{'fraction_non_alphanumeric': 0.06046112742025926, 'fraction_numerical': 0.019658937229936998, 'mean_word_length': 3.597816196542311, 'pattern_counts': {'":': 0, '<': 7, '<?xml version=': 0, '>': 10, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 18, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Moment sequences and difference equations

We recall the definition and the properties of a moment sequence and recall that all real sequences that have a finite rank of its Hankel matrix (see definition in the sequel) satisfy a homogeneous linear equation with constant coefficients. Then we analyze cases when a difference equation with constant coefficients and suitably chosen initial conditions and having as an input a positive moment sequence has a solution that is a positive moment sequence. We give one general simple result and give many examples illustrating the theory. The main simple result states that the roots of the odd multiplicity of the characteristic equation must lie outside the support of the measure that produces the moment sequence that is in the input and the initial conditions suitably chosen.


Introduction
The purpose of this note is to examine the situation when there exists an intersection of a certain important class of real sequences and homogeneous difference equations with constant coefficients and also the cases when the difference equation 'excited' by a positive moment (pm) sequence produces a positive moment sequence. It will turn out that firstly, all sequences having a finite rank of their so-called Hankel matrices satisfy certain homogeneous difference equation with constant coefficients and secondly, that not every pm sequence produces in this way a pm sequence. The answer depends on the fact that the roots of the characteristic equation lie in the support of the measure that produces the input pm sequence and also on the fact that the initial conditions are suitably chosen. In this way, one can use the properties of the solutions of the output of the difference equation with given pm input to test if the roots of the characteristic equation lie in the support of the measure that produced the pm input sequence.
Another purpose of this note is to provide a probabilistic interpretation of the results presented in the excellent paper of Bennett [4] and thus present their sometimes much simpler proofs. We also aim to generalize some of these results and also provide a connection between moment sequences and difference equations.
The paper is organized in the following way. In the next section, we recall the definition of the moment sequence and formulate sufficient conditions for a moment sequence to identify measures that generated this moment sequence. Then we analyze the set of positive moment (pm) sequences. Give several examples and formulate a series of operations on pm sequences that result in another pm sequence. The following two sections present results concerning the question of when a difference equation with constant coefficients with the input being a pm sequence produces a pm sequence.

Moment sequences
Let us agree that if the area of integration is not given explicitly, then we assume that it is equal to the support of the measure with respect to which the integral is calculated. Let us start with the following definition of the moment sequence.
Let µ be a certain measure on the real line. Let µ + and µ − be the elements of the Lebesgue decomposition of µ, i.e., that µ + and µ − are two positive measures such that µ(B) = µ + (B) − µ − (B) for every µ-measurable subset B of the real line. Let us denote |µ| the sum of the two positive measures µ + and µ − . Let us consider a (signed in general) measure µ such that: Then a sequence {m n } n≥0 of reals defined by: is called a moment sequence of the measure µ. Boas theorem (see [2]) states that every sequence is a moment sequence of some signed measure satisfying (2.1). In fact, Boas's theorem asserts some more precise knowledge of the signed measure µ. Namely, we have: Hence, for any real sequence s := {s j } j≥0 we can define the measure µ that generates this sequence according to (2.2). Any measure generating sequence s will be called generating measure of this sequence, briefly gm(s).
Further Let us note, that there can be many real sequences having the same Hankel transform. The in-depth analysis of the relationship between a real sequence and its Hankel transform was done e.g. in [17](chapter 5). For our purposes it will be enough, to notice that following the Theorem of Kronecker for a sequence {a n } n≥0 with the sequence of mlpm {H n } n≥0 and Hankel transform {D n } n≥0 to satisfy condition it is necessary and sufficient that D r−1 = 0 and ∀n ≥ r : D n = 0. Sequences s satisfying condition (2.3) will be called of finite rank briefly fRsequences.
