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Chapter 6: Problem 1

A Markov sequence is realizable if and only if its generating series is rational.

Short Answer

Expert verified
A Markov sequence is realizable if and only if its generating series is rational. To prove this, we showed two directions: 1. If a Markov sequence is realizable, then its generating series is rational. This was done by assuming the realizable Markov sequence satisfies the recurrence relation, multiplying both sides by \(x^n\) and summing over all n, recognizing the left side as a shifted generating function and the right side as a product of coefficients and generating functions. We then observe a rational function equation for the generating function A(x). 2. If a generating series is rational, then the corresponding Markov sequence is realizable. We assumed the generating function A(x) is rational, and wrote it as a fraction of two polynomials. By multiplying both sides by \(Q(x)\), expanding the generating function A(x) by definition, and multiplying the sum term by term by the polynomial Q(x), we deduced a linear recurrence relation for the sequence \((a_n)\). Therefore, a Markov sequence is realizable if and only if its generating series is rational.

Step by step solution

01

Definition of a Markov sequence

A Markov sequence is a sequence of numbers \((a_n)\) that satisfies a linear recurrence relation of the form: \[a_{n+k} = \sum_{j=0}^{k-1} c_j a_{n+j}\] for some fixed integers \(c_0, c_1, ..., c_{k-1}\), and \(k \ge 1\).
02

Definition of a generating function

Given a sequence \((a_n)\), its generating function is defined as: \[A(x) = \sum_{n=0}^{\infty} a_nx^n\]
03

Proving that realizable Markov sequence implies rational generating series

Assume that the Markov sequence \((a_n)\) is realizable, which means it satisfies the recurrence relation: \[a_{n+k} = \sum_{j=0}^{k-1} c_j a_{n+j}\] 1. Multiply both sides of the equation by \(x^n\) and sum over all n: \[\sum_{n=0}^{\infty} a_{n+k}x^n = \sum_{n=0}^{\infty}\sum_{j=0}^{k-1} c_j a_{n+j}x^n\] 2. Recognize the left side as a shifted generating function and the right side as a product of coefficients and generating functions: \[\frac{A(x) - \sum_{n=0}^{k-1} a_n x^n}{x^k} = \sum_{j=0}^{k-1} c_j x^j A(x)\] 3. Observe that the right side is a polynomial in x multiplied by the generating function A(x) and the left side is a rational function. Combining these, we have a rational function equation for A(x).
04

