J Geom Anal
DOI 10.1007/s12220-014-9489-y
Control of Cancellations that Restrain the Growth
of a Binomial Recursion
Magnus Aspenberg · Rodrigo Pérez
Received: 18 August 2010
© Mathematica Josephina, Inc. 2014
Abstract We study a recursion that generates real sequences depending on a parameter x. Given a negative x the growth of the sequence is very difficult to estimate due
to canceling terms. We reduce the study of the recursion to a problem about a family
of integral operators, and prove that for every parameter value except −1, the growth
of the sequence is factorial. In the combinatorial part of the proof we show that when
x = −1 the resulting recurrence yields the sequence of alternating Catalan numbers,
and thus has exponential growth. We expect our methods to be useful in a variety of
similar situations.
Keywords
Catalan numbers · Factorial growth · Integral operators
Mathematics Subject Classification
05A10 · 45P05
1 Introduction
Fix an arbitrary real number x = 0, and consider the sequence defined by the recursive
expression1
1 Indices r less than
n
2
are omitted, as they return vanishing binomials.
Communicated by Jeffrey Diller.
M. Aspenberg
Centre for Mathematical Sciences, Lund University, Box 118, 221 00 Lund, Sweden
e-mail: [email protected]
R. Pérez (B)
Department of Mathematical Sciences, LD-224R IUPUI, 402 N. Blackford St.,
Indianapolis, IN 46202, USA
e-mail: [email protected]
123
M. Aspenberg, R. Pérez
a1 = x,
an = x
n−1 r
ar .
n −r
n
r = 2 (1)
For x = 1, for instance,
we obtain the sequence 1, 1, 2, 7, 34, 214, 1652, . . .. Note
that the last summand n−1
1 an−1 guarantees that an ≥ (n − 1)!. This means that {an }
grows very fast since n! > (n/e)n . We prove
Theorem 1 For any real x = −1, 0 the sequence {an } defined by (1) grows superexponentially.
This is an interesting behavior, and not altogether obvious because when x < 0,
there are a lot of cancellations. In fact, when x = −1, the positive and negative terms
exactly balance out to yield a surprising contrast:
Theorem 2 When x = −1, formula (1) produces the sequence (−1)n Cn of Catalan
numbers with alternating signs, and therefore grows exponentially.
This seems to be a brand new formula for Catalan numbers, which is compelling
all by itself. However, our main interest is in calling attention to our method of proof
for Theorem 1. Here is why:
Let f : C → C be an analytic function with f (0) = 0 and f (0) = eπ iθ , where θ
is an irrational angle. A simple computation shows that there is always a formal power
series ϕ that conjugates f and its linear part z → eπ iθ z, but the coefficients of this
series can grow fast enough for its radius of convergence to be 0.
Under what conditions on θ is it possible to control the growth of the coefficients
so that ϕ results analytic? Since each coefficient is a sum of products of the form
1
(see [6]), the answer depends on the number-theoretic properties of
π
ikθ
e
− eπ iθ
θ . This is because when θ is nearly rational, some of these quotients can become very
large. This is an example of a small denominator problem.
In a famous 1942 paper [7], C. L. Siegel gave a lower bound on the radius of
convergence of the conjugating map when θ is Diophantine. In such situation it was
enough to estimate the growth of the largest product that appears in the formula for the
coefficients of ϕ. Now, it is possible for ϕ to have positive radius of convergence even
if the largest product grows super-exponentially. But for this to happen it is necessary
to detect large cancellations between products, so the coefficients of ϕ have a chance
to grow at a slower exponential rate. Our method of proof for Theorem 1 is tailored
for the detection of these kinds of cancellations.
Formula (1) is a modified version of the recursion that generates the coefficients
of the conjugating map ϕ when f is quadratic. The main simplification is that the
constant x replaces the small denominator terms. This allows us to concentrate on the
structural properties of the conjugating equation that make possible the presence of
cancellations.
The phenomenon at play is very interesting. In the first part of the paper some
combinatorial constructions will allow us to prove the results when x > 0 and when
x ≤ −1. However, when x ∈ (−1, 0), the cancellations and the small size of x
conspire to render elementary arguments ineffective. In the second part we introduce
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Control of Cancellations that Restrain...
functional analytic methods to control the effect of the cancellations in that case. An
application to Siegel disks will be investigated in [1]; also, we expect these ideas to
be useful in a variety of similar situations.
1.1 Structure
The paper has two parts. In the first (Sects. 2 to 4) we prove Theorem 2 and the situation
x∈
/ [−1, 0] of Theorem 1. Section 2 presents some basic facts about hypercube graphs
and the Catalan numbers; Section 3 defines the combinatorial structure we use; and
Sect. 4 contains the proofs.
The second part (Sects. 5 to 7) tackles the case x ∈ (−1, 0) of Theorem 1. Section 5
uses the combinatorial knowledge gained in the first part to derive an alternative
expression for an as the sum of a sequence of numbers Sn (1), . . . , Sn (n−1) constructed
recursively. This sequence is translated into a function sn ∈ L 2 [0, 1], and the recursion
is interpreted as an integral operator. Section 6 contains the proof of the theorem
assuming the statement of Lemma 11, and Sect. 7 is devoted to the proof of Lemma 11.
2 Basic Combinatorial Facts
2.1 Hypercubes
The hypercube graph Hn is the graph whose set of vertices Vn consists of all n-vectors
with coordinates 0 or 1 [3]. Two vertices are adjacent whenever they differ in one
coordinate. There is a natural stratification of Vn by the number of coordinates of each
value in a vertex; accordingly, let Vn,r ⊂ Vn denote the vertices with r coordinates
equal to 1. There are other equivalent definitions of hypercube graphs. The advantage
of the definition in terms of binary coordinates is that the following facts become
obvious; compare Fig. 1.
(H1) |Vn | = 2n. (H2) |Vn,r | = nr .
(H3) If n = m 1 + m 2 , then Hn = Hm 1 × Hm 2 .
items (H1) and (H2) give a succinct proof of the binomial identity
n nIncidentally,
n (Fig. 1) . If instead of just counting vertices, they are assigned weight
=
2
r =0 r
x r , item (H3) furnishes a recursive proof of Newton’s binomial formula:
n
n r
n
(H4)
r x = (1 + x) .
r =0
000
00
0
010
001
01
10
100
011
110
1
11
101
111
Fig. 1 The hypercube graphs H1 , H2 , H3 , and a decomposition of the latter as H1 × H2
123
M. Aspenberg, R. Pérez
Fig. 2 The C3 = 5 monotone paths from (0, 0) to (3, 3)
2.2 Catalan Numbers and Lattice Paths
One of many equivalent definitions of the Catalan numbers 1, 1, 2, 5, 14, 42, 132, . . .
[8, A000108] is via the formula
2n Cn+1 =
n
(n+1) .
The exponential rate of growth of the sequence {Cn } follows easily from Stirling’s
formula:
Cn+1
2n
√
2π(2n) 2n
(2n)!
4n
22n
e
∼ √
=
.
∼
=
√
√
2
2
n n
(n!) (n + 1)
π n 3/2
π n(n + 1)
2π n e
(n + 1)
Definition A lattice path is a path in the lattice Z × Z that moves one horizontal
or vertical unit at every step without self-intersections. We consider monotone paths,
which never move left nor down (Fig. 2). Note that a monotone path from (0, 0) to
(m, n) requires m + n steps. Choosing one such path is tantamount to deciding which
of these steps will be the m horizontal
steps, so the number of monotone lattice paths
.
from (0, 0) to (m, n) is m+n
m
Lemma 3 The number of monotone paths from (0, 0) to (n, n) that do not cross over
the diagonal {y = x} is equal to Cn [9].
Proof of Lemma 3 A monotone path γ from (0, 0) to (n, n) that crosses over the
diagonal will pass through a point ( j, j + 1). Let P be the first such point, and γ the portion of γ going from P to (n, n). Reflecting γ on the diagonal {y = x + 1}
transforms γ into a monotone path from (0, 0) to (n − 1, n + 1). This operation is
bijective because such paths must cross over the diagonal. Therefore, the number of
monotone paths from (0, 0) to (n, n) that do not cross over the diagonal equals the
number of all monotone paths from (0, 0) to (n, n), minus the number of monotone
paths from (0, 0) to (n − 1, n + 1); i.e.,
2n 2n
2n
2n
2n
n
= n .
−
=
−
n+1 n
n+1
n
n−1
n
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Control of Cancellations that Restrain...
3 Structure of an
3.1 Signatures and Binomial Products
The internal structure of the expression an is better understood by separating the
different contributions of weight x r . After expanding the recursive expressions in (1),
the first few terms are
a1 =
a2 =
a3 =
a4 =
a5 =
= x,
= 11 x 2 ,
x 11 a1
= 21 11 x 3 ,
x 21 a2
x 22 a2 + 31 a3 = 22 11 x 3 + 31 21 11 x 4 ,
x 23 a3 + 41 a4 = 23 21 11 x 4 + 41 22 11 x 4
+ 41 31 21 11 x 5 .
(2)
This symbolic manipulation makes it clear that an is the sum of all products of the
form
bs
bs−1
b2
b1
···
· x s+1 ,
(3)
n − bs bs − bs−1
b3 − b2
1
such that
n =: bs+1 > bs > bs−1 > · · · > b1 = 1 , and b j+1 ≤ 2b j ( j = 1, . . . , s). (4)
The last condition is a consequence of the fact that the sum in (1) starts at r =
Note that this condition forces b2 = 2.
n
2
.
Definition A tuple σ = (n, bs , . . . , b1 ) satisfying (4) is called an n-signature. The
n-signature that contains all the numbers from 1 to n is called canonical.
