T0
Numerical Analysis 2, lecture 5:
Chebyshev polynomials
!1
Chebyshev polynomials
are defined by a three-term recursion
T0
1
!1
Tn+1 (x) = 2xTn (x) ! Tn!1 (x) (n = 1, 2,…)
T1
(textbook sections 9.7 & 9.9)
T1
T2 (x) = 2x 2 ! 1, T3 (x) = 4 x 3 ! 3x, T4 (x) = 8x 4 ! 8x 2 + 1, …
• LS approximation
• interpolation
• minimax approximation
T2
T2
Basic facts
degTn = n
T3
T2n is an even function, T2n+1 is an odd function
1-n
T0 (x) = 1, Tn (x) = 2
T4
T3
Tn (x) (for n ! 1) are monic polynomials
Clenshaw’s algorithm (exercise E5)
n
" ckTk (x) = u0 ! xu1 where un = cn ,
k =0
T4
un!1 = cn!1 + 2xun ,
and u j = c j + 2xu j +1 ! u j +2 for j = n ! 2, n ! 1, …, 0.
Numerical Analysis 2, lecture 5, slide! 2
A change of variables converts a Chebyshev
polynomial into a trigonometric polynomial
Chebyshev polynomials can be used
for least-squares approximation
example (p. 272, modified)
Tn (x) = cos(n arccos x) (!1 " x " 1)
1
Find the polynomial p of degree ! 2 that minimizes
$
"1
! n
max Tn (x) = 1
2i ! 1
" ) (i = 1, 2,…, n)
2n
k
Tn ( x! k ) = 1 at x! k = cos( ! ) (k = 0,1, 2,…, n)
n
1
!1
2 !1 2
)
1 " x2
1
c0 = 0, c2 = 0, c1 =
!1" x"1
" (1 ! x
( p(x) " sin(# x 2))2 dx
Solution: p(x) = c0T0 (x) + c1T1 (x) + c2T2 (x)
more basic facts
the n zeros of Tn are cos(
%'0
T j (x)Tk (x) dx = &$ 2
'($
j#k
j=k#0
j=k=0
1
T0 (x) = 1, T1 (x) = x,
( f ,T1 )
1
x " sin(! x 2)
=
$ 1 # x 2 dx = 2J1(! 2) % 1.1336
(T1,T1 ) ! 2 #1
0.15
Chebyshev nodes
f (x) ! c1T1 (x)
extrema
0
orthogonality
!0.15
!1
Numerical Analysis 2, lecture 5, slide! 3
0
1
Numerical Analysis 2, lecture 5, slide! 4
Chebyshev nodes are good for interpolation
Interpolation with Chebyshev nodes
is easily computed in Matlab
polynomial interpolation error
>>
>>
>>
>>
>>
>>
>>
1
f ( x̂) ! P( x̂) =
( x̂ ! x0 )( x̂ ! x1 )!( x̂ ! xn ) f (n+1) (" )
(n + 1)!
max f (x) ! P(x) "
!1" x"1
1
max (x ! x0 )(x ! x1 )!(x ! xn ) max f (n+1) (x)
!1" x"1
(n + 1)! !1" x"1
" 2 j +1 %
x j = cos $
!
# 2n + 2 '&
(
max f (x) ) P(x) *
)1* x*1
2 )n
max f (n+1) (x)
(n + 1)! )1* x*1
f = @(x) 1./(1+25*x.^2);
n = 11;
xe = linspace(-1,1,n);
xc = cos((2*(1:n)-1)*pi/2/n);
t = -1:.01:1;
plot(t,f(t),'b',t,neville(xe,f(xe),t),'r',xe,f(xe),’o’)
plot(t,f(t),'b',t,neville(xc,f(xc),t),'r',xc,f(xc),’o’)
1
1
“waterbed” theorem p.286
For any monic polynomial p of degree n + 1,
0
!1
max p(x) # max 2 !n Tn+1 (x) = 2 !n
!1" x"1
!1" x"1
0
0
!1
1
0
Numerical Analysis 2, lecture 5, slide! 5
Numerical Analysis 2, lecture 5, slide! 6
Chebyshev-node interpolant
can be computed using Clenshaw’s formula
theorem (thm 9.8.1, page 288)
