American Journal of Computational and Applied Mathematics
p-ISSN: 2165-8935 e-ISSN: 2165-8943
2018; 8(3): 47-49
doi:10.5923/j.ajcam.20180803.01

Masoud Saravi
Emeritus Professor of Islamic Azad University of Iran, Iran
Correspondence to: Masoud Saravi, Emeritus Professor of Islamic Azad University of Iran, Iran.
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Copyright © 2018 The Author(s). Published by Scientific & Academic Publishing.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/

Usually the methods based on Taylor expansion series for
have better convergence [1]. But, nearly, all of them contain one or more derivatives of
. The purpose of this paper is to introduce a technique to obtain free from derivatives which works better than methods others that been considered in most text book for solving nonlinear equations by providing some numerical examples.
Keywords: Nonlinear equations, Order of convergence, Euler’s equation, Iteration formulae
Cite this paper: Masoud Saravi, A New Method for Solving Nonlinear Equations Based on Euler’s Differential Equation, American Journal of Computational and Applied Mathematics , Vol. 8 No. 3, 2018, pp. 47-49. doi: 10.5923/j.ajcam.20180803.01.
![]() | (1) |
arise frequently in science and engineering. In most cases it is difficult to obtain an analytical solution of (1). Hence the exploitation of numerical methods for solving such equations becomes a main subject of considerable interests. Usually in all text books the methods split into two sections, namely methods without derivatives and methods with derivatives [2, 3, 4, 6, 8, 9, 10]. Probably the most well-known and widely used algorithm to find a root of
without derivative is the fixed point iteration method. In next section, we introduce a new algorithm and by expressing weak and strong aspect of this method, it will be deduced that the order of convergence is more than other methods without derivatives if the equation (1) contains simple roots.
in (1) by Taylor's series about the point
, we get
By approximating this series we may write
That is,![]() | (2) |
![]() | (3) |
![]() | (4) |
That is![]() | (5) |

![]() | (6) |
can be found by two choices for
in (4). For example, let
, then
and
Therefore equation (6) becomes![]() | (7) |
and
Remark 1: It should be noted that our starting value cannot be a or b, i.e.,
It would be better to start with
where
Remark 2: The sign,
of the square root term is chosen to agree with the sign
to keep
close to
Remark 3: To find the order of convergence of this method we need some difficult square root computations, hence we avoid these computations. But the following examples in next section show that the order of convergence of this method must nearly be quadratic.
This equation has been taken from [5] and has a real root on (0,1). If we wish to approximate this root with accuracy 10-4 by the bisection method, we need 14 iterations to obtain an approximation accurate to 10-4. Because with
and
we get
But if we apply our method, we obtain
so that
.Even we use chord (modified regula-falsi) method given by
starting with
we come to following results:
Example 2: Equation
has a root in (1/4, 1/3). This equation been considered in [2]. The correct value to (4D) is
The authors used fixed point iteration formula showed that if we write
and start with mid-point of [1/4, 1/3] then
Since
is continue there is an interval within [1/4, 1/3] over which
But by our method with plus sign (since f(1/3)>0) we obtained:
Although if they used a new scheme given by
, with
but with the same starting vale this scheme requires fifteen iterations to converges to the root 0.2872. Let's consider another example. This example was chosen from [3].Example 3: Approximate a zero of
In this book only mentioned that this equation has not real root. The roots correct to (4d) are
We used with starting value
for fixed point iteration method and get
But by (7) we obtained
It seems this equation has only two conjugate complex roots, because we examined several numbers and every time we reached to this result. This fact may be examined by considering complex equation
Example 4: Consider
This equation has a root on (1, 2) and is given in [2,4, 5,6]. Let
By fixed point iteration method we obtained following results:
Of course in this example
Now we apply our method. We have
Since
hence we use (7) with minus sign and we get
We also used Newton method with
and get
Note 1: Although by Newton method we had the same result on first iteration but this is not always true. See following example [4].Example 5: Consider
has a root in (1, 2). We used scheme given by (7) and get
But by Newton’s method with
we obtained
But with initial value
, in third iteration we get
In general, the Newton method works better, in particular when the equation has complex roots. Let’s consider polynomial equations with all complex roots.Example 6: Approximate all roots of equation given by
This example was chosen from [7]. In this book mentioned that this equation has not real roots and with starting value
found
It is clear that a second root will be
The other two roots are
If we start with
with
after 8 iterations we obtain
but by Newton method, with
we need only two iterations.