Timeline for Questions on analytic representations of the Kronecker delta function $\delta(x-1)$ and the Moebius function $\mu(n)$
Current License: CC BY-SA 4.0
60 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2021 at 22:14 | vote | accept | Steven Clark | ||
| Jun 9, 2021 at 17:03 | comment | added | Steven Clark | @მამუკაჯიბლაძე Were you not satisfied with the proof of $\tilde{f}_a'(x)=\underset{N\to\infty,\,M(N)=0}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^\infty\cos\left(\frac{2\pi k x}{n}\right)\right)=\delta(x+1)+\delta(x-1)$ at mathoverflow.net/q/362566 which also implies $\tilde{f}_a(x)=\underset{N\to\infty,\,M(N)=0}{\text{lim}}\left(\frac{1}{\pi}\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^\infty\frac{\sin\left(\frac{2\pi k x}{n}\right)}{k}\right)=-1+\theta(x+1)+\theta(x-1)$? | |
| Jun 9, 2021 at 17:02 | comment | added | Steven Clark | @მამუკაჯიბლაძე With respect to your comment "As for (d) and (e), sorry, I don't see how they have been proven", I only said they were proven for the case $a(n)=\delta_{n-1}$ where $b(n)=\mu(n)$, $\tilde{f}_a'(x)=\delta(x+1)+\delta(x-1)$, and $\tilde{f}_a(x)=-1+\theta(x+1)+\theta(x-1)$. | |
| Jun 7, 2021 at 23:28 | comment | added | Steven Clark | @მამუკაჯიბლაძე I believe the closed form representations of formula (g) for $\tilde{f}_a'(x)$ related to $\zeta(s)$ and $\eta(s)$ at $f=1$ are $\tilde{f}_a'(x)=1+\cos(2 \pi x)-\frac{\sin(2 \pi x)}{\pi x}$ and $\tilde{f}_a'(x)=\frac{\sin (2 \pi x)}{\pi x}-2\cos (\pi x)$. Can we generalize these closed form representations as a function of the evaluation frequency $f$? | |
| Jun 7, 2021 at 23:28 | comment | added | Steven Clark | @მამუკაჯიბლაძე For the cases $a(n)=1$ and $a(n)=(-1)^{n-1}$ where $F_a(s)=\zeta(s)$ and $F_a(s)=\eta(s)$, I suspect it would be easier to illustrate convergence over much larger ranges of $f$ and $x$ if we could come up with generalized closed form representations of formula (g) for $\tilde{f}_a'(x)$ which then could be integrated to derive generalized closed form representations of formula (f) for $\tilde{f}_a(x)$. | |
| Jun 7, 2021 at 22:51 | comment | added | Steven Clark | @მამუკაჯიბლაძე For the case $\delta_{x-1}$ I can only get get formula (f) for $\tilde{f}_a(x)$ to converge at $x=2.01$ for $f\in\{1,2,3\}$ which I suspect is related to the fact that I can only get it to converge at $f=1$ for $|x|<~4.5$. As I said before, the convergence range decreases each time you increase the value of $f$, but the decrease in the convergence range is not quite as linear as I originally indicated. | |
| Jun 7, 2021 at 21:38 | comment | added | Steven Clark | @მამუკაჯიბლაძე For the case $a(n)=\delta_{n-1}$ you should be using $F_a(s)=1$, and note there's an extra $x$ in the numerator in formula (f) in my answer below. | |
| Jun 7, 2021 at 21:16 | comment | added | Steven Clark | @მამუკაჯიბლაძე I'm not sure what you're doing. Here's a plot of formula (f) for $\tilde{f}_a(x)$ for the case of $a(n)=\delta_{x-1}$ where formula (f) is evaluated at $f=2$ and $K=200$: i.sstatic.net/5ggFE.jpg. Formula (f) is shown in orange overlaid on the blue reference function $f(x)=-1+\theta(x+1)+\theta(x-1)$. Are you sure you have formula (f) entered correctly? | |
| Jun 7, 2021 at 20:50 | comment | added | მამუკა ჯიბლაძე | Actually I now looked at the $\delta$ case for a specific $x=2.01$, and, unless I am doing something wrong, the limit in (f) is $\infty$ no matter how I increase $f$ and $K$. As for (d) and (e), sorry, I don't see how they have been proven. | |
| Jun 7, 2021 at 20:05 | comment | added | Steven Clark | @მამუკაჯიბლაძე Formulas (d) and (e) in my answer below have already been proven for the case $a(n)=\delta_{n-1}$ where $b(n)=\mu(n)$ and $\tilde{f}_a'(x)=\delta(x+1)+\delta(x-1)$. I think what's important going forward is a proof of the validity of formulas (f) and (g) in my answer below for all definitions of $a(n)$ for which $F_a(s)=\sum\limits_n a(n)\,n^{-s}$ converges for $\Re(s)\ge 2$. | |
