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Approximation of Rational Functions

Project by: Robera Wubie

Submitted to: Ato Yohannes T.

Addis Ababa University

Department of Applied Mathematics

Submitted to the Department of Applied Mathematics in partial fulfillment of a

B.Sc. Degree

MiliyonNew Stamp

userAccepted set by user

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Acknowledgment

Above all else, Id like to thank the Almighty God and his mother the Virgin Mary for

helping me come this far in life. Next, I would like to thank my family: My mother

Sinafikish Lema, my Father Wube Gemeda, my Sister Selamawit Teshale, my little

brother Moa Wubie, and especially my twin brother Roba Wubie. I would also like to

thank all my teachers in the math department, for teaching me all the knowledge that

they have. It would be an injustice to forget to mention all my friends, especially:

Milliyon Tilahun, Fresenbet G/yohanes, Abener Tewodros, Misganaw Nega and

Eyob Mekonnen, for helping me throught hard times. Finally, I would like to thank

my advisor, Ato Yohannes T, for the guidance and valuable comments he provided

on my project.

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Contents

Introduction....Page 4

1. Approximations of Functions..Page 5

2. Approximations of Rational FunctionsPage 5

3. Constant Approximation.Page 6

4. Linear Approximation.Page 8

5. Chebyshev Polynomials...Page 10

5.1.1 Properties of Chebyshev Polynomials....Page 10

5.1.2 Orthogonality of Chebyshev Polynomials..Page 12

5.1.3 Chebyshev ApproximationPage 12

6. Pade Approximation ...Page 14

7. References....Page 18

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Introduction

In this paper, I am going to be addressing the topic of approximation as it applies to

rational functions. A rational function is a function of the form ()

(), where () 0,

and where both () and () are polynomials.

The Chebyshev polynomials develop the theory of a class of orthogonal polynomials

that are the basis for fitting non-algebraic functions with polynomials of maximum

efficiency. Chebyshev polynomials, can be used to create polynomial approximations

that are significantly, more efficient than Maclaurin series. The Pade approximation as

it applies to rational functions is one of the best efficient tools. In addition, Pade

approximants are excellent mathematical tools that are useful for analyzing non-linear

problems.

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1. Approximation of functions

One fundamental problem which occurs in many orients is approximating a function f

by using a member of functions in , which are a class of functions very easy to

work with. An example could be made of Polynomials, Rational functions or

Trigonometric polynomials, where each particular function in the class is specified by

a numerical value of a number of parameters.

It is to note that there are two types of short comings to take into account here:

a) Error while inputting the data

b) Short comings in the particular model which one intends to adopt for the

purpose of data entry.

2. Approximation of Rational functions

Suppose () is a rational function. An approximating function (), is another

function usually much simpler than f, whose values are approximately the same as

the values of f. The difference () () = () is the error, and measures how

accurate or inaccurate our approximation is. Thus:

() = () + (), () =

() =

The two simplest functions are constant functions and linear functions and if () is

constant or linear function, then we will perform a constant or linear approximation

of f.

When we choose our approximation, we want the error to be as small as possible.

Now, we know what it means for a number to be small, but our error () is a

function, and is therefore less clear what it means for a function to be small. In fact,

there are different notions of smallness for functions, and depending on which one

we use, we can obtain different types of approximations.

In this section, we shall fix , and we shall say that our error is small

if () is very small for all values that are close to a.

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Accordingly, we want to find an approximation for a rational function () on a

small interval [ , + ] by some simpler function (). Then we

want to determine the accuracy of the approximation by bounding the absolute value

of the error |()| = |() ()|.

3. Constant Approximation

The best constant approximation of f is () = () we write the error term as

() = () ()

= () ()

The Mean-value theorem tells us that

() = ()()

, for some z between a and x

()( ) = () ()

= ()

Note: So to bound the error on some interval [ , + ] around a, we need to

bound |()| = |()|| | on the interval [ , + ]. Observe that

| | and so |()| max |()| . We can thus bound the error |()|

as soon as we can bound |()| on [ , + ].

Given that a and to find h, we have the following steps to follow.

1. Choose so that [ , + ] dom (f).

( is an initial guess for h)

2. Find M so that |()| on [ , + ]

3. Choose h in (0, ] so that .

Remember that we want > 0 thus in (0, )

Example: Let () = 4 + 5 and = 2

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Find () and a number h so that

|()| 1

20= 2 2 +

Solution

By the above steps, here is how we solve it.

