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Consider the Fibonacci polynomials
$$F_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\binom{n-j}{j} x^{n - 2j} $$ and the Lucas polynomials
$$L_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\frac{n}{n-j }\binom{n-j}{j} x^{n - 2j} .$$

Let $X$ be the multiplication operator $Xp(x)=xp(x)$ on the polynomials, $D_q$ the $q$-differentiation operator defined by $$D_q p(x)=\frac{p(qx)-p(x)}{qx-x}$$ and $A=X+(q-1)D_q$.

Applying the operator $F_n(A)$ to the constant polynomial $1$ has the curious effect that all binomial coefficients $\binom{n}{j}$ are changed to $q$-binomial coefficients ${n\brack j}$ together with some $q$-power. More specifically, we get $$F_n(A)1=F_n(x|q)= \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor } {q^{j+1\choose 2}}{n-j\brack j} x^{n - 2j} $$ and $$ L_n(A)1=L_n(x|q)= \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor } {q^{j\choose 2}}{\frac{[n]}{[n-j]}}{n-j\brack j} x^{n - 2j} .$$

Is this an isolated phenomenon or do there exist similar formulae for polynomials? In other words, are there polynomials $p_n(x)$ in one variable $x$ and operators $B(q)$ satisfying $B(1)=X$ such that $p_n(B(q))1$ is a simple or beautiful $q$-analogue of $p_n(x)$?

A trivial example would be $p_n(x)=x^n$ and $B(q)=\epsilon(q)X$ with $\epsilon(q)p(x)=p(qx)$.

Edit

Recently I noticed that the Rogers-Szegö polynomials $$R_n(x,s)= \sum_{j = 0}^{n }{n\brack j} x^{j}s^{n-j} $$ can also be represented in this way: $$R_n(x,s)=(x+s+(q-1)xsD_q)^n1.$$ Since this is the most natural $q-$ analogue of the binomial theorem I would be astonished if nobody has as yet seen this fact. The proof is the same as that for the well-known recursion $R_{n+1}(x,s)=xR_{n}(x,s)+(q^n-1)xs R_{n-1}(x,s).$

It follows immediately from the recurrence of the $q-$ binomial coefficients:

$(x+s+(q-1)xsD_q) \sum_{j = 0}^{n }{n\brack j} x^{j}s^{n-j}$ $=\sum_{j = 0}^{n }{n\brack j} x^{j+1}s^{n-j}$ $+\sum_{j = 0}^{n }{n\brack {j}} x^{j}s^{n-j+1}$ $+\sum_{j = 0}^{n }(q^{j}-1){n\brack j} x^{j}s^{n-j+1}$ $= \sum_{j = 0}^{n+1 }({n\brack j}+{n\brack {j-1}}+(q^{j}-1) {n\brack {j}}) x^{j} s^{n-j+1}$ $=\sum_{j = 0}^{n +1}{{n+1}\brack j} x^{j}s^{n-j+1}.$

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    $\begingroup$ A $q$-rious $q$-estion! I love it. Do you have a reference for your "quantisation" of the Fibonacci and Lucas polynomials? This should not be isolated examples: one could try next the binomial expansion and, if it works, the hypergeometric series. $\endgroup$ – Wadim Zudilin Jul 21 '10 at 10:37
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    $\begingroup$ Wadim: See e.g. A new class of q-Fibonacci polynomials in Electr.J. Comb. 10(2003),R19 $\endgroup$ – Johann Cigler Jul 21 '10 at 10:56
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    $\begingroup$ Thanks, Johann! By the way, it's not mentioned that $[n]=(1-q^n)/(1-q)$ and ${m\brack n}=[1][2]\dots[m]/([1]\dots[n]\cdot[1]\dots[m-n])$. $\endgroup$ – Wadim Zudilin Jul 21 '10 at 11:11
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    $\begingroup$ Re your edit, isn't $1+(q-1)xD_q$ just the $\epsilon(q)$ operator you had earlier ? So $R_n$ arises from the (q-binomial) expansion $(x+s\epsilon)^n$. $\endgroup$ – dke Apr 11 '11 at 17:30
  • $\begingroup$ @dke: yes you are right. $\endgroup$ – Johann Cigler Apr 11 '11 at 19:25
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This is not a definitive answer, since it doesn't contain any more examples, but more of an observation.
If you have a sequence $P_n(x)$ for polynomials defined recursively by the formula $$P_n(x)=x^{a_1}P_{n-1}(x)+x^{a_2}P_{n-2}(x)+\cdots+x^{a_k}P_{n-k}(x),$$ and an operator $B(q)$ with $\lim_{q \to 1} B(q)=X$, we get: $$P_n(B)1=B^{a_1}P_{n-1}(B)1+B^{a_2}P_{n-2}(B)1+\cdots+B^{a_k}P_{n-k}(B)1.$$

Denoting $P_n(B)1$ as $\hat P_n(x)$ and supposing that $B$ is linear (which it is, if it depends only on $X$ and $D_q$, like all your examples suggest), we get: $$\lim_{q\to 1}\hat P_n(x)=\lim_{q\to 1}x^{a_1}\hat P_{n-1}(x)+x^{a_2}\hat P_{n-2}(x)+\cdots+x^{a_k}\hat P_{n-k}(x),$$

Thus, $\lim_{q\to 1}\hat P_n(x)$ satisfies the same recursion as $P_n(x)$ does and if the first $k$ terms are equal, then $\lim_{q\to 1}\hat P_n(x) = P_n(x)$ for all $n$. This is precisely the definition of a $q$-analog. Whether it will be interesting is another matter, but I think this must be decided case-by-case.

I noticed that all of your examples have the form $B(q)=X+a(q-1)D_q$. This looks somewhat like a linear term of a series in $D_q$ of the form $$B(q)=X+a_1(q-1)D_q+a_2(q-1)^2D_q^2+\cdots,$$ so maybe if you could find examples with $B$ not linear in $D_q$, it would be very interesting.

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