# Linear Curve Fitting in Visual Basic QuickStart Sample

Illustrates how to fit linear combinations of curves to data using the LinearCurveFitter class and other classes in the Extreme.Mathematics.Curves namespace in Visual Basic.

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``````Option Infer On

' The curve fitting classes reside in the
' Extreme.Mathematics.Curves namespace.
Imports Extreme.Mathematics.Curves
' Vectors and function delegates reside in the Extreme.Mathemaics
' namespace
Imports Extreme.Mathematics

' Illustrates least squares curve fitting of polynomials and
' other linear functions using the LinearCurveFitter class in the
' Extreme.Mathematics.Curves namespace of Extreme Numerics.NET.
Module LinearCurveFitting

Sub Main()
' The license is verified at runtime. We're using
' https://numerics.net/trial-key

' This QuickStart sample illustrates linear least squares
' curve fitting using polynomials and linear combinations
' of arbitrary functions.

' Linear least squares fits are calculated using the
' LinearCurveFitter class:
Dim fitter As New LinearCurveFitter

' We use data from the National Institute for Standards
' and Technology's Statistical Reference Datasets library
' at http:'www.itl.nist.gov/div898/strd/.

' Note that, due to round-off error, the results here will not be exactly
' the same as the NIST results, which were calculated using 500 digits
' of precision!

' We use the 'Pontius' dataset, which contains measurement data
' from the calibration of load cells. The independent variable is the load.
' The dependent variable is the deflection.
Dim deflectionData = Vector.Create(0.11019, 0.21956,
0.32949, 0.43899, 0.54803, 0.65694, 0.76562, 0.87487, 0.98292,
1.09146, 1.20001, 1.30822, 1.41599, 1.52399, 1.63194, 1.73947,
1.84646, 1.95392, 2.06128, 2.16844, 0.11052, 0.22018, 0.32939,
0.43885999999999997, 0.54798, 0.65739, 0.76596, 0.87474, 0.983, 1.0915, 1.20004,
1.30818, 1.41613, 1.52408, 1.63159, 1.73965,
1.84696, 1.95445, 2.06177, 2.16829)
Dim loadData = Vector.Create(150.0, 300, 450, 600, 750, 900,
1050, 1200, 1350, 1500, 1650, 1800,
1950, 2100, 2250, 2400, 2550, 2700,
2850, 3000, 150, 300, 450, 600,
750, 900, 1050, 1200, 1350, 1500,
1650, 1800, 1950, 2100, 2250, 2400,
2550, 2700, 2850, 3000)

' You must supply the curve whose parameters will be
' fit to the data. The curve must inherit from LinearCombination.
'
' Here, we use a quadratic polynomial:
fitter.Curve = New Polynomial(2)

' The X values go into the XValues property:
' ...and Y values go into the YValues property:
fitter.YValues = deflectionData

' The Fit method performs the actual calculation:
fitter.Fit()

' A Vector containing the parameters of the best fit
' can be obtained through the
' BestFitParameters property.
Dim solution = fitter.BestFitParameters
' The standard deviations associated with each parameter
' are available through the GetStandardDeviations method.
Dim s = fitter.GetStandardDeviations()

Console.WriteLine("Calibration of load cells")
Console.WriteLine("    deflection = c1 + c2*load + c3*load^2 ")
Console.WriteLine("Solution:")
Console.WriteLine("c1: {0,20:E10} {1,20:E10}", solution(0), s(0))
Console.WriteLine("c2: {0,20:E10} {1,20:E10}", solution(1), s(1))
Console.WriteLine("c3: {0,20:E10} {1,20:E10}", solution(2), s(2))

Console.WriteLine("Residual sum of squares: {0}", fitter.Residuals.Norm())

' Now let's redo the same operation, but with observations weighted
' by 1/Y^2. To do this, we set the WeightFunction property.
' The WeightFunctions class defines a set of ready-to-use weight functions.
fitter.WeightFunction = WeightFunctions.OneOverYSquared
' Refit the curve:
fitter.Fit()
solution = fitter.BestFitParameters
s = fitter.GetStandardDeviations()

' The solution is slightly different:
Console.WriteLine("Solution (weighted observations):")
Console.WriteLine("c1: {0,20:E10} {1,20:E10}", solution(0), s(0))
Console.WriteLine("c2: {0,20:E10} {1,20:E10}", solution(1), s(1))
Console.WriteLine("c3: {0,20:E10} {1,20:E10}", solution(2), s(2))
Console.WriteLine()

'
' Fitting combinations of arbitrary functions
'

' The following example estimates the two parameters, c1 and c2
' in the theoretical model for conductance:
'     k(T) = 1 / (c1 / T + c2 * T*T)

Dim temperature = Vector.Create(12.29, 13.75, 14.82,
16.12, 18.04, 18.67, 20.52, 22.68, 25.15,
27.72, 30.24, 33.21, 36.48, 39.86, 50.4)
Dim conductance = Vector.Create(25.35, 27.88, 29.93,
30.42, 31.0, 31.96, 32.47, 30.33, 31.14,
27.46, 23.29, 20.72, 17.24, 14.71, 9.5)

' First, we transform the dependent variable:
Dim y = Vector.Reciprocal(conductance)

' y is a linear combination of basis functions 1/T and T*T.
' Create a function basis object:
Dim basisFunctions As Func(Of Double, Double)() = {AddressOf f1, AddressOf f2}
Dim basis As New GeneralFunctionBasis(basisFunctions)

' Create a LinearCombination curve using this function basis:
Dim myCurve As New LinearCombination(basis)

' Set the curve fitter properties:
fitter.Curve = myCurve
fitter.XValues = temperature
fitter.YValues = y
' Reset the weights
fitter.WeightFunction = Nothing
fitter.WeightVector = Nothing

' Now compute the solution:
fitter.Fit()
solution = fitter.BestFitParameters
s = fitter.GetStandardDeviations()

' Print the results
Console.WriteLine("Conductance of copper: k(T) = 1 / (c1/T + c2*T^2)")
Console.WriteLine("Solution:")
Console.WriteLine("c1: {0,20:E10} {1,20:E10}", solution(0), s(0))
Console.WriteLine("c2: {0,20:E10} {1,20:E10}", solution(1), s(1))

Console.WriteLine("Residual sum of squares: {0}", fitter.Residuals.Norm())

Console.Write("Press Enter key to exit...")