Timeline for Orthogonal polynomials with quadratic recurrence coefficients
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2018 at 21:37 | comment | added | Twi | @ChristianRemling: You are right. The spectrum is purely discrete. There is a criterion applicable to this case. It also seems that the operator is semi-bounded but I didn't try to prove it. I also didn't get an explicit result in any special case. | |
| Jan 5, 2018 at 20:28 | comment | added | Christian Remling | It should be possible though to obtain some qualitative information. I suspect the spectrum will be purely discrete and bounded below (and obviously it's unbounded above). | |
| Jan 5, 2018 at 20:03 | comment | added | Christian Remling | Yes, you are right, the recursion can be rewritten to bring it to this form. So you want to compute the spectral measure of this Jacobi matrix, and I'd be quite surprised if that could be done explicitly (maybe in friendly special cases such as $a=0$). | |
| Jan 5, 2018 at 8:29 | comment | added | Twi | @ChristianRemling: There is always a positive measure wrt which the sequence of polynomials generated by the three term recurrence is orthogonal providing that the first coefficient, in this case $n(n+b)$, is real for all $n\geq0$ and the second coefficient, in this case $n(n+a)$, is positive for all $n>0$. One explanation relies on the Spectral theorem for self-adjoint operators. Here, the operator is the Jacobi operator $J$ with diagonal $n(n+b)$ and off-diagonal $\sqrt{(n+1)(n+1+a)}$, $n\geq0$. The meas. of orthogonality for the above family is closely related to the spec. meas. of $J$. | |
| Jan 4, 2018 at 18:09 | history | asked | Twi | CC BY-SA 3.0 |