Timeline for Curious identity between the two kinds of Chebyshev polynomials
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2020 at 0:12 | comment | added | MannyC | I was about to show the details of my proof in a self-answer but I did not have time today. Essentially I want to express the left hand side as a contraction of $d$ dimensional, rank $n$ symmetric traceless tensors and the right hand side as traces of products of matrices belonging to the Clifford algebra in $d$ dimensions. While I believe these concepts can be generalized to any $d$, I have no rigorous way of doing so. | |
| Feb 15, 2020 at 23:38 | comment | added | Abdelmalek Abdesselam | Why only for the canonical scaling dimension $\alpha$? There is no reason to be afraid of fractional free fields. | |
| Feb 15, 2020 at 10:27 | history | edited | darij grinberg | CC BY-SA 4.0 |
added 1066 characters in body
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| Feb 15, 2020 at 8:23 | comment | added | MannyC | After your proof I realized that this could be trivially generalized to general Gegenbauer polynomials by letting $U_n(x)\to G_n^{(\alpha)}(x)$ and $2T_{\lambda_j}(x) \to 2\alpha T_{\lambda_j}(x) $. Furthermore, for $\alpha = (d-2)/2$, $d\in\mathbb{N}$ I can prove it in my way too. | |
| Feb 15, 2020 at 7:18 | vote | accept | MannyC | ||
| Feb 15, 2020 at 2:22 | history | answered | darij grinberg | CC BY-SA 4.0 |