Remark 1. Notice that if the support of measure µ X is finite say consisting of , then each moment m n has the following form : and thus the sequence of mlpm has the following form Now notice, that each matrix of the form [α i+j k ] i,j=0,...,n has rank 1, hence for every n ≥ m matrix [EX i+j ] i,j=0,...,n has at most rank m. Thus the sequences of such, finitely supported measures, have their Hankel transforms consisting of zeros for elements with indices greater than m. In other words, we have shown that all moment sequences of finitely supported measure are fR-sequences. Proposition 1. Let s be a real sequence. Assume that its gm(s) is finitely supported and the cardinality of the support is r, then there exist r constants b 0 , . . . , b r−1 such that for m ≥ 0 : In other words all elements of sequence s satisfy a homogeneous linear equation with constant coefficients (see the definition and properties below in Section 3).
Proof. Assume that the support of gm(s), that we will denote by µ, consists of points x 1 , . . . , x r ∈ R . Let coefficients b i , i = 0, . . . , r − 1 be defined by the relationship.
is the zero measure. Hence, in particular, we have ∀m ≥ 0: x m v(dx) = 0. But this means that equation (2.4) is satisfied by the sequence s.
The converse statement also holds. Namely, we have the following theorem that have been formulated and proved in [17](Chapter 5 Thm. 2.3). This theorem can be traced back to Kronecker (from 1881)..

Theorem 2.
Rank of Hankel transform of a real sequence s is finite iff its generating measure µ(dx) satisfies the following condition: There exist a polynomial P (x) such that the measure v(dx) = P (x)µ(dx) satisfies the following condition: ∀m ≥ 0 : More precisely, let the rank in question be r, then there exist r numbers b 0 , . . . , b r−1 such that for m ≥ 0 : In other words there exists a homogeneous difference equation of order r such that all elements of s with indices grater that r−2 are defined by this equation with initial conditions s 0 , . . . , s r−1 . Moreover, coefficients {b i } r−1 i=0 are defined by the elements s r , . . . , s 2r−1 of the sequence s. In other words the sequence is uniquely defined by its first 2r values. Besides, denoting P (x) = x r − r−1 j=0 b j x j and v(dx) = P (x) we see that the condition (2.5) is satisfied.

Remark 2.
Notice that if we assume that gm(s) is identifiable by moments, then the condition (2.5) means that the measure P (x)µ(dx) is the zero measure and iff polynomial P (x) has all different and real roots, then from Proposition 1 it follows that the measure is finitely supported and their support consists of the root of the polynomial P .
From now on we will be interested in the cases when measure µ is a positive measure, that is µ − = 0. The exhaustive study of conditions guaranteeing identifiability by moments of the positive measure was done recently in [9].
Let supp µ denote its support and then let us define a sequence of numbers defined in the following way: for n ≥ 0. We will call such sequence a p(ositive)m(oment) sequence. It is easily seen that by multiplying a pm sequence by a positive number we get another pm sequence. Hence the set of pm sequences forms a cone M in the space of all sequences. To simplify notation we will denote for simplicity integrals over supp µ by simply by . Obviously, we have m 2i ≥ 0, for i = 0, 1, 2 . . .. Using Cauchy-Schwarz inequality, we immediately have: for all 0 ≤ j ≤ n. Remembering that function |x| a is convex for all α > 1 and after applying Jensen's inequality we get for all 0 < α < β Consequently, for the case m 0 = 1, we get for all j ≥ n Additionally, if supp µ ⊂ [0, ∞) then obviously we also have seen that m 1/n n constitutes an non-decreasing sequence.
Sometimes, switching to the so-called random variables is more straightforward and intuitive. Namely, it is known that for every positive such measure µ that µ(dx) = 1 one can define another measurable space (Ω, B, P ) (so-called probability space) and measurable mapping X : Ω → R such that: for every measurable subset B of the real line. By the way, measure µ is called the distribution of the random variable X and we obviously have: for every real, integrable (mod µ) function f on the real line. The distribution of any random variable is called sometimes a probability measure on the real line. Traditionally, the integral over Ω with respect to the probability measure P is simply denoted by E (-expectations). So we will exchangeable use the notation We have important and simple result.
Theorem 3. A real sequence is a pm sequence iff the sequence of its mlpm is positive semi-definite.
Proof. Let X be a random variable with a moment sequence {a n } , i.e., a k = EX k , k ≥ 0. Then for any real sequence {α n } then for any N we have E N j=0 α j X j 2 ≥ 0. But this quantity can be written in the following way: . , X N ) and x T denotes transposition of the vector x. Now, this inequality means that matrices EX N X T N N ≤0 are nonnegative definite, hence their major determinants must be nonnegative.
