Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $d(x)=\prod_{s=1}^{n}(x-a_s)$ and $c(x)=\prod_{s=1}^{n}(x-a_s+b_s h)$ be polynomials, where $a_s, b_s$ are some complex numbers. What are the polynomial solutions of the difference equation $W(x+h)=\frac{c(x+h)}{d(x+h)}W(x)$ for $W(x)$? Thank you very much.

Edit: $b_s$ are positive integers.

share|cite|improve this question
up vote 1 down vote accepted

It is instructive to look at this from the stand-point of general solution, which is $$ W(x) =\kappa \cdot \prod_{s=1}^n \frac{\Gamma\left(\frac{x-a_s}{h} + b_s + 1 \right)}{\Gamma\left(\frac{x-a_s}{h} +1 \right)} $$

Using the recurrence equation for $\Gamma(z)$, i.e. $G(z+n) = (z+n-1) \cdots (z+1)z \Gamma(z)$, we get that $W(x)$ is polynomial iff if $b_s$ are non-negative integers.


The necessary condition is that there exists a permutation of the tuple $\left[ b_1 +1 - \frac{a_1}{h}, \ldots, b_n +1 - \frac{a_n}{h}\right]$ so that it equals to $\left[m_1 + 1 - \frac{a_1}{h}, \ldots, m_n + 1 - \frac{a_n}{h}\right]$ for non-negative integers $m_1,\ldots,m_n$.

A non-trivial example was provided by @DidierPiau, with $n=2$, $h=1$, $[a_1,a_2] = [1,1/2]$, $[b_1,b_2] = [1/2,1/2]$. Then $[b_1+1-a_1/h,b_2+1-a_2/h] = [1/2,1]$, and $[1-a_1/h,1-a_2/h] = [0,1/2]$. The ratio of $\Gamma$-function, thus simplifies to $x$, and $W(x) = \kappa x$.

share|cite|improve this answer
That the $b_s$ must be nonnegative integers for $W$ to be a polynomial function is not so clear. Consider the case $n=2$, $h=a_1=1$, $a_2=b_1=b_2=\frac12$, then $W(x)=\kappa x$ hence $W$ is a polynomial function. – Did Nov 23 '11 at 17:11
@Didier, thank you for your comment. – LJR Nov 23 '11 at 17:14
@user9791, Ahhh, post podified without notification. (As a result, the $b_s$ are complex numbers and, on the next line, positive integers...) – Did Nov 23 '11 at 17:26
@DidierPiau Thanks for the counterexample. I have expanded my answer to covert the case when $\Gamma$-s in numerator are shifts of those in the denominator up to a permutation. – Sasha Nov 23 '11 at 17:37
@Sasha, is the solution unique? – LJR Nov 23 '11 at 19:26

When the $b_s$ are nonnegative integers, the general solution is $$ W(x)=V(x)\cdot\prod\limits_{s=1}^n\,\prod\limits_{k=1}^{b_s}\,(x-a_s+kh), $$ where $V$ has period $h$. Thus $W$ is a polynomial function if and only if $V$ is constant.

Edit The proof is direct: call $V(x)$ the ratio of $W(x)$ by the product over $(s,k)$ on the RHS. Check that the functional equation $d(x+h)W(x+h)=c(x+h)W(x)$ (which is not a difference equation) is equivalent to $V(x)=V(x+h)$. Then the condition that $W$ is a polynomial implies that $V$ is a rational function. The only periodic rational functions are constant hence the proof is complete.

share|cite|improve this answer
can $V(x)$ be a polynomial? – LJR Nov 26 '11 at 22:53
Polynomial and periodic? This does not leave a lot of choice, don't you think? – Did Nov 27 '11 at 6:58
thank you very much. – LJR Nov 27 '11 at 13:21
how do you get this formula? What is the ralation between $\Gamma(hz)$ and $\Gamma(z)$? – LJR Nov 27 '11 at 14:19
See edit. $ $ $ $ – Did Nov 27 '11 at 16:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.