Generalized Chebyshev
polynomials and plane trees
Anton Bankevich
St. Petersburg State University
Jass’07
Chebyshev
Pafnuty Lvovich Chebyshev( May 26, 1821 –
December 8, 1894)
One of nine children, he was born in the village of
Okatovo, the district of Borovsk, province of Kaluga
into the family of landowner Lev Pavlovich Chebyshev.
In 1832 the family moved to Moscow.
In 1837 Chebyshev started the studies of mathematics
at the philosophical department of Moscow University
and graduated from the university as “the most
outstanding candidate”.
In 1847, Chebyshev defended his dissertation “About integration with the help of
logarithms” at St Petersburg University. Chebyshev lectured at the university from
1847 to 1882. In 1882 he left the university and completely devoted his life to
research.
Chebyshev is known for his work in the field of probability, statistics and number
theory. Chebyshev is considered to be one of the founding fathers of Russian
mathematics. Among his students were Aleksandr Lyapunov and Andrey Markov
Chebyshev polynomials
There exist a lot of definitions of Chebyshev
polynomials. We shall now consider several of them.
cos( nx)  Tn (cos( x))
Minimal norm
For any given n, among the polynomials of degree n with the
leading coefficient 1, Tn/2n-1 is the one in which the maximal
absolute value on the interval [− 1,1] is minimal. This maximal
absolute value is 1/2n-1, and |Tn(x)| reaches this maximum exactly
n + 1 times: with x = −1, 1 and with the other n − 1 extremal
points of f.
Chebyshev polynomials are important in the approximation theory
because the roots of the Chebyshev polynomials (which are also
referred to as Chebyshev nodes) are used as nodes in polynomial
interpolation.
Basis of the space of polynomials
• Chebyshev polynomials appear to be an orthonormal basis
for the space of polynomials in one variable with respect to
the following scalar product.
1
 f , g 

f ( x) g ( x)
dx
1 x
So, for any polynomial f(x) of degree n the coefficients in
the formula
n
2
1
f ( x)   c k Tn ( x)
k 0
can be found in the following way:
1
ck 

1
f (t )Tn (t )
1 t
2
dt
Formulas for Chebyshev polynomials
Tn1 ( x)  2 xTn ( x)  Tn1 ( x)
 n  n2 j 2
Tn ( x)     x
( x  1) j
2 j  n 2 j 
1
(1) n 1  x 2 d n
2 n 2
Tn ( x) 
(1  x )
n
(2n  1)!! dx
Composition
• Th Chebyshev polynomials form an Abelian group with
respect to the operation of composition
Proof :
Tn(Tm( cos (x)))  Tn( cos (mx))  cos (mnx) 
 Tmn ( x)  cos( nmx) 
 Tm( cos (nx))  Tm(Tn( cos (x)))
Generalizations
• A lot of attempts were made to
generalize the notion of
Chebyshev polynomials.
|| F ( x) || max{| F ( x) |: x  [1, 1]}
n
|| F ( x) || max{| F ( x) |: x   [ k ,  k ]}
k 1
Polynomials in several variables
Motivation
Let f(z) be a polynomial of degree n with complex coefficients
Let us consider an inverse image of a random point w:
f
1
( w)  {z | f ( z )  w}
Usually, this set consists of n distinct points.
What is the minimal size of the inverse image of the point w?
The answer is 1. The only examples are a +(b+cz)n.
Motivation
And what is the minimal common size of the inverse
image of two points w1 and w2?
The answer is n+1. But why?
Motivation: minimal size of inverse
image
k
p ( z )  w1    ( z  ai )
i
l
p ( z )  w2    ( z  bi )
i 1
i 1
k
p' ( z )  ( z  ai )
l
p ' ( z )  ( z  bi )  i 1
 i 1
i 1
i 1
k
l
a  
i 1
i
i 1
i
i
 deg( f )  n
k
 (a
i 1
l
i
k
l
k
l
i 1
i 1
i 1
i 1
 1)   (  i  1)  deg( f ' )  n  1
i 1
k  l    i    i  ( ( i  1)   (  i  1))  n  n  (n  1)  n  1
Motivation
But when does this degenerate case take place? What
are the examples?
