12 Interpolation Spline

8/10/2019 12 Interpolation Spline

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Curve-Fitting

Spline Interpolation


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Curve Fitting

RegressionLinear Regression

Polynomial Regression

Multiple Linear Regression

Non-linear Regression

Interpolation

Newton's Divided-Difference InterpolationLagrange Interpolating Polynomials



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For some cases, polynomialscan lead to erroneous results

because of round off error and

overshoot.

Alternative approach is to apply

lower-order polynomials to

subsets of data points. Suchconnecting polynomials are

called spline functions.


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(a)Linear spline

Derivatives are not

continuous Not smooth

(b) Quadratic spline

Continuous 1

st

derivatives

(c) Cubic spline

Continuous 1st

& 2nd

derivatives

Smoother


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Quadratic Spline


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Quadratic Spline

Spline of Degree 2

A function Qis called a spline of degree 2if The domain of Qis an interval [a, b].

Qand Q'are continuous functions on [a, b]. There are pointsxi(called knots) such that

a = x0< x1< < xn= band Qis a polynomial ofdegree at most 2on each subinterval [xi,xi+1].

A quadratic spline is a continuously differentiablepiecewise quadratic function.


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Exercise Which of the following is a quadratic spline?

]2,1[21

]1,0[

]0,2[

)( 2

2

xx

xx

xx

xB

]2,1[1

]1,0[2

]0,2[

)(

2

2

2

xxx

xxx

xx

xA


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Observations

n+1points

nintervals

Each interval is connected by a 2nd-order

polynomial Qi(x) = aix2 + bix + ci, i = 0, , n1.

Each polynomial has 3unknowns

Altogether there are 3nunknowns

Need 3nequations (or conditions) to solve for 3n

unknowns

Quadratic Interpolation


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1. Interpolating conditions On each sub interval [x

i

,xi+1

], the function Qi

(x)mustsatisfy the conditions

Qi(xi) =f(xi)and Qi(xi+1) =f(xi+1)

These conditions yield 2nequations

Quadratic Interpolation (3nconditions)

1...,,0

)(

)(

11

2

1

2

ni

xfcxbxa

xfcxbxa

iiiiii

iiiiii


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2. Continuous first derivatives The first derivatives at the interior knots must be

equal. This adds n-1more equations:

1...,,122 11 nibxabxa iiiiii

We now have 2n + (n1) = 3n1equations.

We need one more equation.


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3. Assume the 2ndderivatives is zero at the first

point. This gives us the last condition as


002 11 aa

With this condition selected, the first two points areconnected by a straight line.

Note: This is not the only possible choice orassumption we can make.


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Example

Fit quadratic splines to the given data

points.

i 0 1 2 3

xi 3 4.5 7 9

f(xi) 2.5 1 2.5 0.5


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Example (Solution)

1. Interpolating conditions

5.09815.2749

5.2749

0.15.425.20

0.15.425.20

5.239

333

333

222

222

111

111

cbacba

cba

cba

cba

cba

2. Continuous first derivatives

3322

2211

1414

99

baba

baba

3. Assume the 2nd derivatives is zero at the first point.

01a


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00

0

5.0

5.2

5.2

1

15.2

00000000101140114000

000019019

1981000000

1749000000

0001749000

00015.425.20000

00000015.425.20000000139

3

3

3

2

2

2

1

1

1

cb

a

c

b

a

c

ba

Example (Solution)

We can write the system of equations in matrix form as

Notice that the coefficient matrix is sparse.


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Example (Solution)

The system of equations can be solved to yield

3.916.246.1

46.1876.664.05.510

333

222

111

cba

cbacba

]9,7[3.916.246.1

]7,5.4[46.1876.664.0

]5.4,3[5.5

)(2

2

xxx

xxx

xx

xQ

Thus the quadratic spline that interpolates thegiven points is


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Efficient way to derive quadratic spline

2

1

1

1

11

1

11

)()(2

)()()(

as)(formto

rearrangedandresolved,,integratedbeturnincanwhich

)('

asformLagrangein)('writecanWe

).,(and),(throughpassinglinestraightais)('

hatimpliest tconditionsderivativefirstcontinuousThe

line.straingtais)('functionquadraticais)(

).('Let

i

ii

iiiiii

i

i

ii

ii

ii

ii

i

iiiii

ii

iii

xxxx

zzxxzxfxQ

xQ

zxx

xxz

xx

xxxQ

xQ

zxzxxQ

xQxQ

xQz


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2

1

1 )()(2

)()()( iii

iiiiii xx

xx

zzxxzxfxQ

11)('and

)('),()(verifycorrect,isthat thisseeTo

iii

iiiiii

zxQ

zxQxfxQ

s.'ofvaluethedeterminetoneedstillWe iz

ii

iiii

iii

xx

xfxfzz

nxfxQ

1

11

11

)()(2

:equationsfollowingthe

obtaincanweequations,resultingesimpify thand

1,-...,0,ifor)()(settingBy


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nixx

xfxfzz

xf"z

xxxx

zzxxzxfxQ

ii

iiii

i

ii

iiiiii

...,1,0,)()(

2

0))(assumeweif(0

where

)()(2

)()()(

1

11

00

2

1

1


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Cubic Spline

Spline of Degree 3

A function Sis called a spline of degree 3if

The domain of Sis an interval [a, b].

S, S'and S"are continuous functions on [a, b].

There are points ti(called knots) such that

a = t0< t1< < tn= band Qis a polynomial ofdegree at most 3on each subinterval [ti, ti+1].


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Cubic Spline (4nconditions)

1. Interpolating conditions (2nconditoins).

2. Continuous 1stderivatives (n-1conditions)

The 1stderivatives at the interior knots must be equal.

3. Continuous 2ndderivatives (n-1conditions) The 2ndderivatives at the interior knots must be equal.

4. Assume the 2ndderivatives at the end points are

zero (2conditions). This condition makes the spline a "natural spline".


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Efficient way to derive cubic spline

1,...,1for

)()()()(

6)(2

0,0

solvingfromobtainedbecans'unknowntheandwhere

)(6

)()(

6

)(

)(6

)(6

)(

1

11

1111

0

1

111

3

1

31

ni

h

xfxf

h

xfxf

zhzhhzh

zz

zxxh

xxzh

h

xfxxz

h

h

xf

xxh

zxx

h

zxS

i

ii

i

ii

iiiiiii

n

i

iii

iii

i

iii

i

i

i

ii

i

ii

i

i


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Summary

Advantages of spline interpolation overpolynomial interpolation

The conditions that are used to derive thequadratic and cubic spline functions

Characteristics of cubic spline Overcome the problem of "overshoot"

Easier to derive (than high-order polynomial)

Smooth (continuous 2nd-order derivatives)

12 Interpolation Spline

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