Note Compare the recursion (1) with the similar looking α1 = x, αn = x rn−1
=n/2 αr
in which the binomial coefficients have been removed. From the above discussion we
see that αn is the sum of weights x s+1 , taken over all signatures (n, bs , . . . , b1 ).
Replacing x with 1 shows that the number of distinct n-signatures is given by the
recursion
N1 = 1,
Nn =
n−1
r =
n
2
Nr .
The numbers {Nn } = {1, 1, 1, 2, 3, 6, 11, 22, 42, . . .} form the Narayana–Zidek–
Capell sequence [8, A002083].
123
M. Aspenberg, R. Pérez
5
5
5
5
5
5
4
4
4
4
4
4
3
3
3
3
3
3
2
2
2
2
2
2
1
1
1
1
1
1
(6,3,2,1)
(6,4,2,1)
(6,4,3,2,1)
(6,5,3,2,1)
(6,5,4,2,1)
(6,5,4,3,2,1)
Fig. 3 The towers associated with all 6-signatures. The second tower, for instance, has blocks at positions
1, 2, and 4. The rightmost signature is the canonical one
3.2 Arrays and Blocks
When faced with an expression made of binomial coefficients, the natural thing to
ask is what kind of combinatorial object is being counted. To a given signature
σ = (n, bs, . . . ,b1 ) we
b1 will assign a tower that can be filled with an array of numbers
bs
·
·
·
in exactly n−b
1 ways.
s
Definition Given σ = (n, bs , . . . , b1 ), consider a tower of n −1 square cells split into
blocks of lengths (n − bs ), (bs − bs−1 ), . . . , (b3 − b2 ), (b2 − b1 ) from top to bottom
as in Fig. 3. The position of a block is the height b j of its lowest cell, so the signature
condition b j+1 ≤ 2b j implies that a block is never taller than its position. An array
associated with σ is an assignment of numbers to every cell in the tower of σ such that
the numbers in the j-th block (at position b j ) are chosen from the set {1, 2, . . . , b j }
and appear in descending order. An array associated with the canonical signature is
also called canonical.
Lemma 4
(a) The number
b1 of n-arrays associated with a signature σ = (n, bs , . . . , b1 ) is
bs ·
·
·
n−bs
1 .
(b) The number of canonical n-arrays is (n − 1)!
(c) The total number of n-arrays is given by formula (1) when x = 1.
Proof The tower associated with σ has s blocks. The j-th block is based at position
b j and its length is b j+1 − b j (for the topmost block the length is n − bs ). Therefore,
the j-th block can be filled with an arbitrary choice of b j+1 − b j numbers between
j
possibilities. This proves (a), from which item (b) follows
1 and b j ; i.e., b b −b
j+1
j
immediately. To prove (c), assign the weight x s to every n-array with s blocks, and
note from (3) that an represents the sum of weights over all n-arrays. Thus, when
x = 1, an simply counts the number of n-arrays as claimed.
Definition The sequence of numbers that specifies an array is called a pattern. Our
convention is to read patterns from the bottom up; thus, for instance, the rightmost
array in Fig. 4 has pattern [1232].
Lemma 5 A tuple of numbers [t1 , . . . , tn−1 ] is realized as the pattern of some array
if and only if
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Control of Cancellations that Restrain...
4
1
4
1
4
2
4
1
4
1
4
2
3
2
3
3
3
3
3
2
3
3
3
3
2
1
2
1
2
1
2
2
2
2
2
2
1
1
1
1
1
1
1
1
1
1
1
1
Fig. 4 The 5-signature (5, 3, 2, 1) has six associated arrays: The block at position 2 can hold either a 1 or
a 2, while the block at position 3 can hold any descending combination of the numbers 1, 2, 3
t j ≤ j for all 1 ≤ j ≤ n − 1.
Proof In a canonical array every block has length one. This means that the position
of the j-th block is b j = j, and the number in this block is t j . Thus, in this case, the
pattern condition is equivalent to t j ≤ b j = j, proving the result for canonical arrays.
In a non-canonical array, the cell at position j belongs to a block at position i ≤ j.
The pattern condition states that the number t j in that cell must be at most i, and the
result follows.
3.3 Array Hypercubes
Definition If an array has a block at position p with more than one cell, the block
can be split into two shorter blocks. The result is a valid array since the blocks have
positions p and p + η > p (η is the location of the split within the original block),
and the numbers in both blocks are all at most p. We call this operation on arrays a
split. Note that an array can usually be split in several ways, all of which commute.
Moreover, repeated splitting eventually results in a canonical array.
The reverse operation is also well defined. If an array has two consecutive blocks
at positions p and p + η, and the numbers contained in both blocks run together in
descending order, the two blocks can be combined into a single one. This is because
the new block is at position p and contains numbers in descending order, which
means that the length of the new block cannot exceed its position. In other words,
condition (4) is satisfied. This operation on arrays is called a merge. As with splitting,
merge operations are commutative, and repeated merging must terminate. An array
where no pair of blocks can be merged is called primitive.
Definition The graph Gn on the set of n-arrays is defined by joining any two arrays
related by a single split/merge operation.
Note that arrays with the same pattern of numbers are connected by successive
splits to the same canonical array, and thus this condition is equivalent to belonging in
the same connected component of Gn . Moreover, a split/merge is possible at a given
position if and only if the numbers at that position are in descending order. In particular,
splitting/merging does not depend on the structure of blocks in an array, but only on
the pattern of numbers. This yields the following lemma.
Lemma 6 Every connected component of G n is isomorphic (in the graph-theoretical
sense) to a hypercube graph.
123
M. Aspenberg, R. Pérez
Fig. 5 The pattern [12142] has
2 descents, and thus determines
four 6-arrays connected by
split/merge operations into a
square H2
5
2
4
4
3
1
2
2
5
2
1
1
4
4
(6,4,2,1)
3
2
1
1
5
(6,4,3,2,1)
5
2
4
4
1
3
1
2
2
2
2
1
1
4
4
(6,5,4,2,1)
3
1
2
2
1
1
(6,5,4,3,2,1)
Proof Consider an array A ∈ Gn . The connected component C of A consists of all
arrays with the same pattern of numbers as A. This pattern has descents (locations
where the numbers are in descending order). Now view such locations as placeholders
for a symbol 1 or 0 depending on whether two blocks of A meet at that location or
not. This puts the arrays of C in correspondence with vertices of the hypercube graph
H ; see Fig. 5. Since a split/merge depends only on the pattern of numbers, all edges
of H are included and C is isomorphic to H .
Observation Every hypercube C ⊂ Gn has a unique primitive array and a unique
canonical array. Since canonical arrays are in one-to-one correspondence with valid
patterns, the numbers of hypercubes in Gn and of primitive n-arrays are both equal
to (n − 1)! by Lemma 5. Also, if is as in the proof above, the primitive array has
s = n − − 1 blocks (because the canonical array has n − 1), so C is isomorphic to
Hn−s−1 .
4 The First Proofs
Each an is a polynomial in x. When written with the monomials ξr x r in ascending
order by degree, we say an is in basic format. Since every n-array with r − 1 blocks
contributes x r to the total an , the coefficient ξr counts the number of such arrays.
Example The first few an in basic format are (compare (2)):
a1 = x,
a2 = x 2 ,
a3 = 2x 3 ,
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Control of Cancellations that Restrain...
a4 = x 3 + 6x 4 ,
a5 = 10x 4 + 24x 5 ,
a6 = 8x 4 + 86x 5 + 120x 6 .
(5)
When n = 6, for instance, we know that the towers with 4 blocks correspond to
the three
signatures
(6, 4,3,2,1),(6,
5, 3, 2, 1), and
(6,
5,
4, 2, 1) (see Fig. 3). These
have 24 31 21 11 = 36, 51 23 21 11 = 30, and 51 41 22 11 = 20 associated arrays
respectively, and we see that each of these 86 arrays contributes x 5 to the value of a6 .
Recall that the array graph Gn consists of isolated hypercube components. We can
also break an down as a sum of contributions by hypercubes. Let A be a primitive narray with s = r − 1 blocks. We know that A contributes x r to an . Since the connected
component C of G containing A is homeomorphic to Hn−r , items (H2) and (H4) in
Sect. 2.1 give
a∈C
(H2)
x (#blocks of a)+1 =
n−r n −r
j=0
j
(H4)
x r + j = x r (1 + x)n−r .
(6)
Let primn (r ) be the number of primitive arrays with r − 1 blocks; this is also the
number of components of Gn homeomorphic to Hn−r . Equation (6) shows that the
total contribution to an of all arrays in all such hypercubes is primn (r ) · x r (1 + x)n−r ,
so
n
(7)
an =
primn (r ) · x r (1 + x)n−r .
r =log2 n (Condition (4) implies that the least possible number of blocks is log2 n .)
When an is written in that form, we can read that each of the primn (r ) components
of Gn with dimension n − r contributes x r (1 + x)n−r to the total sum an . We will say
that an is in binomial format.
Example Let us compute the binomial format of a6 . In Eq. (5) we found that the lowest
degree monomial of a6 is 8x 4 . This indicates that the least number of blocks in a 6array is 3, and that there are 8 arrays with 3 blocks. All of these are clearly primitive,
so prim6 (4) = 8. These arrays (together with those obtained by splitting) determine 8
copies of H2 in G6 . Gathering the monomials of all arrays in these hypercubes gives
a6 = 8x 4 + 86x 5 + 120x 6
= (8x 4 + 16x 5 + 8x 6 ) + 70x 5 + 112x 6
= 8x 4 (1 + x)2 + 70x 5 + 112x 6 .