An obvious example is a polynomial zn,where w1=0, w2
is a nonzero complex number. But there are some more
examples.
The other example is Tn(z) Chebyshev polynomials,
where w1=-1, w2=1. It can be easily seen that the inverse
k
{cos(
) | k  {0,1,..., n}}.
image of these points is
n
But there are much more examples of this kind. And they
are called generalized Chebyshev polynomials.
Motivation: minimal size of inverse
image
k
p ( z )  w1    ( z  ai )
i
l
p ( z )  w2    ( z  bi )
i 1
i 1
k
p' ( z )  ( z  ai )
l
p ' ( z )  ( z  bi )  i 1
 i 1
i 1
i 1
k
l
a  
i 1
i
i 1
i
i
 deg( f )  n
k
 (a
i 1
l
i
k
l
k
l
i 1
i 1
i 1
i 1
 1)   (  i  1)  deg( f ' )  n  1
i 1
k  l    i    i  ( ( i  1)   (  i  1))  n  n  (n  1)  n  1
Definition
A complex point z is called a critical point of polynomial f if
f’(z)=0.
A point w is called critical value of f if there exists a critical
point z such that f(z)=w.
Polynomial f(z) is called generalized Chebyshev polynomial
(GCP) if it has at most two critical values.
These polynomials are sometimes called Shabat polynomials
Certain properties of GCP
The following can be easily seen:
If f(z) is a GCP, then f(az+b) is GCP
If f(z) is a GCP, then af(z)+b is GCP
We shall call polynomials f and g equivalent if there exist
constants a, b, A, B, such that:
f ( z )  A  Bg (az  b)
All the polynomials equivalent to certain GCP appear to be
GCP too. So since this moment we shall observe only the
polynomials with critical values equal to 0 and 1.
Geometric point of view
Let us consider the inverse image of a segment. We
colored the ends of the segment in black and white colors.
Let [C0; C1] be the segment free of critical values of f.
Then inverse image would be a set of disjoint sets, each of
them being homeomorphic to a segment.
f
Geometric point of view
Then we suppose that there are no critical points inside a
segment, but they may appear at one or both ends. Then
some curvilinear segments may glue together with the
monochrome vertexes.
f
Inverse image of a segment
But what picture can we see in the case where f is GCP?
The answer is that we will see a bicolored plane tree
It is a star for zn
- and a chain for Chebyshev polynomials
Inverse image of a segment
Th For each GCP f the inverse image of a segment is a
bicolored plane tree.
Proof:
In case of existence of a circuit our polynomial g takes
only real values on the boundary of domain bounded by
our circuit.
Proof of theorem
Then in case of existence of a circuit our polynomial g takes only real
values on the boundary of domain bounded by our circuit.
Then the harmonic function
Im(g) equals zero on the
boundary.
But it means that Im(g)=0 on the whole domain, which is a
contradiction.
The inverse theorem
Th For each bicolored plane tree T there exists a GCP
such that the corresponding inverse image is T.
Moreover, this GCP is unique up to the equivalence
introduced above.
Plane tree
What is plane tree?
• Plane tree is a picture of tree
• Plane tree is a structure of a tree and for every vertex
v a cyclic order on the vertices adjacent to v.
Canonical geometric form
For each bicolored plane tree, there exists a GCP. And we
have presented a way to construct a geometrical form from
the GCP. So for each tree we can construct its canonical
geometric form.
Once more:
f ( z )  z 3 ( z  1)
Examples of geometric forms
Examples of geometric forms
Construction of GCP
The next question is:
Given a tree, how can we construct a polynomial?
