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I'm trying to understand this lemma from my course notes:

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It's my understanding that a linear recurrence relation is one of the form $x_n = p(n, x_{n - 1}, \dots, x_{n - k}) = q(n) + b_1 x_{n - 1} + b_2 x_{n - 2} + \dots + b_k x_{n - k}$ for $n \geq k$. This recurrence relation is homogeneous if $q = 0$.

I find the above proof very terse, so below I give my understanding along with comments/questions where I don't understand things.

First part: Rational generating function defines a homogeneous linear recurrence relation

Start with $f(x) = a_0 + a_1 x + \dots = \frac{g(x)}{h(x)}$ where $g(x) = b_l x^l + \dots + b_0$ and $h(x) = c_k x^k + \dots + c_0$. Bring the denominator over to the left so you get $(c_k x^k + \dots + c_i x_i) (a_0 + a_1 x + \dots ) = b_l x^l + \dots + b_0$. We want to set up something like $a_j = \alpha_1 a_{j - 1} + \dots + \alpha_n a_{j - n}$. Because we are trying to set up a linear homogeneous recurrence relation, we clearly need to set up something that will work for all $j > \text{[something]}$, but it's not clear at this stage what that something should be.

Clearly based on the product $(c_k x^k + \dots + c_i x_i) (a_0 + a_1 x + \dots )$ we are going to get products of $a$ coefficients with $c$ coefficients. This suggests that we might have to divide an expression involving $a_j c_{\text{something}}$ by $c_{\text{something}}$ in order to isolate $a_j$. The only two $c_{\text{something}}$s that we know are not equal to zero are $c_i$ and $c_k$ (which may or may not be distinct).

If we try to work with $c_k$ then we would be looking at the coefficient for the $x^{k + j}$ term, which would be something like $c_k a_j + c_{k - 1} a_{j + 1} + \dots + c_i a_{j + (k - i + 1)}$. This doesn't look like the type of expression we're looking for ($a_j = \alpha_1 a_{j - 1} + \dots + \alpha_n a_{j - n}$) so we'll try using $c_i$.

If we use $c_i$ then we are looking at the coefficient for the $x^{i + j}$ term, which would be something like $c_i a_j + c_{i + 1} a_{j - 1} + \dots + c_k a_{j - (k - i)}$. Comparing both sides of $(c_k x^k + \dots + c_i x_i) (a_0 + a_1 x + \dots ) = b_l x^l + \dots + b_0$, we see that $c_i a_j + c_{i + 1} a_{j - 1} + \dots + c_k a_{j - (k - i)} = b_{i + j}$. We want $b_{i + j} = 0$ so that we can just divide by $c_i$ and move terms over and get our desired recurrence relation. This implies a restriction $i + j > l$. We also see that we need $j - (k - i) \geq 0$, so our restrictions on $j$ are that $j > l - i$ and $j \geq k - i$.

Given this, we can divide both sides of $c_i a_j + c_{i + 1} a_{j - 1} + \dots + c_k a_{j - (k - i)} = 0$ by $c_i$ and we get that $a_j = - \frac{1}{c_i} ( c_{i + 1} a_{j - 1} + \dots + c_k a_{j - (k - i)})$, which is our desired recurrence relation.

Why is my bound for $j$ different from that given in the proof ($j > \max \{ k, l \}$)?

Second part: linear recurrence relation (having function $q(n)$ that has its own rational generating function) has a rational generating function

This part doesn't make much sense to me at all. I see that $q$ essentially defines a sequence starting at index $k$, as opposed to zero, so I can see that $q$ could have an associated generating function. However the generating function $g(x)$ given for $q$ is not even a polynomial, much less rational. Is the assumption that $g(x)$ can be expressed in a rational form?

I guess $f(x)$ is supposed to be the generating function for the linear recurrence relation? I suppose a linear recurrence relation defines a sequence so there is guaranteed to be an associated generating function.

