MIN-MAX POLYNOMIALMATH 174 – Numerical Analysis I

INTRODUCTION

The goal of Min-Max Approximation is to find an

approximating polynomial, called Min-Max

Polynomial, that has the smallest maximum

absolute error from the true function.

Min-Max Polynomials are hard to find, but they

can be approximated by the Chebyshev

Polynomials of the first kind. (Min-Max

Approximation is also called Chebyshev

Approximation)

Comparing the computational expense, it is more

practical to use Min-Max Polynomials than to use

Taylor Series.

PROPERTIES OF MIN-MAX

POLYNOMIALS

Existence and Uniqueness Property – there

exists a unique Min-Max Polynomial that can

approximate any function.

Equal-Error Property – given the Min-Max

Polynomial P(x) that approximates Y(x),

P(x)–Y(x) takes the extreme values of size E,

with alternating signs, at least n+2 times. (E is

the maximum error)

*The equal-error property is the identifying feature

of a min-max polynomial.

THE EXCHANGE METHOD

This is an algorithm for finding the Min-Max

Polynomial through its equal-error property.

However, producing a Min-Max Polynomial (a

polynomial with equal-error behavior) is not

usually within reach, so exchange method can be

used just to obtain a polynomial with an

acceptable closeness to the Min-Max polynomial.

Theoretically, the method assures convergence to

the Min-Max Polynomial.

DISCRETE CASE

The Min-Max Line (also called Equal-Error or

Chebyshev line):

Given any three points (xi,yi) and xi is distinct

(i=1,2,3):

The line must pass through (x1, y1+h1), (x2,

y2+h2) and (x3, y3+h3); where h1=h, h2= –h

and h3=h. This shows the equal-error

property, which is missing all three points by

equal amounts with alternating signs.

DISCRETE CASE

Other Formulas:

• B1= x3–x2 , B2= x3–x1 , B3= x2–x1

• h= –(B1y1–B2y2+B3y3)/(B1+B2+B3)

Example 1:

Find the equal error line for the data points (0,0),

(1,0) and (2,1).

Solution:

B1=2–1=1 B2=2–0=2 B3=1–0=1

h= –[1(0) –2(0)+1(1)]/[1+2+1]= –1/4

DISCRETE CASE

Continuation of Example 1:

Therefore the line passes through (0,–1/4),

(1,1/4) and (2,3/4). By point-slope formula

using any two of the points, the equation of the

line is P(x) = (1/2)x –1/4.

DISCRETE CASE

Example 2: The Exchange Method

Find the equal error line for the following points:

(0,0), (1,0), (2,1), (6,2), (7,3).

Step 1. Select an initial triple and find the

Chebyshev line for this triple.

Let’s choose (0,0), (1,0) and (2,1); and from

Example 1, the Chebyshev line is P(x) = (1/2)x –

1/4, where h= –1/4.

Step 2: Compute the errors at all data points. Call

the absolute value of the largest of these errors as

H. If the absolute value of h=H, then the search is

over, else go to Step 4.

DISCRETE CASE

Continuation of Example 2:

The errors at all five data points are –1/4, 1/4, –1/4, 3/4, 1/4, respectively.

This makes H=3/4, which is not equal to the absolute value of h=1/4.

Step 4: Exchange step: Choose a new triple by adding to the old triple a data point at which H occurs. Then disregard one of the former points, in such a way that the remaining three have errors of alternating sign.

The new triple is (1,0), (2,1) and (6,2). By doing Step 1 to 3 again, we will find out that the three points will lead us to the min-max line of the entire data set which is P(x)=(2/5)x – (1/10).

DISCRETE CASE

The Min-Max Parabola

The algorithm will start by choosing an initial quadruple.

The equal error parabola of the quadruple is P(x)=a+bx+cx2, where P(xi)–y(xi)=+h.

The same Steps 1-4 of the exchange method will be followed in obtaining the Min-Max Parabola of the entire given data set.

Example 3:

Find the min-max parabola for (-2,2), (-1,1), (0,0), (1,1) and (2,2).

