A New Algorithm for Solving Nonlinear Equations by Using Least Square Method
Nasr Al Din Ide
Faculty of Science, Department of Mathematics, Aleppo University, Aleppo, Syria
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To cite this article:
Nasr Al Din Ide. A New Algorithm for Solving Nonlinear Equations by Using Least Square Method. Mathematics and Computer Science. Vol. 1, No. 3, 2016, pp. 44-47. doi: 10.11648/j.mcs.20160103.12
Received: August 22, 2016; Accepted: August 31, 2016; Published: September 18, 2016
Abstract: Finding the roots of nonlinear algebraic equations is an important problem in science and engineering, later many methods have been developed for solving nonlinear equations. These methods are given [1-27], in this paper, a new Algorithm for solving nonlinear algebraic equations is obtained by using least square method by fitting a polynomial form of degree two (or parabolic form). This paper compares the present method with the method given by Jutaporn N, Bumrungsak P and Apichat N, 2016 [1], which was used nonlinear regression method in form of logarithm function. We verified on a number of examples and numerical results obtained show that the present method is faster than the method, which used the logarithm function given by [1].
Keywords: Nonlinear Algebraic Equations, Least Square Method, Logarithm Function
1. Introduction
There are several well-known methods for solving nonlinear algebraic equations of the form
f(x)=0 (1)
Where f denote a continuously differentiable on and has at least one root α, in
Such as Newton method, Bisection method, Regula Falsi method and several another methods see for example [2-27].
Here we describe a new method by using least square method as a parabola polynomial form of degree two:
(2)
Where A, B and C are the unknown constant.
We used three points (a, f(a)), (b, (f(b)) and (c, f(c)) where , then we find that, this procedure lead us to the root α of equation (1).
2. The Present Method
In beginning, we define three initial points (a, f(a)), (b, (f(b)) and (c, f(c)), now by using least square method we fit the polynomial (2) as suit, let is the error or the different value between the true value and he estimated value , therefore,
(3)
And the sum of square error,
(4)
Or,
(5)
To find A, B, C we will minimize this function, taking the derivative of (5) equal zero with respect to A, B and C we find the three normal equations:
3A+ (a+b+c).B+ ().C= f(a)+f(b)+f(c) (6)
(a+b+c).A+ ().B+ ().C= a.f(a)+b.f(b)+c.f(c) (7)
().A+().B+().C= .f(a)+.f(b)+.f(c) (8)
Hence, we find A, B and C. Now by solving the equation we find the two roots and of (2), we choose or which verify or
3. Algorithm. 1
The present method has 6 steps:
1. Take is an initial interval, which has at least a root in this interval.
2. Compute .
3. Determine the constants A, B and C by solving the system of three linear algebraic equations (6), (7) and (8).
4. Solve the equation of second degree for determine the root of (1), or which verify or
5. Replace the interval with which Contains the root.
6. Return step (2) until the absolute error .
4. Examples
Example 1. Consider the equation: f(x) == 0. Applying present method, by using Maple program we find the approximate equation of second degree:
In addition, we find for the initial interval [0.1, 1] after 4 iterations, the approximate solution x=0.345954815671666 and the error .
This result is better than the result given by [1] which give the root x for 6 iterations with error . Figure 1 illustrate the plots of approximate solution of the equation
Moreover, exact solution of the equation for example 1, in the interval [0.1, 1].
Example 2. Consider the equation: f(x) == 0. Applying present method, by using Maple program we find the approximate equation of second degree:
In addition, we find for the initial interval [0.1, 1] after four iterations, the approximate solution x=0.5671432901053765 and the error .
This result is better than the result given by [1] which give the root x for seven iterations with error . Figure 2 illustrate the plots of approximate solution of the equation
Moreover, exact solution of the equation = 0 for example 1, in the interval [0.1, 1].
Figure 2 illustrate that, in the interval [0, 1] we have the same graph of the exact solution and the solution given by present method.
Example 3. Consider the equation: f(x) == 0. Applying present method, by using Maple program we find the approximate equation of second degree:
In addition, we find for the initial interval [0.1, 1] after four iterations, the approximate solution x=0.5671432904097837 and the error .
This result is better than the result given by [1] which give the root x for six iterations with error . Figure 3 illustrate the plots of approximate solution of the equation
Moreover, exact solution of the equation = 0 example 1, in the interval [0.1, 1].
5. Conclusion
The present paper suggests a new algorithm for solving nonlinear algebraic equations; this method presented by equations (6), (7), (8) and by the algorithm 1. Numerical examples show that this method is remarkably effective for solving nonlinear algebraic equations and it is much faster than the method given by [1], 2016, which was used nonlinear regression method in form of logarithm function.
References