curve fitting, interpolation · 2012. 4. 23. · relevance of curve fitting relevance of...

Curve Fitting, Interpolation

Mike Renfro

February 27, 2008

Mike Renfro Curve Fitting, Interpolation

IntroductionSpline Interpolation

Part I

Curve Fitting, Interpolation



Relevance of Curve FittingRelevance of Interpolation

Relevance of Curve Fitting

Extracting parameters from experimental data:

A standard tensile test involves pulling on opposite ends of amaterial specimen and recording the amount of stretch(strain) in the specimen as the load and stress on thespecimen is gradually increased.

The initial portion of the stress-strain graph is theoretically astraight line, and its slope represents the modulus of elasticityof the specimen material.





Realistically, finite amounts of experimental error will beintroduced into both stress and strain measurements, causing thedata to not fall along a single straight line.





One method of curvefitting attempts to find asingle “best-fit” slope andintercept on a line thatwill minimize the overallamount of error betweenthe line and the datapoints. The slope of thatline is then used as a bestestimate of the specimen’selastic modulus.




Relevance of Interpolation

The behavior of many engineering systems is very complex anddifficult or impossible to describe analytically. Experiments can berun for a large set of test cases, but not an infinite set.Interpolation provides a method to estimate the results of untestedcases based off the results of known cases.





For example, the specific heat property of a given material can befound via calorimetry experiments, but is not a constant for allpossible temperatures.





Interpolation provides a way for us to make close estimates of thematerial’s specific heat at other temperatures, in this case, byfitting a series of simple compatible curves to each segment of theoriginal data.



Step Function SplineLinear SplineQuadratic Spline

Piecewise Spline Interpolation

Some interpolation methods shown early in Chapter 5 attempt tofit a single function (normally a polynomial) to a wide selection ofdata points. This can cause several problems shown in Figure 5.8.




Problem 1

Some functions simply aren’t accurately described with a singlepolynomial function.




Problem 2

Most such interpolations yield an nth order polynomial when givenn + 1 points — the result is a function with large oscillations notconfirmed by further experiments. Even if the theoretical functionis a close match for a single polynomial, small errors in the fitpoints can drastically throw off the calculations.




Terminology

Spline: a piecewise polynomial of low order (normally linear,quadratic, or cubic).

Piecewise: the particular function is defined only on a smallrange of x values, not across the entire function domain.

Knots: the data points that the splines must pass through.




Trivial Case: Step Function Spline

The step function spline is the simplest case. The interpolatedfunction value is identical to the function value at the previousknot. Since this has the potential to display large discontinuities, itis almost never used.

fi (x) = f (xi−1)




Linear Spline

A linear spline connects adjacent knots with straight lines.




Development of Linear Spline

For example, the first two knots in a set of data points are(x0, f (x0)) and (x1, f (x1)). The equation of the line connectingthem is

f1(x) = f (x0) +

(f (x1)− f (x0)

x1 − x0

)(x − x0)




Development of Linear Spline

For the general case of knots i − 1 and i , the equation would be

fi (x) = f (xi−1) +

(f (xi )− f (xi−1)

xi − xi−1

)(x − xi−1)




Linear Spline Procedure

Given: a table of xi and f (xi ) data, and a value x to interpolate af (x) value for.

Locate the interval of xi points that contains the required x .

Plug the appropriate values of xi−1, xi , f (xi−1), f (xi ), and xinto the previous formula and calculate the correspondingf (x).




Linear Spline Example

(5.10) Find a linear spline to fit the following data:

i 0 1 2 3 4

xi 2.0 3.0 6.5 8.0 12.0

f (xi ) 14.0 20.0 17.0 16.0 23.0

Use the results to estimate the value of f at x = 7.0.

x = 7.0 lies on the interval [6.5, 8.0].

Use i = 3, xi−1 = 6.5, xi = 8.0, f (xi−1) = 17.0, andf (xi ) = 16.0.

The spline equation becomes

f3(x) = f (x2) +

(f (x3)− f (x2)

x3 − x2

)(x − x2)




Linear Spline Example (continued)

Substitute the values for i , xi , xi−1, f (xi ), and f (xi−1):

f3(x) = 17.0 +

(16.0− 17.0

8.0− 6.5

)(x − 6.5)

Collect the terms:

f3(x) = 21.3333− 0.6667x

Substitute x = 7.0 into the equation:

f (7.0) = 21.3333− 0.6667(7.0) = 16.6664




Quadratic Spline

One drawback to the linear spline is that even though it is acontinuous function (unlike the step function spline), it does notexhibit a continuous first derivative. To overcome this limitation,we use higher-order polynomials instead of straight lines.




Quadratic Spline Development

Assuming we have n + 1 knot coordinates (xi ,f (xi )),i = 0, 1, 2, · · · , n, we’ll need to calculate the equations for ndifferent quadratic functions that connect them. Each of thesefunctions can be expressed in the form

fi (x) = ai + bix + cix2, i = 1, 2, · · · , n

where ai , bi , and ci are unknown constants. Since we have 3nunknowns, we can solve for them if we can find 3n independentequations concerning them.





