Ch10 Interpolation

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Interpolation

“Numerical Methods with MATLAB”, Recktenwald, Chapter 10and

“Numerical Methods for Engineers”, Chapra and Canale, 5th Ed., Part Five, Chapter 18and

“Applied Numerical Methods with MATLAB”, Chapra, 2nd Ed., Part Four, Chapters 15 and 16

PGE 310: Formulation and Solution in Geosystems Engineering

Dr. Balhoff

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How Fast are you going?

What time is it?

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Basic Ideas of Interpolation

Mathematical equivalent of “reading between the lines” Data - discrete samples of some function, f(x)

Uses an interpolating function between points

Data might… Exist as an experiment

Analytic function that is difficult to evaluate

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Interpolation versus Curve Fitting (Regression)

Interpolation Data assumed exact

Interpolating function between points

Curve passes EXACTLY through the data points

Curve Fitting Uncertainty in data

Uses a regression curve to fit data

Curve passes “near” the points

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Types of Interpolation

Single Polynomial Divided Difference

Lagrange Identical

Multiple polynomials Hermite

Lagrange

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Newton’s Divided-Difference is a Useful Form

For “n” points, I can always fit an “n-1” degree polynomial Line b/w 2 points

Parabola b/w 3 points

15th order polynomial b/w 16 points

Find coefficients of polynomial

2 10 1 2 1( ) n

nP x a a x a x a x

1 2 1 1 2 1( ) ( ) ( )n nP x b b x x b x x x x x x

Standard Form

Newton’s Divided Difference

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Linear Interpolation (Simplest Divided-Difference)

2 11 1

2 1

( ) ( )( ) ( )

f x f xf x f x x x

x x

1 1 2 1( )P x b b x x

Imagine Similar Triangles:

Newton-Form:

x1 x2x

f(x1)

f(x2)

f(x)12

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1

1 )()()()(

xx

xfxf

xx

xfxf

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General Form- Newton’s Interpolating Polynomials

1 2 1 1 2 1( ) ( ) ( )n nP x b b x x b x x x x x x

1 1

2 1 2

3 1 2 3

1 2 1

( )

[ , ]

, ,

, , , ,n n n

b f x

b f x x

b f x x x

b f x x x x

[ , ] [ , ], , i j j ki j k

i k

f x x f x xf x x x

x x

Where the b’s are:

2 1 2 11 2

1

[ , , ] [ , , , ], , , n n

nn

f x x f x x xf x x x

x x

( ) ( ), i ji j

i j

f x f xf x x

x x

The divided differences are:

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Linear Interpolation for f(x)=ln(x)

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Quadratic Interpolation example

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Higher Order Estimate

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Lagrange Interpolating Polynomials

Lagrange is simply a reformulation of the Newton polynomial !!!! Avoids computation of Divided Differences

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( ) ( ) ( )n

n j jj

P x L x f x

1

( )n

kj

k j kk j

x xL x

x x

1 1 2 2( ) ( ) ( ) ( )n nP x L f x L f x L f x

n

n

xx

xx

xx

xx

xx

xxL

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Lagrange Form:

2 10 1 2( ) n

nP x a a x a x a x

1 2 1 1 2 1( ) ( ) ( )n nP x b b x x b x x x x x x

Standard Form:

Newton’s Form:

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Problems with Single Polynomial Interpolation

“N-1” Order polynomial through “N” points

Forces oscillations which probably don’t exist naturally

Adding More points can make it worse

How do we fix the problem? Connect a series of interpolating functions

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Review…

Unlike Curve Fitting, Interpolation goes exactly through the data points

Interpolating Polynomials: fit ‘n-1’ order polynomial to ‘n’ data points

Newton Divided Difference

Lagrange Polynomials

Divide Difference and Lagrange are equivalent!

1 2 1 1 2 1( ) ( ) ( )n nP x b b x x b x x x x x x

1 1 2 2( ) ( ) ( ) ( )n nP x L f x L f x L f x 1

( )n

kj

k j kk j

x xL x

x x

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Piecewise Polynomials

Not always a good idea to fit a single polynomial to “n” data points Especially when “n” gets really big (like 10 or so) Lots of oscillations can result in error

Connect together a series of piecewise polynomials Splines

A few different types of splines Linear – functions match at “knot” points Quadratic – derivatives ALSO match at “knot” points Cubic – 1st and 2nd derivatives match at knot points

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Piecewise Polynomial Interpolation

Practical solution to high-degree polynomial interpolation

Use set of lower degree interpolants Each defined on sub-interval of domain Used instead of single function

approximation

Relationship b/w adjacent piecewise functions is of fundamental importance Shape is affected by constraints on

continuity

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Piecewise Linear Interpolation

Simplest scheme uses linear interpolants

Continuity In function at breakpoints

But NOT in the derivatives

Calculate linear interpolants using Lagrange

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222111 )( y

xx

xxy

xx

xxyLyLxf

Linear Lagrange polynomial

For each (n-1) sub-section

1,...,2,1)( 1

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1

niyxx

xxy

xx

xxxf i

ii

ii

ii

ii

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Cubic Spline

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Cubic Spline Interpolation

Basic Idea: Force continuity in 1st and 2nd derivatives at knots

(n-1) splines & 4 coefficients each = 4*(n-1) unknowns1. 2(n-1) known function values

2. n-2 derivatives must be equal at INTERIOR knots

3. n-2 second derivatives must be equal at INTERIOR knots

4n-6 constraints, but 4n-4 unknowns – need 2 more constraints!

iiiii dxcxbxaxf 23)(Standard Form:

Lagrange Form: We will derive….

Find the coefficients for each n-1 polynomials

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Options for Additional Constraints on End Points?

1. Fixed-slope end conditions – constrain the first derivative on the 2 end points

2. “Not-a-knot” end conditions – requires continuity in third derivative at first internal knot. Most accurate condition

3. “Natural” spline end condition – force second derivative equal to zero (We’ll use this one)

0)("0)(" 1 nxfandxf

(These are the last two constraints I need!)

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Cubic Spline Derivation

Forcing continuity in 2nd derivative gives:

Integrating equation twice gives cubic spline formula

" "1

"

3 3

11 1

11

1

1 1

1

"1

( ) ( )( )

6( ) 6( )

( ) ( )

( )

( )

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( ) ( )

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i i ii i i i

i i ii

i i

i i ii

i i i i

i

i i

i

i

i

f x f x

f x

f x x x x xx x x x

f x x xx x

x x

f x xf x xx x

x x

Can plug in xi’s and f(xi)’s, but still don’t know the 2nd derivatives

" "11

1 1

"( ) ( ) ( )i ii i i i i

i i i i

x x x xf x f x f x

x x x x

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Calculate Second Derivatives for Cubic Spline

"1 21 2 2 1

2 1 12 1 1

( ) "( ) "( )( ) 2( ) ( )

6 6[ ( ) ( )] ( ) ( )

i i i i i i

i i i ii i

i

i i

i ix x xf x f xx x x

f x f x f x f x

f

x x

x

x x

But First derivates must be equal at the internal knots

Differentiate the cubic spline equation and equate:

Write the above equation for each internal knot and substitute second derivative = 0 for end points (natural end condition) results in as many equations as unknowns.

Solving for second derivatives, I can then plug back into cubic spline equation

' '1 1 1( ) ( )i i i if x f x

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Conclusion

Curve Fitting: Best Fit curve to data

Interpolation: Function passes THROUGH all the data Newton/Lagrange: n-1 order polynomial for n points

Hermite and Spline: Piecewise polynomials

Cubic Spline Does not require knowledge of derivatives a priori

Forces continuity of first and second derivative

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