Curve Fitting and Interpolation - Examples & Applications

This section provides practical examples of data analysis using curve fitting techniques (such as linear regression, polynomial regression, and linearization) and interpolation methods (such as Newton's Divided-Difference, Lagrange polynomials, and Splines).

Linear Regression

Linear least-squares regression aims to find the "best fit" line y=a0+a1xy = a_0 + a_1x through a set of data points by minimizing the sum of the squares of the residuals.

Basic: Linear Least-Squares Regression

Fit a straight line to the following set of data points:
x=[1,2,3,4,5]x = [1, 2, 3, 4, 5]
y=[0.5,2.5,2.0,4.0,3.5]y = [0.5, 2.5, 2.0, 4.0, 3.5]
Determine the equation of the best-fit line.

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Linearization of Nonlinear Relationships

Many nonlinear relationships, such as exponential models (y=αeβxy = \alpha e^{\beta x}) or power models (y=αxβy = \alpha x^\beta), can be transformed into a linear form using logarithms.

Intermediate: Linearizing a Power Equation

Fit a power model y=αxβy = \alpha x^\beta to the following data:
x=[1,2,3,4,5]x = [1, 2, 3, 4, 5]
y=[0.5,1.7,3.4,5.7,8.4]y = [0.5, 1.7, 3.4, 5.7, 8.4]

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Intermediate: Linearizing an Exponential Model

Fit an exponential model y=αeβxy = \alpha e^{\beta x} to data. The transformation is ln⁡(y)=ln⁡(α)+βx\ln(y) = \ln(\alpha) + \beta x. Given transformed coefficients a0=ln⁡(α)=1.5a_0 = \ln(\alpha) = 1.5 and a1=β=−0.2a_1 = \beta = -0.2, state the final model.

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Polynomial Regression

Advanced: Second-Order Polynomial Regression Setup

Set up the system of normal equations to fit a parabola y=a0+a1x+a2x2y = a_0 + a_1x + a_2x^2 to n=5n=5 data points where ∑x=15\sum x = 15, ∑x2=55\sum x^2 = 55, ∑x3=225\sum x^3 = 225, ∑x4=979\sum x^4 = 979, ∑y=12.5\sum y = 12.5, ∑xy=45\sum xy = 45, and ∑x2y=180\sum x^2y = 180.

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Newton's Divided-Difference Interpolating Polynomials

Newton's divided-difference method creates an interpolating polynomial based on divided differences of the data points.

Advanced: Newton's Divided-Difference

Estimate ln⁡(2)\ln(2) using a second-order Newton interpolating polynomial based on the following three points:
x0=1,f(x0)=0x_0 = 1, \quad f(x_0) = 0
x1=4,f(x1)=1.386294x_1 = 4, \quad f(x_1) = 1.386294
x2=6,f(x2)=1.791759x_2 = 6, \quad f(x_2) = 1.791759

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Lagrange Interpolating Polynomials

The Lagrange interpolating polynomial is simply a reformulation of the Newton polynomial that avoids the computation of divided differences.

Intermediate: First-Order Lagrange Interpolation

Estimate ln⁡(2)\ln(2) using a first-order Lagrange interpolating polynomial with the points:
x0=1,f(x0)=0x_0 = 1, \quad f(x_0) = 0
x1=4,f(x1)=1.386294x_1 = 4, \quad f(x_1) = 1.386294

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Spline Interpolation

Spline interpolation applies lower-order polynomials to subsets of data points, ensuring that the curves connect smoothly. This avoids the wild oscillations (Runge's phenomenon) characteristic of high-degree polynomial interpolation.

Basic: Linear Splines

Given three points (1,0),(2,5)(1, 0), (2, 5), and (4,1)(4, 1), formulate the linear splines.

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