most recent 30 from http://mathoverflow.net 2013-05-24T15:01:17Z http://mathoverflow.net/feeds/user/7540 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx Herman Tulleken 2010-10-04T12:24:21Z 2012-01-28T11:49:48Z

<p>What is the indefinite sum of the tangent function, that is, the function $T$ for which </p> <p>$\Delta_x T = T(x + 1) - T(x) = \tan(x)$</p> <p>Of course, there are infinitely many answers, who all differ by a function of period 1. Ideally, I would like the solution to be of the form</p> <p>$T(x) = $ nice_function$(x)$ + possibly_ugly_periodic_function$(x)$, where nice is at least piece-wise continuous.</p> <p>If any of the following sums can be found, the sum of tan can also be found:</p> <ul> <li><p>$\sum \sec x$</p></li> <li><p>$\sum \csc x$</p></li> <li><p>$\sum \cot x$</p></li> <li><p>$\sum \frac{1}{e^{ix} + 1}$</p></li> </ul> <p>I have tried several methods without success, including using a newton series (which does not converge for non-integer $x$), and trying to guess possible functions. </p> <p>I would also appreciate lines of attack if a solution is not known.</p>

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42432#42432 Herman Tulleken 2010-10-22T09:43:49Z 2010-10-22T09:43:49Z

Thank you Anixx!

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42432#42432 Herman Tulleken 2010-10-22T07:30:27Z 2010-10-22T07:30:27Z

@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.

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42432#42432 Herman Tulleken 2010-10-22T07:04:06Z 2010-10-22T07:04:06Z

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.

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42975#42975 Herman Tulleken 2010-10-21T14:28:43Z 2010-10-21T14:28:43Z

Indeed, it is easy to prove that $T(z + \pi) - T(z) = -[\Psi(1 - (\pi/2 + z)) -\Psi(\pi/2 + z)] = -\pi \cot \pi(\pi/2 + z)$.

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42975#42975 Herman Tulleken 2010-10-21T13:08:34Z 2010-10-21T13:08:34Z

Yes, thanks for the additional information. Interestingly, it looks like $T(x + \pi) - T(x)$ is also periodic, with period 1.

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42903#42903 Herman Tulleken 2010-10-20T18:46:32Z 2010-10-20T18:46:32Z

So far, this answer is more useful than the other one, so I will accept it (even though it is not immediately clear that it will converge, it is something that we can work with).

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42903#42903 Herman Tulleken 2010-10-20T17:31:21Z 2010-10-20T17:31:21Z

I have made the periodic assumption you speak of (and of all the possible solutions, does this not hold for 1 of them?) to obtain a plot of certain points (several thousand) in the interval $[0, 1)$ In this case the function can have any value in any interval, however small (if I reason correctly). My initial hope was that we could add a periodic function (period 1) to this and get something like your blue function.

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42903#42903 Herman Tulleken 2010-10-20T16:52:43Z 2010-10-20T16:52:43Z

Am I reading the graph right? The blue graph ($T$) cuts the $x$-axis near (just before) 3, and near (just before) 4, $T$ is very large. But just before 3, $\tan x$ is close to 0. Can this be? $T(x+1)−T(x) = LARGE - 0 \neq SMALL = \tan x$ where x is just below 3. (Sorry, I know this is clumsy and inaccurate, but I hope you get my point).

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42432#42432 Herman Tulleken 2010-10-18T06:04:33Z 2010-10-18T06:04:33Z

@Anixx Very interesting, thank you! How did you discover this?

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42144#42144 Herman Tulleken 2010-10-15T10:48:49Z 2010-10-15T10:48:49Z

Thanks, interesting; could you please also post the link to the article containing that expression?

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/41098#41098 Herman Tulleken 2010-10-08T08:53:58Z 2010-10-08T08:53:58Z

It's interesting that a simple setup can lead to a complicated function like this. Thanks for your answer. (I am still interested if this is somehow a studied function).

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/41034#41034 Herman Tulleken 2010-10-05T07:49:41Z 2010-10-05T07:49:41Z

Ok, a few quick tests in Matlab shows that even in the range $[0 \pi/2]$ that the indefinite sum of Taylor series does not converge. In light of what Oleg Eroshkin said below, this makes sense, so this method cannot be used for finding an approximation.

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/41034#41034 Herman Tulleken 2010-10-04T16:05:00Z 2010-10-04T16:05:00Z

The indefinite integral (typically) requires the use of substitution, which exists because of the &quot;nice&quot; chain rule for differentiation. Differencing does not have a nice chain rule (there are some variants, but they do not help in this case). I'll play with truncated Taylor series and see where it gets me (I guess I should have tried that already).

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/41031#41031 Herman Tulleken 2010-10-04T15:29:07Z 2010-10-04T15:29:07Z

Yes, I am pretty sure if there is a simple expression, it is not an elementary function, which is why these symbolic programs don't know it. But it might still be a studied special function...

http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/41021#41021 Herman Tulleken 2010-10-04T14:53:28Z 2010-10-04T14:53:28Z

Thank you. I know the solution to $T(x + \pi) - T(x)$, but I am hoping for a solution to the $T(x+1) - T(x)$ case. But if there is no &quot;simple&quot; solution, what avenues can I explore to get some sort of solution?