Comp Phys3

8/11/2019 Comp Phys3

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Interpolation, Extrapolation&

Polynomial Approximation

November 13, 2012

Interpolation, Extrapolation & Polynomial Approximation
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Introduction

In many cases we know the values of a function f (x ) at a set of pointsx 1, x 2, ..., x N , but we dont have the analytic expression of the functionthat lets us calculate its value at an arbitrary point. We will try toestimate f (x ) for arbitrary x by drawing a curve through the x i andsometimes beyond them.

The procedure of estimating the value of f (x ) for x [x 1, x N ] is calledinterpolation while if the value is for points x /[x 1, x N ] extrapolation.

The form of the function that approximates the set of points should be a

convenient one and should be applicable to a general class of problems.Interpolation, Extrapolation & Polynomial Approximation
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Polynomial Approximations

Polynomial functions are the most common ones while rational andtrigonometric functions are used quite frequently.

We will study the following methods for polynomial approximations:

Lagranges PolynomialHermite Polynomial

Taylor Polynomial

Cubic Splines

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Lagrange Polynomial

Lets assume the following set of data:

x 0 x 1 x 2 x 3x 3.2 2.7 1.0 4.8f (x ) 22.0 17.8 14.2 38.3

f 0 f 1 f 2 f 3

Then the interpolating polynomial will be of 4th order i.e.

ax 3

+ bx 2

+ cx + d = P (x ). This leads to 4 equations for the 4 unknowncoefficients and by solving this system we get a = 0.5275, b = 6 .4952,c = 16.117 and d = 24.3499 and the polynomial is:

P (x ) = 0.5275x 3 + 6 .4952x 2 16.117x + 24 .3499

It is obvious that this procedure is quite laboureus and Lagrangedeveloped a direct way to nd the polynomial

P n (x ) = f 0L0(x ) + f 1L1(x ) + .... + f n Ln (x ) =n

i =0

f i Li (x ) (1)

where Li (x ) are the Lagrange coefficient poly no m ialsInterpolation, Extrapolation & Polynomial Approximation
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L j (x ) = (x x 0)(x x 1)...(x x j 1)(x x j +1 )...(x x n )(x j x 0)(x j x 1)...(x j x j 1)(x j x j +1 )...(x j x n )

(2)

and obviously:L j (x k ) = jk =

0 if j = k 1 if j = k

where jk is Kroneckers symbol.

ERRORSThe error when the Lagrange polynomial is used to approximate acontinuous function f (x ) it is :

E (x ) = ( x x 0)(x x 1).... (x x n )f (n+1) ( )

(n + 1)!

where [x 0, x N ] (3)

NOTELagrange polynomial applies for evenly and unevenly spaced points. Stillif the points are evenly spaced then it reduces to a much simpler form.

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Lagrange Polynomial : Example

Find the Lagrange polynomial thatapproximates the function y = cos( x ).We create the table

x i 0 0.5 1f i 1 0.0 -1

The Lagrange coeffiecient polynomials are:

L1(x ) = (x x 2)(x x 3)(x 1 x 2)(x 1 x 3)

= (x 0.5)(x 1)(0 0.5)(0 1)

= 2x 2 3x + 1 ,

L2(x ) = (x x 1) (x x 3)

(x 2 x 1)(x 2 x 3) =

(x 0)(x 1)(0.5 0)(0 .5 1)

= 4(x 2 x )

L3(x ) = (x x 1)(x x 2)(x 3 x 1)(x 3 x 2)

= (x 0)(x 0.5)(1 0)(1 0.5)

= 2x 2 x

thus

P (x ) = 1 (2x 2 3x + 1) 0.4 (x 2 x ) + ( 1) (2x 2 x ) = 2x + 1

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The error will be:

E (x ) = x (x 0.5) (x 1)3 sin( )

3!

e.g. for x = 0 .25 is E (0.25) 0.24.

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Newton Polynomial

Forward Newton-Gregory

P n (x s ) = f 0 + s f 0 + s (s 1)2! 2f 0 + s (s 1)( s 2)3!

3f 0 + ...

= f 0 + s 1 f 0 +

s 2

2f 0 + s 3

3f 0 + ...

