[MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - Printable Version +- Tetration Forum ( https://tetrationforum.org)+-- Forum: Tetration and Related Topics ( https://tetrationforum.org/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://tetrationforum.org/forumdisplay.php?fid=3)+--- Thread: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) ( /showthread.php?tid=1732) |

RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - JmsNxn - 04/04/2023
OMG! I SEE WHAT YOU ARE DOING. THIS WAS A HUGE PROBLEM I COULD NEVER SOLVE. I ACTUALLY HAVE AN OLD MO POST ABOUT IT THAT GOT DOWNVOTED TO OBLIVION, LMAO! Let me spit ball for a moment to remind myself. I'm just going to try to give some context to my understanding of this problem. If I'm way off base let me down easy Okay; let's refer to a function \(f(z)\) which is holomorphic on the unit disk \(\overline{\mathbb{D}}/1\). This means not only is it holomorphic for \(|z| < 1\) but also holomorphic for \(|z| = 1, z \neq 1\). Holomorphy on a boundary is taken to mean it is analytic on the boundary. You are constructing a contour that looks like: \[ \int_{|z| = 1} f(z)\,dz\\ \] This type of contour and integration problem can be related to Mellin transforms (or fourier transforms in general). Let's map \(1^+ \to \infty\) and \(1^- \to -\infty\) (where this is as a path approaching \(1\) along \(\partial{\mathbb{D}}\)). And then additionally we map the unit circle to \(\mathbb{R}\). This is a perfectly legal construction; and looks like: \[ h(z) = \frac{az + b}{cz+d}\\ \] I'll find \(a,b,c,d\) but, know first, they exist prima facie. We are mapping a clircle to a clircle and a point to a point (a clircle is just a fancy word for a circle on the Riemann Sphere projection); and therefore this defines a unique LFT. Since \(h(1) = \infty\) the bottom terms are \(c = 1, d = -1\). We now map: \[ \Im\left(\frac{az + b}{z-1}\right) = 0\,\,\text{if}\,\,|z| =1\\ \] Which is solvable as \(a = i\) and \(b=i\); and we just end up with the good old fashion automorphism \(h(z) = i\frac{z+1}{z-1}\). We're going to write \(h'(z) = \frac{-2i}{(z-1)^2}\). Additionally; we should know that \(h(h(z)) = z\); so changing variables is pretty standard. Let's also parameterize the contour \(|z| = 1\) with \(e^{i\theta}\) for \(-\pi \le \theta \le \pi\). Where \(z= e^{i\theta}\) and \(dz = ie^{i\theta}d\theta\). Where then we are at two new expressions for this integral: \[ \begin{align} \int_{|z| = 1} f(z)\,dz &= \int_{\mathbb{R}} f(h(x)) h'(x)\,dx = -2i\int_{\mathbb{R}} \frac{f(h(x))}{(x-1)^2}\,dx\\ &= i\int_{-\pi}^\pi f(e^{i\theta})e^{i\theta}\, d\theta\\ \end{align} \] You can do more substitutions if you'd like; (to get the mellin transform); but this is enough to talk about Fourier; and I'll stick to that. Let's assume now that \(f(h(x))h'(x) = g(x)\). The function \(g(\xi)\) is holomorphic for \(\Im(\xi) > 0\) by assumption. When we integrate this we get: \[ \int_{\mathbb{R}} g(x)\,dx = \int_{|z| = 1} f(z)\,dz \] Now equally so we have the integral, for \(t > 0\): \[ \int_{\mathbb{R}} g(x+it)\,dx = \int_{\mathbb{R}} g(x)\,dx \] Which follows by Cauchy's theorem; and the assumption that for \(r < 1\): \[ \int_{|z| = r} f(z)\,dz \] Is integrable; which we know to be true........ Now this doesn't look like your problem; not at first glance. But it is your problem. You have a single singularity at a point on the boundary; and you are holomorphic for all intents and purposes away from this singularity on the boundary. I am mapping it to the upper half plane. Now additionally; I have assumed that \(g(\xi)\) is holomorphic on \(\mathbb{R}/1\). I'm going to take an additional assumption (which was part of my MO question originally, that I probably phrased terribly). Let's assume that \(\mathcal{L}\) is a contour from \(-\infty\) to \(\infty\), where \(\Im(\mathcal{L}) \neq 0\) and \(\Im(\mathcal{L}) < -\delta(x)\), for \(\delta>0\) appropriately small which goes to zero as \(x \to \pm \infty\). A great example of this technique can be seen in the Newman proof of the prime number theorem. I'm only taking some small notes from there; and not bothering with the rigour yet; just painting a picture. But this allows us to pull out a much cleaner (better bounded, absolute convergence) Laplace