| Jun 7, 2021 at 20:05 | comment | added | Steven Clark | @მამუკაჯიბლაძე The exact equivalence of formulas (4) and (5) for $\tilde{a}(x)$ in my question above at non-integer values of $x$ is no longer as important to me now that I understand the higher-level context revealed in formulas (f) and (g) in my answer below. | |
| Jun 7, 2021 at 19:49 | comment | added | Steven Clark | @მამუკაჯიბლაძე But beware when evaluating formula (f) each time you increase the evaluation frequency $f$ while holding $K$ constant you decrease the range of convergence over $x$. For example, if formula (f) diverges for $|x|>12$ when evaluated at $f=1$, it will diverge for $|x|>6$ when evaluated at $f=2$, and it will diverge for $|x|>4$ when evaluated at $f=3$ (assuming all three evaluation frequencies are evaluated with the same value of $K$). | |
| Jun 7, 2021 at 19:44 | comment | added | Steven Clark | @მამუკაჯიბლაძე Formulas (e) and (g) for $\tilde{f}_a'(x)$ in my answer below diverge to $\infty$ as $f\to\infty$ when $x=n\in\mathbb{Z}\land a(n)\ne 0$ because they're analytic representations which converge in a distributional sense to $f_a'(x)=\sum\limits_n a(n)\,\delta(x-n)$ where $\delta(x)$ is the Dirac delta function. Are you not familiar with the Fourier series for the Dirac comb? Try evaluating formulas (d) and (f) for $\tilde{f}_a(x)$ and you'll see they converge to $f_a(x)=\sum\limits_n a(n)\,\theta(x-n)$ where $\theta(x)$ is the Heaviside step function. | |
| Jun 7, 2021 at 19:09 | comment | added | მამუკა ჯიბლაძე | Sorry, I don't understand these (f) and (g) at all. It seems the sum converges for each fixed $f$ but goes to infinity as $f$ increases. Why do you include this $f$? With $f=1$ this is very similar to the series by Risomar Sousa but for large $f$ seems to be unbounded. And in what sense could they be equivalent to (d) and (e) I also don't understand at all. | |
| Jun 7, 2021 at 18:49 | comment | added | Steven Clark | @მამუკაჯიბლაძე For the case $a(n)=1$ associated with $F_a(s)=\zeta(s)$, I believe the analytic formula for $\tilde{f}_a'(x)$ defined in formula (g) in my answer below has the closed form representation $\tilde{f}_a'(x)=1+\cos(2 \pi x)-\frac{\sin(2 \pi x)}{\pi x}$ when evaluated at $f=1$. Do you have any insight on how to generalize this closed form representation to all positive integer values of $f$ for the case $a(n)=1$ associated with $F_a(s)=\zeta(s)$? | |
| Jun 7, 2021 at 16:38 | comment | added | Steven Clark | @მამუკაჯიბლაძე By the way, reuns provided a proof of formula (e) in my answer below for the case $a(n)=\delta_{x-1}$ where $b(n)=\mu(n)$ and $\tilde{f}_a'(x)=\delta(x+1)+\delta(x-1)$ in the answer posted to my Math Overflow question at mathoverflow.net/q/362566. Formulas (e) and (g) in my answer below only converge in a distributional sense, and I'm not sure what that implies about convergence of either formula when evaluated at the finite limit $f=1$ associated with the question above. | |
| Jun 7, 2021 at 16:00 | comment | added | Steven Clark | @მამუკაჯიბლაძე But formula (4) in my answer above is simply formula (g) in my answer below normalized by $\frac{1}{2f}$ and then evaluated at $f=1$, just as formula (5) in my answer above was simply formula (e) in my answer below normalized by $\frac{1}{2f}$ and then evaluated at $f=1$. | |
| Jun 7, 2021 at 15:56 | comment | added | Steven Clark | @მამუკაჯიბლაძე I recommend you just ignore my original formulas and their conjectured relationship with the formulas in the PDF and focus on formulas (f) and (g) in the answer I posted below. Whereas my original formulas (d) and (e) in my answer below require evaluation at specific values of N (which you're not comfortable with), the corresponding formulas (f) and (g) in my answer below eliminate this requirement. | |