Remark 3. Let us recall that from the matrix theory it follows that: 1) the symmetric matrix is positive definite iff all its leading principal minors are positive, 2) the symmetric matrix is positive semi-definite iff all its principal minors are non-negative.
Remark 4. Consequently, we can notice that if the sequence of Hankel transforms of a sequence is positive, then the sequence of mlpm is positive definite, consequently, the sequence is a pm sequence.

Remark 5.
On the other hand, we have sequences having non-negative Hankel transforms that are not positive moment sequences. One of them is the following sequence {1, 1, 1, 1, 0, 0, . . . .}. One can see that the Hankel transform of this sequence is {1, 0, 0, 1, 0, . . .} . The sequence {1, 1, 1, 1, 0, 0, . . . .} cannot be a pm sequence since EX 3 = 1 and EX 4 = 0! Recently Berg&Szwarc in [5] have made a contribution that clarifies the case of non-negativity of the sequence of Hankel transform of a sequence s and the existence of a finitely supported positive measure generating s. Namely, they proved the following result: be such a real sequence that its sequence of Hankel transforms {D j } ∞ j=0 satisfy the following condition: ∃r > 0 : D j > 0 for j ≤ r − 1 and D j = 0 for j ≥ r, then there exists a positive measure µ supported on exactly r points such that s is a pm sequence and gm(s) = µ.
Following the paper by Bennett [4], we have yet another way of deciding if a given sequence is a moment sequence. Namely, we have the following criterion (compare [1]).

Definition 2.
Let I be a segment of a real line. A given sequence {m j } j≥0 is a pm sequence on I iff for every polynomial n k=0 c k x k that is nonnegative on I we have n k=0 c k m k ≥ 0. Remark 6. It is well-known that if the cardinality of the support of the measure µ is infinite, then the sequence of Hankel transforms of the moment sequence of the measure µ is strictly positive. Now, having random variables we can simply utter some rules concerning moment sequences. Most of them were formulated in the paper of Bennett [4]. Some of them had quite complicated proofs. Due to probabilistic interpretation, we can substantially simplify these proofs. Assertions 6. 7. and 8. of the Proposition below seem to be unknown to Bennett. Namely, we have: Proposition 2. Let {a n } n≥0 and {b n } n≥0 be two pm sequences. Then, so are the following sequences: 4. If a pm sequence {a n } n≥0 is nonnegative, then also pm is the following 5. Suppose {a n } n≥0 is a pm sequence, then for all k ≥ 1 the following sequences {a 2k+n } n≥0 are the pm sequences.
6. Suppose that for a given pm sequence {a n } n≥0 we have a 0 = 1 and a 2 = a 2 1 , then ∀n ≥ 0 : a n = a n 1 . 7. Suppose that a sequence {b n } n≥0 is a pm sequence and i) suppose further that b 0 = 1, (1 − p). 8. Suppose that a sequence {b n } n≥0 is a pm sequence then for all n ≥ 1 : 2n . Further, if the measure generating sequence {b n } has infinite support, then the sequence {b 2n } is log-concave. In particular the following sequence {b 2n+2 /b 2n } n≥0 is non-decreasing.
Proof. 1. Let us define the following matrices A n = [a i+j ] i,j=0,1,...,n and B n = [b i+j ] i,j=0,1,...,n . Since the sequences {det A n } n≥0 and {det B n } n≥0 are nonnegative by assumption, then so the matrices {A n } and {B n } are nonnegative defined hence so are their convex combinations, that is the sequence {det[αA n + βB n } n≥0 is nonnegative.
2. Obviously, the sequence {E (αθX + β(1 − θ)Y ) n } where independent random variables X and Y and independent also of the random variable θ are such that a n = EX n and b n = EY n for all n ≥ 0. But we have 3. If two independent random variables X and Y are such that EX n = a n and EY n = b n , the E(XY ) n = EX n EY n is also a pm sequence. Notice, that a kn = E(X k ) n . Note, that if X is such a random variable that a n = EX n, then (−1) n a n = E(−X) n . Now, we apply assertion 1 with α = β = 1, b n = (−1) n a n . 4. We apply assertion 1. with α = β = 1, a n = E( |X|) n , b n = (−1) n a n for some random variable X. Now the sequence {a n + b n } contains zeros for odd indices and 2E |X| k for n = 2k. But we can always divide elements of a pm sequence by a positive number. See also [7], p. 40.
