The continuous $q-$Hermite polynomials are defined by

$${H_{n + 1}}(x|q) = 2x{H_n}(x|q) +( {q^n}-1){H_{n - 1}}(x|q)$$

with initial values ${H_{ - 1}}(x|q) = 0$ and ${H_0}(x|q) = 1.$

Cf. e.g. http://aw.twi.tudelft.nl/~koekoek/askey/ch3/par26/par26.html

The only simple special values I know of are $$\sqrt {{q^n}} {H_n}\left( {\frac{1}{2}\left( {\sqrt q + \frac{1}{{\sqrt q }}} \right)|{q^2}} \right) = (1 + q)(1 + {q^2}) \cdots (1 + {q^n}).$$ I have seen this formula in the literature but do not remember where. Could you please give me some references? Are there other simple special values known?

  • $\begingroup$ Why are there 3 variables inside $H_n$ in your special-value formula ? $\endgroup$ – F. C. Jun 7 '13 at 15:27
  • $\begingroup$ Sorry, I have corrected it. $\endgroup$ – Johann Cigler Jun 7 '13 at 19:50
  • $\begingroup$ The value at $x=q-1$ is divisible by a power of $q-1$. $\endgroup$ – F. C. Feb 10 '14 at 16:20

Half a year late, but here is a solution. I don't know any good references to your special value, but it can easily be derived using other literature.

From Koekoek and Swarttouw take the Askey-Wilson polynomials $$ p_n(x;a,b,c,d|q) = \frac{(ab, ac, ad | q)_n}{a^n} {}_4 \phi_{3} (q^{-n}, abcdq^{n-1}, ae^{i\theta}, ae^{-i\theta}; ab, ac, ad; q, q). $$ It is easily seen that $$ p_n(\frac{1}{2}(a + a^{-1}); a, b, c, d | q) = a^{-n}(ab,ac,ad;q)_n, $$ and since the Askey-Wilson polynomials are symmetric in $a$, $b$, $c$ and $d$ you have a similar result for $x = \frac{1}{2}(b + b^{-1})$, $x = \frac{1}{2}(c + c^{-1})$ and $x = \frac{1}{2}(d + d^{-1})$.

Now let's rewrite $H_n(x|q^2)$ in Askey-Wilson polynomials. From exercise 9.10.ii from Basic Hypergeometric Series from George Fasper and Mizan Rahman you have quadratic transformation $$ H_n(x|q^2) = Q_n(x;q^{1/2},-q^{1/2}|q) $$ where $Q_n$ are the Al-Salam-Chihara polynomials.

Then following (limit) transformations (http://aw.twi.tudelft.nl/~koekoek/askey/ch4/ch4.html) we rewrite: $$ Q_n(x;q^{1/2},-q^{1/2}|q) = p_n(x;q^{1/2},-q^{1/2},0,0|q). $$ And therefore $$ H_n(\frac{1}{2}(q^{1/2} + q^{-1/2})|q^2) = \frac{(-q;q)_n}{q^{n/2}}. $$

I hope this helps.


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