Step 1: Let us choose =

, then [ , + ],

[2.01, 2 + 0.1] = [1.9,2.1]

Step 2: We want to bound the absolute value of () = 3 4 for [1.9, 2.1],

we have:

1.9 2.1

3.61 4.41

10.83 3 13.23

6.83 3 4 9.23

Thus |()| = |3 4| 9.23 = [1.9,2.1]

Step 3: Find h in (0,

] so that 9.23 = =

. We want 0 <

= 0.1

and

()(.)=

. 0.0054.

Let = 0.005. From the above steps, we finally get:

2 2 +

1.995 2.005

|()| . = 0.04615 0.05 =1

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The constant approximation () = 5 to () = 4 + 5 is accurate to within

an error |()| at most

as long as [2 0.05, 2 + 0.05].

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4. Linear Approximation

The best linear approximation for f near a is:

() = () + ()( )

The function whose graph is the tangent line to the graph of f through (, ()). It is

called the linear approximation to f at a and its error is

() = () () ()( )

The Intermediate man value theorem tells us that

() ( )

2= () () ()( )

= () () = (),

.

Therefore, we can control the error () as soon as we can find an upper bound

|()| for the second derivative.

How do we find h in linear approximation?

Step 1: Choose so that [ , + ] ().

.

Step 2: find M so that |()| on [ , + ].

Step 3: Choose h in (0, ] so that

. At last, since we have succeeded with

the above steps, if [ , + ] then:

|()| =|()|()

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Example: Given a function () = 4 + 5 and = 2, find () and a

number h, so that

|()|

= when 2 2 +

Solution:

First, we find out the value of ().

() = (2) + (2)( 2)

= 5 + 8( 2)

= 8 11

Now, we find the value of h

1. Choose so that [ , + ] ()

Let =

, [ , + ] = [1.9,2.1]

2. We want to bound the absolute value () = 6, [1.9,2.1], we have

1.9 2.1

()| 6(2.1)

= 12.6 =

Thus |()| 12.6 = [1.9,2.1].

3. Find h in (0,0.1] so that (.)

=

=

We want 0 < 0.1

from

=

(.)

.

0.0079365

0.08909

Let = 0.08

We finally get: 2 2 +

2 0.08 2 + 0.08

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1.92 2.08

This implies that |()|

5. Chebyshev Polynomials

We now turn our attention to the problem of representing a function with a minimum

error. This is a central problem in the software development of digital computers because

it is more economical to compute the values of the common functions using an efficient

approximation than to store a table of values and employ interpolation techniques

because digital computers are essentially only as arithmetic devices, the most elaborate

function they can compute is rational function, which is a ration of polynomials.

5.1 Properties of Chebyshev polynomials

Property 1: Recurrence relation

Chebyshev polynomials can be generated in the following way.

Set () = 1 () = , and by using the recurrence relation

2() = () + ()

The above equation can be arranged as such to give the recurrence relation:

() = () () Eqn 1

Note that, for any [1,1] there exists a such that = .

Let us define the set of polynomials () = where = for 1 1.

These polynomials, called Chebyshev polynomials, exist for all but the definition of

() only makes sense for [1,1].

To determine the form of these polynomials we recall the trigonometric formula.

( + 1) + ( 1) = 2

Again, we can rewrite the trigonometric identity like the following:

() + () = (),

which is a similar equation with Eqn 1.

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The Chebyshev polynomials can be generated by equation 1, and they give:

() = 0 = 1

() = =

() = 2 = 2 1

() = 3 = 2(2 1) = 4 3

() = 4 = 2(4 3) (2 1)

= 8 8 + 1

() = 5 = 2(4 3) (2 1)

= 16 20 + 5

Observe that if n is even () contains even powers of x and if n is odd () contains

odd powers of x.

Property 2: Leading Coefficient

The coefficient of is () is 2where 1.

Property 3: Symmetry

When = 2, () is an even function. That is:

() = ()

When = 2 + 1, () is an odd function, that is:

() = ()

Property 4: Trigonometric representation on [, ].

() = ( arccos()) for 1 1

Property 5: |()| 1, 1 1,

Proof: Since we know that 1 1, then by definition 1 () 1 for

[1,1].

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5.2 Orthogonality of Chebyshev polynomials

Two polynomials (), () are Orthogonal on an interval [, ] if their inner product

< , >= ()() () = 0, () 0

Set of polynomials is orthogonal if each polynomial is orthogonal to each other.

