Timeline for What is the indefinite sum of tan(x)?
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36 events
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| May 27, 2014 at 15:21 | history | edited | Anixx | CC BY-SA 3.0 |
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| Jun 25, 2013 at 5:23 | history | wiki removed | Kim Morrison | ||
| Oct 22, 2010 at 8:24 | comment | added | Anixx | @Herman Tulleken Here is it: $$\prod _x \tan(x) = \frac{e^{\frac{1}{2} i (-2 x+\pi +2) x} \left(-1;e^i\right){}_x \left(e^i;e^i\right){}_{x-1}}{\left(-1;e^{-2 i}\right){}_x}$$ | |
| Oct 22, 2010 at 7:30 | comment | added | Herman Tulleken | @Gerald, I hope it can, especially since more generally $\sum \frac{a}{1-q^x} = a\frac{\psi_{q}(x)}{\ln q} - ax$ whenever everything is nicely defined. With $a = i$ and $q = e^{2i}$, we get the $\tan$ (translated) case. | |
| Oct 22, 2010 at 7:04 | comment | added | Herman Tulleken | Also, Annix, could you please give the link to the indefinite product of $\tan x$ you had here somewhere again? There is a nice relationship between $\int T(x)$ and $\prod \tan x$ which I wish to investigate. | |
| Oct 21, 2010 at 22:50 | comment | added | Anixx | functions.wolfram.com/ElementaryFunctions/Cot/06/02/0001 | |
| Oct 21, 2010 at 18:58 | comment | added | Gerald Edgar | QUESTION: Is there some sense in which Anixx's other solution, the one I liked, can be interpreted as $ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)$ ?? | |
| Oct 20, 2010 at 18:06 | comment | added | J. M. isn't a mathematician | If you're trying to subtract two divergent quantities to get a meaningful answer, shouldn't you be trying to figure out first why exactly those divergences cancel out? | |
| Oct 20, 2010 at 15:44 | comment | added | Anixx | See my new answer with the plot of this function. | |
| Oct 18, 2010 at 21:40 | comment | added | Anixx | Gerald Edgar, I did best of what I could. I think this is the most close answer to the question. | |
| Oct 18, 2010 at 21:11 | comment | added | Gerald Edgar | @Anixx: But, so far, it has no known definition. So should we cite it as the solution to a problem? | |
| Oct 18, 2010 at 18:27 | comment | added | Anixx | Yes, it is not supported by Mathematica. This does not mean it does not exist. | |
| Oct 18, 2010 at 17:25 | comment | added | Gerald Edgar | Now do $q$-Polygamma with $q=e^{2i}$, which is needed here! For example: QPolyGamma[0.1234,Exp[2*I]] // N | |
| Oct 18, 2010 at 16:22 | comment | added | Anixx | @ Gerald Edgar, on this page mathworld.wolfram.com/q-PolygammaFunction.html they use q-polygamma with |q|=e are they wrong? Also Mathematica evaluates q-polygamma with |q|>1 quite well. reference.wolfram.com/mathematica/ref/QPolyGamma.html | |
| Oct 18, 2010 at 13:19 | comment | added | Gerald Edgar | Note that $\psi_q(x)$ is defined only if $|q|<1$, which fails for $q=e^{2i}$. The series (or product) involved diverges even for complex $x$. | |
| Oct 18, 2010 at 6:04 | comment | added | Herman Tulleken | @Anixx Very interesting, thank you! How did you discover this? | |
| Oct 18, 2010 at 5:57 | vote | accept | Herman Tulleken | ||
| Oct 20, 2010 at 18:46 | |||||
| Oct 18, 2010 at 5:13 | comment | added | Gerry Myerson | @Anixx, I don't know what you mean by "the question caption." I see no infinite sum anywhere in the question, nor any sum over an infinite range (although I see several sums over an unspecified range), so I regret to say that pointing to the question has not helped me to deconfuse the two kinds of sum. Please be patient with me and try again. Thanks. | |
| Oct 17, 2010 at 23:41 | comment | added | Anixx | @ Gerald Edgar This function has wider domain of definition than the real axis and the values on the real axis may be can be assumed using limit technique. Even if not, then this is a solution for the complex plane except real axis. | |
| Oct 17, 2010 at 23:18 | comment | added | Gerald Edgar | @Anixx: See (1) and (2) in the Mathworld page mathworld.wolfram.com/q-PolygammaFunction.html Note that (2) involves an INFINITE series, and (1) -- if you follow the link involved -- involves an INFINITE product. Both of these are DIVERGENT for almost all real x in the case of $e^{2i}$. | |
| Oct 17, 2010 at 23:17 | comment | added | Gerald Edgar | By the way: ask Wolfram Alpha to plot the function $ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)$ or to numerically evaluate it for some real number $x$. | |
| Oct 17, 2010 at 22:47 | comment | added | Anixx | Maybe by absence of the sign of infinity? And also this is clearly stated in the question caption. | |
| Oct 17, 2010 at 22:40 | comment | added | Gerry Myerson | I have always confused an infinite sum with a sum over an infinite range. How may I deconfuse the two? | |
| Oct 17, 2010 at 21:52 | comment | added | Anixx | Gerald, you are confusing the infinite sum with sum over infinite range. There is no infinite series here. | |
| Oct 17, 2010 at 21:18 | comment | added | Gerald Edgar | And the $q$-Pochammer symbol $(a,q)_\infty$ with $q=e^{2i}$ in the definition (1) also diverges for almost all $a$. | |
| Oct 17, 2010 at 21:11 | comment | added | Gerald Edgar | @Mariano: of course not. The series (2) that "defines" it diverges for almost all real $x$ | |
| Oct 17, 2010 at 18:02 | history | edited | Anixx | CC BY-SA 2.5 |
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| Oct 17, 2010 at 15:34 | comment | added | Mariano Suárez-Álvarez | Is the q-polygamma well-defined at $q=e^{2i}$? | |
| Oct 17, 2010 at 8:56 | history | edited | Anixx | CC BY-SA 2.5 |
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| Oct 17, 2010 at 8:51 | history | edited | Anixx | CC BY-SA 2.5 |
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| Oct 17, 2010 at 0:02 | history | edited | Anixx | CC BY-SA 2.5 |
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| Oct 16, 2010 at 23:48 | history | edited | Anixx | CC BY-SA 2.5 |
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| Oct 16, 2010 at 23:40 | history | edited | Anixx | CC BY-SA 2.5 |
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| Oct 16, 2010 at 23:01 | history | edited | Anixx | CC BY-SA 2.5 |
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| Oct 16, 2010 at 22:55 | history | edited | Anixx | CC BY-SA 2.5 |
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| Oct 16, 2010 at 22:47 | history | answered | Anixx | CC BY-SA 2.5 |