Transform (in Newman's case)/Fourier Transform (general case)/Mellin Transform (Ramanujan's raw intuition case). Then we can also write: \[ \int_{\mathcal{L}} g(\xi)\,d\xi = \text{Res}(g(\xi),\xi = 1) + \int_{|z| = r} f(z)\,dz\\ \] What this is saying; is that the "residue at infinity" can be found just like a normal residue; if you change your domains and do some riemann mapping magic. Because; going back to our discussion with \(f\); we've defined: \[ \text{Res}(f,z=1) = \text{An integral on the real line}\\ \] But this integral actually equals another integral on the real line that's more manageable. You can pull out the fourier coefficients if you'd like--you're probably getting something pretty ugly; but it's still a nice idea. A lot of this discussion is mostly just to motivate you that your "residues at infinity" are mappable to normal infinite integrals; and then the application of residue calculus atop that. And it's incredibly fucking advanced. I can't remember my dumb question exactly; but it was essentially the statement that in this restricted scenario we should have some nice mellin transform results--which relate to Ramanujan--which relate to all the beautiful shit you keep seeing Also, Caleb. Go easy on me. I'm just spit balling here, and trying to give my two cents. I apologize if I'm being stupid, lmao. But if you go backwards from here; you should be able to find a function \(q(z)\) such that: \[ \int_{|z| = 1} f(z)\,dz = \int_{|z| = r} f(z) q(z)\,dz\\ \] And it could be more manageable. I apologize I can't be of more help! These are things I would exclusively wrestle with using the Mellin transform; and what you're detailing relates to an arc \(\gamma\) such that \(\gamma(0) = 0\) and \(\gamma(\infty) = \infty\). And similarly \(\gamma*(0) =0\) and \(\gamma*(\infty) = \infty\). Then if a function is differintegrable we have: \[ \int_\gamma \vartheta(w)\,dw = \int_{\gamma^*}\vartheta (w)\,dw \] Which is just a fancy cauchy's integral theorem. At infinity; if this isn't differintegrable, I conjectured that there was an equivalence: \[ \lim_{R\to \infty} \int_{\gamma_R}\vartheta(w)\,dw - \int_{\gamma^*_R}\vartheta (w)\,dw = 0\\ \] Then this function was differintegrable... Where \(\gamma_R\) is \(|\gamma| \le R\). This largely went nowhere but would mean fantastic and huge things for determining when a function is differintegrable. Which would again; talk about how we handle fourier transforms, and fourier coefficients of your residues. Have you considered: \[ \int_{|z| = 1} f(z) z^k\,dz\\ \] Which are fourier coefficients at their core. Either way, don't shoot me. I'm mostly just casually spitballing here. I love your work and I just like to talk. Regards, James RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - tommy1729 - 04/04/2023
(04/04/2023, 12:23 PM)JmsNxn Wrote: ... First, I want to say I believe in that conjecture. I think it can be proven for smooth curves. Fractal curves might be the tricky part. In fact I bet someone already has, then again I did assume in the past alot has been done that turned out to be surprisingly never formally considered. Im talking about constructible curves here, so do not shoot me with nonconstructible exotic set theoretical curves that require AC. **Ok BELOW I will talk about poles and residues now , but it is actually more general like log singarities and essential singularities etc ** But anyways, what I really wanted to say is that we are considering residues at the BOUNDARY. That is another matter than residues on the interior. The boundary does not take the full range or domain. For instance the domain may contain + oo , but not - oo. When mapping infinity to 0 , we get that the point zero is not completely considered but just the positive or negative direction (corresponding to +oo or -oo). So most theorems in complex analysis are for connected open sets or interiors of connected open sets on the riemann sphere. But that is not so much the case here. The angle that cuts the point of the domain is crucial. If that is even well-defined ( fractal , nonanalytic curves , curves based on AC etc ) Furthermore, one can not simply say f(z) is analytic at the reals and for Im(z) > 0 ** but not analytic for Im(z) < 0 ** and then take