5. Let X be such a random variable that EX n = a n /a 0 . Let µ(.) denote the distribution of X and let us denote by ν k the probability measure x 2k dµ(x)/a 2k . Let us denote Y k the random variable that has distribution ν k .
is a pm sequence and by assertion 1. also {a 2k+n } ns>0 is a pm sequence. 6. Let X denote a random variable whose moments are α n , that is EX n = α n . We have var(X) = EX 2 − (EX) 2 = ρ 2 − ρ 2 = 0. But this equality means that the distribution of X is a one-point distribution, i.e. P (X = ρ) = 1.
7. For the proof see Lemma 2 of [25]. 8. Since by the fact that {b n } n≥0 is a pm sequence, then all central minors of the matrix [b i+j ] i,j≥0 must be nonnegative and moreover, in particular, we must have , for all n ≥ 1. Now, if the measure that produces the sequence {b n } has infinite support, then ∀n ≥ 0 : b 2n > 0. Hence, it proves log-concavity. On the other hand, we can easily deduce from (2. Let us give some examples. i) Take measure χ to be a one-point probability measure concentrated at t ∈ (0, 1). Then we have h i,n (χ) = n i t i (1−t) n−i . Let is take b n = c, and α = β = 1, then from assertion 2. it follows that c n j=0 n j t j a j (1 − t) n−j is a pm sequence provided {a j } is. Further taking α = 1/2, t = 1/2, we see that the so-called binomial transform of the sequence {a j } , , i.e., n j=0 n j a j is also a pm sequence. If we take α = −β = −1 = −2t, then also the following sequence n j=0 (−1) j n j a j which is called inverse binomial transform is a pm sequence.
ii) Take In particular, if β = 1 we have h i.n (χ) = 1 n+1 . Hence, following assertion 2. Proposition 2 we get Hence we see that sequence of arithmetic means of the convolution of two pm sequences is also a pm sequence.
As a corollary from the interesting results of Layman (see [10]) we have the following result. Proof. First if a 1 = 0 then the assertion is true. So let us assume that a 1 = 0. Let us notice that we have (2.8) n j=0 n j (−a 1 ) j a n−j = (−a 1 ) n n j=0 n j a n−j /(−a 1 ) n−j .

Now notice that
n j=0 n j a n−j /(−a 1 ) n−j is the binomial transform of the sequence a j /(−a 1 ) j j≥0 . By the Layman's Theorem 1, these two sequences have the same Hankel transforms. Now, the sequence a j /(−a 1 ) j j≥0 has the Hankel transform equal to det[a i+j ] 0≤i,j≤n /(−a 1 ) 2n . Hence the sequence n j=0 n j a n−j /(−a 1 ) n−j has the same Hankel transform. Now by (2.8) we deduce that the sequence n j=0 n j (−a 1 ) j a n−j n≥0 has the Hankel transform equal to (−a 1 ) 2n det[a i+j ] 0≤i,j≤n /(−a 1 ) 2n = {det[a i+j ] 0≤i,j≤n }. Remark 8. From the above-mentioned lemma it follows that the random variables X and X − EX have the same Hankel transforms. In other words, the sequence of (ordinary) moments of a random variable and the sequence of its central moments have the same Hankel transforms. Lemma 2. If {a n } n≥0 is a pm sequence, then the following sequence of polynomials indexed by n ≥ 0 2n j=0 a j x j /j! assumes only nonnegative values for x ∈ R.
Proof. We consider sequence of Hausdorff means with {a n } , {b n = x −n n!}, α = 1/d, β = 1/(1 − d), χ being the one point measure concentrated at point d ∈ (0, 1), Consequently the following sequence is a pm sequence: Now since all elements of a pm sequence with even indexes are positive and since 1/x 2n > 0 for all x = 0 and (2n)! is positive we get our assertion.