=

n

i =0

s i

i

f 0 (4)

where x s = x 0 + s h and

f i = f i +1 f i (5)

2f i = f i +2 2f i +1 + f i (6)

3f i = f i +3 3f i +2 + 3 f i +1 f i (7)

n f i = f i + n nf i + n 1 + n(n 1)

2! f i + n 2

n(n 1)(n 2)3!

f i + n 3 + (8)

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Newton Polynomial

ERROR The error is the same with the Lagrange polynomial :

E (x ) = ( x x 0)(x x 1).... (x x n )f (n+1) ( )

(n + 1)! where [x 0, x N ] (9)

x f (x ) f (x ) 2 f (x ) 3 f (x ) 4 f (x ) 5 f (x )0.0 0.000

0.2030.2 0.203 0.017

0.220 0.0240.4 0.423 0.041 0.020

0.261 0.044 0.0320.6 0.684 0.085 0.052

0.246 0.0960.8 1.030 0.181

0.5271.0 1.557

Table: Example of a difference matrix

Backward Newton-Gregory

P n (x ) = f 0+ s 1 f 1+

s + 12

2f 2+ ...+ s + n 1

n n f n

(10)Interpolation, Extrapolation & Polynomial Approximation
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x F (x ) f (x ) 2 f (x ) 3 f (x ) 4 f (x ) 5 f (x )0.2 1.06894

0.112420.5 1.18136 0.01183

0.12425 0.001230.8 1.30561 0.01306 0.00015

0.13731 0.00138 0.0000011.1 1.44292 0.01444 0.00014

0.15175 0.01521.4 1.59467 0.01596

0.167711.7 1.76238

Newton-Gregory forward with x 0 = 0 .5:

P 3(x ) = 1 .18136 + 0 .12425 s + 0 .01306 s 2 + 0 .00138

s 3

= 1 .18136 + 0 .12425 s + 0 .01306 s (s 1)/ 2 + 0 .00138 s (s 1) (s 2)/ 6= 0.9996 + 0 .3354 x + 0 .052 x 2 + 0 .085 x 3

Newton-Gregory backwards with x 0 = 1 .1:

P 3(x ) = 1 .44292 + 0 .13731 s + 0 .01306 s + 1

2 + 0 .00123 s + 2

3= 1 .44292 + 0 .13731 s + 0 .01306 s (s + 1) / 2 + 0 .00123 s (s + 1) ( s + 2) / 6= 0.99996 + 0 .33374 x + 0 .05433 x 2 + 0 .007593 x 3

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Hermite Polynomial

This applies when we have information not only for the values of f (x )but also on its derivative f (x )

P 2n 1(x ) =n

i =1

Ai (x )f i +n

i =1

B i (x )f

i (11)

where

Ai (x ) = [1 2(x x i )Li (x i )] [Li (x )]2 (12)B i (x ) = ( x x i ) [Li (x )]2 (13)

and Li (x ) are the Lagrange coefficients.

ERROR : the accuracy is similar to that of Lagrange polynomial of order2n!

y (x ) p (x ) = y (2n+1) ( )(2n + 1)!

[(x x 1)(x x 2) . . . (x x n )]2 (14)


l l l
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Hermite Polynomial : Example

Fit a Hermite polynomial to thedata of the table:

k x k y k y k 0 0 0 01 4 2 0

The Lagrange coefficients are:

L0(x ) = x x 1

x 0 x 1=

x 40 4

= x 4

4 L1(x ) =

x x 0x 1 x 0

= x 4

L0(x ) = 1x 0 x 1 =

14 L1(x ) =

1x 1 x 0 =

14

ThusA0(x ) = 1 2 L0(x x 0) L

20 = 1 2

1

4(x 0)

x 4

4

2

A1(x ) = 1 2 L0(x x 1) L21 = 1 2

1

4(x 4)

x

4

2= 3

x

2

x

4

2

B 0(x ) = ( x 0) x 4

4

2= x

x 4

4

2B 1(x ) = ( x 4)

x

4

2

And the Hermite polynomial is:

P (x ) = (6 x )x 2

16.


T l P l i l
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Taylor Polynomial

It is an alternative way of approximating functions with polynomials. Inthe previous two cases we found the polynomial P (x ) that gets the same

value with a function f (x ) at N points or the polynomial that agrees witha function and its derivative at N points. Taylor polynomial has the samevalue x 0 with the function but agrees also up to the Nth derivative withthe given function. That is:

P (i ) (x 0) = f (i ) (x 0) and i = 0 , 1,..., n

and the Taylor polynomial has the well known form from Calculus

P (x ) =N

i =0

f (i ) (x )i !

