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
40 events
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| Jul 15, 2021 at 20:46 | history | edited | Steven Clark | CC BY-SA 4.0 |
Updated formula (s) for Zeta(s) with one that converges for Re(s)<2 instead of 1<Re(s)<2.
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| Jun 16, 2021 at 19:07 | comment | added | Steven Clark | @მამუკაჯიბლაძე Real analytic formulas for $\delta(x-1)$ can be used to derive complex analytic formulas for a variety of functions via Mellin convolutions such as $f(y)=\int\limits_0^\infty\delta(x-1)\ f\left(\frac{y}{x}\right)\ \frac{dx}{x}$ and $f(y)=\int\limits_0^\infty\delta(x-1)\ f(y\ x)\ dx$ which generally converge for $Re(y)>0$ (e.g. see formulas (52) to (70) in the answer I posted to my own question on Math StackExchange at math.stackexchange.com/q/2380164). | |
| Jun 16, 2021 at 19:00 | comment | added | Steven Clark | @მამუკაჯიბლაძე As the simplest possible example, I consider a real analytic formula for the Dirac delta function $\delta(x-1)$ far more important than a complex analytic formula for the Kronecker delta function $\delta_{n-1}$ derived from partial evaluation of the $\delta(x-1)$ formula. | |
| Jun 15, 2021 at 22:14 | vote | accept | Steven Clark | ||
| Jun 15, 2021 at 22:11 | history | edited | Steven Clark | CC BY-SA 4.0 |
Corrected lower limit k=0 to k=1 in formulas (f), (g), and (j) and added formulas (m) to (t).
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| Jun 14, 2021 at 20:34 | comment | added | Steven Clark | @მამუკაჯიბლაძე Whereas von Mangoldt's explicit formula illustrates the connection between the primes and the zeros of $\zeta(s)$ (which are the poles of $-\frac{\zeta'(s)}{\zeta(s)}$), formula (f) for $\tilde{f}_a(x)$ illustrates the connection between the primes and the values of $F_a(s)=-\frac{\zeta'(s)}{\zeta(s)}$ at even positive integer values of $s$. | |
| Jun 14, 2021 at 20:22 | comment | added | Steven Clark | @მამუკაჯიბლაძე Formula (f) for $\tilde{f}_a(x)$, when evaluated with $a(n)=\Lambda(n)$ where $F_a(s)=-\frac{\zeta'(s)}{\zeta(s)}$, gives an alternate real analytic formula for $\psi(x)$. The von Mangold explicit formula only converges for $x>1$ whereas formula (f) for $\tilde{f}_a(x)$ converges for $x\in\mathbb{R}$. Real analytic formulas for $f(x)=\sum\limits_{n\le x} a(n)$ are hard to come by. | |
| Jun 14, 2021 at 20:22 | comment | added | Steven Clark | @მამუკაჯიბლაძე I'm sorry my point seems to be lost on you. The significance of formulas (f) and (g) is that they're real analytic formulas for $\tilde{f}_a(x)$ and ${f}'(x)$. Are you also uninterested in real analytic formulas such as von Mangoldt's explicit formula for the second Chebyshev function: $\psi_o(x)=x-\sum\limits_\rho\frac{x^\rho}{\rho}-\log(2\,\pi)-\frac{1}{2}\,\log\left(1-\frac{1}{x^2}\right)$ where $\psi(x)=\sum\limits_{n\le x}\Lambda(n)$? | |
| Jun 14, 2021 at 19:38 | comment | added | მამუკა ჯიბლაძე | While I see some interest in finding entire functions with given values at natural numbers and explicit arithmetically interesting coefficients with potential applications for number theory, I don't see any significance in these (f) and (g). | |
