Timeline for Possible bug in finite sum over inverse squares $\sum\limits_{i=1}^n \frac{1}{(x (n-i)+i)^2}$
Current License: CC BY-SA 4.0
16 events
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| May 19, 2020 at 7:40 | comment | added | Dr. Wolfgang Hintze | @ Max1 See my edit with the official rejection of the bug. | |
| May 19, 2020 at 7:39 | history | edited | Dr. Wolfgang Hintze | CC BY-SA 4.0 |
Official response added
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| May 11, 2020 at 9:03 | comment | added | Dr. Wolfgang Hintze | @ Max1 Thanks for your comment. The question of uniqueness is indeed a valid one, Normally it is additionally requested that the continuation has a simple analytic form. You can add any term of the form $\sin (n p)$ to the result. But this extension has exponential growth in the complex $n$-plane. As a useful and typical exercise you can try to derive an integral representation for the harmonic sum $H_n = \sum_{k=1}^n \frac{1}{k}$. | |
| May 11, 2020 at 8:23 | history | edited | Dr. Wolfgang Hintze | CC BY-SA 4.0 |
added 1007 characters in body
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| May 10, 2020 at 20:44 | comment | added | Max1 | @ Dr. Wolfgang Hintze Thanks for your comment, I now understood the way you derived your analytic continuation. I'm not sure about uniqueness since there might be more than one integral representation, but I see that you only need one which yields correct results for integer n. If you find the time it would surely be very interesting to get an update about the result of the bug report. | |
| May 10, 2020 at 20:28 | comment | added | Dr. Wolfgang Hintze | 1) Don't worry and save your time, Soler's answer is plainly wrong as I have explained now to full extent. He should correct or withdraw it. 2) I recommend you to look into this topic: to find an integral representation is a normal way to extend a sum as a function of n analytically. A famous example is the harmonic sum. There are sometimes even simpler methods like $\sum_1^n k = \frac{1}{2}n(n+1)$. 3) And finally, I think you are making things more complicated than necessary: there's a bug in Mathematica but I have obtained expressions for all real values of the parameter $x>0$ and any $n$. | |
| May 10, 2020 at 20:02 | comment | added | Max1 | @ Dr. Wolfgang Hintze I have totally not understood the reasoning behind soner's method, so I cannot really comment on it. I am skeptical about considering real n, tho. The sum only allows integer n as it's limit. Hence the sum cannot be regarded as an analytic function w.r.t n. Since the sum is not an analytic function it is not accessible to analytic continuation as far as I can see. I have expanded my own answer to give reasoning, why n should not be considered as the independent variable and how to correct the limiting process with regard to the polygamma expression given by mathematica. | |
| May 10, 2020 at 7:51 | comment | added | Dr. Wolfgang Hintze | @ Max1 Please see my updated comment on Soler's wrong answer. There I am considering real n, and show that his formla is wrong except for n=Integer+-1/3 using the reflection formula for the polylog functions. We have understood more now but still there is a bug in Mathematica (I have filed a bug report). | |
| May 10, 2020 at 7:45 | history | edited | Dr. Wolfgang Hintze | CC BY-SA 4.0 |
elaborate more on the wrong answer of Soler
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| May 9, 2020 at 22:08 | comment | added | Max1 |
One very important point. We employ different limiting procedures. I had suggested the limit Limit[PolyGamma[1,1-2n+c]-PolyGamma[1,1-n+c], c->0]. This limit exists and yield the expected value for n=10. This limit is different from the limit in the OP which could be written as Limit[PolyGamma[1,1-2(n+c)]-PolyGamma[1,1-(n+c)], c->0] and does not exist (divergent). In my opinion the limiting procedure is the key point to this question and I'll try to elaborate on that in my answer tomorrow.
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| May 9, 2020 at 18:11 | comment | added | Max1 |
Comment regarding "Comment on Soler's answer": According to my analysis. The expression PolyGamma[1, 1 - 2 n] - PolyGamma[1, 1 - n] does not diverge for integer n. It is ill defined for integer n. This is similar to how $f:\mathbb{R}\setminus\{0\}\to\mathbb{R}$ with $f(x)=1/x-1/x$ would be ill defined at 0. Nevertheless $f(x)$ has a natural extension to all of $\mathbb{R}$, namely $g:\mathbb{R}\to\mathbb{R}$ with $g(x)=0$. So I think "divergent" is incorrect here (to be proofed).
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| May 9, 2020 at 17:43 | history | edited | Dr. Wolfgang Hintze | CC BY-SA 4.0 |
Comment on the most appreciated answer
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| May 9, 2020 at 17:20 | comment | added | Dr. Wolfgang Hintze | @Max1 Thanks for you opinion but I don't agree: IMHO is it sufficient to state that a wrong result provided by Mathematica is wrong, and that this should be considered as a bug. I shall file a bug report so that the Mathematica experts can search for reasons of the bug and try to remove it in the next version. This is not the first time I have filed a bug report. Let's see. | |
| May 9, 2020 at 13:59 | comment | added | Max1 |
This is an interesting analysis, but it fails to address the point why mathematicas result sg[2,n] is to be considered wrong. It is just states that as a fact. The expression sg[2,n] is ill defined for $n\in\mathbb{N}>0$ (as far as I can see), but I do not see a problem with that as GenerateConditions -> False is the default setting of Sum. The GenerateConditions point is addressed in the answer by Soner. Without GenerateConditions -> True I would not expect mathematica to accompany the solution with conditions on x where the solution is applicable.
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| May 9, 2020 at 8:12 | history | edited | Dr. Wolfgang Hintze | CC BY-SA 4.0 |
added 413 characters in body
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| May 9, 2020 at 7:51 | history | answered | Dr. Wolfgang Hintze | CC BY-SA 4.0 |