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 "Curve Fitting Solutions" Straight Line Fit Method
 f(x) = (1.72857142857143, 0.928571428571429, 0.190862703084106)
 (constant)  (x)   (standard deviation)

 x-array = { , , , , , } y-array = { , , , , , } [ Initial x-array: { 0.0, 1.0, 2.0, 3.0, 4.0, 5.0 } ] [ Initial y-array: { 1.9, 2.7, 3.3, 4.4, 5.5, 6.5 } ]

IMPLEMENTATION
Curve Fitting Linear Fit

The most popular curve fitting technique is the least squares method. The simplest linear regression is the straight line fit, which attempts to fit a straight line using the least squares technique.

Algorithm Creation

The model function has the following simple form:

 f(x;α) = a + bx

where the sum of the linear regression becomes,

 S(a,b) = ∑ni=0[yi - f(xi;α]2 = ∑ni=0(yi - a - bxi)2

The standard deviation α can be expressed by,

 α = √ S / n - m Testing the Straight Line Fit Method

In order to test theStraight Line Fit Method as defined above, a new TestStraightLineFit() static method has been added and executed. Supporting code and methods are not shown.

static void TestStraightLineFit();
{
ListBox1.Items.Clear();
double[] xarray = new double[] { t1, t2, t3, t4, t5, t6 };
double[] yarray = new double[] { t7, t8, t9, t10, t11, t12 };
double[] x = new double[] { t11, t12, t13, t14 };
double[] results = CurveFitting.StraightLineFit(xarray, yarray);
VectorR v = new VectorR(results);
ListBox1.Items.Add(" " + v.ToString());
}

As a sample we provide the input data points using two double arrays x-array and y-array. Running this example generates results (1.729, 0.929, 0.191). Therefore, the regression line is given by:

 f(x) = 1.729 + 0.929x

and the standard deviation is,

 0.191

The user can manipulate all values and try variations on the arrays themselves by specifying new estimate values.

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