(x x 0)i (15)

ERROR : was also estimated in calculus

E N (x ) = ( x x 0)N +1f (N +1) ( )(N + 1)!

(16)


T l P l i l E l
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Taylor Polynomial : Example

We will show that for the calculation of e = 2 .718281828459... with13-digit approximation we need 15 terms of the Taylor expansion.All the derivatives at x = 1 are:

y 0 = y (1)0 = y

(2)0 = ... = y

(n)0 = 1

thusp (x ) =

n

i =1

1i !

x n = 1 + x + x 2

2 + ... +

1n!

x n

and the error will be

|E n | = x n+1 e

(n + 1)! = e

16! < 3

16! = 1 .433 10 13

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Interpolation ith C bic Splines


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Interpolation with Cubic Splines


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Let the cubic for the i th interval, which lies between the points (x i , y i )and (x i +1 , y i +1 ) has the form:

y (x ) = ai (x x i )3 + b i (x x i )2 + c i (x x i ) + d i

Since it ts at the two endpoints of the interval:

y i = ai (x i x i )3

+ b i (x i x i )2

+ c i (x i x i ) + d i = d i y i +1 = ai (x i +1 x i )3 + b i (x i +1 x i )2 + c i (x i +1 x i ) + d i

= ai h3i + b i h2i + c i hi + d i

where hi = x i +1 x i . We need the 1st and 2nd derivatives to relate the

slopes and curvatures of the joining polynomials, by differentiation we gety (x ) = 3 ai (x x i )2 + 2 b i (x x i ) + c i y (x ) = 6 ai (x x i ) + 2 b i


The mathematical procedure is simplied if we write the equations in
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The mathematical procedure is simplied if we write the equations interms of the 2nd derivatives of the interpolating cubics. Lets name S i the 2nd derivative at the point (x i , y i ) then we can easily get:

b i = S i 2 , ai =

S i +1 S i 6hi (17)

which means

y i +1 = S i +1 S i

6hi h3i +

S i 2

h2i + c i hi + y i

and nally

c i = y i +1 y i

hi

2hi S i + hi S i +16

(18)

Now we invoke the condition that the slopes of the two cubics joining at

(x i , y i ) are the same:y i = 3ai (x i x i )2 + 2 b i (x i x i ) + c i = c i y i = 3ai 1 (x i x i 1)2 + 2 b i 1 (x i x i 1) + c i 1

= 3 ai 1h2i 1 + 2b i 1hi 1 + c i 1

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By equating these and substituting a, b , c and d we get:

y i = y i +1 y i

hi 2hi S i + hi S i +1

6

= 3S i S i 1

6hi 1h2i 1 + 2

S i 12

hi 1 + y i y i 1

hi 1

2hi 1S i 1 + hi 1S i 6

and by simplifying we get:

hi 1S i 1 + 2 ( hi 1 + hi ) S i + hi S i +1 = 6y i +1 y i

hi

y i y i 1hi 1

(19)

If we have n + 1 points the above relation can be applied to the n 1internal points. Thus we create a system of n 1 equations for the n + 1unknown S i . This system can be solved if we specify the values of S 0 andS n .


The system of n 1 equations with n + 1 unknown will be written as:
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The system of n 1 equations with n + 1 unknown will be written as:.......

h0 2(h0 + h1) h1h1 2(h1 + h2) h2

h2 2(h2 + h3) h3....... .......

hn 2 2(hn 2 + hn 1 ) hn 1

S 0S 1S 2:S n 2S n 1S n

= 6

....y 2 y 1

h1 y 1 y 0

h0y 3 y 2

h2 y 2 y 1

h1....y n y n

1hn 1 y n

1 y n

2hn 2....