| Jun 7, 2021 at 4:00 | comment | added | მამუკა ჯიბლაძე | There are zillions of entire functions with prescribed values at integers. Among them, the ones given by particular series indicated in that preprint by Risomar Sousa seem to have interesting connection to number theory. I don't see anything significant going on beyond that. In particular, all of your sums require specifying a particular summation method, otherwise it is completely unclear which globally defined function do you mean by them. Different summations might converge to different functions, and so far I don't even see whether any of them provide any well-defined function at all. | |
| Jun 6, 2021 at 14:14 | comment | added | Steven Clark | @მამუკაჯიბლაძე The answer I posted below illustrates that reuns as well as the author of the paper really only understood the tip of the iceberg. | |
| Jun 6, 2021 at 1:40 | answer | added | Steven Clark | timeline score: -2 | |
| Jun 4, 2021 at 2:09 | comment | added | Steven Clark | @მამუკაჯიბლაძე I don't expect reuns to accept the answer, but I provided information related to his formula that perhaps might be of interest to others. The question by reuns was soliciting feedback from the author of the PDF whose account has been terminated. | |
| Jun 4, 2021 at 2:06 | comment | added | მამუკა ჯიბლაძე | Well, frankly speaking I strongly doubt that @reuns will accept this answer - the question was about anything essential from that preprint that has not been addressed. | |
| Jun 3, 2021 at 23:54 | comment | added | Steven Clark | @მამუკაჯიბლაძე Did you see the answer I posted to reuns related question on Math StackExchange at math.stackexchange.com/q/4160465? Also, for some definitions of $a(n)$ (e.g. $a(n)=\delta_{n-1}$ and $a(n)=\mu(n)$) formula (3) above seems to evaluate similar the real part of the generalization of reuns formula I posted in a comment above when the inner sum is started at $m=-1$ instead of $m=1$ except there seems to be an offset related to $a(n)$ (and perhaps the evaluation limit). | |
| Jun 3, 2021 at 21:58 | history | edited | Steven Clark | CC BY-SA 4.0 |
Corrected formula references in paragraphs preceding figures (3) and (5)
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| Jun 3, 2021 at 18:57 | comment | added | მამუკა ჯიბლაძე | Yes, I had in mind analog of (3). It should be easy to derive the analog of (4) too | |
| Jun 3, 2021 at 17:01 | comment | added | Steven Clark | @მამუკაჯიბლაძე Thanks. In the generalized formula I posted in a comment above I started the inner series at $m=1$ (which reuns didn't specify). I see you started the inner sum at $m=0$ instead of $m=1$ which I was also investigating because I noticed it seemed to be more closely related to the PDF formula when it's inner series is started at $j=1$ instead of $j=0$, and $j=1$ gives the correct evaluation of $a(0)$. | |
| Jun 3, 2021 at 16:16 | comment | added | მამუკა ჯიბლაძე | Yes, the coefficients sum up to $\frac{(2 \pi i)^k (k-1)}{2 (k+1)!}$ for $\zeta$ and to $\frac{(2 \pi i)^k \left(2^k-k-1\right)}{2^k (k+1)!}$ for $\eta$; accordingly, the series gives $1+\frac{\left(1-e^{2 i \pi x}\right) (1-i \pi x)}{2 \pi i x}$ for $\zeta$ and $-\frac{1-e^{2 i \pi x}}{2 \pi i x}-e^{i \pi x}$ for $\eta$ | |
| Jun 3, 2021 at 14:47 | comment | added | Steven Clark | @მამუკაჯიბლაძე The PDF formula has closed-form expressions for the cases $a(n)=1$ and $a(n)=(-1)^{n-1}$ associated with $\zeta(s)$ and $\eta(s)$ which are $\frac{1}{2}(1+\cos(2 \pi x))$ and $-cos(\pi x)$. I'm wondering if the formula posted by reuns also has closed-form expressions for the cases $a(n)=1$ and $a(n)=(-1)^{n-1}$. | |