The remaining 70 arrays with 4 blocks must be primitive since they were not
incorporated into an H2 ; thus, prim6 (5) = 70. These arrays (together with those
obtained by splitting) determine 70 copies of H1 in G6 . Thus,
8x 4 (1 + x)2 + 70x 5 + 112x 6 = 8x 4 (1 + x)2 + 70x 5 (1 + x)1 + 42x 6 .
123
M. Aspenberg, R. Pérez
The first few an in binomial format are:
a1 = x 1 y 0 ,
a2 = x 2 y 0 ,
a3 = 2x 3 y 0 ,
a4 = x 3 y 1 + 5x 4 y 0 ,
a5 = 10x 4 y 1 + 14x 5 y 0 ,
a6 = 8x 4 y 2 + 70x 5 y 1 + 42x 6 y 0 ,
where y stands for 1 + x. This convention makes for cleaner looking expressions, and
will be used consistently in the rest of the paper.
Now that the combinatorial structure is in place, the proofs of Theorem 2 and the
case x ∈
/ [−1, 0] of Theorem 1 are straightforward.
Proof of Theorem 2 Since x = −1, the only non-zero term in (7) occurs when r = n.
In other words,
an = (−1)n primn (n).
Now, primn (n) is the number of n-arrays with n − 1 blocks; that is, arrays that are both
primitive and canonical. By definition, these are arrays with non-decreasing patterns
of n − 1 numbers. Any such pattern can be associated with a monotone lattice path
from (0, 0) to (n − 1, n − 1) that does not cross over the diagonal. Simply subtract
1 from each entry in the pattern and interpret the results as heights of the horizontal
steps of a monotone path (compare Fig. 2). This procedure is bijective, so by Lemma 3,
primn (n) = Cn .
Proof of Theorem 1 when x ∈
/ [−1, 0] Depending on the sign of x, one of the two
formats for an displays no cancellations.
x > 0: All monomials ξr x r in the basic format of an are positive because the coefficient ξr counts arrays with r − 1 blocks. In particular, an is larger than the highest
order monomial. The coefficient of this monomial is the number of arrays with the
most blocks; i.e., canonical arrays. By Lemma 4,
an > (n − 1)! · x n .
x < −1: Since y < 0, all terms primn (r ) · x r y n−r in the binomial format of an have
the same sign, and do not cancel each other. Also, |x| > |y|, so
|an | =
r
primn (r ) · |x r y n−r | >
primn (r ) · |y n | = (n − 1)! · |y n |,
r
where the last equality follows from the observation after Lemma 6.
In both cases, |an | is larger than (n − 1)! · w n for some positive w, and the result
holds.
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Control of Cancellations that Restrain...
5 The Case x ∈ (−1, 0)
The situation when x ∈ (−1, 0) is more delicate because the terms in both the basic
and binomial formats of an have alternating signs. Our strategy in this second part
involves a different representation of an (Eqs. (8) and (9)). We will interpret the
sequences {Sn (1), . . . , Sn (n − 1)} as functions sn ∈ L 2 [0, 1] and the recursion (9) as
a sequence of integral operators An : sn → sn+1 . Then we deduce some facts about
the shape of the graph of sn , and about the limit operator T = limn→∞ An . We will
use this information to show that the largest eigenvalue λ of T bounds from below the
exponential rate of decay of an /(n − 2)!
5.1 A New Recursion
So far we have established that an is the sum of contributions of the form x r over
a large set of arrays (for each array, r − 1 is the number of blocks). We grouped
arrays with the same number pattern into hypercube graphs, and showed that an is
the sum of contributions x n− (1 + x) = x n− y over hypercubes C, where is
the dimension of each C. This dimension is the number of descents in the associated
pattern, so we can abandon arrays and express an directly as a sum of contributions over
patterns:
n−log2 n −1
an =
=0
patterns with
descents
x n− y .
This allows us to sort the contributions to an made by individual patterns. To this
end, consider an n-pattern π = [t1 , . . . , tn−1 ]. If the truncated pattern [t1 , . . . , tn−2 ]
contributes x a y b to an−1 , then π contributes x a y b+1 or x a+1 y b to an depending on
whether tn−1 < tn−2 or not (i.e., on whether π has one extra descent or not at the last
position). This motivates the following definition.
Definition For n ≥ 2 let Sn (r ) denote the sum of contributions of all patterns
[t1 , . . . , tn−1 ] such that tn−1 equals r (thus, r can take values in {1, . . . , n − 1}).
In particular, S2 (1) = x 2 , and
n−1
an =
Sn ( j).
(8)
j=1
By the previous argument, Sn+1 (r ) can be computed from the contributions of
n-patterns:
r
n−1
Sn ( j) + y ·
Sn ( j)
(9)
Sn+1 (r ) = x ·
j=1
j=r +1
(when r = n−1 or n, there is
no descent in the last position, so (9) should be interpreted
to mean Sn+1 (n − 1) = x · n−1
j=1 Sn ( j) = Sn+1 (n)).
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M. Aspenberg, R. Pérez
Table 1 Entries of the
sequences Sn for 2 ≤ n ≤ 5,
computed according to the
recursion (9)
n \r
1
2
x2
3
x3
x3
4
x 4 + yx 3
2x 4
2x 4
5
x 5 + 5yx 4
3x 5 + 3yx 4
5x 5 + yx 4
2
3
4
5x 5 + yx 4
5.2 Sinusoidal Shape of Sn
The last term of the left sum in (9) is x · Sn (r ). Substituting this with the equivalent
expression (y − 1) · Sn (r ), we obtain
Sn+1 (r ) = x ·
r −1
Sn ( j) + y ·
r −1
j=1
Sn ( j) + (y − 1) · Sn (r )
j=r +1
j=1
=x ·
n−1
Sn ( j) + y ·
n−1
Sn ( j) − Sn (r )
j=r
= Sn+1 (r − 1) − Sn (r ),
(10)
valid for 2 ≤
r ≤ n − 2. Moreover, Eq. (10) remains valid for r = 1 if we define
Sn (0) = y · n−2
j=1 Sn−1 ( j). In this manner we derive a more practical method of
computing the sequence Sn :
• S2 (1) = x 2 .
For n ≥ 3:
• Sn (0) = y n−2
j=1 Sn−1 ( j),
• Sn (r ) = Sn (r − 1) − Sn−1 (r ), (1 ≤ r ≤ n − 2),
• Sn (n − 1) = Sn (n − 2).
Observation Sn (0) is not part of the original sequence, but including this auxiliary
term makes the definition of the sequence considerably simpler. It is vital to the coming
arguments that Sn (0) = yan−1 and Sn (n − 2) = xan−1 are both 0 or have opposite
signs.
For illustration purposes Table 2 shows the first few sequences Sn (2 ≤ n ≤ 5),
including the auxiliary values Sn (0) (compare with the expressions in Table 1). Also,
Fig. 6 shows a sample plot of values for Sn .
Definition Let n ≥ 3. Given a fixed x, the sequence Sn has
• a sign change at (A, B) if 0 ≤ A < B ≤ n − 1, all terms Sn ( j) are equal to 0 for
every index A < j < B (perhaps an empty set), and either
Sn (A) < 0 < Sn (B) (an up-change) or
Sn (A) > 0 > Sn (B) (a down-change);
123
Control of Cancellations that Restrain...
Table 2 Entries of the sequences Sn for 2 ≤ n ≤ 5, computed according to the recursion (10)
n \r
0
1
2
—
x2
3
yx 2
(y − 1)x 2
(y − 1)x 2
4
(2y 2 − 2y)x 2
(2y 2 − 3y + 1)x 2
(2y 2 − 4y + 2)x 2
(2y 2 − 4y + 2)x 2
5
(6y 3 −11y 2 +5y)x 2
(6y 3 −13y 2 +8y−1)x 2
(6y 3 −15y 2 +12y−3)x 2
(6y 3 −17y 2 +16y−5)x 2
2
3
4
(6y 3 −17y 2 +16y−5)x 2
Substituting y = 1 + x confirms that these values are identical to those in Table 1
Fig. 6 The sequence S20 when x = −2/7
• an extreme at (A, B) if Sn−1 has a sign change at (A + 1, B). Note that there is at
least one index between A and B, and Sn ( j) equals some fixed value M for every
A < j < B;
If Sn (A) < M > Sn (B), the extreme value Mis a local maximum;
If Sn (A) > M < Sn (B), the extreme value Mis a local minimum;
• an inflection at (A, B) if Sn−1 has an extreme at (A + 1, B). Note that there are at
least two indices between A and B.
The sign change/extreme/inflection is located 2 at B − 1.
Observation From (10) we get Sn−1 (r ) = Sn (r − 1) − Sn (r ), so the sequence Sn−1
acts as a “discrete derivative” of the sequence Sn . A corollary of property (ShB)
below is that in our inflections the “slope” is always steeper in the middle than
on the sides. Also, note that Sn (n − 1) cannot be part of a change, extreme, or
inflection because Sn (n − 1) = Sn (n − 2) (we allow this range in the definition solely to make the proofs easier to write). Thus, in practice, a sign change
has B ≤ n − 2 and, since the definition of extreme at level n depends on a sign
change at level n − 1, an extreme has B ≤ n − 3. Similarly, an inflection has
B ≤ n − 4.
2 The proof of (ShE) below illustrates the convenience of this extra definition.
123
M. Aspenberg, R. Pérez
Proposition 7 For all n ≥ 3 the sequence Sn satisfies
For some r in the range 1 to n − 2 we have Sn (r ) = 0.
There are at most one sign change, one extreme, and one inflection.
A local maximum M is positive. A local minimum M is negative.
There is at most one r in the range 1 to n − 2 such that Sn (r ) = 0.
