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Interpolation

Lecture NotesDr. Rakhmad Arief SiregarUniversiti Malaysia Perlis

Applied Numerical Method for Engineers

Chapter 17

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Interpolation

The first-order The second-order

The third-order

nnxaxaaaxf ...)( 2

210

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Interpolation

There are variety of alternative forms for expressing an interpolating polynomial.

The methods are: Newton’s Divided-difference interpolating

polynomial Linear interpolation Quadratic interpolation

Lagrange interpolating polynomial

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Linear interpolation

The simplest form of interpolation is to connect two data points with a straight line

01

01

0

011 )()()()(

xx

xfxf

xx

xfxf

)()()(

)()( 0101

0101 xx

xx

xfxfxfxf

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Linear interpolation

The simplest form of interpolation is to connect two data points with a straight line

01

01

0

011 )()()()(

xx

xfxf

xx

xfxf

)()()(

)()( 0101

0101 xx

xx

xfxfxfxf

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Ex. 18.x

Due to special order, you need to make a special taper pin with standard size of 3.5 mm.

Compute by using linear interpolation for the max. size and min size based on the following data.

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Dimensions at large end

of some standard

Taper Pins

- Metric Series-

Max:3.51

Min: 3.503mm

9(1,0)

(2,0.6931472)

(6,1.791759)

(2,0.3583519), t=48.3%

(4,1.386294)

t=33.3% (2,0.4620981)

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Quadratic Interpolation

To reduce error in Ex. 18.1,we need to introduce curvature into the line.

This is accomplished with a second-order polynomial.

After multiplying

))(()()( 1020102 xxxxbxxbbxf

nnxaxaaaxf ...)( 2

210

12021022

201102 )( xxbxxbxxbxbxbxbbxf

2210)( xaaaxf

1020100 xxbxbba 120221 xbxbba 22 ba

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Quadratic Interpolation

A simple procedure can be used to determine the value of the coefficients.

For b0 with x = x0 can be computed:

Then substitute back, which can be evaluated at x = x1

Finally substitute back b0 and b1

))(()()( 1020102 xxxxbxxbbxf

00 xfb

01

011 xx

xfxfb

12

01

01

12

01

2 xx

xx

xfxf

xx

xfxf

b

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Ex. 18.2

Fit a second-order polynomial to the three points used in Ex. 18.1

x0=1 f(x0)=0 x1=4 f(x1)=1.386294 X2=6 f(x2)=1.791759 Use the polynomial to evaluate ln 2

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Ex. 18.2

Solution

00 xfb

01

011 xx

xfxfb

12

01

01

12

01

2 xx

xx

xfxf

xx

xfxf

b

00 b

4620981.014

0386294.11

b

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4620981.046386294.1791759.1

2

b

0518731.02 b

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Ex. 18.2

Substituting value of b0, b1, b2

For x = 2

The relative percentage error is t = 18.4%

))(()()( 1020102 xxxxbxxbbxf

)4)(1(051873.0)1(4620981.00)(2 xxxxf

5658444.0)2(2 f

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General Form of Newton’s Interpolating Polynomial

The nth-order polynomial

As comparison, below is 2nd-order polynomial

))...()((...)()( 110010 nnn xxxxxxbxxbbxf

))(()()( 1020102 xxxxbxxbbxf

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General Form of Newton’s Interpolating Polynomial

The nth-order polynomial

The coefficient

))...()((...)()( 110010 nnn xxxxxxbxxbbxf

00 xfb

011 , xxfb

0122 ,, xxxfb

011 ,,...,, xxxxfb nnn

.

.

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General Form of Newton’s Interpolating Polynomial

Recursive nature divided differences

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