| Jun 2, 2021 at 19:33 | comment | added | მამუკა ჯიბლაძე | It is closely related. Both entire functions can be obtained from the series $f(x)=\sum\frac{a_n/n^2}{1-x^2/n^2}$ (which only has limited radius of convergence). In the question $f(x)$ is multiplied by $\frac{1-e^{2\pi i x}}{2\pi i}$ while to obtain series from the preprint you multiply $f(x)$ by $\frac{\sin(2\pi x)}{2\pi}$ | |
| Jun 2, 2021 at 18:35 | comment | added | Steven Clark | @მამუკაჯიბლაძე I think reuns is saying that the analytic formula for $\tilde{\mu}(x)$ can be generalized as follows: $\tilde{a}(x)=\underset{K\to\infty}{\text{lim}}\left(-2\sum\limits_{k=2}^K x^k \sum\limits_{m=1}^{\frac{k}{2}-1}\frac{(2 i \pi )^{k-2 m-2}\,F_a(2 m+2) }{(k-2 m-1)!}\right)$. But I don't see how this relates to the PDF formulas. This formula generally evaluates to a complex result for $x\in\mathbb{R}$ except at the integers, whereas the PDF formulas evaluate to a strictly real result for $x\in\mathbb{R}$. | |
| Jun 2, 2021 at 17:25 | history | edited | Steven Clark | CC BY-SA 4.0 |
Corrected n to x in formulas (3) and (4).
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| Jun 2, 2021 at 15:50 | comment | added | მამუკა ჯიბლაძე | In subsequent comments (s?)he claims that the same proof works also for the functions from that preprint. | |
| Jun 2, 2021 at 15:45 | comment | added | Steven Clark | @მამუკაჯიბლაძე Actually the question and comments posted by reuns imply some of the formulas in the PDF don't converge. Note reuns derived a different formula for $\mu(n)$, and note his comment: The pdf is sloppy/wrong, that's why my post starts with "in order to repair". This is actually why I asked if you had investigated convergence of the PDF formulas. | |
| Jun 2, 2021 at 15:34 | comment | added | მამუკა ჯიბლაძე | The same value as you, $2.999$. In the meanwhile I reached 433180, looks more or less the same. Still, here it is: i.sstatic.net/h2gWz.png. I guess the problem is that there are larger and larger gaps between the consecutive $N$ satisfying your condition 2 (I still do not see why it might be necessary). At each such gap the partial sums jump significantly; on my plots these are the straight line segments. As for (3) and (4) they seem to converge always. In fact in math.stackexchange.com/q/3493770/214353 @reuns claims to have a proof that these are indeed entire functions. | |
| Jun 2, 2021 at 13:51 | comment | added | Steven Clark | @მამუკაჯიბლაძე What value of $x$ did you use in your last plot? I know the oscillation in $\tilde{a}(x)$ associated with the oscillation in the offset converges to zero as $N\to\infty$ because the offset itself converges to zero as $N\to\infty$ (which is equivalent to the PNT). But it's possible there are other factors contributing to the oscillation in $\tilde{a}(x)$, and it's also possible formulas (3) and (4) don't converge for the cases $a(n)=\delta_{n-1}$ and/or $a(n)=\mu(n)$. Have you investigated the convergence of formulas (3) and (4) at a specific value of $x$ as $N\to\infty$? | |
| Jun 2, 2021 at 12:53 | comment | added | მამუკა ჯიბლაძე | I think for such cases one must look further. I did run it until $N$ about 380000 or so, and I think the picture does not suggest that the oscillations eventually will settle down. For $\delta$, the subsequence of those $N$ satisfying $\sum_{n\leqslant N}b(n)=0$ goes like $2,39,40,58,65,93,...,385111,385112,385139,385140,...$; looking at the plot of the partial sums for these $N$ i.sstatic.net/hGi7S.png I don't think one can say with confidence that there will be definite limit, no? | |
| Jun 1, 2021 at 22:13 | comment | added | Steven Clark | @მამუკაჯიბლაძე For example, here's a plot analogous to Figure (7) for the case $x=2.999$: i.sstatic.net/XPBjV.jpg. Note $\tilde{a}(2.999)$ seems to evaluate to almost a constant as a function of increasing magnitudes of $N$ where $\tilde{a}(0)=M(N)=0$. | |