Chebyshev interpolation with Clenshaw’s
formula is easily computed in Matlab
orthogonality
%'0
2i ! 1
If xi = cos(
" ) (i = 1, 2,…, n + 1), then # T j ( xi ) Tk ( xi ) = &(n + 1) 2
2n + 2
i=1
('n + 1
zeros of Tn+1
n+1
for 0 ) j, k ) n, and the polynomial p of degree m that minimizes
n+1
2
i=1
is p(x) =
m
1 n+1
2 * n+1
f (xi ) + #
#
# Tk (xi ) f (xi )/.Tk (x).
n + 1 i=1
n + 1 ,+ i=1
k
=1
!#
#"##
$
!###"###$
c0
m=n
interpolation
ck
1
0.15
example
2 3
sin( 3 ! 4)x " 1.129x
3
{
function u = chebpolval(c,x)
n = length(c);
u = c(n)*ones(size(x));
if n > 1
ujp1 = u;
u = c(n-1) + 2*x*c(n);
for j = n-2:-1:1
ujp2 = ujp1;
ujp1 = u;
u = c(j) + 2*x.*ujp1 - ujp2;
end
u = u - x.*ujp1;
end
function [c,x] = chebpolfit(fname,n)
x = cos((2*(1:n)'-1)*pi/2/n);
y = feval(fname,x);
T = [zeros(n,1) ones(n,1)];
c = [sum(y)/n zeros(1,n-1)];
a = 1;
for k = 2:n
T = [T(:,2) a*x.*T(:,2)-T(:,1)];
c(k) = sum( T(:,2) .* y)*2/n;
a = 2;
end
j$k
j=k$0
j=k=0
# ( p(xi ) ! f (xi ))
1
0
}
interpolates sin(! x 2) at ± 3 2, 0 .
!0.15
!1
0
Numerical Analysis 2, lecture 5, slide! 7
1
>>
>>
>>
>>
f = @(x) 1./(1+25*x.^2);
0
[c,x] = chebpolfit(f,11);
!1
t = -1:.01:1;
plot(t,f(t),'b',t,chebpolval(c,t),'r',x,f(x),'ok')
0
1
Numerical Analysis 2, lecture 5, slide! 8
The Chebyshev interpolant is “almost”
the best minimax approximation
minimax approximation
Find a polynomial p of degree ! n that minimizes max p(x) " f (x)
The minimax approximation can be computed
using the alternation principle
approx. of monotone function by straight line
a! x!b
Weierstrass’s theorem
min
max p(x) " f (x) # 0 for any continuous f
!####"####$
deg( p)!n "1! x!1
En ( f )
2 -1/2
Least-squares approximation, w=(1-x )
max f (x) !
!1" x"1
n
+ log n) $ E ( f )
# c jT j (x) < (4
!#"#$ n
j =0
n=20
%
% 2 j +1 ((
1 n
max f (x) ! # c j T j (x) < ' 1 +
# tan &' 4(n + 1) $ )* * + En ( f )
!1" x"1
n
+
1
&
)
j =0
j =0
!#####"#####$
n
!4
Numerical Analysis 2, lecture 5, slide! 9
what happened, what’s next
• Chebyshev polynomials satisfy a 3-term recurrence
and are orthogonal in two ways
•2
-nT
•
•
n
is the monic polynomial with…
smallest weighted continuous 2-norm
smallest discrete maxnorm (with nodes at zeros of Tn+1)
• “almost minimax” :
•
•
example (p. 300-301)
weighted continuous LS approximation
interpolation with zeros of Tn (= discrete LS approx. at zeros of Tn+1)
• minimax approximation is a nonlinear problem
Next: Boundary Value Problems (§10.8, 10.9, 10.11)
Numerical Analysis 2, lecture 5, slide! 11
d
d
Find p(x) = c0 + c1x
that minimizes
d
max p(x) " sin(# x 2)
0! x!1
!7
Chebyshev-node interpolation
n=20
Error extrema should be at end points and at an interior point.
!c0 = d
sin("# 2) ! c0 ! c1# = !d
1 ! c0 ! c1 = d
"
2
cos("# 2) ! c1 = 0
$
&
&
&
%
&
&
&'
c1 = 1
! = 0.5607
c0 = 0.1053
d = "0.1053
Numerical Analysis 2, lecture 5, slide! 10