We define the function $h$ in such a way that all its coefficients beyond $k - 1$ are zero. Since $g(x)$ is presumed to be rational, we can write $g(x) = \frac{r(x)}{s(x)}$ where $r$ and $s$ are polynomials, therefore to be more thorough we can write $f(x) = \frac{h(x)s(x) + r(x)}{s(x)(1 - b_1 x - b_2 x^2 - \dots - b_k x^k)}$.

Does what I'm saying about this second part make sense? I am trying to talk myself through it. What is the intuition behind the definition of $h(x)$? I guess we're trying to make some kind of finite polynomial and that is the definition that works?

I appreciate any help.

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First part: We have a generating function $f(x)=a_0+a_1x+a_2x^2+\cdots$ which is a rational function \begin{align*} f(x)&=\frac{g(x)}{h(x)}\\ a_0+a_1x+a_2x^2+\cdots&=\frac{b_0+b_1x+\cdots+b_lx^l}{c_0+c_1x+\cdots+c_kx^k}\qquad\qquad b_l\ne 0, c_k\ne 0 \end{align*} We consider $h(x)f(x)=g(x)$ and obtain \begin{align*} &\left(c_ix^i+c_{i+1}x^{i+1}+\cdots+c_kx^k\right)\left(a_0+a_1x+a_2x^2+\cdots\right)\\ &\qquad=b_0+b_1x+\cdots+b_lx^l\tag{1} \end{align*} with $c_i\ne 0$ the coefficient of the smallest power of $x$ of $h(x)$ not equal to zero.

In the following we use the coefficient of operator $[x^q]$ to denote the coefficient of $x^q$ of a series. We want to find a recurrence relation in terms of $a_j,$. We take $j>\max\{k,l\}$ which is convenient, since the coefficient of $[x^j]g(x)=0$ and $[x^j]h(x)f(x)$ has to respect all coefficients of $h(x)$.

We calculate $[x^{j+i}]h(x)f(x)$ and obtain from (1) \begin{align*} \color{blue}{[x^{j+i}]h(x)f(x)}&=[x^{j+i}]\left(c_ix^i+c_{i+1}x^{i+1}+\cdots+c_kx^k\right)\\ &\qquad\qquad\qquad\cdot \left(a_0+a_1x+a_2x^2+\cdots\right)\\ &=\left(c_i[x^{j+i}]x^i+c_{i+1}[x^{j+i}]x^{i+1}+\cdots+c_k[x^{j+i}]x^k\right)\tag{2}\\ &\qquad\qquad\qquad\cdot \left(a_0+a_1x+a_2x^2+\cdots\right)\\&=\left(c_i[x^{j}]+c_{i+1}[x^{j-1}]+\cdots+c_k[x^{j+i-k}]\right)\tag{3}\\ &\qquad\qquad\qquad\cdot \left(a_0+a_1x+a_2x^2+\cdots\right)\\ &\,\,\color{blue}{=c_ia_j+c_{i+1}a_{j-1}+\cdots+c_{k}a_{j+i-k}}\tag{4}\\ \end{align*}

Comment:

  • In (2) we use the linearity of the coefficient of operator.

  • In (3) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

  • In (4) we select the coefficient of $[x^q], j+i-k\leq q \leq j$.

Since $j>\max\{k,l\}$ we obtain from (1) \begin{align*} \color{blue}{[x^{j+i}]g(x)}=[x^{j+i}]\left(b_0+b_1x+\cdots+b_lx^l\right)\color{blue}{=0}\tag{5} \end{align*} and we obtain from (4) and (5) \begin{align*} \color{blue}{a_j=-\frac{c_{i+1}}{c_i}a_{j-1}-\cdots-\frac{c_{k}}{c_i}a_{j+i-k}} \end{align*} in accordance with the proof of the lemma.

Second part: Note the Lemma states:

  • ... Furthermore if ... the function $q(n)$ has a rational generating function ...

It follows a representation of the generating function $g(x)=\frac{r(x)}{s(x)}$, with $r, s$ polynomials, wlog relatively prime and $\deg{r}<\deg{s}$ is admissible.

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