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Curve Fitting using Lagrange Interpolation

, 12 Sep 2008
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Fitting a set of points P to a curve using Lagrange Interpolation Polynomial.

Introduction

In this article, I will explain curve fitting using the Lagrange interpolation polynomial. Curve fitting is used in a wide spectrum in engineering applications such as cars and air crafts surface design. The main problem is, given a set of points in the plan, we want to fit them in a smooth curve that passes through these points. The order of the curve f(x) depends on the number of points given. For example, if we have two points, the function will be of the first order, and the curve will be the line that passes through these two points, while if you have three points, the function will be of the second order f(x) = x2 . Let's first explain the Lagrange polynomial, then we will proceed to the algorithm and the implementation. In this article, I am using C# for coding.

Background

Lagrange Polynomial

An interpolation on two points, (x0, y0) and (x1, y1), results in a linear equation or a straight line. The standard form of a linear equation is given by y = mx + c, where m is the gradient of the line and c is the y-intercept.

m = y1 − y0 / x1 − x0
c = y0 − mx0

which results in:

y = (y1 − y0 / x1 − x0) * x + (x1 y0 − x0 y1 )/ (x1 − x0)

The linear equation is rewritten so that the two interpolated points, (x0, y0) and (x1, y1), are directly represented.

With this in mind, the linear equation is rewritten as:

P1(x) = a0(x − x1) + a1(x − x0)

where a0 and a1 are constants . The points x0 and x1 in the factors of the above equation are called the centers. Applying the equation at (x0, y0), we obtain:

y0 = a0(x0 − x1) + a1(x0 − x0)

or

a0 = y0/ x0−x1 

At (x1, y1), we get:

y1 = a0(x1 − x1) + a1(x1 − x0), or a1 = y1/ x1−x0

Therefore, the linear equation becomes:

P1(x) = y0 (x− x1) /(x0 − x1) + y1 (x− x0) / (x1 − x0)

The quadratic form of the Lagrange polynomial interpolates three points, (x0, y0), (x1, y1), and (x2, y2). The polynomial has the form:

P2(x) = a0(x− x1)(x− x2) + a1(x− x0) (x− x2) + a2(x− x0)(x− x1)

with centers at x0, x1, and x2. At (x0, y0):

y0 = a0(x0 − x1)(x0 − x2) + a1(x0 − x0)(x0 − x2) + a2(x0 − x0)(x0 − x1),

or

a0 = y0/ (x0 − x1)(x0 − x2)
a1 = y1 / (x1 − x0)(x1 − x2)
a2 = y2/ (x2 − x0)(x2 − x1)
P2(x) = y0 (x− x1)(x− x2) / (x0 − x1)(x0 − x2) + y1 (x − x0)(x − x2)/ 
       (x1 − x0)(x1 − x2) + y2 (x− x0)(x− x1)/ (x2 − x0)(x2 − x1)

In general, a Lagrange polynomial of degree n is a polynomial that is produced from an interpolation over a set of points, (xi , yi ) for i = 0, 1, . . ., n, as follows:

Pn(x) = y0L0(x) + y1L1(x) + ··· + ynLn(x)

Using the Code

The Algorithm

Given the interpolating points (xi , yi ) for i =0, 1, . . . ,n;

for i = 0 to n
  //the cumulative multiplication from k = 1 (and k not equal i) to n 
  Evaluate Li (x) = ∏nk=1,k != i (x−xk ) / (xi−xk ) ;
endfor 

Evaluate Pn(x) = y0L0(x) + y1L1(x) + ··· + ynLn(x);

Download the source code from the top of this page to view the code.

Points of Interest

Numerical computing is a very interesting field in software development. This article is related to that field ... And, I will post more articles soon about Computational Geometry ...such as Convex Hull algorithm and Triangulation.

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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About the Author

Fady Aladdin
Software Developer
Egypt Egypt
No Biography provided

Comments and Discussions

 
GeneralA compact version Pinmembergustavtr20-Feb-12 11:24 
GeneralMy vote of 5 PinmemberMarkDaniel5-Jan-12 12:21 
Questionhow to get all pixels coordinate from curve fitting using Lagrange interpolationan on an image PinmemberMrKyaw24-Oct-11 16:18 
Dear Fady,
 

 
The topic here is quite useful for me and I am tryig to get the x, y coordinates in terms of pixels form the user drawn curve on the image.
 
My targeted output was show in this link
http://imageshack.us/photo/my-images/4/currentdetectedboundary.png/ http://imageshack.us/photo/my-images/196/targetedoutput.png/Seems http://imageshack.us/photo/my-images/98/originalboundarymanualh.png/
 
and integrate or using it tough for me as well as took so much time to do that.
 
Please advise me any alternative simple ways since I was stuck on this stage one week plus and rushing on it.
 

 

 
Thanks and best regards
 
Kyaw
GeneralRadius of curve PinmemberM i s t e r L i s t e r25-Aug-10 9:48 
Generalneed help plz Pinmemberd_e_e_o6-Aug-10 4:33 
GeneralCorrected an Overflow error PinmemberSBGTrading22-Jul-10 5:05 
GeneralContact Pinmembericae27-Apr-09 5:42 
GeneralRe: Contact PinmemberFady Aladdin27-Apr-09 7:19 
GeneralArea under the curve PinmemberDorisvaldo16-Sep-08 15:06 
Generalpiecewise bezier PinmemberSarath.14-Sep-08 21:54 

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