contour integral or path integrals on the real line and pretend everything is fine. Again the point 0 is only partially in the upper plane and the same for its neighbourhood , again a case of edge or boundary. Not only are you then working on the boundary, what may be an obstacle to the conditions for the theorems , but saying that the boundary of analyticity is analytic itself with a straight poker face is dubious magic math. That is like saying the prime zeta function is analytic at its boundary of analyticity. And then integrate around some of the poles of the prime zeta near the boundary , even though they are only partially used as poles + there is a dense set of poles around all poles , and pretend the theorems for contour integration around poles are valid. And ignore that we did not take the whole neighbourhood of those points. And then conclude the Riemann Hypothesis is true. I actually have seen similar "proofs" like that for the prime twin conjecture. So a big warning here. ( remember in the other thread were you thought every integral around the origin with path a circle at origin with any radius ( even the one corresponding to the natural boundary ) was 0 as extension to cauchy ? (it was false) This seems a similar problem ) Dont get me wrong you had some good ideas and the Riemann mapping is a good idea as we mentioned before. But the R mapping is more of a " restatement " then a solution. Comparable to the formulas of cauchy linking zero's and integrals of a function , they give alot of insight and be useful , but actually solving NEITHER of them in general. ( in particular the generalized argument principle in mind ) Look at Calebs pictures again that will probably help. Not trying to be rude or belittle but the problem is complex. Let me restate a crucial part of the Argument principle as written in most books : ... Specifically, if f(z) is a meromorphic function inside and on some closed contour C, and f has no zeros or poles on C, then ... The problem is NOT magically solved by making a riemann mapping and fourier series and assuming the condition has been altered. Even stronger the condition is NESSESSARY and is not a weakness of proving power ; counterexamples exist that defy the relaxation of conditions. regards tommy1729 RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - Caleb - 04/04/2023
(04/04/2023, 12:23 PM)JmsNxn Wrote: OMG!I spent some time thinking about what you've said-- and I think I've come up with some preliminary results. This post is really only some early thoughts, so it probably won't be very well organized. Anyway, first, we need a lemma Cauchy's Integral theorem for open curves Cauchy's integral theorem can be derived very directly for circles. In particular, if we have \[\int_C z^k dz = i \int_{0}^{2 \pi} r^{k+1} e^{i {k+1} \theta} d\theta = \begin{cases} 2\pi i & k = -1 \\ 0 & \text{otherwise} \end{cases} \] Then, by homotopy, this works for all closed loops. However, really, a circle is only special because it loops back on itself when we multiply by any integer. But, for instance, a half circle loops on itself whenever we mutliply by an even integer. So, for a half circle, we will have removed half of every term. In particular, if \(C\) is a half circle, then \[\int_C z^k dz = i \int_{-\pi/2}^{\pi/2} r^{k+1} e^{i {k+1} \theta} d\theta = \begin{cases} \pi i & k = -1 \\ 0 & k \text{ odd } \\ r^{k+1}\frac{2 (-1)^{k/2}}{k+1} i & \text{k even} \end{cases}\] By homotopy again, this means that every open curve with those endpoints evaluates to the same value. A similar sort of thing applies if we take \(C\) to be a third of a circle, in which case every 3rd term gets removed, etc. So, if we compute an integration with open curves, and the endpoint of those open curves lie on a circle and the two endpoints are a rational multiple of pi angle apart, then this weakened version of Cauchy's integral theorem can apply. Now that we have this out the way, let us return back to the original problem, computing \( \int_{-i \infty}^{i \infty} e^{z^2} \). Let me share some intuition on where I see this approach going. In my view, \( e^{z^2} \) actually have a residue at \( + \infty \), and a different residue at \( - \infty \). NOTICE, this is not the typical way that people