1+3
] (−3) = 1 −
3
4
= 0.25
𝑓2 = 𝑓( 𝑥2) = 𝑓(0.25) = −0.234375
𝑥3 = 𝑥2 − [
𝑥1−𝑥2
𝑓1−𝑓2
] = 0.186441
∴ 𝑓3 = 𝑓( 𝑥3) = 0.074276
𝑥4 = 𝑥3 − [
𝑥3−𝑥2
𝑓3−𝑓2
] 𝑓3 = 0.201736
∴ 𝑓4 = 𝑓( 𝑥4) = −0.000470
𝑥5 = 𝑥4 − [
𝑥4−𝑥3
𝑓4−𝑓3
] 𝑓4 = 0.201640
6.3 Advantages of secantmethod:
1. It converges at faster than a linear rate, so that it is more rapidly
convergent than the bisection method.
2. It does not require use of the derivative of the
function, something that is not available in a number
of applications.
3. It requires only one function evaluation per iteration,

as compared with Newton’s method which requires
two.
6.4 Disadvantagesofsecantmethod:
1. It may not converge.
2. There is no guaranteed error bound for the computed
iterates.
3. It is likely to have difficulty if f 0(𝛼) = 0. This
means the x-axis is tangent to the graph of 𝑦 = 𝑓 (𝑥)
at 𝑥 = 𝛼.
4. Newton’s method generalizes more easily to new
methods for solving simultaneous systems of nonlinear
equations.

EXERCISE
Que—Solve the following Examples:
1. Find a positive rootof the following equation by bisection method : 𝑥3
−
4𝑥 − 9
(𝐴𝑛𝑠𝑤𝑒𝑟: 2.7065)
2. Solve the following by using Newton – RaphsonMethod :𝑥3
– 𝑥 − 1
(𝐴𝑛𝑠𝑤𝑒𝑟 ∶ 1.3247 )
3. Solve the following by method of false position (RegulaFalsi Method) :
𝑥3
+ 2𝑥2
+ 10𝑥 – 20 (𝐴𝑛𝑠𝑤𝑒𝑟 ∶ 1.3688)
4. Solve the following equation by Secant Method up to four iteration.
𝑐𝑜𝑠𝑥 − 𝑥𝑒 𝑥
= 0 in interval (0,1).
(𝐴𝑛𝑠𝑤𝑒𝑟: 0.5317058606)
REFERENCES

1. Numerical Method for Science and Computer science by M.K. Jain, S.R.K.
Iyenger, R.K. Jain
2. http://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&v
ed=0CCMQFjAB&url=http%3A%2F%2Fwww.b-
u.ac.in%2Fsde_book%2Fbca_numer.pdf&ei=uKenVMatGceduQTt9ILwBw
&usg=AFQjCNH7oooXkwN7zusFPiJS7MksUkmZaw&bvm=bv.82001339,
d.c2E&cad=rja
3. http://1.bp.blogspot.com/_BqwTAHDIq4E/TCqWRUqn7qI/AAAAAAAAA
CE/oe-W-JZaQRI/s400/Dibujo.bmp
4. http://www.google.co.in/imgres?imgurl=http%3A%2F%2Fimage.tutorvista.
com%2Fcms%2Fimages%2F39%2Fnewton%252513raphson-
method.JPG&imgrefurl=http%3A%2F%2Fmath.tutorvista.com%2Fcalculus
%2Fnewton-raphson-method.html&h=283&w=500&tbnid=VCvTr-
sjzqmjDM%3A&zoom=1&docid=o_YQeEzCa1CVHM&ei=B6SnVMCED
omlNqbsgaAN&tbm=isch&ved=0CB0QMygBMAE&iact=rc&uact=3&dur
=4533&page=1&start=0&ndsp=17
5. http://www.google.co.in/url?sa=i&rct=j&q=&esrc=s&source=images&cd=
&cad=rja&uact=8&ved=0CAcQjRw&url=http%3A%2F%2Fmathumatiks.c
om%2Fsubpage-373-Regula-Falsi-Method.htm&ei=6aWnVK-
1Eo3muQSuqILABg&bvm=bv.82001339,d.eXY&psig=AFQjCNGE5QOy
UayJysWeiBMOzx7QBybl_w&ust=1420359120049725
6. http://www.numericmethod.com/About-numerical-methods/roots-of-
equations/secant-method
7. http://www.dummies.com/how-to/content/how-to-solve-nonlinear-
systems0.html
8. http://studyhelpszone.blogspot.in/2009/07/advantages-and-disadvanteges-of-
secant.html

SECANT METHOD
The Newton-Raphson algorithm requires the evaluation of two functions (the function and its
derivative) per each iteration. If they are complicated expressions it will take considerable
amount of effort to do hand calculations or large amount of CPU time for machine calculations.
Hence it is desirable to have a method that converges (please see the section order of the
numerical methods for theoretical details) as fast as
Newton's method yet involves only the evaluation of the
function. Let x0 and x1 are two initial approximations for
the root 's' of f(x) = 0 and f(x0) & f(x1) respectively, are
their function values. If x 2 is the point of intersection of x-
axis and the line-joining the points (x0, f(x0)) and (x1,
f(x1)) then x2 is closer to 's' than x0 and x1. The equation
relating x0, x1 and x2 is found by considering the slope 'm'
m =
f(x1) - f(x0) f(x2) - f(x1) 0 - f(x1)

= =
x1 - x0 x2 - x1 x2 - x1
x2 - x1 =
- f(x1) * (x1-x0)
f(x1) - f(x0)
x2 = x1 -
f(x1) * (x1-x0)
f(x1) - f(x0)
or in general the iterative process can be written as
xi+1= xi -
f(xi) * (xi - xi-1 )
i = 1,2,3...
f(xi) - f(xi-1)
This formula is similar to Regula-falsi scheme of root bracketing methods but differs in the
implementation. The Regula-falsi method begins with the two initial approximations 'a' and 'b'
such that a < s < b where s is the root of f(x) = 0. It proceeds to the next iteration by
calculating c(x2) using the above formula and then chooses one of the interval (a,c) or (c,h)
depending on f(a) * f(c) < 0 or > 0 respectively. On the other hand secant method starts with two
initial approximation x0 and x1 (they may not bracket the root) and then calculates the x2 by the
same formula as in Regula-falsi method but proceeds to the next iteration without bothering
about any root bracketing.

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