Y

From the solution of this linear systems we get the coefficients ai , b i , c i and d i via the relations:

ai = S i +1 S i 6hi , b i = S i 2

(20)

c i = y i +1 y i

hi

2hi S i + hi S i +16

(21)

d i = y i (22)

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Choice I Take, S 0 = 0 and S n = 0 this will lead to the solution of thefollowing (n 1) (n 1) linear system:

2(h0 + h1) h1h1 2(h1 + h2) h2

h2 2(h2 + h3) h3....hn 2 2(hn 2 + hn 1)

S 1S 2S 3

S n 1

= Y

Choice II Take, S 0 = S 1 and S n = S n 1 this will lead to the solution of the following (n 1) (n 1) linear system:

3h0 + 2 h1 h1h1 2(h1 + h2) h2

h2 2(h2 + h3) h3... ...

hn 2 2hn 2 + 3 hn 1

S 1S 2S 3

S n 1

= Y

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Choice III Use linear extrapolation

S 1 S 0h0= S 2 S 1h1

S 0 = (h0 + h1)S 1 h0S 2h1S n S n 1

hn 1=

S n 1 S n 2hn 2

s n = (hn 2 + hn 1)S n 1 hn 1S n 2

hn 2this will lead to the solution of the following (n 1) (n 1) linearsystem:

(h0+ h1)( h0 +2 h1)h1

h21 h20h1

h1 2(h1 + h2) h2h2 2(h2 + h3) h3

.... ....h2n 2

h2n 1hn 2

(hn 1+ hn 2)( hn 1+2 hn 2)

hn 2

S 1S 2S 3

S n 1

= Y

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Condition III :

6 0 01 4 10 0 6

S 1S 2S 3

=3672108

S 0 = 0 S 1 = 6S 2 = 12 S 3 = 18S 4 = 24

Condition IV :

2 1 0 0 01 4 1 0 00 1 4 1 00 0 1 4 10 0 0 1 2

S 0S 1S 2S 3S 4

=

6367210866

S 0 = 0S 1 = 6S 2 = 12S 3 = 18S 4 = 24


Interpolation with Cubic Splines : Problems


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p p

1 The following data are from astronomical observations andrepresent variations of the apparent magnitude of a type of variablestars called Cepheids

Time 0.0 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1.0Apparentmagnitude

0.302 0.185 0.106 0.093 0.24 0.579 0.561 0.468 0.302

Use splines to create a new table for the apparent magnitude forintervals of time of 0.5.

2 From the following table nd the acceleration of gravity atTubingen (48o 31) and the distance between two points withangular separation of 1 of a degree.

Latitude Length of 1 of arcon the parallel

local accelerationof gravity g

00 1855.4 m 9.7805 m/sec 2

150 1792.0 m 9.7839 m/sec 2

300 1608.2 m 9.7934 m/sec 2

450 1314.2 m 9.8063 m/sec 2

600 930.0 m 9.8192 m/sec 2

750 481.7 m 9.8287 m/sec 2

900 0.0 m 9.8322 m/sec 2


Rational function approximations


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pp

Here we introduce the notion of rational approximations for functions .We will constrain our discussion to the so called Pade approximation .A rational approximation of f (x ) on [a, b ], is the quotient of twopolynomials P n (x ) and Q m (x ) with degrees n and m1

f (x ) = R N (x ) P n (x )Q m (x )

= a0 + a1x + a2x 2 + ... + an x n

1 + b 1x + b 2x 2 + ... + b m x m , N = n + m

i.e. there are N + 1 = n + m + 1 constants to be determined.The method of Pade requires that f (x ) and its derivatives are continuousat x = 0 . This choice makes the manipulation simpler and a change of variable can be used to shift, if needed, the calculations over to aninterval that contains zero.We begin with the Maclaurin series for f (x ) (up to the term x N ), thiscan be written as :

f (x ) c 0 + c 1x + ... + c N x N

where c i = f ( i ) (0) / (i !).1

Sometimes we write R

N (x

) R

n, m (x

).Interpolation, Extrapolation & Polynomial Approximation

Pade approximation


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pp

Then we create the difference

f (x ) R N (x ) c 0 + c 1x + ... + c N x N

a0 + a1x + ... + a n x n

1 + b 1x + ... + b m x m

=c 0 + c 1x + ... + c N x

N (1 + b 1x + ... + b m x m ) (a0 + a1x + ... + an x n )

1 + b 1

x + ... + b m

x m

If f (0) = R N (0) then c 0 a0 = 0 .In the same way , in order for the rst N derivatives of f (x ) and R N (x )to be equal at x = 0 the coefficients of the powers of x up to x N in thenumerator must be zero also.