At least one of the two values minr Sn (r ) and maxr Sn (r ) (r = 0, . . . , n − 2)
is attained at an endpoint: Sn (0) or Sn (n − 2).
(ShE) Assume Sn has an extreme and an inflection. Let their locations be E ≥ 1 and
I ≥ 2, respectively. Then
(ShO)
(ShA)
(ShB)
(ShC)
(ShD)
(a) E < I ⇒ |Sn−1 (1)| ≤ |Sn−1 (0)| and |Sn−1 (n − 3)| ≥ |Sn−1 (n − 2)|,
(b) E ≥ I ⇒ |Sn−1 (1)| ≥ |Sn−1 (0)| and |Sn−1 (n − 3)| ≤ |Sn−1 (n − 2)|.
Proof of (ShO) Suppose that Sn (r ) = 0 for 1 ≤ r ≤ n − 2. Then Sn (0) = xy Sn (n −
2) = 0 too, and we find from (10) that Sn−1 (r ) = 0 for every 1 ≤ r ≤ n − 3. This
leads to a contradiction since S3 (1) = x 3 < 0.
Proof of (ShA) By the definition of extreme and inflection we only need to prove that
Sn has at most one sign change for n ≥ 3. From Table 2 it is clear that S3 has a sign
change (indeed, a down-change) for any value of x. Let us assume then that n ≥ 4.
If Sn has multiple sign changes, they must alternate between up- and down-changes.
Say Sn has an up-change followed by a down-change (the other case is analogous).
In between these two sign changes there is a (positive) local maximum. Now, at least
one of Sn (0) and Sn (n − 2) is non-negative (both are zero or have opposite signs), so
Sn also has a (negative) local minimum.
Two extremes for Sn mean two sign changes for Sn−1 . Thus, we have found that
multiple sign changes for Sn would imply multiple sign changes for Sn−1 . But S4 can
only have one sign change between S4 (0) and S4 (2) = xy S4 (0).
Proof of (ShB) Suppose Sn has a local minimum M ≥ 0 at (A, B) (a local maximum
M ≤ 0 is ruled out in an analogous way). Since there are no other extremes, both Sn (0)
and Sn (n − 2) must be strictly larger than Sn (A + 1) = M ≥ 0, which contradicts the
fact that they have opposite signs.
Proof of (ShC) The statement is trivially true for n = 3 and follows from (ShO) when
n = 4, so we can assume n ≥ 5.
Note that it is possible for Sn to have two zeros in the range 0 to n − 2; namely,
when Sn (0) = Sn (n − 2) = 0. In this situation (ShO) forces Sn to have a non-zero
extreme at E (1 ≤ E ≤ n − 3), and consequently, Sn−1 has a sign change at E. It
follows that Sn−1 (0) and Sn−1 (n − 3) are different from 0; otherwise, the sign change
of Sn−1 would imply two extremes, contradicting (ShA).
Still assuming Sn (0) = Sn (n − 2) = 0, consider what would happen if Sn had one
more zero:
– If Sn (n − 3) = 0, then Sn−1 (n − 2) = 0 (using (10)), and we saw above that this
contradicts the sign change of Sn−1 at E.
123
Control of Cancellations that Restrain...
– If Sn (1) = 0, we would have Sn−1 (1) = 0 (again, using (10)). As above, Sn−1 (0)
cannot be 0, so let us compare the sign of Sn−1 (0) with the sign immediately to the
left of the sign change at E:
The signs cannot be the same because Sn−1 (1) = 0 would be an extreme, contradicting (ShB). But the signs cannot be opposite either because that would imply a
second sign change to the left of E, contradicting (ShA).
– If Sn (1) = 0 and Sn (n − 3) = 0, a zero in between would imply two extremes,
giving again a contradiction.
When Sn (0) and Sn (n − 2) are different from 0 the signs of these endpoints are
opposite and Sn has (exactly) one sign change. It remains to consider the possibility
that this sign change consists of a long run of consecutive zeros. Accordingly, assume
that Sn (r ) = · · · = Sn (r +k) = 0 (r, k ≥ 1), with Sn (r −1) < 0 and Sn (r +k +1) = 0
(the case Sn (r − 1) > 0 is analogous). By (ShB) Sn (r + k + 1) must be positive. But
then
Sn−1 (r ) < 0 = Sn−1 (r + 1) = · · · = Sn−1 (r + k) > Sn−1 (r + k + 1),
so Sn−1 would have a local maximum M = 0, contradicting (ShB).
Proof of (ShD) Assume Sn−1 (1) ≤ 0 so Sn (1) ≥ yan−1 (the case Sn−1 (1) > 0 is
analogous). It follows that Sn−1 cannot have a down-change at (A, B) unless A = 0,
and therefore Sn cannot have a local minimum. As a consequence, the value minr Sn (r )
is attained at either r = 0 or r = n − 2, and the conclusion follows.
Proof of (ShE) We can assume that the extreme of Sn is a local maximum (the local
minimum case is analogous). Then Sn−1 has an up-change at E, and an extreme at I .
Note in particular that yan−2 = Sn−1 (0) < 0. There are two cases:
(a) The up-change of Sn−1 is located before its extreme. Hence the extreme is a local
maximum, and therefore it lies above Sn−1 (n − 2) = xy Sn−1 (0) > 0. By (ShD),
the minimum of Sn−1 is Sn−1 (0), so
Sn−1 (0) ≤ Sn−1 (1) ≤ Sn−1 (E) ≤ 0.
(The last inequality holds because E is the last point before Sn−1 becomes
positive.)
(b) The up-change of Sn−1 is located at, or after its extreme. Hence the extreme is
a local minimum, and therefore it lies below Sn−1 (0). Now, all values Sn−1 (r )
for 0 ≤ r ≤ I are negative because the up-change is located at E ≥ I . We
claim further that these values are non-increasing; otherwise, Sn−1 would have a
negative local maximum. Since Sn−1 has an up-change, Sn−1 (n − 2) = xan−2 is
non-negative, and we have
0 ≥ yan−2 = Sn−1 (0) ≥ Sn−1 (1).
The proofs of the inequalities at the right extreme of Sn−1 are analogous.
123
M. Aspenberg, R. Pérez
5.3 Sn Becomes a Step Function
Formula (9) induces a linear operator An : Rn−1 −→ Rn . Here we will embed An as
an integral operator An : L 2 [0, 1] −→ L 2 [0, 1], and find an operator T which is the
limit of {An } in the operator norm. The goal will be to link the growth of {an } to the
spectral properties of T .
For n ≥ 2, the r -th entry of the column vector
⎤
⎡
Sn (1)/(n − 2)!
⎥
⎢
..
n−1
sn = ⎣
⎦∈R
.
Sn (n − 1)/(n − 2)!
represents the average contribution to an of patterns with last entry tn−1 = r (of
which there are (n − 2)!). With this notation, Eq. (9) can be interpreted as a linear
transformation
sn+1 = (An · sn )/(n − 1),
(11)
where An is the n × (n − 1) matrix whose (i, j)-entry is x if i ≥ j, and y otherwise.
the standard basis vector e j
Let E n : Rn−1 −→ L 2 [0, 1] be the linearmap that sends
j−1
j
n−1 , n−1
. The vector sn maps to the step
j−1
j
.
function sn = E n (sn ), such that sn (u) = Sn ( j)/(n − 2)! whenever u ∈ n−1
, n−1
to the characteristic function of the interval
The maps {E n } embed the linear operators An : Rn−1 −→ Rn into linear operators
An : L 2 [0, 1] −→ L 2 [0, 1]. In particular, Eq. (11) takes the form
1
sn+1 (u) = [An (sn )] (u) =
αn (u, v) · sn (v)dv,
(12)
0
where the kernel αn is a piecewise constant function whose value at (u, v) ∈
j−1
j
× n−1
is (Fig. 7)
, n−1
x if i ≥ j
αn (u, v) =
i.e., equal to (An )i j .
y otherwise
i−1
n
, ni
Observation The factor 1/(n − 1)in (11) is hidden as a normalization factor in (12).
i
Indeed, when u lies in the interval i−1
n ,n ,
1
[An (sn )] (u) =
αn (u, v) · E n (sn )(v)dv
0
=
n−1
(An )i j
j=1 j−1
j
n−1 , n−1
123
Sn ( j)
dv
(n − 2)!
Control of Cancellations that Restrain...
Fig. 7 The kernel α10 (the
shaded region is 10 ). Note that
the unit square is divided into
1 by 1
rectangles of size 10
9
v
y
x
u
n−1
Sn ( j)
1
=
(An )i j
·
,
(n − 2)! n − 1
j=1
which is the i-th entry of sn+1 . In particular, (compare Eq. (8)):
1
an = (n − 1)!
sn (v)dv.
0
To prove the theorem we need to show that there is a sequence of indices {n k } so
1
the rate of exponential decay of the integrals 0 sn k (v) dv is bounded from below. The
bound will be dictated by the largest eigenvalue λ of the limit operator of An .
5.4 The Limit Operator T
Here we define the limit operator T of the sequence {An }, and establish some of its
basic properties [4], [5, Sect. 11].
Let T : L 2 [0, 1] −→ L 2 [0, 1] by
1
κ(u, v) · f (v)dv,
(T f )(u) =
0
with kernel (Fig. 7)
κ(u, v) =
x
y
if u ≥ v
otherwise.
123
M. Aspenberg, R. Pérez
Lemma 8 The operator T is the limit of {An } in the operator norm:
1
T − An 2 ≤ √ .
n
Proof The kernel of T − An is the function κ − αn . Since y is just shorthand for 1 + x,
we see that κ − αn is the characteristic function of the staircase region n in the unit
square, consisting
of the upper triangle {0 ≤ u, v ≤ 1 | u < v}, minus those rectangles
i−1 i j−1 j n , n × n−1 , n−1 such that i < j. Then the Lebesgue measure of n is
μ(n ) =
1
1
1 (n − 2)(n − 1)
−
·
= .