| Jun 14, 2021 at 17:19 | comment | added | Steven Clark | @მამუკაჯიბლაძე I don't think evaluating $\tilde{a}(x)$ at non-integer values of $x$ is of much value since there are an infinite number of functions that evaluate correctly at integer values of $x$ but which generally evaluate very differently at non-integer values of $x$. Therefore I don't really see the point in extending $a(n)$ to a complex analytic function $\tilde{a}(x)$ where $x\in\mathbb{C}$. I think the real value in the paper is it lead to formulas (f) and (g) above for $\tilde{f}_a(x)$ and $\tilde{f}_a'(x)$. | |
| Jun 14, 2021 at 17:02 | comment | added | მამუკა ჯიბლაძე | I don't understand. If $f=1$ gives what you want why do you need anything else? | |
| Jun 14, 2021 at 16:24 | comment | added | Steven Clark | @მამუკაჯიბლაძე I believe the relationship $\tilde{a}(x)=\frac{1}{2}\tilde{f}_a'(x)$ when $\tilde{f}_a'(x)$ is evaluated at the frequency $f=1$ generalizes to $\tilde{a}(x)=\frac{1}{2\,f}\tilde{f}_a'(x)$ when $\tilde{f}_a'(x)$ is evaluated at a finite frequency $f>1$. | |
| Jun 14, 2021 at 15:58 | comment | added | Steven Clark | @მამუკაჯიბლაძე I suspect $\tilde{f}_a'(x)$ and $\tilde{f}_a(x)$ are analytic when evaluated at any finite evaluation frequency $f$, but only converge as $f\to\infty$ when $x\in\mathbb{R}$. | |
| Jun 14, 2021 at 15:50 | comment | added | Steven Clark | @მამუკაჯიბლაძე Yes the original question was about extending $a(n)$ to the analytic function $\tilde{a}(x)$ where $x\in\mathbb{C}$. My revelation was that $\tilde{a}(x)$ is just $\frac{1}{2}\tilde{f}_a'(x)$ evaluated at the frequency $f=1$. At this point I really don't understand if or why not $\tilde{f}_a'(x)$ and/or $\tilde{f}_a(x)$ extend to $x\in\mathbb{C}$ when evaluated at frequencies $f>1$. | |
| Jun 14, 2021 at 15:38 | comment | added | Steven Clark | @მამუკაჯიბლაძე Also note in $F_a(s)=s\int\limits_0^\infty f_a(x)\,x^{-s-1}\,dx=\sum\limits_{n=1}^\infty a(n)\ n^{-s}$, the integral is over real $x$. The inverse relationship is $f_a(s)=\frac{1}{2 \pi i}\int\limits_{c-i\,\infty}^{c+i\,\infty} F_a(s)\frac{x^s}{s}\,ds$ where $F_a(s)=\sum\limits_{n=1}^\infty a(n)\ n^{-s}$ converges for $Re(s)>b$ and $c>b$. | |
| Jun 14, 2021 at 15:33 | comment | added | მამუკა ჯიბლაძე | Is not the question about extending $a(n)$ to an analytic function of $n$? Risomar Sousa extends it to an entire function, i. e. finds a series that converges for all complex $n$. | |
| Jun 14, 2021 at 14:56 | comment | added | Steven Clark | @მამუკაჯიბლაძე The summatory function $f(x)=\sum\limits_{n\le x} a(n)$ only makes sense for $x\in\mathbb{R}$. However it can evaluate to a complex result when $a(n)\in\mathbb{C}$ (e.g. $a(n)=\chi _{5,2}(n)=\{0,1,i,-i,-1,...\}$. | |
| Jun 14, 2021 at 5:59 | comment | added | მამუკა ჯიბლაძე | Yes I see now, and confirm the convergence. Except still the result is only real-analytic, I could not use it for complex $x$. | |
| Jun 14, 2021 at 0:01 | comment | added | Steven Clark | @მამუკაჯიბლაძე For proofs of the validity of formulas (f) and (g), please see the answer I posted at mathoverflow.net/q/395266 to my more recent question on closed form expressions.. | |
| Jun 13, 2021 at 2:27 | comment | added | Steven Clark | @მამუკაჯიბლაძე Ok, thanks. I think it would help me to see your point if you could quantify how the evaluation increases as the evaluation frequency $f$ increases and explain how this quantification is consistent with the evaluations of formula (f) for $\tilde{f}_a(x)$ that I illustrated in Figures (1) to (4) of my new question at mathoverflow.net/q/395112 and Figures (5) to (8) of my updated answer at math.stackexchange.com/q/4160465. | |