First, the quadratic functions must be continuous at the knots.That is, the function value at an interior knot xi must be the sameregardless of whether it is calculated using piecewise function fi (x)or fi+1(x):

fi (x = xi ) = ai + bixi + cix2i = f (xi ), i = 1, 2, · · · , n − 1

fi+1(x = xi ) = ai+1 + bi+1xi + ci+1x2i = f (xi ), i = 1, 2, · · · , n− 1

The above equations provide 2n − 2 conditions, out of a 3n totalrequired.

Warning: the x values are known here, the a, b, and c values arethe unknowns!





Second, the first and last quadratic functions must pass throughthe endpoints x0 and xn, respectively:

f1(x = x0) = a1 + b1x0 + c1x20 = f (x0)

fn(x = xn) = an + bnxn + cnx2n = f (xn)

The above equations provide 2 conditions, making the total so far2n.





Third, we need continuous first derivatives. The quadraticfunctions themselves have continuous first derivatives, but we haveto ensure that the function slopes are continuous at each interiorknot:

f ′i (x) = bi + 2cix , i = 1, 2, · · · , n

f ′i (x = xi ) = f ′i+1(x = xi )

which yields

bi + 2cixi = bi+1 + 2ci+1xi , i = 1, 2, · · · , n − 1

The above equations provide n − 1 conditions, making the total sofar 3n − 1.





Finally, we need one more condition, and have some leeway in thecondition we can select. One common condition is to specify thatthe second derivative (curvature) of the last quadratic function iszero at xn:

f ′′n (x = xn) = 2cn = 0

cn = 0

We now have a 3n set of equations, with 3n unknowns. Use any ofthe methods from Chapter 3 to solve them.




Quadratic Spline Example

(5.11) Find a quadratic spline to fit the data given in Example5.10. Use the results to estimate the value of f at x = 7.0.

i 0 1 2 3 4

xi 2.0 3.0 6.5 8.0 12.0

f (xi ) 14.0 20.0 17.0 16.0 23.0




Quadratic Spline Example: Step 1

The quadratic functions must be continuous

fi (x = xi ) = ai + bixi + cix2i = f (xi ), i = 1, 2, · · · , n − 1

fi+1(x = xi ) = ai+1 + bi+1xi + ci+1x2i = f (xi ), i = 1, 2, · · · , n− 1

f1(x1) = a1 + 3b1 + 9c1 = 20

f2(x2) = a2 + 6.5b2 + 42.25c2 = 17

f3(x3) = a3 + 8b3 + 64c3 = 16

f2(x1) = a2 + 3b2 + 9c2 = 20

f3(x2) = a3 + 6.5b3 + 42.25c3 = 17

f4(x3) = a4 + 8b4 + 64c4 = 16




Quadratic Spline Example: Step 2

The first and last functions must pass through the endpoints x0

and xn

f1(x0) = a1 + 2b1 + 4c1 = 14

f4(x4) = a4 + 12b4 + 144c4 = 23




Quadratic Spline Example: Steps 3 and 4

The first derivatives must match at each interior knot

b1 + 6c1 − b2 − 6c2 = 0

b2 + 13c2 − b3 − 13c3 = 0

b3 + 16c3 − b4 − 16c4 = 0

Finally, set the second derivative at x4 to 0

c4 = 0

and we have 12 equations and 12 unknowns.




Quadratic Spline Example: Matrix Form

1 3 9 0 0 0 0 0 0 0 0 00 0 0 1 6.5 42.25 0 0 0 0 0 00 0 0 0 0 0 1 8 64 0 0 00 0 0 1 3 9 0 0 0 0 0 00 0 0 0 0 0 1 6.5 42.25 0 0 00 0 0 0 0 0 0 0 0 1 8 641 2 4 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 1 12 1440 1 6 0 −1 −6 0 0 0 0 0 00 0 0 0 1 13 0 −1 −13 0 0 00 0 0 0 0 0 0 1 16 0 −1 −160 0 0 0 0 0 0 0 0 0 0 1

a1b1c1a2b2c2a3b3c3a4b4c4

=

20171620171614230000

Which can be solved for the unknown a, b, and c values:a1 = −25.7858, a2 = 10.1683, a3 = 105.1112, a4 = 2.0000,b1 = 29.1548, b2 = 5.1854, b3 = −24.0278, b4 = 1.7500,c1 = −4.6309, c2 = −0.6361, c3 = 1.6111, c4 = 0.0000.




Quadratic Spline Example: Conclusion

The quadratic spline functions can then be written as:

f1(x) = −25.7858 + 29.1548x − 4.6309x2; 2.0 ≤ x ≤ 3.0

f2(x) = 10.1683 + 5.1854x − 0.6361x2; 3.0 ≤ x ≤ 6.5

f3(x) = 105.1112− 24.0278x + 1.6111x2; 6.5 ≤ x ≤ 8.0

f4(x) = 2.0000 + 1.7500x ; 8.0 ≤ x ≤ 12.0

and the value of f at x = 7.0 is then

f3(7.0) = 105.1112− 24.0278(7.0) + 1.6111(7.0)2 = 15.8605


curve fitting, interpolation · 2012. 4. 23. · relevance of curve fitting relevance of...

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