TAKING THE CONVOLUTED OUT OF BERNOULLI CONVOLUTIONS: A DISCRETE APPROACH

In this paper we consider a discrete version of the Bernoulli convolution problem traditionally studied via functional analysis. We discuss several innovative algorithms for computing the sequences with this new approach. In particular, these algorithms assist us in gathering data regarding the maximum values. By looking at a family of associated polynomials, we gain insight on the local behavior of the sequence itself. This work was completed as part of the Clemson University REU, an NSF funded program 1 .

(2) Consider the finite sequences recursively given by B 0 = (1) and B n+1 = dup n (B n ) + shf n (B n ).B n is called the n th level Bernoulli sequence.The names "dup" and "shf" reference the duplication and shifting of the coordinates.
In this paper, we are primarily interested in the rate at which the maximum value of B n is growing with n.We develop two independent algorithms to do this.The first gives a recursive method for computing the entire sequence B n when n is small.By encoding the sequence as coefficients of a polynomial, the second gives a method for computing specified entries of B n when n is larger.In addition, we provide numerical data concerning B n .
1.1.Motivation.Classically, Bernoulli convolutions have been studied as a problem in functional analysis.A Bernoulli convolution is obtained as an infinite convolution of Bernoulli measures [1].The Bernoulli measure, denoted by b(X), is the measure corresponding to the discrete probability density function on the real line with value 1/2 at 1 and −1.The Bernoulli convolution with parameter q for 0 < q < 1 is the measure µ q (X) = b(X) * b(X/q) * b(X/q 2 ) * ... .This measure was first studied by Jessen and Wintner [4].They showed that µ q is continuous for any q.
A different perspective on Bernoulli convolutions is obtained through a functional equation.For 0 < q < 1, consider the functional equation Date: March 9, 2010.This research was supported by NSF grant DMS-0552799.
for t on the interval I q := [−1/(1 − q), 1/(1 − q)].It can be shown that there is a unique bounded solution F q (t) to the above equation.Moreover, F q (t) is the distribution function of µ q , that is F q (t) = µ q ((−∞, t]).For an introduction to this, refer to Chapter 5 in Experimental Mathematics in Action [1].For a more in depth analysis, refer to Sixty years of Bernoulli convolutions [6].
Jessen and Winter, in [4], showed that F q (t) is either absolutely continuous or purely singular.The major question regarding the solutions of ( 3) is to determine the values of q which make F q (t) absolutely continuous.If F q (t) is absolutely continuous, rather than considering the function F q (t), one may consider its derivative f q (t) := F q (t).Upon differentiating, the functional equation for F q (t) yields the following equation: The existence of an absolutely continuous solution F q (t) to ( 3) is equivalent to the the existence of an L 1 (I q ) solution f q (t) to (4).When 0 < q < 1/2, Kershner and Wintner [5] proved that F q (t) is always singular.For these values of q, the solution F q (t) is an example of a Cantor function, a function that is constant almost everywhere.It can be shown that if q = 1/2, then the solution F q (t) is absolutely continuous.
The case when q > 1/2, however, is significantly harder and more interesting.In 1939, Erdős [3] showed that F q (t) is again singular for q of the form q = 1/θ with θ a Pisot number.There is little else that is known for other values of q > 1/2.One interesting result due to Solomyak [7] is that almost every q > 1/2 yields a solution F q (t) that is absolutely continuous.Hence it is surprising that no actual example of such a q is known.Specifically, the case when q = 2/3 remains a mystery.
In [1], Girgensohn asks the question of computing f q (t) for various values of q.The author starts an arbitrary initial function f 0 (t) ∈ L 1 (I q ) and iterates the transform to gain a sequence of functions f 0 , f 1 , f 2 , ....He shows that if this sequence of functions converges to a bounded function, then it converges to the unique solution of (4).
Calkin [2] specifically looked at the above process for q = 2/3.Rather than working on the interval I q , we shift the entire interval to [0, 1] for simplicity.The transform T q now becomes the map T : Geometrically, this transform (6) takes two scaled copies of f (x): one on the interval [0, 2/3] and the other on [1/3, 1], and adds them.The scaling factor of 3/4 gives us that In other words, the average value of T f (x) equals that of f (x).In this setting, the question now reads: starting with the constant function f 0 (x) = 1, does the iteration determined by the transform in (6) converge to a bounded function?
1.2.Discrete version.Rather than viewing T as a transform on [0, 1], we consider the discrete analogue mentioned earlier.The sequences dup(B n ) and shf(B n ) defined in (1) and ( 2) are analogous to the two shifted copies of the function on [0, 1] in (6).We start with the sequence B 0 = (1).We recursively generate sequences given by We are interested in the global behavior of B n .In particular, we investigate the growth rate of the maximum element of B n as n grows large.Before we discuss our results, we present some notations that are useful throughout this paper.We call B n the Bernoulli sequence on level n.We refer to the map (dup n + shf n ) : R n −→ R 3n seen in ( 7) as the process of duplicate, shift, add or DSA for short.We usually write the n th level Bernoulli sequence as B n = (b 0 , b 1 , ..., b t ) where t = 3 n − 1.The fact that B n has a total of 3 n terms follows directly from the definition of dup n and shf n in ( 1) and ( 2).We define m n := max(B n ). 1 1 2 2 3 3 2 3 4 4 3 4 3 4 4 3 2 3 3 2 2 1 1 1 1 Figure 1.This shows the process of DSA at three low levels.The figure demonstrates computation of the Bernoulli sequence on level n + 1 from the Bernoulli sequence on level n for n = 0, 1 and 2 using the definition.
It is immediate from the definition that the maximum value of B n must occur on the middle third of the Bernoulli sequence.Furthermore, because the sequence is palindromic, the maximum must occur on the first half of the middle third.
The following observation can be proved using induction on n and the definition of B n in (7).
Observation 1.1.The mean of the elements of B n is (4/3) n .
The major question that we considered is whether m n also grows like (4/3) n .A positive result here is equivalent to the existence of a bounded solution to the functional equation ( 4) for q = 2/3.In turn, this is equivalent to the solution F q (t) of (3) being absolutely continuous for q = 2/3.Beginning with a computational approach, we address this and other related questions throughout this paper.

