minmax polynomial, min-max approximation
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Numerical Analysis, Polynomial Approximation, Min-max approximation, minmax approximationTRANSCRIPT
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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