MATH 4300, FALL 2011:
CHEBYSHEV POLYNOMIALS
PR HEWITT
The great Russian mathematician Pafnuty Lvovich Chebyshev1 introduced a
family of delightful polynomials. Consider the trig identity
cos(2θ) = 2 cos2 (θ) − 1.
This tells us that cos(2θ) is a quadratic polynomial in cos(θ), namely 2x2 − 1. This
is the degree-2 Chebyshev polynomial Q2 .2 More generally, we define the degree-n
Chebyshev polynomial Qn by the rule
(1)
cos(nθ) = Qn (cos(θ)).
It is not immediately obvious that this definition even gives us polynomials. If you
play with the trig identities a bit then you soon see the pattern. You can prove
rigorously that definition (1) does indeed yield a polynomial of degree n using
induction and the addition formulas:
cos((n + 1)θ) = cos(θ) cos(nθ) − sin(θ) sin(nθ)
sin((n + 1)θ) = sin(θ) cos(nθ) + cos(θ) sin(nθ)
These lead to a recursive formula for Qn . If we set x = cos(θ) and y = sin(θ), and
note that y 2 = 1 − x2 , then we can quickly complete a table of values:
cos(nθ)
sin(nθ)
n
0
1
0
1
x
y
2
2x2 − 1
2xy
3
4x3 − 3x
(4x2 − 1)y
4 8x4 − 8x2 + 1 (8x3 − 4x)y
For example, to get line n = 3 from line n = 2 we compute that
x · (2x2 − 1) − y · 2xy = x · (2x2 − 1) + 2x · (x2 − 1) = 4x3 − 3x
y · (2x2 − 1) + x · 2xy = (4x2 − 1)y
Extend the table a few more lines — it’s quite relaxing.
This is only the beginning of the fun. Exploring the Chebyshev polynomials
reveals an embarrassing wealth of treasure. They are beautiful.
Date: 3 October 2011.
1
You can read about Chebyshev at the MacTutor website for the history of math: http:
//www.gap-system.org/~history/Biographies/Chebyshev.html. Even more fun is The Thread,
by Philip Davis, which traces Chebyshev’s first name around the globe.
2The Cyrillic letter Q, pronounced “che” is the first letter in the Russian spelling of Chebyshev’s
name. Davis’ journey is launched by a dispute over the transliteration of this letter!
1
2
PR HEWITT
1
0.5
0
−0.5
−1
−1
−0.5
0
0.5
1
There are n zeroes of Qn in [−1, 1], at the points
(2k + 1)π
(2)
cos
, k = 0, ±1, ±2, . . .
2n
The n − 1 extrema are all ±1, at the points
kπ
(3)
cos
, k = 0, ±1, ±2, . . .
n
Like all fun mathematics, they turn out to be extraordinarily useful. They
are useful in numerical approximation, in large part because of their orthogonality
relations:


Z 1
0 if m 6= n,
2
dx
Qm (x)Qn (x) √
(4)
= 1 if m = n 6= 0,

π −1
1 − x2

2 if m = n = 0.
These can be proved using the substitution x = cos(θ).
Assignment 9: due Monday, 10 October
Let V = P5 (R). Set
β = (1, x, x2 , x3 , x4 , x5 )
γ = (Q0 , Q1 , Q2 , Q3 , Q4 , Q5 )
Thus, β is the standard (ordered) basis for V . We explore the relationship between
β and γ.
(1) Prove that γ is a basis. (You may use without proof the fact that Qn has
degree n.)
(2) Compute the change-of-basis matrix [IV ]βγ . By definition, the columns of
this matrix give the coordinates of the Qn in terms of the standard basis.
(3) Compute the change-of-basis matrix [IV ]γβ . By definition, the columns of
this matrix give the coordinates of the xn in terms of the Chebyshev polynomials.
CHEBYSHEV POLYNOMIALS
3
Extra credit: due Friday, 9 December
The remaining problems are extra credit.
(4) Prove that Qn is a polynomial of degree n.
(5) Differentiate cos(nθ) to find a direct relationship between the columns in
the table above.
(6) Prove that the n zeroes of Qn are given by formula (2). Conclude that they
all lie in [−1, 1].
(7) Prove that the n − 1 extrema of Qn are all ±1, and occur at the points
given by formula (3).
(8) Prove the orthogonality relations (4).
(9) Prove that Qn satisfies the differential equation
(1 − x2 ) Q00n (x) − x Q0n (x) + n2 Qn (x) = 0.