Recursive attempts at computing the Bernoulli sequence
2.1.The naive method: DSA.The process of duplicate, shift, add gives a naive method for computing Bernoulli sequences.Despite the simplicity in describing this process, DSA is not computationally feasible for large values of n.The primary shortfall of the DSA method is that each Bernoulli level B n has 3 n terms-each successive computation takes roughly three times as long as the previous.Likewise, at each successive level we must keep track of three times as many entries as on the previous level.Using the DSA method without modifications, we have been able to compute B n for n = 1, 2, ..., 20.However, the fact that one must know all 3 n entries of level n to compute level n + 1 limits the practicality of this algorithm.
Figure 2.This shows some of the low level Bernoulli sequences generated using DSA.In the above plots, the horizontal axis gives the index and the vertical axis gives the entry corresponding to that index in the Bernoulli sequence.Moving clockwise from the top left, we see B n for n = 5, 7, 11 and 9.
2.2.The improvement: DEM.We now consider an alternate approach that addresses some of the issues arising from the DSA algorithm.The process we call double, enlarge, merge, abbreviated DEM, is a way of encoding the Bernoulli sequence B n as a sequence of length 2(2 n − 1).The advantage with DEM is that this auxiliary sequence is in the order of 2 n as opposed to the 3 n size increase required for the DSA process.The DEM algorithm is based on the observation that in a given Bernoulli sequence, many individual entries are consecutively repeated.Rather than keeping consecutive repeats, we only keep the index where the Bernoulli sequence either increases or decreases.For this encoding to be meaningful, it is important to note that when comparing two consecutive entries in a Bernoulli sequence, the difference between these entries will never be greater than one in absolute value.This can be proved inductively from the definition in (7).
Given the Bernoulli sequence B n = (b 0 , b 1 , ..., b t ) with t = 3 n − 1, the DEM representation is (d 1 , ..., d r ) where d i is the index of the i th jump in B n up to a sign (by jump, we mean two consecutive entries which are not equal).Suppose the i th jump occurs at index j in B n , so b j and b j+1 are different.Then For example, the DEM representation of B 2 = (1, 1, 2, 3, 2, 3, 2, 1, 1) is (2, 3, −4, 5, −6, −7).
We now describe how to translate the DSA method to this new representation.The process of duplicate becomes double: each element from the original list is multiplied by two.The process of shift becomes enlarge: each element from the doubled list is modified by 3 n as follows: if the element is positive, we add 3 n , for negative elements we subtract 3 n .The process of add translates to merge: we discard the original sequence and consider the new lists attained in the double and enlarge processes.We then add two additional elements 3 n and −2(3 n ) to these lists, respectively.Finally, we merge sort the elements of the two lists according to their absolute value to obtain the DEM representation of the next Bernoulli level.
2.3.Data.Using the DEM method, we are able to compute the Bernoulli sequence B n up to level n = 27.For each of these levels, the maximum value is of particular interest.Typically, the maximum is attained many times on a given level.Although we found no clear pattern regarding the location of the maximum values, we do see obvious patterns in the convergence of the maximums themselves.Below we provide a table detailing the data we collected using the DSA and DEM algorithms.
Level  1: The maximums m n appear to grow like (4/3) n .The fact that m n (3/4) n seems to be converging, albeit slowly, supports this claim.We describe how these polynomials behave under the process of DSA.We see that the duplication b 0 , b 0 , b 1 , b 1 , ..., b r , b r corresponds to the polynomial (1 + x)p n (x 2 ).Shifting the sequence 3 n places to the right corresponds to multiplication by x 3 n .By adding the duplicate and the shift of the sequence, we arrive at the sequence on the next level.This yields the recursive relation with P 0 (x) = 1.