Proving that rational generating series implies realizable Markov sequence

Now assume that the generating function A(x) is rational, meaning it can be written as the fraction of two polynomials: \[A(x) = \frac{P(x)}{Q(x)}\] 1. Multiply both sides by \(Q(x)\), to obtain an equation relating polynomials P(x) and Q(x): \[Q(x)A(x) = P(x)\] 2. Expand the generating function A(x) by definition: \[Q(x)\sum_{n=0}^{\infty} a_nx^n = P(x)\] 3. Multiply the sum term by term by the polynomial Q(x), resulting in a new sum: \[\sum_{n=0}^{\infty} a_n(Qx^n) = P(x)\] 4. Since P(x) is a polynomial with finite terms, the sum on the left can be truncated at a specific term, say k, giving a finite sum of polynomials: \[\sum_{n=0}^{m} a_n(Qx^n) = P(x)\] 5. Using the properties of polynomials, we can deduce a linear recurrence relation for the sequence \((a_n)\), proving that \((a_n)\) is a realizable Markov sequence. Since we have shown that a realizable Markov sequence implies a rational generating series and a rational generating series implies a realizable Markov sequence, we have proven that a Markov sequence is realizable if and only if its generating series is rational.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Linear Recurrence Relations
A linear recurrence relation is a sequence where each term is defined as a linear combination of its preceding terms. These recurrence relations play a pivotal role in mathematics and computer science because they provide a way to describe complex sequences and systems.
Consider the sequence \(1, 4, 7, 10, ...\). This is a simple linear sequence where each term after the first begins with the previous term plus three. A more complex version involves combining several previous terms, scaled by constants, also known as coefficients. This is the foundation of Markov sequences.
When solving problems with linear recurrence relations, it's key to identify the relation's order, the coefficients, and initial conditions. The 'order' refers to how many previous terms in the sequence are used for the calculations of the following terms.
With linear recurrence relations, unique solutions can often be determined by a given set of initial conditions, which are the first few terms of the sequence. Once these are known, the entire sequence can theoretically be unfolded. Therefore, understanding the nature of these relations is vital for students aiming to solve complex problems involving sequences in mathematics.
Decoding Generating Functions
A generating function is an infinite series that encodes the information of a sequence. The beauty of generating functions lies in their ability to simplify the handling of sequences, particularly when working with combinatorial problems or, as in our case, Markov sequences.
In essence, each term in the sequence is associated with a corresponding term in the power series. The coefficients of \(x^n\) in the power series \(A(x)\) are precisely the terms of the sequence, and \(A(x)\) serves as a compact representation. This can be incredibly powerful, as operations on generating functions can correlate to meaningful operations on the sequences they represent.
Consider the function \(A(x) = a_0 + a_1x + a_2x^2 + ...\) where \(a_n\) are the terms of a sequence. This power series is not just a list of terms; it's a tool that enables us to utilize calculus and algebraic techniques to work with the sequence. For instance, differentiating or integrating \(A(x)\) can yield another generating function that provides insights into the original sequence. Thus, mastering generating functions is crucial for students to solve and understand a range of problems in discrete mathematics and in working with Markov sequences.
Exploring Rational Generating Series
A rational generating series is a generating function that can be expressed as the ratio of two polynomials. This property is highly significant, as sequences represented by rational generating series bear the hallmark of having a predictable and regular structure鈥攐ften indicative of a linear recurrence relation.
Due to their rational nature, these generating functions can be manipulated algebraically to unearth the underlying recurrence relations of the sequence. For example, in the case of a Markov sequence, the generating function yields a polynomial equation that implicitly contains the sequence's recurrence relation.
The practicality of a rational generating series cannot be overstated. By handling these as fractions of polynomials, students can apply long-established polynomial algebra to solve them. This includes tasks such as partial fraction decomposition, polynomial division, and finding roots.
It is also worth noting for students that not all generating series are rational. Determining whether a generating series is rational can be a pivotal decision point in choosing the correct approach to solve a problem involving sequences鈥攆urther emphasizing the importance of fully grasping this concept in their studies.

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Most popular questions from this chapter

Let \(\Lambda\) be any time-invariant complete behavior and let \(\sim\) be the following equivalence relation on \(\Omega\) : $$ \omega \sim \omega^{\prime} \Leftrightarrow \lambda(\omega \nu)=\lambda\left(\omega^{\prime} \nu\right) \quad \forall \nu $$ (to be more precise, one should write the translated version of \(\nu\) ). This is the Nerode equivalence relation. Prove that \(\lambda\) admits a finite- cardinality realization if and only if there are only finitely many equivalence classes under the Nerode relation. (Hint: It only takes a couple of lines, using previous results.)

There is a continuous-time version of the above result. Consider continuous- time polynomial systems $$ \dot{x}=f(x, u), \quad y=h(x) $$ where \(\mathcal{X}, \mathcal{U}, \mathcal{y}, h\) are as above and \(f\) is polynomial. We assume completeness. Show that, again, there exists then a finite set of controls \(\omega_{1}, \ldots, \omega_{k}\) such that, for any pair of states \(x\) and \(z\), these states are distinguishable if and only if one of the controls \(\omega_{i}\) distinguishes them. (Hint: Use analytic controls and argue in terms of the corresponding derivatives \(y(0), \dot{y}(0), \ldots\) that there must exist for each such control a polynomial \(\Delta_{\omega}\) so that \(x\) and \(z\) are indistinguishable by \(\omega\) if and only if \(\Delta_{\omega}(x, z)=0\). The Hilbert Basis Theorem is needed for that. Now apply the Basis Theorem again.)