As before majority of the assertions of the Proposition below are known. We provide their simple, probabilistic proofs. Proposition 3. The following sequences are pm sequences:  , where (α) (n) = α(α + 1) . . . (α + n − 1) is the so called raising factorial of α, 7) ∀α, β > 0 : where n j denotes Stirling number of the second kind. , β) is the value of beta function at α and β, Now taking β = k − α we deduce that (α) (n) (k+n−1)! . 8) Fibonacci numbers are defined as the solution of the following difference equation F n+2 = F n+1 + F n . with F 0 = 0, F 1 = 1. Hence are positive. Now we apply assertion 1 of Proposition 2 We have also Similarly we can consider F n+5 and generally F n+2k+1 . They constitute pm sequence (no probabilistic in general) since > 0 for k ≥ 0. Applying third assertion of the above-mentioned Proposition 2, we deduce that {F 2n+2 } n≥0 . The fact that {F n+1 /(n + 1)} n≥0 and {F 2n+2 /(n + 1)} n≥0 are pm sequences follows third assertion of the above-mentioned Proposition 2, where a n = F n+1 or a n = F 2n+2 and b n = 1/(n + 1). Similarly, using the well known (see e.g. [8]) property of Fibonacci numbers F 1 + . . . + F n = F n+2 − 1 and then (2.7) with a n = F n+1 and b n = 1 we see that {(F n+2 − 1)/(n + 1)} n≥0 is a pm sequence. Now we apply assertion 3 of Proposition 2 and use the fact that 2n + 1. The last case i.e. proof that the sequence {1/F 2k+n } n≥0 is a pm sequence for every natural k is treated and proved in the excellent paper by Berg [3]. 9) We start with the observation that if X has the so-called Poisson distribution with parameter λ ≥ 0 , i.e., 10) Now, we use the fact that the so-called Bell numbers B n are defined as We also use the well-known fact that B n+1 = n j=0 n j B j and apply assertion 2) of the Proposition 2 with h i,n (χ) = n i .

Moment sequences and linear difference equations
First, let us fix the terminology. The best one seems to be taken from the systems theory. So let us consider the following difference equation with constant parameters: with the so-called initial conditions: r 0 = p 0 , . . . , r m−1 = p m−1 .
If ∀n ≥ 0 : c n = 0, then the equation is called homogeneous otherwise it is called non-homogeneous. Recall that we dealt with this type of difference equations, above, when discussing fR-sequences.
For the sake of the completeness of the paper, let us recall the basic properties of the linear difference equations with constant parameters.
Given the so-called system parameters: {d j } m j=0 , (sometimes one says about input-output system parameters) the input sequence {c n } n≥0 and initial conditions, the task is to find the so-called output sequence {r n } n≥0 . The equation (3.1) with the set of initial values is called initial value problem. It is known that every the initial value problem has always a solution, i.e. the sequence {r n } is determined uniquely. In other words, the set of parameters, the input sequence and the set of initial conditions determine the output sequence uniquely.
Moreover, the parameter m is called the order of the difference equation and the following algebraic equation: For the case of multiple roots we direct the reader to the monograph [6].
We start with the homogeneous linear difference equations with constant coefficients. Obviously the following sequence of numbers: for different real b k , and non-negative α k , k = 1, . . . , m is a pm sequence. By the way, the support of the measure that generates this sequence has finite cardinality, equal to m and consists of points of the set {b 1 , . . . , b m }. On the other hand, it is known that this sequence, let's call it {r n } n≥0 is a solution of the following difference equation: The fact that in the case of pm sequences only different roots are concerned can be seen, when studying the following example. Let's take m = 2 and b 1 = b 2 = a i.e. we consider the following equation r n+2 − 2ar n+1 + a 2 r n = 0, with r 0 = 1 and r 1 = r1. It is not difficult to check that the following sequence r n = a n−1 (nr 1 − a(n − 1)), for n ≥ 0 satisfies the equation. Now, as it is also easy to check, the Hankel transform of this sequence is 1, −(r1 − a) 2 , 0, . . . . Hence, unless a = r1, for no real a and r1, {r n } is a pm sequence. If r1 = a we have the obviously trivial sequence r n = a n .
