Timeline for What is the indefinite sum of tan(x)?
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10 events
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| Jan 28, 2012 at 20:24 | comment | added | Zsbán Ambrus | I agree with Anixx: the function can be made sort of nice in that it's piecewise continuous, the set of the places of discontinuitites is discrete and has no accumulation point, though these discontinuities get closer and closer to each other as you go to infinity. You do this by choosing the function on the interval [0, 1) as an arbitrary continuous function, say all zeros, then extend the only possible way. | |
| Oct 21, 2010 at 17:43 | comment | added | Anixx | Besides this is is from general considerations evident that for any continuous analytic function on the interval from $(-\pi/2, \pi/2)$ can be constructed an indefinite sum function, continuous and analytic on the same interval plus a unit range. | |
| Oct 21, 2010 at 17:30 | comment | added | Anixx | Oleg's argument's claims are fairly correct until the final conclusion. He correctly noted that the function should have poles at x+1+k or x-k for any natural k. The key word here is "either". The continuous sum function can be arranged so that for any pole right of zero all consecutive poles would be in x+1+k points, so to the right of the first one and for each pole in the left of zero all poles would be in the points x-k, so to the left of it. Thus in the neighbourhood of zero itself the number of poles is limited. | |
| Oct 21, 2010 at 16:06 | comment | added | Noah Snyder | Oleg's argument would only work if there were more "mixing" of the two cases. For example, if the former case happened for infinitely many negative numbers then the argument would go through. But as long as all sufficiently negative numbers are in the latter case and all sufficiently positive ones are in the former case, then there's a hole in the argument. | |
| Oct 21, 2010 at 15:50 | comment | added | Noah Snyder | Hrm Gerald's explanation looks pretty convincing, so let's sort out what the error is here... Is it just that "dense subset if the reals" doesn't actually follow from the conclusion? Instead what happens is that if m is positive then we have the "former" case while if m is negative we have the "latter" case. Thus the poles get more common as x increases but is always discrete for any any given interval. | |
| Oct 21, 2010 at 15:38 | comment | added | Anixx | For all those who voted for this answer, please notice that there is a counter-example to Eroshkin's claims. Such functions exists, mostly continuous and its plot is shown. See the currently selected answer. | |
| Oct 17, 2010 at 18:09 | history | edited | Qfwfq | CC BY-SA 2.5 |
added 4 characters in body
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| Oct 8, 2010 at 8:53 | comment | added | Herman Tulleken | 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). | |
| Oct 8, 2010 at 8:53 | vote | accept | Herman Tulleken | ||
| Oct 18, 2010 at 5:57 | |||||
| Oct 5, 2010 at 2:38 | history | answered | Oleg Eroshkin | CC BY-SA 2.5 |