When \(\mathbb{K}=\mathbb{R}\) or \(\mathbb{C}\), one may use complex variables techniques in order to study realizability. Take for simplicity the case \(m=p=1\) (otherwise one argues with each entry). If we can write \(W_{\mathcal{A}}(s)=P(s) / q(s)\) with \(\operatorname{deg} P<\) \(\operatorname{deg} q\), pick any positive real number \(\lambda\) that is greater than the magnitudes of all zeros of \(q\). Then, \(W_{\mathcal{A}}\) must be the Laurent expansion of the rational function \(P / q\) on the annulus \(|s|>\lambda\). (This can be proved as follows: The formal equality \(q W_{\mathcal{A}}=P\) implies that the Taylor series of \(P / q\) about \(s=\infty\) equals \(W_{\mathcal{A}}\), and the coefficients of this Taylor series are those of the Laurent series on \(|s|>\lambda\). Equivalently, one could substitute \(z:=1 / s\) and let $$ \widetilde{q}(z):=z^{d} q(1 / z), \widetilde{P}(z):=z^{d} P(1 / z), $$ with \(d:=\operatorname{deg} q(s)\); there results the equality \(\widetilde{q}(z) W(1 / z)=\widetilde{P}(z)\) of power series, with \(\widetilde{q}(0) \neq 0\), and this implies that \(W(1 / z)\) is the Taylor series of \(\widetilde{P} / \widetilde{q}\) about 0 . Observe that on any other annulus \(\lambda_{1}<|s|<\lambda_{2}\) where \(q\) has no roots the Laurent expansion will in general have terms in \(s^{k}\), with \(k>0\), and will therefore be different from \(W_{\mathcal{A} .)}\) Thus, if there is any function \(g\) which is analytic on \(|s|>\mu\) for some \(\mu\) and is so that \(W_{\mathcal{A}}\) is its Laurent expansion about \(s=\infty\), realizability of \(\mathcal{A}\) implies that \(g\) must be rational, since the Taylor expansion at infinity uniquely determines the function. Arguing in this manner it is easy to construct examples of nonrealizable Markov sequences. For instance, $$ \mathcal{A}=1, \frac{1}{2}, \frac{1}{3 !}, \frac{1}{4 !}, \ldots $$ is unrealizable, since \(\mathcal{W}_{\mathcal{A}}=e^{1 / s}-1\) on \(s \neq 0\). As another example, the sequence $$ \mathcal{A}=1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots $$ cannot be realized because \(\mathcal{W}_{\mathcal{A}}=-\ln \left(1-s^{-1}\right)\) on \(|s|>1\).

Let \(\omega_{i}, i=1, \ldots, k\) be \(k\) different positive real numbers. Show that there is some continuous-time time-invariant linear system with outputs and no inputs \(\Sigma=(A, 0, C)\) such that: \- \(\Sigma\) is observable, and \- for each set of \(2 k\) real numbers \(a_{i}, \varphi_{i}, i=1, \ldots, k\), there is some initial state \(x\) so that $$ \lambda_{x}^{0, t}=\eta(t)=\sum_{i=1}^{k} a_{i} \sin \left(2 \pi \omega_{i} t+\varphi_{i}\right) $$ for all \(t \geq 0\). Conclude from the above discussion that, if $$ \frac{1}{\delta}>2 \max _{i=1, \ldots, m}\left|\omega_{i}\right| $$ then the complete function \(\eta(t)\) can be recovered from the values $$ \eta(0), \eta(\delta), \eta(2 \delta), \ldots $$ for every set of \(a_{i}\) 's and \(\varphi_{i}\) 's.

Calculate a canonical realization, and separately calculate the rank of the Hankel matrix, for each of these examples with \(m=p=1\) : 1\. The sequence of natural numbers \(1,2,3, \ldots\) 2\. The Fibonacci sequence \(0,1,1,2,3,5,8, \ldots\)
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