2
2
n(n − 1)
n
It follows that
1/2
1
T − An 2 ≤ μ(n )
=√ .
n
5.5 Eigenfunctions of T
The operator T can be expressed as follows:
1
κ(u, v) · f (v) dv
(T f )(u) =
0
1
=x·
1
f (v)dv + y ·
0
f (v) dv
0
= −F(u) + y · F(1) − x · F(0),
where F is any primitive of f . To find the eigenvalues of T , set
(λ f )(u) = (T f )(u) = −F(u) + y · F(1) − x · F(0),
and differentiate to obtain the ODE
f (u) =
− f (u)
,
λ
with general solution
f (u) = Ce−u/λ .
123
(13)
Control of Cancellations that Restrain...
A primitive of f is F(u) = −λCe−u/λ , so substituting in (13) gives
λ Ce−u/λ = λCe−u/λ − λC y · e−1/λ − x .
Thus, λ is an eigenvalue if and only if
e−1/λ =
Note that
x
y
x
.
y
< 0 exactly when x ∈ (−1, 0). Then we can write the eigenvalues as
λm =
log
x
y
−1
−1
=
for all m ∈ Z,
+ 2mπ i
log xy + (2m + 1)π i
and the eigenfunction corresponding to λm is
u
f m (u) = xy e(2m+1)π iu .
Definition For ease of notation, we write the absolute values of the two largest eigenvalues as λ := |λ−1 | = |λ0 | and μ := |λ−2 | = |λ1 |.
5.6 Weighted 2-Norm
Definition The family of functions { f m }m∈Z is not orthonormal with respect to the
1
standard inner product 0 f m (v)gn (v) dv. To turn it into one (and to simplify our
computations) we introduce the weighted inner product
1
f, g :=
x −2v
f (v)g(v) dv,
(14)
y
0
and its associated 2-norm f 2 := f, f 1/2 . We will not use the standard inner
product and 2-norm in the core of the proof, so this choice of notation should create
no confusion. However, as a constant reminder we will use the term weighted 2-norm.
Lemma 9 There is a finite constant G > 1 such that the standard 2-norm of a function
f is bounded between G1 f 2 and G f 2 (the weighted 2-norm).
u
Proof It follows since the function xy is bounded away from 0 and ∞ when
0 ≤ u ≤ 1.
As a corollary, we get
Lemma 8 The operator T is the limit of {An } in the weighted operator norm:
G
T − An 2 ≤ √ .
n
123
M. Aspenberg, R. Pérez
With the new notation in place, we can interpret the family of eigenfunctions of T
as follows.
Lemma 10 The family { f m } forms a Hilbert basis of L 2 [0, 1].
Proof Let
an arbitrary function in L 2 [0, 1]. Since x ∈ (−1, 0), the function
y fu be
−π
iu
g(u) = x e
is continuous and bounded from above, so f g ∈ L 2 [0, 1]. After
rescaling the standard basis of L 2 [−π, π ], we obtain the representation
( f g)(u) =
cm · e2mπ iu ,
m∈Z
which implies
f (u) =
u
cm · xy e(2m+1)π iu =
cm · f m (u).
m∈Z
m∈Z
Thus, the family { f m } is complete, and since it is orthonormal the claim follows
because L 2 [0, 1] with the weighted product is a Hilbert space (see, e.g., [2, Theorem
4.13]).
Note that for m ≥ 0 the pair of functions f −(m+1) , f m are complex conjugate and
their eigenvalues have the same magnitude. As a consequence, a convenient basis for
the subspace L 2R [0, 1] ⊂ L 2 [0, 1] of real-valued functions is
x u cos (2m + 1)π u , x u sin (2m + 1)π u
y
y
m≥0
.
6 The Functional Approach
In this section we establish the lower bound on the exponential rate of decay of the
sequence an /(n − 2)!. The long proof of Lemma 11 interferes with the flow of logic,
and is consequently deferred to the next section.
Definition The eigenfunctions f −1 and f 0 with largest eigenvalue λ span a complex
two-dimensional subspace of L 2 [0, 1]. Let E ⊂ L 2R [0, 1] denote the real slice of this
u
u
subspace generated by x cos π u , x sin π u . The space E ⊥ spanned by all
y
y
other eigenfunctions is orthogonal to E, so that (by Lemma 10) L 2R [0, 1] = E ⊕ E ⊥ .
The projections onto E and E ⊥ (with respect to our new weighted inner product) are
denoted P and P ⊥ , respectively. By Parseval’s theorem we can define the angle θn (in
the range [0, π/2]) by any of the three equivalent formulas
sin θn :=
123
P ⊥ sn 2
Psn 2
P ⊥ sn 2
, cos θn :=
, tan θn :=
.
sn 2
sn 2
Psn 2
Control of Cancellations that Restrain...
Intuitively, the closer θn is to 0, the better sn resembles a function in E.
Now we can describe the strategy of the proof:
Step 1: We use the shape properties of the sequence Sn to show that the angles θn are
bounded away from π/2.
Step 2: The sequence {θn } converges to 0, so the functions sn become progressively
sinusoidal.
Step 3: There is a sequence of indices {n k } such that {|an k |} is comparable to {sn k 2 }.
Meanwhile, sn 2 ≥ (λ−ε)n for arbitrarily small ε, and the result will follow.
6.1 Step 1 (θn ≤ < π/2)
Fix n. If Sn has a sign change, let (a, b) denote its locus. Then the value z n
:= a/(n − 1) ∈ [0, 1) is such that sn (u) · sin(π(u − z n )) never changes sign. When
there is no sign change, z n := 0 trivially satisfies the same condition. We will show
there is a K > 0 such that for all n,
sn (u), x u sin π(u − z n ) > K sn 2 .
y
(15)
u
The projection Psn is larger than |sn (u), xy sin π(u − z n )|/ √1 , and thus the angle
2
θn is bounded away from π/2 by
√
θn < := arccos( 2K ) < π/2.
The proof of (15) follows from Corollary 12 and Lemma 13 below, using
K = C C .
Lemma 11 The sequences {sn 1 } and {sn ∞ } are comparable in the sense that
there is a constant C such that for all n,
sn ∞ ≥ sn 1 ≥ Csn ∞ .
This is the main technical lemma, and its proof is deferred to Sect. 7.
Corollary 12 The sequences {sn 1 } and {sn 2 } are comparable in the sense that
there is a constant C such that for all n,
1
C sn 2
≥ sn 1 ≥ C sn 2 .
Proof The left side follows from the Cauchy–Schwarz inequality together with
Lemma 9. On the right we have
1
sn 21
≥C
2
sn 2∞
≥C
sn2 (v) dv,
2
0
123
M. Aspenberg, R. Pérez
where C is as in Corollary 12. The integral is the square of the standard 2-norm of sn ,
C 2
sn 22 .
so the last quantity is larger than G
Lemma 13 Let z n be defined as above. Then there is a constant C such that
sn (u), x u sin π(u − z n ) > C sn 1 .
y
Proof For every z ∈ [0, 1], let Mz,ε be the maximum of the function | xy |−u ·| sin π(u −
z)| in a small interval [z − ε, z + ε] ∩ [0, 1]. This quantity is strictly positive and it
varies continuously on both z and ε. Given a fixed z ∈ [0, 1], the Lebesgue measure
of the set
L z,ε := {u ∈ [0, 1] s.t. | xy |−u · | sin π(u − z)| < Mz,ε }
varies continuously on ε because | xy |−u · | sin π(u − z)| is nowhere constant. Then we
can choose ε (depending on z) so that the measure of L z,ε is C/2, where C ≤ 1 is the
constant of Lemma 11. Since [0, 1] is compact, the lower bound M := inf z∈[0,1] Mz,ε
is positive.
Recall that z n is chosen so that sn (u) · sin π(u − z n ) has constant sign. For simplicity,
let
us assume that this sign is positive. Then the weighted inner product
sn (u), x u sin π(u − z n ) equals (compare (14)):
y
1
| xy |−v sn (v) · sin π(v − z n ) dv ≥
| xy |−v sn (v) · sin π(v − z n ) dv >
L z n ,ε
0
M
L z n ,ε
But
⎛
⎞
1
⎜
⎟
|sn (v)| dv = M ⎝ |sn (v)| dv −
|sn (v)| dv ⎠.
0
L z n ,ε
|sn (v)| dv < μ(L z n ,ε )sn ∞ =
C
2 sn ∞
< 21 sn 1
L z n ,ε
by Lemma 11, so we get
sn (u), x u sin π(u − z n ) >
y
M
2 sn 1 .
6.2 Step 2 (θn → 0)
We will show in Lemma 14 that when n is large enough, the sequence {tan θn } enters a
decreasing regime that makes it eventually converge to 0. This establishes the desired
result.
123
Control of Cancellations that Restrain...
First note that
P ⊥ An sn 2
P ⊥ (T sn + (An − T )sn )2
=
P An sn 2
P(T sn + (An − T )sn )2
⊥
μP ⊥ sn 2 + P ⊥ (An − T )sn 2
P T sn 2 + P ⊥ (An − T )sn 2
= ≤ P T sn 2 − P(An − T )sn 2 λPsn 2 − P(An − T )sn 2 tan θn+1 =
because T commutes with the projections P and P ⊥ . Moreover, since θn < , the two
terms in the denominator are comparable to λsn 2 and (An − T )sn 2 , respectively.
In particular, Lemma 8 allows us to remove the absolute value in the denominator by
assuming n is large enough.
