Timeline for Resultant of linear combinations of Chebyshev polynomials of the second kind
Current License: CC BY-SA 4.0
18 events
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| Sep 3, 2022 at 17:27 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Aug 2, 2022 at 0:22 | comment | added | Michael Hardy | @TerryTao I've now edited your answer by adding the part about the slashes. | |
| Aug 2, 2022 at 0:21 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Aug 1, 2022 at 14:54 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Jul 31, 2022 at 23:22 | comment | added | Terry Tao | @darijgrinberg (This identity, by the way, is a close cousin of the standard fact that the Fejer kernel is the average of Dirichlet kernels. Indeed I was led to this argument by inspecting the Fourier transform of $\sum_{k=0}^{n-1} U_k$ and observing its resemblance to the (Fourier transform of the) Fejer kernel.) | |
| Jul 31, 2022 at 23:18 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Jul 31, 2022 at 23:08 | comment | added | Terry Tao | @MichaelHardy OK, I let LaTeX do its thing with the parentheses. | |
| Jul 31, 2022 at 23:07 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Jul 31, 2022 at 23:06 | comment | added | Terry Tao | @darijgrinberg Sort of. From the usual trig addition formulae one gets $\sum_{k=0}^{n-1} U_k(\cos \theta) = U_n(\cos \frac{\theta}{2}) U_{n-1}(\cos \frac{\theta}{2}) / 2 \cos(\frac{\theta}{2})$. | |
| Jul 31, 2022 at 22:17 | comment | added | darij grinberg | "and so the $n-1 = \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n-1}{2} \rfloor$ zeroes of $\sum_{k=0}^{n-1} U_k$ take the form $\cos( \frac{2\pi j}{n+1} )$ for $1 \leq j < (n+1)/2$ and $\cos( \frac{2\pi j}{n} )$ for $1 \leq j < n/2$": This is very nice, but shouldn't this just mean that $\sum_{k=0}^{n-1} U_k$ factors into two smaller Chebyshevs? | |
| Jul 31, 2022 at 21:45 | comment | added | Michael Hardy | I wonder whether you prefer the typography of the first line below to the second or you simply don't consider this sort of thing worthy of attention or what? $$ \begin{align} & U_{n}( \cos( \frac{2\pi j}{n} ) ) = \sin((n+1)\frac{2\pi j}{n}) / \sin(\frac{2\pi j}{n}) = +1 \\ {} \\ & U_n \left( \cos\left( \frac{2\pi j}{n} \right) \right) = \left.\sin\left((n+1)\frac{2\pi j}{n}\right) \right/ \sin\left(\frac{2\pi j}{n}\right) = +1 \\ \end{align} $$ | |
| Jul 31, 2022 at 19:27 | comment | added | W. Wang | @rafik. the first equality follows from the definition of U_{n-1}.the second equality seems not correct. I think the result is -1. | |
| Jul 31, 2022 at 19:17 | comment | added | Terry Tao | Signs corrected (the point is that $\sin(x)$ is $2\pi$-periodic and odd). | |
| Jul 31, 2022 at 19:17 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Jul 31, 2022 at 19:08 | comment | added | W. Wang | Thank you for your beautiful and clear proof. | |
| Jul 31, 2022 at 19:08 | comment | added | zeraoulia rafik | pleas how you got this :$ U_{n-1}( \cos( \frac{2\pi j}{n+1} ) ) = \sin(n\frac{2\pi j}{n+1}) / \sin(\frac{2\pi j}{n+1}) = (-1)^{j+1}$ ? is that for $ n\to \infty $ | |
| Jul 31, 2022 at 19:01 | vote | accept | W. Wang | ||
| Jul 31, 2022 at 18:35 | history | answered | Terry Tao | CC BY-SA 4.0 |