Secant

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03.04 
 
Nonlinear Equations – Secant Method 
 
 
 
After reading this chapter, you should be able to: 
 
1. Understand how to use the Secant method to solve for the roots of nonlinear 
equations. 
 
 
 
What is the Secant method and why would I want to use it versus the Newton-
Raphson method? 
The Newton-Raphson method of solving the nonlinear equation f(x)=0 is given by 
the recursive formula 
)f'(x
)f(x - = xx
i
i
ii 1+ (1) 
From the above equation, one of the drawbacks of the Newton-Raphson 
method is that you have to evaluate the derivative of the function. With 
availability of symbolic manipulators such as Maple, Mathcad, Mathematica and 
Matlab, this process has become more convenient. However, it is still can be a 
03.04.1 
03.04.2 Chapter 03.04 
 
laborious process. To overcome this drawback, the derivative, f’(x) of the 
function, f(x) is approximated as 
1
1 )()()(
−
−
−
−=′
ii
ii
i xx
xfxf
xf (2) 
Substituting Equation (2) in (1), gives 
)()(
))((
1
1
1
−
−
+ −
−−=
ii
iii
ii xfxf
xxxfxx . (3) 
The above equation is called the Secant method. This method now requires two 
initial guesses, but unlike the bisection method, the two initial guesses do not 
need to bracket the root of the equation. The Secant method may or may not 
converge, but when it converges, it converges faster than the Bisection method. 
However, since the derivative is approximated, it converges slower then Newton-
Raphson method. 
 
The Secant method can be also derived from geometry shown in Figure 
(1). Taking two initial guesses, xi and xi-1, one draws a straight line between f(xi) 
and f(xi-1) passing through the x-axis at xi+1. ABE and DCE are similar triangles. 
Hence 
 
DE
DC
AE
AB = 
11
1
1
)()(
+−
−
+ −
=− ii
i
ii
i
xx
xf
xx
xf
 
On rearranging, it gives the secant method as 
)()(
))((
1
1
1
−
−
+ −
−−=
ii
iii
ii xfxf
xxxfxx 
 Nonlinear Equations – Secant Method 03.04.3 
 
 
 f(x) 
 f(xi) 
 f(xi-1) 
xi+1 xi-1 xi 
 X 
 B 
 C 
 E D A 
 
Figure 1: Geometrical representation of the Secant method 
 
 
 
Example 1 
You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for 
ABC commodes. The ball has a specific gravity of 0.6 and has a radius of 5.5 cm. 
You are asked to find the distance to which the ball will get submerged when 
floating in water. 
 
03.04.4 Chapter 03.04 
 
 
Figure 5. Floating ball problem 
 
The equation that gives the depth ‘x’ to which the ball is submerged under water 
is given by 
010993.3165.0 423 =×+− −xx 
Use the secant method of finding roots of equations to find 
a) the depth ‘x’ to which the ball is submerged under water. Conduct three 
iterations to estimate the root of the above equation. 
b) find the absolute relative approximate error at the end of each iteration, 
and 
c) the number of significant digits at least correct at the end of each iteration. 
 
Solution 
( ) 423 10993.31650 −×+−= x.xxf 
Let us assume the initial guesses of the root of ( ) 0=xf as 0201 .x =− and 
 0500 .x =
 
Iteration #1 
The estimate of the root is 
 Nonlinear Equations – Secant Method 03.04.5 
 
=1x ( )( )( ) ( )10
100
0
−
−
−
−−
xfxf
xxxfx 
 
( ) ( )( ) ( )4213142030 10
42
0
3
0
0 10993316501099331650
1099331650
−
−−
−
−
−
+−−+−
−×+−−=
x.x.xx.x.x
xxx.x.xx 
( )
( ) ( ) ]x....[]x....[
]..[]x....[.
423423
423
10993302016500201099330501650050
0200501099330501650050050 −−
−
+−−+−
−×+−−=
 = 0.06461 
 
The absolute relative approximate error, a∈ at the end of iteration #1 is 
 100
1
01 ×−=∈
x
xx
a 
 100
064610
050064610 ×−=
.
.. 
 %61.22 =
The number of significant digits at least correct is 0, as you need an absolute 
relative approximate error of at less than 5% for one significant digit to be correct 
in your result. 
 
Iteration #2 ( )( )
( ) ( ) 01
011
12 xfxf
xxxfxx −
−−= 
( ) ( )( ) ( )4203042131 01
42
1
3
1
1 10993316501099331650
1099331650
−−
−
+−−+−
−×+−−=
x.x.xx.x.x
xxx.x.xx 
( ) ( )
( ) ( ) ]x....[]x....[
..]x....[.
423423
423
10993305016500501099330646101650064610
050064610109330646101650064610064610 −−
−
+−−+−
−×+−−=
06241.0= 
The absolute relative approximate error, a∈ at the end of iteration #1 is 
03.04.6 Chapter 03.04 
 
100
2
12 ×−=∈
x
xx
a 
 100
062410
064610062410 ×−=
.
.. 
 = 3.525% 
The number of significant digits at least correct is 1, as you need an absolute 
relative approximate error of less than 5%. 
 
Iteration #3 ( )( )
( ) ( ) 12
122
23 xfxf
xxxfxx −
−−= 
 
( ) ( )( ) ( )4213142232 12
42
2
3
2
2 10993316501099331650
1099331650
−−
−
+−−+−
−×+−−=
x.x.xx.x.x
xxx.x.xx 
( ) ( )
( ) ( ) ]x....[]x....[
..]x....[.
423423
423
10993306461016500646101099330624101650062410
0646100624101099330624101650062410062410 −−
−
+−−+−
−×+−−=
 06238.0=
The absolute relative approximate error, a∈ at the end of iteration #1 is 
 100
3
23 x
x
xx
a
−=∈ 
 100
062380
062410062380 ×−=
.
.. 
 %0595.0=
The number of significant digits at least correct is 1, as you need an absolute 
relative approximate error is at least 5%. 
 
 
 
 
 Nonlinear Equations – Secant Method 03.04.7 
 
 
Table 1: Secant method results as a function of iterations 
Iteration 
Number 
I 1−ix ix 1+ix %a∈ ( )1+ixf 
1 
2 
3 
4 
0 
1 
2 
3 
0.02 
0.05 
0.06461 
0.06241 
0.05 
0.06461 
0.06217 
0.06238 
0.06461 
0.06241 
0.06238 
0.06238 
22.61 
3.525 
0.0595 
3.64x10-4 
-1.9812x10-5 
-3.2852x10-7 
2.0252x10-9 
-1.8576x10-12 
 
 
NONLINEAR EQUATIONS 
Topic Secant Method for solving Nonlinear Equations. 
Summary Text book notes of Secant method of finding roots of nonlinear 
equations, including examples. 
Major General Engineering 
Authors Autar Kaw 
Date June 25, 20072 
Web Site http://numericalmethods.eng.usf.edu 
 
	What is the Secant method and why would I want to use it versus the Newton-Raphson method?
	Example 1
	Solution
	Iteration #1
	Iteration #2
	Iteration #3
	Table 1: Secant method results as a function of iterations