4. Alternating Sets
Let f be a continuous function on the interval [a,b]
, and let p be a polynomial approximation. An alternating set on f,p
is defined to be a sequence of points x0,…,xn-1
such that:
- a≤x0<x1<⋯<xn-1≤b
,
- f(xi)-p(xi)=(-1)iE
for i=0,1,…,n-1
.
where either E=‖f-p‖
or E=-‖f-p‖
. The number n
is the length of the alternating set.
Example
Let f(x)=x2
on the interval [-1,1]
, and let p(x)
be the constant function p(x)=1/2
.
Clearly ‖f-p‖=1/2
, and the points (-1,0,1)
form an alternating set of length 3.
Given two alternating sets X=(x0,…,xn)
and Z=(z0,…,zm)
for f, p, we say that Z extends X if, as sets, {x0,…,xn}⊆{z0,…,zm}
.
The Chebyshev theorem states that if p is a polynomial of degree ≤n
of best approximation to f, then f, p has an alternating set of length n+2
. We start by proving a simple partial result.
Let f be a continuous real-valued function on [a,b]
, and suppose that p is a polynomial of best approximation of degree n to f. Then there is an alternating set on f, p of length 2.
Proof
Let x0
be a minimum and x1
be a maximum of the function f(x)-p(x)
on [a,b]
, and let m0=f(x0)-p(x0)
and m1=f(x1)-p(x1)
. They must be of opposite sign and equal in magnitude, for otherwise we could add a constant to p(x)
and get a better approximation. Specifically, if q(x)=p(x)+(m1+m0)/2
, then:
m0-m1+m02≤f(x)-q(x)≤m1-m1+m02
m0-m12≤f(x)-q(x)≤m1-m02
So ‖f-q‖≤(m1-m0)/2≤‖f-p‖
, and equality holds if and only if m0=-m1
. If x0<x1
, then (x0,x1)
is the required alternating set; otherwise, (x1,x0)
is the alternating set.