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An error in my presentation

April 24, 2013

I gave two talks regarding the Askey-Wilson Operator and the Structure Equation.  Not to bore the non-mathematics audience who is reading this, the Askey-Wilson Operator is like differentiation in Calculus.  So x^n becomes nx^{n-1} under normal differentiation.  But Askey-Wilson Operator turns T_n to [n]_q U_{n-1}.  Got the pattern?

A certain kind (set) of Orthogonal Polynomials was characterized by a certain three-term recurrence relation, shortened TTRR (e.g. F_n = F_{n-1}+F_{n-2} is a TTRR for the famous Fibonacci sequence that goes, 1, 1, 2, 3, 5, 8, 13, 21, …) and the Structure Equation given by the usual differentiation in Calculus \frac{d}{dx}.  But by replacing the operator \frac{d}{dx} by other linear operators, it gave rise to other interesting quantum orthogonal polynomials.  I put in the Askey-Wilson Operator and consider (of the three different cases ) case 1.  That was already well-known and solved (even verbally by Al-Salam and maybe Chihara), but we sent out a paper claiming another proof, completely different from the one Al-Salam proved and experts knew.

Then I sat down and try yet another proof which is similar but different from that of our paper.  The difference is, instead of using the Chebyshev polynomial of the first and second kind as basis for our polynomials, I used the usual \{1, x, x^2, x^3, x^4, \cdots \}.

So after months of work, I gave the proof in a 50-minute lecture held in the school and another 15 min talk in Indiana University of Pennsylvania in an MAA meeting.  Here is a part of many slides shown in my talks: ({\cal D}_q is the Askey-Wilson operator and P_n is one of the orthogonal polynomials)

ErrorInPresentation

A few weeks later, I worked on the formula again ( for R_{n=1}-R_n which is going to be equal to C_n, which is crucial in the TTRR (three-term recurrence relation), I found a terrible error in my formula.  But in my talk, no one pointed out the error.  The correct formula should be

[n-2]_q (R_n + \frac{n}{4} ) = [n]_q (R_{n-1}+ \frac{n-2}{4})

It may look like just a minor mistake, I multiplied n  by four instead of dividing it by 4.  My apologies to those who got the wrong formula.

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