One has to underline that the case of two complex but conjugate roots of the characteristic equation also leads to the non-pm case. More precisely, if one considers, for example, the difference equation: r n+2 + a 2 r n = 0, with r 0 = 1, r 1 = r1. Then {r n } is not a pm sequence since we have r 2 = −a 2 < 0.
The situation becomes more complicated and generally different when one considers nonhomogeneous linear equations with constant coefficients. That is when one considers the following difference equation: with {c n } being a pm sequence. First, let us consider the case when {c n } is a pm sequence generated by a discrete distribution with finite support.
As an example, let us take m = 1 and c n = d n .
That is, let us consider such an equation with an initial condition r 0 = p 0 . As it is commonly known the solution of such an equation is given by the formula: r n = a n − d n a − d + p 0 a n , for n ≥ 0. Finding Hankel transform of this sequence we get: p 0 , p 0 (d − a) − 1, 0, . . . . Hence {r n } is a pm sequence iff p 0 ≥ 0 and p 0 (d − a) − 1 ≥ 0. In particular, if d > a then p 0 ≥ 1 d−a . Let us note, that if we consider a second-order nonhomogeneous equation of the following form: r n+2 + a 2 r n = b n , with r 0 = r0 and r 1 = r1. One can easily split solving this equation into two parts. First to consider the case of even n and then the odd n case. Anyway, solving this equation doesn't cause any difficulty. With the help of Mathematica, one obtains the following sequence of Hankel transform of the sequence {r n } . Namely, we get: Now, it is enough to notice that the polynomial is −b 2 r1 2 + a 2 r0 2 + 2r1b − 1 + 2a 2 r0 − a 2 (a 2 r0 2 + r1 2 ) negative for all b. Consequently sequence {r n } cannot be a pm sequence for any b different from zero. We will see in a moment that this is not the case when one considers sequence {c n } being a moment sequence of absolutely continuous measure.
To avoid unnecessary complications we will consider signed measures dµ defined on the real line that satisfy the so-called Cramer's condition, that is that there exists δ > 0 such that It is known, that if a measure satisfies this condition, then it can be identified by its moments. Let us call the set of such measures Cra. Now let us consider a positive measure dA ∈ Cra and a polynomial P (x) both such that dB(x) = 1 P (x) dA(x) ∈ Cra. Note that following Proposition 1 of [22] if only (3.6) 1 P (x) 2 dA(x) < ∞ and 1/P (x) ≥ 0 on the supp A, then dB ∈ Cra and it is a positive measure.
Theorem 5. Let sequences, respectively {a n } n≥0 and {b n } n≥0 be pm sequences generated by the measures dA and dB. Then, the sequences are related to one another by the following difference equation: with initial conditions Coefficient c j is of x j in the expansion of the polynomial P (x).
Proof. We start with an obvious equality: ∀n ≥ 0: a n = x n dA(x). Now, recalling that P (x) = m j=0 c j x j we see that a n = P (x)x n dB(x) = m j=0 c j b j+n . Now to get unique solution of this equation, we need to set m initial conditions that can be found by calculating m numbers defined by (3.8).
Remark 10. In other words the difference equation (3.7) with an input of the form of a pm sequence {a n } n≥0 generated by a positive measure dA ∈ Cra has the solution that is also a pm sequence generated by the measure dB also belonging to Cra iff its characteristic polynomial P (x) satisfies condition (3.6). In particular, it has roots of odd multiplicity outside the support of the measure dA.

Remarks and examples
Let us return to the examples that were analyzed above.
Example 1. We start with the example with two complex conjugate roots of the characteristic equation Now, let us consider the following equation: with r 0 = 1 a 2 +b 2 , r 1 = r1. It turns out that the Hankel transform of r n is equal to 1 a 2 +b 2 , (b−r1(a 2 +b 2 ))(b+r1(a 2 +b 2 )) (a 2 +b 2 ) 2 , −(b−r1(a 2 +b 2 )) 2 a 2 +b 2 , 0, . . . , . Hence, unless r1 = b a 2 +b 2 the solution of the above-mentioned equation cannot be a pm sequence. Thus, the fact that the sequence {r n } is a pm sequence heavily depends on the initial conditions.