The first way
is to write a system of equations:
k
p( z )    ( z  ai ) 
α and β are the degrees of black
i
i
i 1
l
p( z )  1    ( z  bi )  i
i
and white vertices of the graph.
i 1
Now we have n algebraic equations in n+2 variables. This is
sufficient to construct the polynomial.
Tree families
While constructing our GCP, we only used αi and βi, which
are the degrees of black and white vertices of the graph.
But for each multiset of degrees there are a lot of trees with
the same multisetset. All such trees form a family <α, β>,
where α=(α1, α2, …αk), β =(β1, β2, … βn+1-k).
By solving the equations from the previous slide we can
find all the polynomials which correspond to the trees of
this family.
Several trees and their polynomials
z ( z  2 z  a)
3
2
34  6 21
a
7
z ( z  2 z  a)
3
2
34  6 21
a
7
Several trees and their polynomials
z ( z  1) ( z  a)
3
2
25a  12a  24a  16  0
3
2
Computation of GCP
The computation of coefficients of GCP for a tree with a big number of
vertices becomes a complicated problem since we have to solve a system of
algebraic equations of a high degree, so another way had to be found and it has
been found.
The idea of the new method is the following:
First we calculate coefficients not precisely. It can be done by using the class of
polynomials with the following conditions:
There exist complex numbers C and w such that
g(w) = C, f’(w)=0, f’’(w)≠0
If f’(z) = 0, then f(z) =±1 or z = w
In the second step we calculate precise values of the coefficients.
Computation of GCP: a small example
Galois group action on trees
It can be easily seen that the group Gal(Q/Q) acts on the
set of Chebyshev polynomials with algebraic coefficients.
k
p( z )  w1    ( z  ai )  i
i 1
l
p ( z )  w2    ( z  bi )  i
i 1
So we can define an action of Gal(Q/Q) of bicolored
plane trees.
This leads us to the definition of the field of definition of
a tree: this is a field which corresponds to the subgroup
of Gal(Q/Q) that fixes the given tree.
GCP coefficients
But what is the connection between the field of definition
and generalized Chebyshev polynomials? The answer is
given in the following theorem:
Th For any bicolored plane tree there exists a generalized
Chebyshev polynomial whose coefficients belong to the field
of definition of the tree.
Orbits vs families
It can be easily seen that the action of Galois group on the
tree does not change the multiset of valences of the tree.
So the orbit of a tree is a subset of its family. But do they
coincide?
The answer is: not always.
Orbits vs families
z ( z  1)
4
2
25
z (z  2z  )
9
4
2
Composition of polynomials
It is well known that the composition of Chebyshev polynomials is Chebyshev
polynomial, too. And is this the case for generalized Chebyshev polynomials?
f ( g ( x))'  0 ?  ? f ( g ( x))  {0,1}
g ' ( x)  0
f ' ( g ( x))  0
f ({0,1})  {0,1}
Th Let f, g be GCP such that {f(0), f(1)} lie in {0, 1}. Then the
composition f(g(x)) is also GCP.
Composition of trees
Let us imagine that we have two trees T1 and T2. Their
polynomials are f1 and f2. And we need to construct a tree
which corresponds to f1◦f2. Do we have to calculate f1 and f2,
find their composition and then construct a tree or there is a
direct way?
Composition of trees
tail
body
head
Composition of trees
Plane maps: Belyi functions
Till this moment we have been talking about plane trees,
but what about plane maps?
Let f be a rational function on the Riemann sphere that satisfies the
following conditions:
• f has only three critical values: 0, 1, infinity
• All points of f-1(1) are critical with the degree exactly equal to 2.
Such functions are called Belyi functions.
Plane maps: map construction
But how are Belyi functions connected with maps?
f-1(∞) are vertices.
f-1(1, ∞) are edges ( f-1(1) are points on edges )
f-1(0) are faces
1
f ( z) 
1  g ( z) 2