Note: Chebyshev polynomials are orthogonal on interval (1,1] with () =

.

Example: Verify that the Orthogonality of the Chebyshev Polynomials () =

(), () = ().

Proof:

< (), () >= 1(4 3)

1

1 = 0

5.3. Chebyshev Approximation

The Chebyshev approximation polynomial () or degree for () over [1,1] can

be written as a sum of {()}.

() () = ()

The coefficients {} are computed with the formulas:

=1

+ 1 ()()

=1

+ 1 ()

and

=2

+ 1 ()()

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=2

+ 1 () (

(2 + 1)

2 + 2 = 1,2,3,

Example: Find the Chebyshev polynomials () that approximate the function

() = over [1,1].

Solution:

The Coefficients can be calculated using the above formulas, and the nodes

= ((2 + 1))

8 = 0,1,2,3

=1

4 ()

=1

4 =

1.26606568

=1

2 ()

=1

2

= 1.13031500

=1

2 ()

=1

2 (

2(2 + 1)

8)

= 0.27145036

=1

2 ()

14

=1

2 (

3(2 + 1)

8)

= 0.04379392

Therefore, the Chebyshev Polynomials () for is:

() = 1.26606568() + 1.13031500() + 0.27145036()

+ 0.04379392()

If we expand in powers of X, of the Chebyshev polynomial, then:

() = 0.99461332+ 0.99893324 + 0.54290072 + 0.1751768

6. Pade Approximation

A Pade rational approximation to () on the interval [, ] is the quotient of two

polynomials () and () of degree and respectively. We use the notation ,()

to denote the quotient.

,() =()

()

Let us assume that assume that , and that () is a Maclaurin polynomial

expansion of degree + at least, then:

() ,() =()

()

Where () and () are polynomials of degree and respectively.

Note: Our goal is to make the maximum error as small as possible. For a given amount of

computational effort, one can usually construct a rational approximation that has a smaller

over all error on [, ], then a polynomial approximation.

The method of Pade requires () and its derivative be continiuous at = 0. First, it makes

the manipulation simpler. Second, a change of variable can be used to shift the calculations

over to an interval that contains zero.

Theorem: Every pade form of type (, ) for () yields the same rational function.

Proof:

(, ) (, ) Are Pade forms

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= 0()

= 0()

So

+ = 0())

Thus, =>

=

Example: Find the Pade approximation ,() for [] =

Solution:

First set up the equation ()() () = 0 , where = = 2.

() = + +

() = 1 + +

() = 1 + +

2+

6+

24

Show from

()() () = 0

(1 ) + (1 + ) + 1

2 + +

+ 1

6+

2

+

+ 1

24

6

+2

= 0

Secondly, solve the equation ()() () = 0

1 = 01

1 + = 0.2

+ + = 03

+

+ = 0.4

+

+

= 05

After solving the above equations, we will get the following results:

16

= 1, =1

2, =

1

12 =

1

2 =

1

12

Show [] =

[] = ,() =()

()=

1 +12

+1

12

1 12

+1

12

=12 + 6 +

12 6 +

[] [] = (12 + 6 +

12 6 + )

The maximum error is |[] []| 0.00399611

Example 2: Find the pade approximation ,() for [] = [].

Solution:

First, set up the equation ()() () = 0

We have given = = 4

() = + + +

+

() = 1 + + +

+

() = 1

2+

24

720+

40320

Now from

()() () = 0

(1 ) + ( + ) + 1

2 +

+ 2

+

+ 1

24

2

+ +

24

2

+ 1

720+

24

2

+

720+

24

+ 1

40320

720

+24

Second, solve the equation ()() () = 0

1 = 0....1

+ = 02

17

+ = 03

+ = 0..4

+ = 0.5

= 06

+

= 0.7

+

= 0............8

+

= 09

After solving, the equations we will get

= 1, = 0, =

, = 0, =

= 0, =11

252, = 0, =

13

15120

[] = []

[] =15120 6900 + 313

15120 + 660 + 13

[] [] = [] (15120 6900 + 313

15120 + 660 + 13)

The maximum error is

|[] []| 3.5987 10

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References

Applied Numerical Analysis, 6th edition by Gerald/Weathey

Prentice Hall- Numerical Methods using MATLAB, 3rd edition

Approximation of Functions, G. Lorentz

Fundamental Numerical Methods and Data analysis, G. Collins