We will need two variations (inequalities (16) and (17)) on this basic estimate for
the tangent. Using Lemma 8 again we find that
√
μP ⊥ sn 2 + sn 2 G/ n
√
λPsn 2 − sn 2 G/ n
√
nμ sin θn + G
= √
nλ cos θn − G
⎛
⎞
μ + √n G
sin θn
⎠ tan θn .
=⎝
G
λ − √n cos
θ
tan θn+1 ≤
(16)
(17)
n
Lemma 14 Let n be larger than both constants Ca
9G 2
(λ−μ)2
G2
,
λ2
=
9G 2
(λ−μ)2 cos2 and
+
and suppose further that θn < . Then there are constants
Cb =
0 < ε < 1 and R > 0 such that
√
3G
a) If n sin θn ≥ λ−μ
, then
tan θn+1 < (1 − ε) tan θn .
b) If
√
n sin θn ≤
3G
λ−μ ,
then
R
tan θn+1 < √ .
n
In other words, when n is sufficiently large, each step in the sequence {tan θn }
affords a definite relative decrease, or a slower but absolute decrease.
Proof of Lemma 14
a) The assumptions θn < and n > Ca imply
G
λ−μ
G
.
>λ− √
>λ−
λ− √
3
n cos θn
n cos 123
M. Aspenberg, R. Pérez
The condition
√
n sin θn ≥
3G
λ−μ
is equivalent to
G
λ−μ
μ+ √
<μ+
,
3
n sin θn
so (17) is smaller than
#
μ+
λ−
λ−μ
3
λ−μ
3
$
tan θn =
2μ + λ
tan θn .
2λ + μ
2μ+λ
2λ+μ
< 1, this case is proved.
%
√
√
3G
b) When n sin θn ≤ λ−μ
, we also have n cos θn ≥ n −
(16) gives
Since
9G 2
.
(λ−μ)2
Substituting in
3G
+G
μ λ−μ
tan θn+1 < %
,
9G 2
λ n − (λ−μ)
−
G
2
where the denominator is positive since n > Cb . The result follows.
6.3 Step 3 (| sn k | > cst (λ − ε)n k )
We are ready to prove that there is a constant such that
1
sn (v) dv ≥ (λ − ε)n
(18)
0
along a subsequence of indices. The immediate consequence is factorial growth of |an |,
1
since an = (n − 2)! · 0 sn (v) dv. Equation (18) follows at once from Propositions 15
and 16.
Proposition 15 There is an infinite integer sequence n 1 < n 2 < · · · , and a constant
0 < W ≤ 1 such that for all k,
1
sn (v) dv ≥ W sn 2 .
k
k
0
Proposition 16 For every ε > 0 there are constants H, N > 0 such that for n > N ,
sn 2 > H (λ − ε)n .
The proof of Proposition 15 uses the following auxiliary result:
123
Control of Cancellations that Restrain...
Lemma 17 There is a constant 0 < K < 1, and an infinite integer sequence n 1 <
n 2 < · · · such that for all k,
1
1
Psn (v) dv ≥ K |Psn (v)| dv.
(19)
k
k
0
0
We prove Lemma 17 first, and then Propositions 15 and 16.
Proof of Lemma 17 The projection Psn (u) lies in the eigenspace E generated by the
functions | xy |u · sin(π u) and | xy |u · cos(π u). Therefore, it has the form
Psn (u) = Hn | xy |u · sin(π u + ωn ),
(20)
for some positive constant Hn and phase shift ωn ∈ R/(2π Z). We need some facts
about the weighted sine function f ω (u) := | xy |u sin(π u + ω) (ω ∈ R/(2π Z)):
a) Since
1
0
| f ω (u)| du > 0, the function
1
0 f ω (u) du h(ω) := 1
0 | f ω (u)| du
is well defined for all ω and satisfies |h(ω)| ≤ 1. The equality is attained only at
ω = 0, π ; i.e., at the angles for which f ω (u) is a strictly non-positive or strictly
non-negative function.
b) A two-step integration by parts shows that
log | xy | sin(π u + ω) − π cos(π u + ω)
x u
+ c.
f ω (u) du = | y |
π 2 + log2 | xy |
The linear combination of trigonometric functions in the numerator can be written
as
%
± π 2 + log2 | xy | · sin(π u + ω + ϕ),
(21)
x (or
where the sign is negative exactly when log | xy | < 0, and ϕ = arctan −π
log y &%
ϕ = π/2 when x = −1/2) depends only on | xy |. Since |λ| = 1
π 2 + log2 | xy |,
one antiderivative of f ω (u) is thus given by
±|λ|| xy |u · sin(π u + ω + ϕ).
It follows that there are exactly two phase angles '
ω and '
ω + π for which h(ω) = 0.
The angle '
ω is different from 0, π because h(0) = h(π ) = 1. Thus, for ε0 sufficiently small we can find closed intervals I1 and I2 = I1 + π , both of length π/2, so
◦
◦
that 0, π ∈ I1 ∪ I2 , but '
ω and '
ω + π are more than a distance ε0 away from I1 ∪ I2 . Let
123
M. Aspenberg, R. Pérez
I be the union of I1 , I2 , and the four closed intervals of length ε0 at their endpoints.
Clearly, there is a uniform bound h(ω) > K > 0 for every ω ∈ I. We will prove the
lemma by finding arbitrarily large indices n k for which ωn k ∈ I.
First, let us compute the effect of T on any function in the eigenspace E. Recall
that E is generated by the eigenfunctions f −1 and f 0 , and that
| xy |u cos(π u) = ( f −1 + f 0 )/2 , | xy |u sin(π u) = ( f 0 − f −1 )/2i,
so
#
T (| xy |u
sin(π u)) = (λ0 f 0 − λ−1 f −1 )/2i =
=
=
=
where as in (21), ϕ = arctan
−π
x
log | y |
−| xy |u eiπ u
−| xy |u e−iπ u
$
(
−
2i
log | xy | + π i log | xy | − π i
|λ|2 | xy |u Im (− log | xy | + π i)eiπ u
|λ|2 | xy |u − log | xy | sin(π u)+π cos(π u)
±|λ|| xy |u sin(π u + ϕ),
(or ϕ = π/2 when x = −1/2). A similar
computation yields the same phase ϕ when T is applied to | xy |u cos(π u). By linearity,
T effects the same phase shift on every function f ω ∈ E.
In particular, consider the following. Since the operators T and P commute,
Psn+1 = T Psn + P(An − T )sn = ±Hn |λ|| xy |u sin(π u + ωn + ϕ) + P(An − T )sn ,
where Hn is the constant from (20) modulo a choice of sign as in (21). Now,
P(An − T )sn ∈ E, so the above can be written as
Psn+1 = Hn |λ|| xy |u sin(π u + ωn + ϕ + εn ).
Comparing this with (20), we see that Hn = Hn+1 and
ωn+1 = ωn + ϕ + εn .
√
But θn < < π/2, so Hn = 2Psn 2 (by (20)) is comparable to sn 2 , and
P(An − T )sn 2 is not larger than (An − T )sn 2 . Since Lemma 8 implies that
(An −T )sn 2
approaches 0, we see that the error εn will become arbitrarily small.
sn 2
We have |ϕ| ≤ π/2, so there is a number N , depending only on ϕ, so that the angles
{ϕ, 2ϕ, . . . , N ϕ} cut the circle in N intervals of length ≤ π/2. It follows that for any
x ∈ R/(2π Z), at least one of the angles {x, x + ϕ, x + 2ϕ, . . . , x + N ϕ} must lie in
I1 ∪ I2 .
Now, given any n large enough for
|εn | + · · · + |εn+N | < ε0
123
Control of Cancellations that Restrain...
to hold, choose 0 ≤ r ≤ N so that ωn + r ϕ ∈ I1 ∪ I2 . Then
ωn+r = ωn + r ϕ + (εn + · · · + εn+r )
is in the set I of good angles.
With the above result we are ready to prove Propositions 15 and 16, establishing
(18) and our main result.
Proof of Proposition 15 Consider the sequence {n k } from Lemma 17, truncated in the
beginning so that the last expression in
1
1
1
sn (v) dv = Psn (v) dv + P ⊥ sn (v) dv k
k
k
0
0
0
1
1
⊥
≥ Psn k (v) dv − P sn k (v) dv 0
(22)
0
is positive. Let us evaluate both terms. By Lemma 17,
1
Psn (v) dv ≥ K Psn 1 ≥ K W Psn 2 ,
k
k
k
0
where 0 < W ≤ 1 is a lower bound on the quotient of the 1-norm and weighted
u
2-norm of xy sin(π u + φ) (for an arbitrary phase shift φ). On the other hand, the
triangle and Cauchy–Schwarz inequalities give
1
P ⊥ sn (v) dv ≤ P ⊥ sn 2 .
k
k
0
If n k is sufficiently large, then tan θn k < K W/2 ≤ 1/2, where W is as before, and we
get
|P ⊥ sn k 2 = tan θn k Psn k 2 ≤
KW
2 Psn k 2 .
Plugging these estimates back into (22) gives
1
sn (v) dv ≥ K W Psn 2 −
k
k
KW
KW
KW
2 Psn k 2 = 2 Psn k 2 = 2
cos θn k sn k 2 .
0
But K 2W cos θn k has a positive lower bound W > 0 since tan θn k < 1/2, thus
proving the result.
123
M. Aspenberg, R. Pérez
Proof of Proposition 16 For large enough n, the right-hand side of
sn+1 2 = T sn + (An − T )sn 2 ≥ T sn 2 − (An − T )sn 2
is positive. In fact, the two terms on the right have the bounds
T sn 2 ≥ P T sn 2 = T Psn 2 = λ cos θn sn 2
and
(An − T )sn 2 ≤
G
√
s n n 2
by Lemma 8 , so we arrive at
sn+1 2 ≥ λ cos θn −
G
√
n
sn 2 .