view residues. In complex analysis, normally, one thinks of there as being ONLY ONE INFINITY. In particular, there is only a single infinity on the Riemann sphere, so we can't exactly seperate out \( + \infty\) and \( -\infty \) by disjoint neighborhoods. However, my idea is the following. We think about taking a contour like this This contour is viewed as enclosing \( + \infty \) but not enclosing \( - \infty \). Then, we map this contour under \( dz \to d\frac{1}{z}\). This is the intuition for the steps I'm trying to achieve, its basically an attempt to seperate out different residues. You can ignore this section if you're not interested in the intution, since this maybe makes no sense at all, but it is nonetheless the background for why I'm doing the stuff I'm doing in the following steps. Ok, now to the actual stuff. Let's map do the map \( dz \to d \frac{1}{z} \). Now we are dealing with the integration \[-\int_C \frac{e^{\frac{1}{z^2}}}{z^2} dz \] (the extra 1/z^2 comes from the change of variables). The contour I'm looking at looks like this. So there is an infinitesimal indent on the side of \( + \infty \). First, observe that \[ \frac{e^{1/z^2}}{z^2} = \sum_{n=1}^\infty \frac{1}{z^{2n}(n-1)!}\] So \[\frac{e^{1/z^2}}{z^2} = \sum_{n=1}^\infty \frac{a_n}{z^n}, a_n = \begin{cases} (n-1)! & \text{n even} \\ 0 & \text{n odd} \end{cases}\] Now, we apply the Cauchy's open integral theorem on the infinitesimal contour (call it D). It tells us that the value of this contour integral is \[\int_D \sum_{n=1}^\infty \frac{a_n}{z^{n}}dz = 2i \sum_{n=1}^\infty \frac{(-1)^n a_{2n}}{\varepsilon^{2n-1}(2n-1)} = \frac{(-1)^n }{(n-1)!}\frac{1}{\varepsilon^{2n-1}(2n-1)} \] Now, this is obviously terrible to compute as \( \varepsilon \to 0 \). However, notice that we can write this as a contour line integral, in particular, we can write it as \[ \int_{3/4 - i \infty}^{ 3/4 + i \infty} \frac{1}{2i} \csc(\pi n)\frac{(-1)^n }{(n-1)!}\frac{1}{\varepsilon^{2n-1}(2n-1)} dn\] BUT, THEN WE CAN SWITCH THE DIRECTION OF THIS CONTOUR!!! It picks up the pole at \( 2n-1 = 0\), i.e. at \( n= 1/2 \), and then everything else goes away, since the rest of the integral picks up the terms \[-2i \sum_{n=1}^\infty \frac{(-1)^n a_{2(-n)}}{\varepsilon^{2(-n)-1}(2(-n)-1)}\] Which is simply zero as \( \varepsilon \to 0 \). Notice that this means the integral over \( D\) is equal to \( \frac{1}{2i} a_{2n} \) evaluated at \( n = \frac{1}{2} \). This means the residue is equal to \( 2 \pi i \frac{a_{2 \frac{1}{2}}}{2i} = \pi a_{2 \frac{1}{2}} \) Lets check that this right. We have that \( a_{2n} = \frac{1}{(n-1)!} \). So \( a_{2 \frac{1}{2}} = \frac{1}{ (1/2-1)!} = \frac{1}{\Gamma(1/2)} = \frac{1}{\sqrt{\pi}} \). So we have that the integral is equal to \( \frac{\pi}{\sqrt{\pi}} = \pi \). OKAY, THIS IS PRETTY INSANE. It means that we can compute half these residues by looking at the half derivative of the function. It also says something much more interesting. When I see \(a_{2 \frac{1}{2}} \), I'm very strongly tempted to cancel out the 2 and the 1/2. But, in the case of e^z^2, you can do this, and its because there is some kind of incoherence in the derivatives of the function. The odd order derivatives make it look like the function should have derivative 0, but its even order derivatives make it look like it should have a derivative of 1/(n!). I think that in general, whenever fractional integration makes sense, it means that a_n is given by one single form, not a piecewise formula like with the exponential. In that case, we shoudl have \( a_{2 \frac{1}{2}} = a_1\). In that case, everything reduces back down into regular residues, since \( a_1 \) is the residue term. So, I think that we have that this integration method gives the usual residue precisely when the differential integral makes sense and is 'coherent' in the sense that it isn't defined in the kind of piecewise way that e^z^2 is defined. Anyway, I'll look at all this stuff later in detail and rewrite this in a coherent way but for now I have to go, hopefully these initial thoughts are still helpful though. RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - tommy1729 - 04/04/2023