Let us consider one more example. . That is, consider the following difference equation: r n+2 + r n = 1 n + 1 .

Now let us recall, that we have:
1 0 x n dx = 1 n + 1 .
So the measure dA has the density equal to 1 for x ∈ [0, 1] and zero otherwise. According to the theorem above, the initial conditions should be: Remark 11. This example suggests that this sensitivity either for the initial conditions or for the fact if 1 P (x) dA(x) is a positive measure can be used in the numerical calculation to test if, for example, the roots of the odd multiplicity of the polynomial P (x) lie in the support of dA or finding the first values of the integral (3.8).
Let us return now to the question if a convolution of two pm sequences is a pm sequence. From the formula (2.7) it follows that the sequence of arithmetic averages of a convolution of two pm sequences is a pm sequence. However, if we consider the simple case of one sequence, say, {a n } n≥0 is a moments sequence of the distribution dA and the sequence say {b n } n≥0 then the solution of the difference equation r n+1 − br n = a n , with r 0 = r0 is given by the formula Hence for r0 = 0 sequence {r n } is a sequence of convolutions of {b n } and {a n }. Moreover, we know from the Theorem 5 that it is a moments sequence if only point b lies outside the support the measure dA, more precisely if dA(x)/(x − b) is a positive measure as it follows from Theorem 5.
Remark 12. Now, let us notice that the scheme in that we have two positive measures absolutely continuous with respect to one another is a general situation. One considers it in the series of papers [19], [21], [20], [23], [18]. These papers provide many consequences of such assumptions, including infinite expansions of the Radon-Nikodym derivative: and {α i (x)} i≥0 is the sequence of polynomials orthogonal with respect to the measure dA. The expansion (4.1) converges in mean-square mod dA provided dB dA (x) 2 dA(x) < ∞. Knowing sequences {α i } and {β i } one is able to find a numerical sequence {c i } and thus get the expansion (4.1). Moreover, following the above mentioned positions of literature, there exists a finite linear relationship between two sets of polynomials {α i } and {β i } orthogonal respectively to dA and dB, provided dB dA (x) = 1 P (x) . This observation was first made by Pascal Maroni in a more general but more confining context, not necessarily concerning measures. Maroni's approach, followed by his associates in the case of measures concerns mostly polynomials of order at most 4. For details see [11]- [16]. More precisely, there exists a table {w i,n } N α≥n≥0,0≤i≤n of real numbers such that: where N is the order of the polynomial P (x). This might lead to a new difference equation, this time with non-constant coefficients.
Remark 13. Note, also, that we could have calculated the moment sequence of dB by expanding 1 P (x) in an infinite power series and integrating term by term getting: where, of course, j≥0 d j x j is the postulated expansion of 1 P (x) . If the series on the right-hand side of the above-mentioned formula is convergent, we have the other relationship combining moments of the measures dA and dB. Such an approach of calculating the sequence of moments of some measure in two different ways can lead to discovering new, interesting relationships, not only in the moment sequences in question but also between sequences describing these moment sequences. This was done for example in the paper [24].
As an example, let us consider two distributions : the semicircle with the density 1 2π √ 4 − x 2 for |x| ≤ 2 and zero otherwise and the so-called arcsine distribution with the density 1 π √ 4−x 2 for |x| < 2 and 0 otherwise. It is elementary to notice that since these distributions are symmetric their odd moments are equal to zero. Further, it is well-known that even elements of the pm sequence generated by these distributions are respectively the so-called Catalan numbers C n = 1 n+1 2n n and the so-called central binomial coefficients. Further, we have Hence, taking into account that 2 (4 − x 2 ) = 1 2 j≥0 x 2 /4 j , for |x| < 2, the (4.2) takes the following form: Note that the convergence here is very slow since C n ∼ = 4 n /n 3/2 . On the other hand, since we have: we get an obvious relationship: We also see that the sequence 2n n n≥0 is the solution of the following difference equation: s n+1 − 4s n = −2C n , with an initial condition : s 0 = 1. Taking into account general solution of the above-mentioned equation, we end up with the following identity: 2n n = 4 n − 1 2 n−1 j=0 4 n−j C j .