Since θn → 0, the result follows.
7 Proof of Lemma 11
The inequality sn 1 ≤ sn ∞ is trivial, but the opposite direction requires estimates
based on the particular shape of the sequences Sn . Recall that sn is defined as a step
function using the rescaling
sn
j+1/2
n−1
=
Sn ( j)
.
(n − 2)!
For technical reasons we will work instead with a continuous piecewise linear function
sn that approximates sn . First, let
2(n−1)−1
2(n−1) ,
n :=
1
2(n−1)
, rn := 1 − n =
and define sn at the points
2 j−1
2(n−1)
( j = 1, . . . , n − 1) by
sn
2 j−1
2(n−1)
=
Sn ( j)
.
(n − 2)!
The function sn is defined in [n , rn ] by linear interpolation. Note that the interval
[n , rn ] covers almost all of [0, 1], but not quite. In particular, while the interval of
sn is [n , rn ].
integration used to compute sn 1 is [0, 1], the corresponding interval for
However, even if the 1-norms are not necessarily the same, we do have
rn
rn
sn =
n
123
sn .
n
Control of Cancellations that Restrain...
Lemma 11 There exists some N > 0 (only depending on x) such that if n ≥ N then
sn 1 ≥ C
sn ∞ .
This implies Lemma 11 because the 1-norms of sn and sn differ by at most
> 0. Indeed, the integrals of |s | and |
C
sn | in any
C sn ∞/n for some constant
n
2 j−1
2 j+1
interval 2(n−1) , 2(n−1) coincide, except at the left and right endpoints, and around
the (unique) sign-change. Since the discrepancy between integrals in each of these
sn follows. The
three intervals is of the order of n1 , the comparison between sn and sn ∞ , gives
above, together with Lemma 11 , and the fact that sn ∞ = sn 1 ≥ sn 1 − C sn ∞
sn ∞
≥ C
sn ∞ − C = sn ∞ C −
n
n
C
n
,
establishing Lemma 11.
To prove Lemma 11 we will move back and forth between the language of
sn . All statements about the shape of
sequences Sn and the language of functions sn .
the sequence Sn can be interpreted as applying to the function Definition Let the sign change of Sn be located at Z , the extreme at E, and the
inflection at I (these are integer indices between 0 and n − 3). To avoid carrying
1
we follow the convention that indices from 1 to n − 1 are represented
factors of n−1
by capital letters, and their counterparts in the interval [0, 1] by the corresponding
I +1/2
lowercase letter. In particular, we let e := E+1/2
n−1 , i := n−1 , and we let z correspond
Z +1/2
to the zero of sn near n−1 . If Sn has no sign change, no extreme, or no inflection,
we let z = n , e = n , or i = rn as necessary.
Idea of the proof Our definition of inflection yields a natural notion of intervals
of concavity for sn . We will find a suitable lower bound for the integral of |
sn | within
each of these intervals as illustrated in Fig. 8. These bounds are linear combinations of
sn (e)|, and |
sn (rn )|, and they suffice to obtain a bound of the form sn 1 >
|
sn (n )|, |
const.
sn ∞ because
y
|
sn (rn )|,
– |
sn (n )| = |x|
sn (n )|, |
sn (rn )|},
– property (ShD) implies |
sn (e)| ≥ min{|
sn (e)|, |
sn (rn )|} = sn ∞ (= sn ∞ ).
– max{|
sn (n )|, |
For later reference let us call this the Maximum-is-Equivalent-to-Extreme Principle,
or meep for short.
Note The proof of Lemma 18 below uses property (ShE) in an essential way. This
assumes that Sn does have an extreme and an inflection. Sequences Sn with no extreme
or no inflection are taken care of before Lemma 18, in the observation preceding
inequality (23).
Basic assumption If there is an extreme, we will assume that it is a local minimum
(a local maximum is treated in an analogous manner). If there is also an inflection,
we assume that E < I . This yields the three cases of Fig. 8, and elicits Lemma 18
123
M. Aspenberg, R. Pérez
eiz
ezi
zei
Fig. 8 These continuous functions are caricatures of the piecewise linear function sn in the different
situations of cases 1, 2, 3. The 1-norm is bounded from below by the shaded areas limited by the linear
functions n . We show that these areas are comparable to the maximum sn ∞
in the form presented here, which discusses inflections near the far left (this uses the
inequalities near r = 0 in (ShE)). The situation E ≥ I involves the mirror image of
Fig. 8 and a discussion of inflections near the far right (this uses the inequalities near
r = n −2 in (ShE)). It is entirely analogous to the discussion for E < I . (The situation
E = I is included in the other case simply because of how we defined the location of
inflections and extremes.)
Note from the basic assumption on Sn that Sn−1 has a down-change at E and an
extreme at I . In particular, Sn−1 is negative to the right of E (so the extreme at I is a local
minimum by (ShB)), and it is positive to the left of E, so Sn−1 (0) = xy Sn−1 (n−2) > 0.
Then Sn−1 (1) < Sn−1 (0) by property (ShE), and Sn−1 (1) > 0 because it is at, or to
the left of, the down-change.
Definition For given n and a, b ∈ [0, 1] we denote by n [a, w 1 , b,
w2 ] : R −→ R
the linear function whose graph is the straight line from a, w1 to b, w2 . Also, let
μ(a, b) be the length b − a of the interval [a, b].
The function sn can adopt one of three forms depending on the order of e, i, and z
(keep in mind that e = z).
In each case we split [0, 1] into the same three intervals [n , e], [e, i], and [i, rn ],
and describe linear functions on these intervals that bound |
sn | from below (compare
Fig. 8).
Case 1 (e < i ≤ z): The linear functions are
(a) On [n , e]: [n , 0, e,
sn (e)]. The area of the triangle is
1
2
· |
sn (e)| · μ(n , e)
(b) On [e, i]: [e,
sn (e), i, 0]. The area of the triangle is
1
2
123
· |
sn (e)| · μ(e, i)
Control of Cancellations that Restrain...
(c) On [i, rn ]: [z, 0, rn ,
sn (rn )]. The slope is
triangles are
|
sn (rn )|
·
μ(z, rn )
μ(i, z)2
μ(z, rn )2
+
2
2
|
sn (rn )|
μ(z,rn )
≥
so the areas of the two
|
sn (rn )| μ(i, rn )2
·
μ(z, rn )
4
≥
1
4
· |
sn (rn )| · μ(i, rn )
The first inequality above follows from a 2 + b2 ≥ (a + b)2 /2.
Case 2 (e < z < i): The linear functions are
sn (e)]. The area of the triangle is
(a) On [n , e]: [n , 0, e,
· |
sn (e)| · μ(n , e)
1
2
(b) On [e, i]: [e,
sn (e), z, 0]. The slope is
are
|
sn (e)|
·
μ(e, z)
|
sn (e)|
μ(e,z) .
μ(e, z)2
μ(z, i)2
+
2
2
≥
The areas of the two triangles
|
sn (e)| μ(e, i)2
·
μ(e, z)
4
≥
1
4
· |
sn (e)| · μ(e, i)
The first inequality above follows as in Case 1(c).
sn (rn )]. The area of the triangle is
(c) On [i, rn ]: [i, 0, rn ,
1
2
· |
sn (rn )| · μ(i, rn )
Case 3 (z < e < i): The linear functions are
|
sn (e)|
sn (e)]. The slope is μ(z,e)
so the areas of the two triangles
(a) On [n , e]: [z, 0, e,
are
|
sn (e)|
·
μ(z, e)
μ(n , z)2
μ(z, e)2
+
2
2
≥
|
sn (e)| μ(n , e)2
·
μ(z, e)
4
≥
1
4
· |
sn (e)| · μ(n , e)
(b) On [e, i]: [e,
sn (e), i, 0]. The area of the triangle is
1
2
· |
sn (e)| · μ(e, i)
n−2+1/2
(c) On [i, rn ]: [a,
sn (a), b,
sn (b)], where a := n−3+1/2
n−1 , b :=
n−1 .
The region bounded by this consists of a rectangle of base μ(i,
r
n
) and
n−1
b , where
·
sn−1 '
height sn (rn ), plus a triangle of base μ(i, rn ) and slope n−2
123
M. Aspenberg, R. Pérez
'
b :=
n−3+1/2
n−2 .
The area of sn is at least
|
sn (rn )| · μ(i, rn ) +
n−1
n−2
sn−1 '
b ·
· μ(i,rn )2
2
In each of the three cases, sn 1 is bounded from below by a sum of three expressions. A trivial weakening of these expressions allows us to consolidate cases Case 1
and 2 into one inequality:
sn (e)| · μ(n , i) + |
1&2 : If (e < z), then sn 1 ≥ 41 |
sn (rn )| · μ(i, rn ) .
n−1
sn (e)| · μ(n , i) + |
3 : If (z < e), then sn 1 ≥ 41 |
sn (rn )| · μ(i, rn ) + n−2
·
sn−1 '
b · μ(i, rn )2 .
Observation Cases 1 and 2 allow the possibility that Sn has no extreme. In that
sn (n )| and |
sn (rn )|. Here,
situation e = n , and the bound for both cases depends on |
the conclusion of Lemma 11 follows directly from the meep (Idea of the proof). For
the rest of the proof we can assume now that Sn has an extreme.
sn 1 ≥
If Sn has no inflection then i = rn and the consolidated estimates give 1
follows as well. For the rest of the proof
|
s
(e)|.
