6/9/2020

Computational Science: Computational Methods in Engineering Fitting to & /

https://empossible.net/academics/emp4301_5301/

Outline

• Exact Fit Methods – Fitting Polynomials • Interpolation & Extrapolation

• MATLAB Implementation https://empossible.net/academics/emp4301_5301/

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Fitting Polynomials

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Statement of the Problem

Suppose it is desired to fit the following Nth‐order to a set of points.

2 N f xaaxax01  2  axN

There exists N+1 unknown coefficients so N+1 points are needed to calculate the coefficients exactly.

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Calculating the Unknown Polynomial Coefficients

First, write the polynomial at the N+1 points.

2 N fx101121 a axax   axN 1 2 N fx201222 a axax   axN 2  2 N f xaaxaxaxNNNNN101121   1

Since this is an exact fit, there are no residual terms.

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Calculating the Unknown Polynomial Coefficients

Second, put the equations into matrix form.

2 N f  xa10 1 xx11 x 1 2 N 2 N  fx101121 a axax   axN 1 f xa211 xx22 x 2 2 N  fx a axax   ax 201222N 2  2 N f xa32 1 xx x  33 3 f xaaxaxax 2  N  NNNNN101121 1  2 N  f xaNN1 1 xxNN11 x N  1

This can be written more compactly as Aside: f  Xa  This is a Vandermonde matrix and is usually ill‐conditioned for large matrices. See Lagrange interpolation.

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Calculating the Unknown Polynomial Coefficients

Last, the matrix equation is solved for [a] to find the polynomial coefficients.

a0  a1 1  aXf      aa 2    aN

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Example 1 (1 of 2)

Fit the following points to a polynomial..

f 00  f(x) f 1.5 1.5 f 4.0 1.0 x Step 1 –Determine order of polynomial

Since there are three points, a quadratic polynomial can be fit.

2 f  xaaxax 01  2

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Example 1 (2 of 2)

Step 2 –The matrix equation is

22 f101121aaxax   f 110100 xx 11  a 0    a 0  f aaxax 22   f  1 xx  a   1.5   1 1.5 2.25  a 201222 2 221     1 22     f301323aaxax   f 311.01416 xx 33  a 2    a 2

Step 3 –Calculate polynomial coefficients.

1 a0 10 0  0  0  a 1 1.5 2.25  1.5  1.45  1    a2 1 4 16  1.0  0.3 

Step 4 –Write the final polynomial. f(x) f  xxx 1.45 0.3 2

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Interpolation & Extrapolation

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What is Interpolation & Extrapolation?

Interpolation –Sometimes an intermediate value located between measured values is needed.

Extrapolation –Sometimes a value located outside of the measured values is needed.

Why? Measuring the new value may be difficult, expensive, time‐consuming or impossible.

Interpolation and extrapolation can be thought of as two steps: Step 1 – Fit data to a curve. Step 2 –Use the curve fit to calculate the new value(s).

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Linear Interpolation

Given two data points, a line can be fit from which it is possible to interpolate or extrapolate anything else.

Given two data points 𝑥, 𝑦 and 𝑥, 𝑦 , the equation for the line is

yy21 yy11 xx  xx21 y 2 error This equation can be used to interpolate 𝑦 from any yi desired position 𝑥. y1 yy yy21xx  ii11 x1 xi x2 xx21

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Polynomial Interpolation (1 of 3)

Interpolation using a quadratic polynomial is likely the most common method of interpolation.

This requires three points.

f  x

2 f  xaaxax 01  2

x

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Polynomial Interpolation (2 of 3)

Write the matrix equation 𝑓𝑋 𝑎 .

2 f11101 xx  a  f  1 xx2  a 2221  2  f33321 xxa  Solve for the polynomial coefficients 𝑎 . xx xx xx 23 13 12 1 xxxx xxxx  xxxx  axxf1 2 1213 1223 1323f 0111  1 2 xx23 xx 13 xx12  axxf 1   f 1222  xxxx xxxx  xxxx 2 2 1213 1223 1323 axxf23331  f3 111  xxxx1213 xxxx 1223  xxxx 1323  14

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Polynomial Interpolation (3 of 3)

The final equations are

xx23 xx 13 xx12 afff0123  xxxx1213 xxxx 12  23 xxxx 1323 

xx23 xx 13 xx12 afff1123   xxxx1213 xxxx 1223  xxxx 1323  11 1 afff2123  xxxx1213 xxxx 12  23 xxxx 1323 

Typically, these are not calculated using these equations. Instead, the matrix inversion is performed numerically.

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Example 1 – Quadratic Interpolation

Given the following points, interpolate the value at 𝑥3.

f 00 f 1.5 1.5 f 4.0 1.0

Solution Step 1 –These points were previously fit to a quadratic polynomial. The answer was

f  xxx 1.45 0.3 2 Interpolated point f(x) Step 2 –This is evaluated at 𝑥3.

f 3 1.45  3 0.3  32 1.65 x 16

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Example 2 – Quadratic Extrapolation

Given the following points, extrapolate the value at 𝑥0.5.

f 00  f 1.5 1.5 f 4.0 1.0

Solution Step 1 –These points were previously fit to a quadratic polynomial. The answer was

2 f  xxx 1.45 0.3 f(x)

Step 2 –Evaluate this at 𝑥0.5. x f 0.5 1.45   0.5 0.3  0.52  0.8 Extrapolated point

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