| Jun 13, 2021 at 2:17 | comment | added | მამუკა ჯიბლაძე | Well, I see it from the coefficients which are polynomials of increasing degrees in $f$, and also I saw it experimentally in some cases. I will show them to you, and I will also think about proving it. | |
| Jun 13, 2021 at 0:49 | comment | added | Steven Clark | @მამუკაჯიბლაძე I'm not sure why you think formula (f) for $\tilde{f}_a(x)$ will converge to larger magnitudes as the magnitude of $f$ increases. The figures in my new question and updated answer illustrate formula (f) for $\tilde{f}_a(x)$ converges closer and closer to the reference function $f(x)=\sum\limits_{n\le x} a(n)$ as the evaluation frequency $f$ increases for $1\le f\le 4$ when $x$ is within the convergence range of formula (f) for $\tilde{f}_a(x)$. | |
| Jun 13, 2021 at 0:46 | comment | added | მამუკა ჯიბლაძე | I don't know about that, I just want to say that I don't see how your display (f) can define anything that does not go to infinity everywhere. | |
| Jun 13, 2021 at 0:36 | comment | added | Steven Clark | @მამუკაჯიბლაძე When $x=n\in\mathbb{Z}$ formula (g) for $\tilde{f}_a'(x)$ should converge to $2\,f\,a(n)$ as $K\to\infty$, so yes formula (g) for $\tilde{f}_a'(x)$ goes to infinity as $f\to\infty$ when $x=n\in\mathbb{Z}\land a(n)\ne 0$, but this is consistent with the convergence of the Fourier series for the Dirac comb of period $1$ at integer values of $x$ (which corresponds to $a(n)=1$). | |
| Jun 12, 2021 at 22:48 | comment | added | მამუკა ჯიბლაძე | I think what I said subsumes your $K\gg fx$ condition and is actually much stronger. I propose to fix $f$ and take the limit $K\to\infty$ for this fixed $f$. For each fixed $f$ you will indeed obtain a well-defined analytic function of $x$. The problem is that, I believe, these functions will attain bigger and bigger values when you will increase $f$. | |
| Jun 12, 2021 at 22:02 | comment | added | Steven Clark | @მამუკაჯიბლაძე I believe what you say is true which is why I modified the limits in formulas (f), (g), and (j) above to specify $K\gg f\,x$. It's not unusual for a series evaluation to require $K\gg x$, but I understand you're uncomfortable with extending it to $K\gg f\,x$. I suspect the only way to prove (or disprove) formulas (f) and (g) above is to derive a generalized closed form representation as a function of the evaluation frequency $f$. Hopefully someone will come up with an answer to my new question on this topic. | |
| Jun 12, 2021 at 21:53 | comment | added | მამუკა ჯიბლაძე | But how do you know what will happen to these graphs when, say, $f$ is one million? Because you do need arbitrarily large $f$, right? For each such $f$ you will get a convergent series, i. e. will obtain definite limit as $K\to\infty$, but at each given $x$ this limit will be bigger and bigger as $f$ gets bigger. | |
| Jun 12, 2021 at 18:03 | comment | added | Steven Clark | @მამუკაჯიბლაძე Figures (1) to (4) in my new question on closed form representations at mathoverflow.net/q/395112 and Figures (5) to (8) in my updated answer at math.stackexchange.com/q/4160465 are perhaps the best illustrations to date of why I believe formulas (f) and (g) in my answer above are valid. | |
| Jun 12, 2021 at 13:47 | history | edited | Steven Clark | CC BY-SA 4.0 |
Corrected delta_{x-1} to delta_{n-1} in a couple of places and modified the limits in formulas (f), (g), and (j).