| Jun 1, 2021 at 22:05 | comment | added | Steven Clark | @მამუკაჯიბლაძე When I evaluate formula (5) for $\tilde{a}(x)$ for the case $a(n)=\delta_{n-1}$ at non-integer values like $x=0.001$, $x=1.001$, and $x=2.999$ in plots analogous to the evaluation at $x=1.5$ illustrated in Figure (7) above, I get virtually no variation in the evaluation result which is why I selected $x=1.5$ for Figure (7) above. So I don't understand why you claim formula (5) doesn't converge when evaluated in the proximity of integer values of $x$. | |
| Jun 1, 2021 at 20:46 | comment | added | Steven Clark | @მამუკაჯიბლაძე I added Figure (7) and a couple of paragraphs to illustrate why I believe formula (5) for $\tilde{a}(x)$ converges as $N\to\infty$. I don't really understand your point. There are many functions which can only be approximated by partial sums of an infinite series, and it's not unusual to have to increase the evaluation limit $N$ as the magnitude of $x$ increases. You can't just stop at a finite value of $N$ and expect series like those defined in formulas (3), (4), and (5) above to converge for all values of $x$. | |
| Jun 1, 2021 at 20:41 | history | edited | Steven Clark | CC BY-SA 4.0 |
Added Figure (7) and a couple of related paragraphs in an attempt to clarify the relationship of the evaluations of formula (5) at non-integers and the offset defined in formula (6).
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| Jun 1, 2021 at 18:52 | history | edited | Steven Clark | CC BY-SA 4.0 |
Correct n to x at the end of second to last paragraph.
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| Jun 1, 2021 at 18:26 | history | edited | Steven Clark | CC BY-SA 4.0 |
Added case for a(n)=(-1)^(n-1) following question (2).
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| Jun 1, 2021 at 17:31 | comment | added | მამუკა ჯიბლაძე | Of course for $N$ with the partial sum at 0 equal to what I want you will get at 0 what you want. But I doubt you can obtain any definite value in a neighborhood of 0 except at 0 itself. And in any case you cannot stop at any $N$ since you need agreement with $a(n)$ for all $n$, so I don't see any point in considering these partial sums. It can also happen that for some subsequences of $N$ satisfying (2) you get some limiting values at some points, and these values can be different for different subsequences. | |
| Jun 1, 2021 at 16:13 | comment | added | Steven Clark | @მამუკაჯიბლაძე For the case $a(n)=\mu(n)$, when $\sum\limits_{n=1}^\infty\frac{b(n)}{n^s}=\frac{1}{\zeta (s)^2}$ is evaluated at $s=0$, the left-hand side once again corresponds to formula (7) for $\tilde{a}(0)$ and in this case the right-hand side corresponds to $\frac{1}{\zeta(0)^2}=4$ which is consistent with Figure (4). | |
| Jun 1, 2021 at 16:13 | comment | added | Steven Clark | @მამუკაჯიბლაძე For the case $a(n)=\delta_{n-1}$, when $\sum\limits_{n=1}^\infty\frac{\mu(n)}{n^s}=\frac{1}{\zeta (s)}$ is evaluated at $s=0$, the left-hand side corresponds to formula (7) for $\tilde{a}(0)$ and the right-hand side corresponds to $\frac{1}{\zeta(s)}=-2$ which is consistent with Figure (2). | |
| Jun 1, 2021 at 16:13 | comment | added | Steven Clark | @მამუკაჯიბლაძე I'm working on a plot that illustrates why I believe formula (5) for $\tilde{a}(x)$ converges when conditions (1) and (2) are met. In the mean time, here are a couple of interesting observations. | |
| Jun 1, 2021 at 15:27 | comment | added | მამუკა ჯიბლაძე | Seems to diverge for $\delta$ too. Actually I do not understand why your condition (2) is necessary. If, say, the sum of $b(n)$ converges, you do have a definite value for $\tilde a(0)$ while condition (2) might be violated. On the other hand it might happen that condition (2) holds but the sum of $b(n)$ does not converge, which means that $\tilde a(0)$ is undefined. So if that sum does not converge then it is not clear whether the function defined by (5) is analytic at 0, and you need some other means to ensure its analyticity there. | |