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This gives additionally N equations for the as and b s

b 1c 0 + c 1 a1 = 0

b 2c 0 + b 1c 1 + c 2 a2 = 0b 3c 0 + b 2c 1 + b 1c 2 + c 3 a3 = 0

...b m c n m + b m 1c n m +1 + ... + c n an = 0

b m c n m +1 + b m 1c n m +2 + ... + c n+1 = 0 (23)b m c n m +2 + b m 1c n m +3 + ... + c n+2 = 0

...b m c N m + b m 1c N m +1 + ... + c N = 0

Notice that in each of the above equations, the sum of subscripts on thefactors of each product is the same, and is equal to the exponent of thex -term in the numerator.


Pade approximation : Example


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For the R 9(x ) or R 5,4(x ) Pade approximation for the function tan 1 (x )

we calculate the Maclaurin series of tan 1(x ):

tan 1 (x ) = x

1

3x 3 +

1

5x 5

1

7x 7 +

1

9x 9

and then

f (x ) R 9 (x ) =x 13 x

3 + 15 x 5 1

7 x 7 + 19 x

9 1 + b 1 x + b 2x 2 + b 3x 3 + b 4 x 4 a0 + a1x + a2x 2 + ... + a5x 5

1 + b 1x + b 2x 2 + b 3x 3 + b 4x 4

and the coefficients will be found by the following system of equations:

a0 = 0 , a1 = 1 , a2 = b 1 , a3 = 13

+ b 2 , a4 = 13

b 1 + b 3 , a5 = 15

13

b 2 + b 4

1

5b 1

1

2b 3 = 0 ,

1

7+

1

5b 2

1

3b 4 = 0 ,

1

7b 1 +

1

5b 3 = 0 ,

1

9

1

7b 2 +

1

5b 4 = 0 (24)

from which we geta0 = 0 , a1 = 1 , a2 = 0 , a3 = 79 , a4 = 0 , a5 =

64945 , b 1 = 0 , b 2 =

109 , b 3 = 0 , b 4 =

521 .

tan 1 x R 9(x ) = x + 79 x

3 + 64945 x 5

1 + 109 x 2 + 521 x 4

For x = 1 , exact 0.7854, R 9(1) = 0 .78558 while from the Maclaurinseries we get 0.8349!


Rational approximation for sets of data


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If instead of the analytic form of a function f (x ) we have a set of k points (x i , f (x i )) in order to nd a rational function R N (x ) such that for

every x i we will get f (x i ) = R N (x i ) i.e.

R N (x i ) = a0 + a1x i + a2x 2i + ... + an x

ni

1 + b 1x i + b 2x 2i + ... + b m x mi

= f (x i )

we will follow the approach used in constructing the approximatepolynomial. In other words the problem will be solved by nding thesolution of the following system of k m + n + 1 equations:

a0 + a1x 1 + ... + an x n1 (f 1x 1) b 1 ... (f 1x m1 ) b m = f 1

::

a0 + a1x i + ... + an x ni (f i x i ) b 1 ... (f i x

mi ) b m = f i ::

a0 + a1x k + ... + an x nk (f k x k ) b 1 ... (f k x mk ) b m = f k

i.e. we get k equations for the k unknowns a0, a1,..., an and b 1, b 2,..., b m .


Rational approximation for sets of data : Example


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We will nd the rational function approximations for the following set of data (-1,1), (0,2) and (1,-1).It is obvious that the sum of degrees of the polynomials in the nominatorand denominator must be (n + m + 1 3). Thus we can write:

R 1,1 (x ) = a0 + a1x 1 + b 1x

which leads to the following system

a0 + ( 1) a1 ( 1) b 1 = 1a0 + 0 a1 0 b 1 = 2a0 + 1 a1 ( 1) b 1 = 1

a0 = 2a1 = 1b 1 = 2

and the rational fuction will be:

R 1,1 (x ) = 2 x 1 2x .

Alternatively, one may derive the following rational function:

R 0,1 (x ) = a0

1 + b 1x + b 2x R (x ) =

21 2x x 2


Rational approximation: Problems


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1 Find the Pade approximation R 3,3(x ) for the function y = e x .Compare with the Maclaurin series for x = 1.

2 Find the Pade approximation R 3,5(x ) for the functions y = cos( x )and y = sin( x ). Compare with the Maclaurin series for x = 1.

3

Find the Pade approximation R 4,6(x ). for the functiony = 1 / x sin(x ). Compare with the Maclaurin series for x = 1.

4 Find the rational approximation for the following set of points:

0 1 2 4

0.83 1.06 1.25 4.15

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