3.2.
Explicit formula for p n (x).This formula (8) allows us to explicitly solve for p n (x).
Theorem 3.1.For n ≥ 1, the polynomials p n (x) satisfy Then α j is the number of ways that j can be written as a sum of distinct elements from S [8].This idea is applicable to the coefficients of our polynomial because our polynomial is a product of terms of the form (1 + x a ).Applying this idea to our situation, we find that the coefficient b j of x j in p n (x) (which is the j th entry on the n th Bernoulli sequence) is precisely the number of ways that j can be written as a sum of distinct terms in the sequence The values in the sequence S are just the exponents of the factors of p n (x) in (9).We now outline an algorithm that can be used to calculate the entry b j for a fixed level n.Let S = {a 1 , ..., a n } where each a i > 0. Let N S (k) denote the number of ways k can be written as a sum of elements from S. Our entire algorithm is based on the following observations: • For any i ∈ {1, ..., n}, we have • If k < 0, then N S (k) = 0.These three relations provide an algorithm for computing the j th entry on a given Bernoulli level.However, our sequence S in (10) takes on a particularly nice form-taking advantage of this we can increase the efficiency of our algorithm.Let S be as in (10).Suppose 0 < k < 2 n−1 .Any element of S containing a positive power of 3 is strictly larger than k itself, so these terms become useless when writing k as a sum of elements in S. Hence, we are left with the distinct powers of 2 in S.But k can now be written uniquely; this is simply the binary expansion of k.So N S (k) = 1 in this case.
Based on the above ideas, we implement a tree branching algorithm to compute N S (k) for various values of k and n.See Figure 3 for a specific example of the PIP algorithm.
3.5.Data.Our PIP algorithm is used to compute isolated points on B n for a given n.It is infeasible to compute entire levels using PIP, so we are not able to calculate the global maximum of a given level with this method.Instead, PIP provides us with local information on the Bernoulli sequence.In particular, the PIP algorithm gives evidence that the Bernoulli sequence B n is converging uniformly as n grows large.Furthermore, we can gather local data on much higher levels than we were able to with DSA or DEM.
We now describe one way in which PIP can be used to gather local information on the Bernoulli sequence.Fix α in [0, 1].Consider the index k = α(3 n − 1) .We now use PIP to compute the entry b k at index k for levels n = 1, 2, ..., 50 (we have used PIP to compute individual entries on levels as high as n = 70).Finally, we take the quotient of b k by (4/3) n .This data is presented in Table 2  The boxes containing the word "cutoff" account for the fact that if 0 < k < 2 n−1 , then there is a unique way to write k as a sum of elements in S (the binary expansion of k).As the diagram suggests, N S (k) = 3, corresponding to the fact that there are three boxes containing the word answer++.Hence the 12 th entry on the 3 rd Bernoulli level is 3.

Conclusion
Studying Bernoulli convolutions through a discrete lens sheds much new insight on the subject.Many of our algorithms would not have been discovered without combinatorial thinking-for example, the PIP algorithm is made possible by the fact that the coefficients of a polynomial can be thought of as counting the number of representations of integers as sums from a certain sequence.The discrete point of view is a very simple way to think about Bernoulli convolutions (the duplicate, shift, add method could be explained to a small child), but a computer has trouble computing more than a handful of Bernoulli sequences.In particular, studying Bernoulli convolutions via combinatorics has led to the discovery and development of two elegant algorithms (DEM and PIP).Using these algorithms we were able to generate the entire Bernoulli sequences at many new levels (up to 27) and also were also able to calculate individual entries of B n at levels as high as 70.
We conjecture that m n = O((4/3) n ), or equivalently that F 2/3 (t) is absolutely continuous.Our two algorithms provide much data to support this claim.

3 .
A Polynomial Approach to DSA 3.1.Translating DSA into a polynomial recursion.By encoding these sequences as coefficients of polynomials, the process of duplicate, shift, add gives a particularly nice recursive relation among the polynomials.Let B n = (b 0 , b 1 , ..., b t ) be the Bernoulli sequence on level n where t = 3 n − 1.Consider the polynomial p n (x) := b 0 + b 1 x + ... + b t x t .

Figure 3 .
Figure 3.The above flowchart illustrates the PIP algorithm.It shows the computation of N S (k) for k = 12 and S = {1, 2, 4, 4, 6, 9}.The boxes containing the word "cutoff" account for the fact that if 0 < k < 2 n−1 , then there is a unique way to write k as a sum of elements in S (the binary expansion of k).As the diagram suggests, N S (k) = 3, corresponding to the fact that there are three boxes containing the word answer++.Hence the 12 th entry on the 3 rd Bernoulli level is 3.

Table 2 :
. We consider various values for α (appearing in the top row), and compute associated entry of the Bernoulli sequence.We chose these specific α values because the only portion of the sequence of interest is the first half of the middle third.