(04/04/2023, 09:28 PM)Caleb Wrote:(04/04/2023, 12:23 PM)JmsNxn Wrote: OMG!I spent some time thinking about what you've said-- and I think I've come up with some preliminary results. This post is really only some early thoughts, so it probably won't be very well organized. Anyway, first, we need a lemma Do you (implicitly) claim to have reduced the problem to fractional differentiation ? I think you forgot a summation symbol around the third equation. Am I correct in thinking if there is only 1 pole and it is at the boundary then only the angle of the curve at that point matters ? Now wait, make that the 2 angles made by the corner , that touches the poles ; the 2 angles are with respect to the cartesian axes. Also what about singularities and branch cuts at the boundary ? Is this really only about poles ? You often talk about doing analytic number theory, but how does this relate ? Sorry if I am a bit slow. regards tommy1729 RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - Caleb - 04/05/2023
(04/04/2023, 10:34 PM)tommy1729 Wrote: Do you (implicitly) claim to have reduced the problem to fractional differentiation ?The connection between fractional integration is the following. We have that \[f(x) = \sum_{n=0}^\infty a_n x^n\] Where \( a_n = \frac{f^{(n)}(0)}{n!} \). My work has reduced the evaluation of certain integrals into what is essentially the evaluation of \( a_\frac{1}{2} \). Some earlier conversations I've had with James suggests to me that the DifferIntegral is the uniquely correct way to interpolate the sequence \( a_n \). Therefore, evaluation of \( a_\frac{1}{2} \) should be equal to \( \frac{f^{(\frac{1}{2})}}{(\frac{1}{2})!} \), which is the half derivative. Also, you are right about the missing summation sign-- I'll try to fix everything later today, and redo my explanation a bit, and maybe try to do some computations to see if the math checks out in other cases. The post is actually only about essential singularities. It won't work for poles, or branch cuts. Actually, for poles, the natural extension of this idea gives something rather trivial. If there is a pole after doing the \( dz \to d \frac{1}{z} \) map, then that means you must have started with a polynomial. In which case, this method gives exactly what you would expect. For odd powers of \(k\), it evaluates to zero (which is what you would expect, since, morally, \( \int_{-\infty}^\infty x^k dx = 0 \). For functions that include an even power, the integral evaluates to infinity. James has a better grasp of how this connects to analytic number theory (he mentioned a bit on this idea when he talked about the prime number theorem in his last post). For now, I don't have a good enough idea on how these ideas work out to actually claim that this method has applications for anything whatsoever-- it might turn out to be completely useless. Actually, one case I'm interested in trying to apply this is to compute the integral \[ \int_{-\infty}^\infty (\eta(x^2) - 1) dx \] Where eta is the Dirichlet Eta function (which is the alternating version of the zeta function). Presumably, the answer to this integral should be given in terms of the half integral of the Eta function, which seems interesting to me (if its true that is). RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - Caleb - 04/05/2023
It looks like proving the connection between fractional integrals and these residues is actually quite easy. Take \[ \int_{-\infty}^\infty f(x^2) dx\] Then by a change of variables this becomes \[\int_{0}^\infty \frac{f(t)}{\sqrt{t}} dt \] Which is nothing more than the half order fractional derivative (after diving by gamma) evaluated at zero, exactly as predicted. So for instance, this shows that the results I've derive for \( e^{-x^2} \) are valid. This also solves that \( \eta \) case illustrated before, and says that its value is just the half order fractional derivative. RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - JmsNxn - 04/06/2023
OH MY FUCKING GOD! I LOVE YOU GUYS! I HAVE NOTES ALL AROUND EXACTLY THIS, BUT I NEVER GOT IT, OR UNDERSTOOD IT! YOU GUYS ARE AWESOME! Beautiful work, both of you! I'll compile some thoughts afterwards, just working through your guys' comments. RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - JmsNxn - 04/06/2023