DISCRETE CASE

Continuation of Example 3:

Solution:

Let’s choose (-2,2), (-1,1), (0,0) and (1,1).

P(xi) –y(xi)=+h will result to the following

systems of equation:

(a–2b+4c)–2 = h

(a–b+c)–1 = –h

a – 0 = h

(a+b+c)–1 = –h

Then solve for a, b, c and h. Then do the

Exchange Method.

DISCRETE CASE

Continuation of Example 3:

We will find out that the min-max parabola

would be P(x)=(1/4)+(1/2)x2 ,

where the maximum error on our quadruple (or

the absolute value of h) = 1/4,

and the maximum error on the entire set (or H)

= 1/4 also.

CONTINUOUS CASE

Chebyshev Polynomials, when truncated, often yield

approximations having almost equal-error behavior

(or almost min-max).

The first four Chebyshev Polynomials are:

T0(x) = 1

T1(x) = x

T2(x) = 2x2–1

T3(x) = 4x3–3x

Its recurrence relation can be written as

TN(x)=2xTN–1(x) –TN–2(x), for N=2,3,…

Note that the coefficient of xN in TN(x) is 2N–1 when

N>1.

CONTINUOUS CASE

Trigonometric Representation on [-1,1]

TN+1(x)+TN–1(x) =2xTN(x)

cos(N+1)θ+cos(N–1)θ=2cosθcosNθ , θ=arccos(x)

Because of the relation TN(x)=cosNθ, it is apparent

that the Chebyshev polynomial have a succession

of maximums and minimums of alternating signs,

each of magnitude 1,

N+1 times on the

interval [-1,1].

CONTINUOUS CASE

On the interval [-1,1], the minimum value of the

error bound of the Lagrange Interpolation is achieved

when the nodes are the Chebyshev abscissas/nodes.

CONTINUOUS CASE

For [-1,1], Chebyshev abscissas (zeros of TN(x)) are:

Lagrange Interpolation using Chebyshev abscissas:

There are possibilities that the maximum of the error

term grows as N→∞ when using equally-spaced

nodes (Runge Phenomenon). Because of this, wild

and large oscillations may occur.

The use of Chebyshev nodes will solve this

phenomenon – the error term will go to zero as

N→∞.

CONTINUOUS CASE

Exercise: Try to compare the maximum absolute errors

of the Lagrange Interpolation for f(x)=ex when the

following nodes were used:

a. equally-spaced: -1, -1/3, 1/3, 1

b. Chebyshev nodes on [-1,1] where N=4

Maximum Absolute Error on [-1,1]:

max│ex–P(x)│, -1<x<1, where P(x) is the Lagrange

Interpolating Polynomial

CONTINUOUS CASE

Transforming the Interval [a,b] to [-1,1]

The interpolating nodes on [a,b] are obtained using:

for k=0,1,…,N

2 2

2 1 2

k k

k

b a a bx t

where t cos ( N k )

CONTINUOUS CASE

LAGRANGE-CHEBYSHEV APPROXIMATION POLYNOMIAL

Assume that PN(x) is the Lagrange polynomial that is

based on the Chebyshev nodes given in the previous slide,

If f belongs to CN+1[a,b], then

Example: For f(x)=sin(x) on [0,π/4], find the Chebyshev

nodes and the error bound for the Lagrange Polynomial

P5(x).

11

1

2

4 1

N( N )

N N a x b

( b a )f ( x ) P ( x ) max{ f ( x ) }

( N )!

5

11 2

12

0 00000720

k

( k )x cos k

sin( x ) P ( x ) .

CONTINUOUS CASE

A direct approach (with the use of the Orthogonal Property

of Chebyshev Polynomials):

The Chebyshev approximation polynomial PN(x) of

degree<N for f(x) over [-1,1], can be written as

CONTINUOUS CASE

Exercise:

Find the Chebyshev polynomial P3(x) that approximates

the function f(x)=ex over [-1,1].

You should get

P3(x)

=0.99461532+0.99893324x+0.54290072x2+0.17517568x3

as your answer.

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