| Jun 1, 2021 at 14:59 | comment | added | Steven Clark | @მამუკაჯიბლაძე A proof of convergence of formula (5) for $\tilde{a}(x)$ for the case $a(n)=\mu(n)$ requires a proof that condition (2) is met, and it may turn out condition (2) is not met. This motivated me to include the simpler case $a(n)=\delta_{n-1}$ where condition (2) is met as well as condition (1). | |
| Jun 1, 2021 at 14:58 | comment | added | Steven Clark | @მამუკაჯიბლაძე In the case of $a(n)=\mu(n)$, if it turns out $\tilde{a}(0)$ defined in formula (7) can only be evaluated to $4$ or $1$ as it was in Figures (4) and (5) for a finite number of values of $N$, then formula (5) for $\tilde{a}(x)$ doesn't converge for the case $a(n)=\mu(n)$ since there are no values of $N$ for which formula (5) for $\tilde{a}(x)$ can be evaluated correctly both at $x=0$ and at an arbitrarily large integer value of $x$. | |
| Jun 1, 2021 at 14:57 | comment | added | Steven Clark | @მამუკაჯიბლაძე The limit in formula (6) associated with condition (1) is met for both $a(n)=\delta_{n-1}$ and $a(n)=\mu(n)$, and condition (2) is also met for $a(n)=\delta_{n-1}$ since in this case $\tilde{a}(0)$ defined in formula (7) corresponds to the Mertens function $M(N)$ which can be evaluated to any integer at an infinite number of values of $N$. | |
| Jun 1, 2021 at 4:01 | comment | added | მამუკა ჯიბლაძე | Oh I see sorry. So does it hold for $\mu$ or $\delta$? And how does it help? You need the limit, if the partial sum attains (even for any other fixed $x$, not just zero) the correct value for infinitely many $N$, you still cannot stop since until you pass to the limit, infinitely many values at (large) integers will be wrong, no? And it might happen that the limit simply does not exist. | |
| May 31, 2021 at 22:40 | comment | added | Steven Clark | @მამუკაჯიბლაძე Perhaps you're looking at formula (2)? Condition (2) is formula (7) where $N$ is the evaluation limit. | |
| May 31, 2021 at 22:13 | comment | added | მამუკა ჯიბლაძე | Sorry I don't understand. There is no $N$ in (2). | |
| May 31, 2021 at 21:55 | comment | added | Steven Clark | @მამუკაჯიბლაძე You're ignoring condition (2). First you must select $N$ such that condition (2) is met. But even if you just look at values of $N$ that meet condition (2), I expect the evaluation at non-integer values of $x$ to bounce around a bit since the offset defined in formula (6) bounces around a bit. I believe this oscillation decreases in amplitude as $N\to\infty$ since formula (6) converges to zero as $N\to\infty$, but it may not be practical to evaluate formula (6) at large enough magnitudes of $N$ to convince yourself. | |
| May 31, 2021 at 21:41 | comment | added | მამუკა ჯიბლაძე | I am just taking sequences of your (5) for $N=1,2,3,...,500$ for various fixed $x$ like $x=1.001$ or $x=2.999$; the values always oscillate with increasing amplitude. | |
| May 31, 2021 at 21:33 | comment | added | Steven Clark | @მამუკაჯიბლაძე Perhaps it may seem that way for magnitudes of $N$ that can be practically evaluated, but I believe it converges as $N\to\infty$ since the offset $\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{b(n)}{n}\right)=0$ for $b(n)=\mu(n)$ and $b(n)=\sum\limits_{d|n}\mu(d)\,\mu\left(\frac{n}{d}\right)$. Try plotting the offset $\sum\limits_{n=1}^N\frac{b(n)}{n}$ and you'll see it bounces around for small magnitudes of $N$ which is perhaps misleading you. | |
| May 31, 2021 at 21:22 | comment | added | მამუკა ჯიბლაძე | Seems like your (5) diverges for any non-integer $x$ for Kronecker delta and for $\mu$ | |
| May 31, 2021 at 21:03 | history | edited | Steven Clark | CC BY-SA 4.0 |
Corrected closed to closely in paragraph preceding Figure (1).
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| May 31, 2021 at 19:31 | history | asked | Steven Clark | CC BY-SA 4.0 |