I AM GOING TO DO SOMETHING I DONT KNOW THE IMPORTANCE OF, JUST FOR CALEB! LISTEN! We are going to define the Dirichlet \(\eta\) function, as the function: \[ \eta(s) = \sum_{n=1}^\infty (-1)^{n-1} n^{-s}\\ \] This function is holomorphic for \(\Re(s) > 0\). Additionally, we know that it is \(O(s)\) bounded for \(\Re(s) > 0\). Which means, there exists a constant \(C_\delta >0\) such that \(|\eta(s)| < C_\delta |s|\) for \(\Re(s) > \delta\) when \(\delta>0\). Now your first instinct might tell you to "fractionally differentiate" \(\eta\). But that's wrong. We instead take: \[ \vartheta(w) = \sum_{n=0}^\infty \eta(1+n) \frac{(-w)^n}{n!} = \frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} \Gamma(s) \eta(1-s) w^{-s}\,ds\\ \] Where: \[ \frac{d^{s-1}}{dw^{s-1}}\Big{|}_{w=0} \vartheta(w) = \eta(s)\\ \] Here is where advanced number theory pokes its head. I am just trying to describe the problem, I am not trying to solve it. For you to describe singularities within the function \(\eta(\infty)\); within these fucking crazy beautiful residues your looking at.... You're solving very deep problems that I don't think you are aware of; and that's why I'm doubtful of some of the shenanigans you're doing. BUT BY GOD IF YOU'RE NOT AT LEAST ON TO SOMETHING! Dirichlet's function \(\eta(s)\) is meromorphic on \(\mathbb{C}\)......... Let's just move some integrals around and pull out some residues This is true and unique and Hilbert Special and fucking JON VON NEUMANNN UNIQUE so long as we assume that these functions are "Ramanujan bounded". Or we subject ourselves to advanced fourier theory. CARLSON WON A FIELDS MEDAL FOR FINDING THE FOURIER RIGOR VERSION OF RAMANUJAN'S RAW. So I Fucking love this Caleb! But just know, the fucking devil's in the details. And you're going to have to go case by case; because this relates to highly advanced analytic number theory. Fucking awesome, regards, James EDIT: Forgot to mention; all it takes to turn your integrals into my integrals is a variable change. I mean, \[ \begin{align} \frac{d^s}{dw^s}\big{|}_{w=0}f(w^2) &= \frac{1}{\Gamma(-s)} \int_{\gamma} f(w^2) w^{-s-1}\,dw\\ &=\frac{1}{2\Gamma(-s)} \int_{\gamma} f(u) u^{-(s+1)/2-1/2}\,du\\ &= \frac{\Gamma(-s/2)}{2\Gamma(-s)} \frac{d^{s/2}}{dw^{s/2}}\big{|}_{w=0}f(w) \end{align} \] I'm still just rapidly posting these things but fuck this is beautiful. RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - Caleb - 04/06/2023
(04/06/2023, 05:30 PM)JmsNxn Wrote: I AM GOING TO DO SOMETHING I DONT KNOW THE IMPORTANCE OF, JUST FOR CALEB! LISTEN!A short thought is that we can re-express fractional derivatives in terms of Cauchy's integral formula. We have that \[ a_n = \int_C \frac{f(z)}{z^n} dz \] The tempting thing is to plug in non-integer values of n. I tried this before, and it didn't work. Now that I know more about complex analysis, its clear why this niave approach doesn't work-- there is a branch cut at non-integer values of \( n \). I believe we get the desired behaviour if we take \( C \) be a contour that goes around the branch cut, for instance, a Hankel contour. So I think this is one way we might start analyzing these things seriously. We can define fractional derivatives in terms of the fractional Cauchy's integral formula, and test if it satisfies the desired behaviour. I guess this also explains why all the integer derivatives of local, whereas the non-integer ones are non-local-- its because this branch cut means the shape of the contour has to change to something that depends on more than just the local behaviour of the function. Also, this is far in the future and certainly jumping the gun, but I think this fractional derivative idea might fit really well with some of the lambert series stuff we did before. Whats interesting is that many lambert series have that 'gateway' along the real line. That gateway should capture enough behaviour to be able to take fractional derivatives, so perhaps even though fractinoal derivatives are non-local, it seems we can still compute them in a reasonable way beyond natural boundaries. I'm just spitballing here, it requires much more careful thought to determine if these thigns are actually meaningful. RE: [MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n)) - JmsNxn - 04/06/2023
Not to sound like a broken record, but: *touches nose* |