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| Jun 10, 2021 at 19:32 | comment | added | Steven Clark | @მამუკაჯიბლაძე And if the closed form expression leads to a proof of formulas (f) and (g) for the case $a(n)=1$ or $a(n)=(-1)^{n-1}$, then it could perhaps lead to a new formula for $\zeta(s)$ or $\eta(s)$. | |
| Jun 10, 2021 at 19:31 | comment | added | Steven Clark | @მამუკაჯიბლაძე Since the convergence range over $x$ seems to decrease each time $f$ increases, it seems to me formulas (f) and (g) need to be evaluated such that $K\gg f\,x$. I believe a generalized closed form expression as a function of $f$ for one of the cases $a(n)=1$, $a(n)=(-1)^{n-1}$, or even $a(n)=\delta_{n-1}$ would perhaps lead to a proof or disproof for the specific case where the closed form expression can be derived. | |
| Jun 10, 2021 at 17:40 | comment | added | მამუკა ჯიბლაძე | Still, each coefficient is a polynomial in $f$, so if you just increase both $K$ and $f$ independently you will always get infinity. In more detail: for each $x$ you have the numbers $S_{K,f}(x)$ (truncating the sum at $K$ for $f$). If you fix $K$ and increase $f$ you get infinity. If you fix $f$ and increase $K$ you get certain finite limit $S_f(x)$. These tend to infinity as $f$ increases. So you need something else to get a finite limit. What is it? | |
| Jun 10, 2021 at 17:02 | comment | added | Steven Clark | @მამუკაჯიბლაძე Correction: The denominators seem to match oeis.org/A156769 which resembles A036279 with the first difference occurring at a(12). | |
| Jun 10, 2021 at 16:33 | comment | added | Steven Clark | @მამუკაჯიბლაძე The coefficients for $k=4$ can be rewritten as $-\frac{4 f}{9}\left(1-\frac{2 \pi^2 f^2}{3}+\frac{2 \pi^4 f^4}{15}-\frac{4 \pi^6 f^6}{315}\right)$ which I believe generalizes to $-\frac{4\, f}{2\,k+1}\left(1-\frac{2 \pi^2 f^2}{3}+\frac{2 \pi^4 f^4}{15}-\frac{4 \pi^6 f^6}{315}+...\right)$ and note the denominators inside of the parenthesis $\{1,3,15,315,2835,155925,6081075,638512875,10854718875,1856156927625,\text{...}\}$ correspond to oeis.org/A036279 which are the denominators in the Taylor series for $\tan(x)$. | |
| Jun 10, 2021 at 16:33 | comment | added | Steven Clark | @მამუკაჯიბლაძე I don't believe you can base convergence on a single coefficient, rather I believe you have to consider what the entire series is doing. You may be right, but I'll have to think about it some more. I appreciate you're feedback and the time you've invested in investigating my question and answer. | |
| Jun 10, 2021 at 15:49 | comment | added | მამუკა ჯიბლაძე | Because when I increase $f$ each of these coefficients goes to infinity. If you have a procedure to obtain a finite limit somehow, it cannot be described as just $K,f\to\infty$, this will just give you infinity for all $x$. | |
| Jun 10, 2021 at 15:03 | comment | added | Steven Clark | @მამუკაჯიბლაძე Your terms seem to exactly match formula (f) for $\tilde{f}(x)$ above for the case $a(n)=\delta_{n-1}$ where $b(n)=\mu(n)$ and $F_a(s)=1$. What is it that you don't understand, and why do you think you get entirely different results? | |
| Jun 10, 2021 at 5:20 | comment | added | მამუკა ჯიბლაძე | I don't understand, I get entirely different results. Could we please check, for $F_a(s)\equiv1$ I get the following series:$$-\frac{4 f}{3} x^3+\left(-\frac{4 f}{5}+\frac{8 \pi ^2 f^3}{15}\right) x^5+\left(-\frac{4 f}{7}+\frac{8 \pi ^2 f^3}{21}-\frac{8\pi ^4 f^5}{105} \right) x^7+\left(-\frac{4 f}{9}+\frac{8 \pi ^2 f^3}{27}-\frac{8 \pi ^4 f^5}{135}+\frac{16 \pi ^6 f^7}{2835}\right) x^9+O\left(x^{11}\right)$$ | |
| Jun 8, 2021 at 19:00 | history | edited | Steven Clark | CC BY-SA 4.0 |
Simpliflied formulas (f), (g) and (j) sllightly and added figures illustrating formula (f).
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| Jun 6, 2021 at 1:40 | history | answered | Steven Clark | CC BY-SA 4.0 |