Then
the
conclusion
of
Lemma
11
n
4
we can assume now that Sn has an inflection. In particular, we are at freedom to use
property (ShE) from now on.
, then we can neglect the
If the interval [n , i] has definite size, say i ≥ min{|x|,y}
6
portion containing μ(i, rn ) in both bounds, and simplify all three cases to
sn 1 ≥ 14 μ(n , i)|
sn (e)| ≥
1 min{|x|,y}
|
sn (e)|,
4
12
(23)
1
since μ(n , i) ≥ min{|x|,y}
− 2(n−1)
≥ min{|x|,y}
, when n is large enough. Equation (23)
6
12
yields the conclusion of Lemma 11 in this situation, again because of the meep.
To finish the proof we have to consider what happens when μ(n , i) is small so
. In this situation we neglect the portion of the bounds 1&2 and 3
that i < min{|x|,y}
6
n−1 sn−1 '
b satisfies one of
that contains μ(n , i), and show that either |
sn (rn )| or n−2
·
the lower bounds (24), (27), (28), all of which have the form constant times |
sn (e)|.
sn ∞ , and we will be
Just as above, this implies a lower bound for sn 1 in terms of done.
As a note of caution, we remark that for this final step we revert to the language
n−1
b in
sn (rn ) and n−2
·
sn−1 '
of sequences Sn . Thus, instead of seeking bounds for terms of sn (e), we get equivalent bounds (with normalizing constants) for Sn (n − 2)
and Sn−1 (n − 3) in terms of Sn (E). Note here that the value μ(i, rn ) is rescaled to
(n − 2) − I .
Lemma 18 below uncovers the key relationship between Sn and Sn−1 . The proof is
delayed till the end of this section so as not to interfere with the proof of Lemma 7.
)
yn *
1
Lemma 18 If I < min |x|n
6 , 6 then |Sn−1 (1)| ≤ 2 |Sn (0)|.
123
Control of Cancellations that Restrain...
Recall from (10) that Sn (1) = Sn (0) − Sn−1 (1), so under the conclusion of
Lemma 18 we get |Sn (0)| ≥ 23 |Sn (1)|. Hence, when |Sn (1)| ≥ |Sn 2(E)| we get
|Sn (n − 2)| =
y
2 y
y
|Sn (0)| ≥
|Sn (1)| ≥
|Sn (E)|.
|x|
3 |x|
3|x|
(24)
It remains only to consider what happens when
|Sn (1)| ≤
|Sn (E)|
.
2
(25)
We will give two estimates for this situation. The first (inequality (27)) holds for
cases 1&2. The second (inequality (28)) holds for case 3.
Case 1 &2: Since there is no change in convexity on (I, n − 2), we have
|Sn (n − 2) − Sn (I )|
≥ |Sn−1 (n − 2)|.
(n − 2) − I
Since the extreme lies lower than the inflection (Sn (E) ≤ Sn (I )), and Sn (E) < 0 <
Sn (n − 2), we can replace the numerator on the left with
y
y
|Sn (n − 2)| + |Sn (E)|
≥ |Sn−1 (n − 2)| =
|Sn−1 (0)| ≥
|Sn−1 (1)|,
(n − 2) − I
|x|
|x|
where the last inequality is due to property (ShE). Again, by constant convexity on
(1, E) (recall E < I ), the above is larger than
y |Sn (E) − Sn (1)|
,
|x|
E
and we have
y (n − 2) − I
y (n − 2) − I
|Sn (E) − Sn (1)| ≥
|Sn (E)|
|x|
E
|x|
2E
(26)
by (25).
To
estimate
the
factor
next
to
|S
(E)|
we
use
n
≥
6
and
the
fact
that
E
<
I <
n
)
*
min |x|n/6, yn/6 . For ease of notation, let m := min{|x|, y}, so E, I < mn/6.
|Sn (n − 2)| + |Sn (E)| ≥
y (1 − 2/n − m/6)
y 4−m
4−m
y (n − 2) − I
≥
≥
≥
|x|
2E
|x|
2m/6
|x| 2m
2|x|
3−x
since y ≥ m. The last expression above equals −2x
when x ∈ (−1, −1/2], and
4+x
equals −2x when x ∈ [−1/2, 0). Both functions are continuous and increasing for all
x < 0, so substituting x = −1 into the first and x = −1/2 into the second shows that
4−m
2|x| > 2. Substituting this back into (26) gives
|Sn (n − 2)| ≥ 2|Sn (E)| − |Sn (E)| = |Sn (E)|.
(27)
123
M. Aspenberg, R. Pérez
Case 3: Here, property (ShE) gives a different bound:
|Sn−1 (n − 3)| =
y
y
y |Sn (E)|
|Sn−1 (0)| ≥
|Sn−1 (1)| ≥
,
|x|
|x|
|x| E − Z
and this concludes the proof of Lemma 11 .
(28)
Proof of Lemma 18 Throughout this section the basic assumption has been that
Sn (E) < 0 and E < I , so that Sn−1 has a local minimum at I and a down-change at E.
Recall that this implies Sn−1 is negative to the right of E, and 0 < Sn−1 (1) < Sn−1 (0).
Moreover, suppose that |x| ≤ y so that the hypothesis of Lemma 18 reads I < |x|n/6
(the complementary case is similar). This inequality, together with E < I , gives the
straightforward
y E + x(n − 1 − I ) < y I + x(n − 1 − I ) = I + x(n − 1) < I +
xn
< 0,
6
(29)
which we will use to derive inequality (31). Since y − x = 1 and n/2 ≤ n − 1,
I <
−xn
−x(n/2)
−x(n − 1)
=
≤
,
6
2+y−x
2+y−x
so
−x(n − 1 − I )
≥ 1.
(2 + y)I
The left expression increases if I is replaced by E in the denominator, giving
− 2E ≥ y E + x(n − 1 − I ).
(30)
As stated in (29), the right expression in (30) is negative, so after multiplying by y,
we get
y
−y E
≤ .
(31)
0<
y E + x(n − 1 − I )
2
Using inequality (31) we proceed to prove Lemma 18 as follows. The sequence
Sn−1 is positive and decreasing from 1 to E, so the average in this span is smaller than
the leftmost term Sn−1 (1). Similarly, Sn−1 is negative and increasing from I to n − 2,
so the average of |Sn−1 | in this span exceeds the rightmost value |Sn−1 (n − 2)|. Thus,
E
j=1
Sn−1 ( j) < E · Sn−1 (1) < E · Sn−1 (0)
n−2
y
yE Sn−1 ( j) ≤
= E |x| |Sn−1 (n − 2)| < |x|'
n j=I
123
n−2
yE Sn−1 ( j) , (32)
|x|'
n j=E+1
Control of Cancellations that Restrain...
yE
where '
n := n − 1 − I . Notice that (29) implies 0 < |x|'
n < 1, so the sum of positive
terms (those to the left of E) is less in absolute value than the sum of negative terms
(those to the right of E):
n−2
E
−
S
(
j)
Sn−1 ( j)
n−1
j=E+1
j=1
n−2
yE ≥ 1 − |x|'
Sn−1 ( j) .
n j=E+1
n−2
Sn−1 ( j) =
j=1
(33)
Now, using (32), (33), and (31),
E
|Sn−1 (1)| < Sn−1 ( j) <
j=1
n−2
−y E = y E−|x|'
Sn−1 ( j) ≤
n j=1
n−2
n−2
yE
|x|'
n
yE <
S
(
j)
S
(
j)
n−1
n−1
|x|'
n yE
j=E+1
1 − |x|'
j=1
n
n−2
y Sn−1 ( j) = 21 |Sn (0)|.
2 j=1
Acknowledgments We wish to give thanks to F. Przytycki and the Polish Academy of Sciences for their
hospitality at the conference center in Be˛dlewo, Poland; to the Institut Mittag-Leffler for its hospitality
during the final phase of this project; to M. Benedicks for conversations that led to the project this work
stems from; and to P. Bleher for suggesting the use of integral operator theory as a tool for estimating
asymptotic growth. We are also highly indebted to the anonymous referee for a very thorough reading
of the manuscript, and in particular for calling our attention to a problem with the original statement of
Proposition 7. The referee spotted many places where mathematical, typographical, and stylistic corrections
enhanced considerably the final version. We are appreciative of the fact that someone took the time to read
our work with such thoroughness. Rodrigo Pérez was supported by NSF grant DMS-0701557.
References
1. Aspenberg, M., Pérez, R.: Catalan, Siegel, and Fibonacci. Work in progress.
2. Conway, John B.: A Course in functional analysis. Graduate texts in mathematics, vol. 96, 2nd edn.
Springer, New York (1990)
3. Harary, Frank, Hayes, John P., Wu, Horng-Jyh: A survey of the theory of hypercube graphs. Comput.
Math. Appl. 15(4), 277–289 (1988)
4. Kato, T.: Perturbation theory for linear operators. Classics in mathematics. Springer, Berlin (1995)
5. Misiurewicz, M., Nitecki, Z.: Combinatorial patterns for maps of the interval. Mem. Am. Math. Soc.
94(456), vi+112 (1991)
6. Pérez, Rodrigo A.: A brief but historic article of Siegel. Not. Am. Math. Soc. 58, 558–566 (2011)
7. Siegel, C.L.: Iteration of analytic functions. Ann. Math. 2(43), 607–612 (1942)
8. Sloane, N. J. A.: The on-line encyclopedia of integer sequences, 2010. Published electronically at www.
research.att.com/~njas/sequences/. Accessed 10 Aug 2012
9. Stanley, R.P.: Enumerative combinatorics. Cambridge studies in advanced mathematics, vol. 2. Cambridge University Press, Cambridge (1999)
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