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Grandi's Series

On the Proof by Integration, you are wrong that it isn't rigorous because of the fact that integrals are only defined up to a constant because this is irrelevant because integrals are not defined up to a multiple of a function and thats what is being compared here the amount of e^xsin(x) and e^xcos(x) in the integral which has to be the same no matter how you carry out the integration.I have also readded the other proof to the page partly because the page seems to really lack content and I think readers will be interested by what methods you can come about what might seem to be a counter intuitive result I certainly was when I read the page in the past and I am disappointed these additional proofs have been removed. Dalek1099 (talk) 01:06, 7 July 2017 (UTC)

@Dalek1099: This is faulty logic, for the following reasons. I will be taking it back out until you can address all of the following:

1. What is the meaning of integrating by parts infinitely many times? A series of functions? Said series is still pointwise divergent and therefore cannot be differentiable, let alone an "antiderivative".
2. As mentioned emphatically in the "Unrigorous Methods" section, one is not allowed to perform the rearrangement of the terms of a divergent series (not even a conditionally convergent one), so the step "factoring out" G is invalid.
3. Even disregarding the above, you are finding an antiderivative using different methods and one must thus consider the possibility of differing by a constant. For any nonzero constant one is not allowed to perform the cancellation step of the proof; two functions that differ by a constant are not equal. Even if we could let $x\to \infty$ on both sides, one is generally not allowed to do that even when the limit exists, let alone when it doesn't.
4. Above all, the series is divergent. Therefore, there can be no "proof" that it has a certain limit. You have to use an alternative method of summation.
5. These "proofs" fail to make any connection to such an alternative method of summation like the Cesaro sum.
6. There are no reliable sources supplied for them.
7. The second "proof" duplicates one of the methods given in the "Unrigorous Methods" section and therefore does not belong anyways.

So sorry, but they must stay out for the time being.--Jasper Deng (talk) 06:09, 7 July 2017 (UTC)

What is effectively being done here is zeta function regularization, the x parameter is effectively acting like the s parameter in that method. That means I can factor out G like that. With regards to the fact that the functions differ by a constant that is irrelevant if I know that an antiderivative is given by -Ge^xcos(x)+Ge^xsin(x)+C or Ge^xsin(x)/2-Ge^xcos(x)/2 +B then the only possibility is that G=1/2 this can be proven simply by the fact that e^x(G-1/2)sin(x)+e^x(-G+1/2)cos(x) has to be a constant function as the difference of two antiderivatives must be constant and that means G=1/2 is the only possibility for that to be true for all x and for G to be constant.

The series can indeed be differentiable depending on how its sum is defined of course if you define the sum based on the method in standard analysis around convergence then it would obviously not be differentiable as it would lack a function value in the first place.Using the methods of summation you mention does indeed make this series differentiable.Although, it is standard usage in zeta function regularization to simply differentiate term by term which makes sense to me as we know that in the case that the series does converge that in its radius of convergence it is entirely valid to differentiate term by term. Thus, if we are to make all series' have a finite sum then one could define the differentiation that way in order to determine what the value for the sum should be as that would be consistent with the definition for convergent series.

You seem to be unaware of summation methods that very much involve manipulating series to determine the result like zeta function regularization or analytic continuation. If these methods are seen as unrigorous you could put the methods under the unrigorous methods section.Dalek1099 (talk) 07:51, 7 July 2017 (UTC)

Heuristic approaches (regularization etc) may be interesting and useful, but should be clearly marked as heuristic. In particular, it should be clear, is this an application of an existing regularization method, or its modification, or a new approach inspired by regularization, or whatever. A (rigorous) proof cannot be started before all relevant definitions are (clearly) formulated. And by the way, it seems Dalek1099 confuses "differentiable" with something else ("convergent"? "summable"? "derivable"?). Boris Tsirelson (talk) 08:05, 7 July 2017 (UTC)

Does the proof as-is mention such regularization? No. This has nothing to do with the Riemann zeta function, and "zeta function" can mean many things, so you will have to elaborate. I still don't think you know what you are talking about; for starters, one cannot fix constants of integration on either side of the equation without initial conditions. If $f(x)g(x)=h(x)g(x)+C$ for $C\neq 0$, then it is invalid to write $f(x)=h(x)$, period. There is nothing ambiguous about that. In this case, the absurdity is clear: you have $g(x)=e^{x}(\sin(x)-\cos(x))$, $f(x)=G$, and $h(x)={\frac {1}{2}}$. Then one can only conclude $G={\frac {1}{2}}+{\frac {C}{g(x)}}$, so one cannot pick any value of G valid for all x! Zeta function regularization is clearly not being applied in the integration proof as-is, and you have not demonstrated you even know how to apply that method to the problem in hand. You aren't even considering the correct problem at hand, since the antiderivative in question is not a linear combination of sine and cosine, but rather, an exponential times that (we're working in the reals, so nothing along the lines of Euler's formula applies here).

Above all, @Dalek1099: we cannot accept your own research (or mine), for that matter. Wikipedia publishes what reliable sources say and original research is not allowed.--Jasper Deng (talk) 08:22, 7 July 2017 (UTC)

@Dalek1099: It would also go a long way if you would not type in run-on sentences.--Jasper Deng (talk) 08:53, 7 July 2017 (UTC)

"Then one can only conclude $G={\frac {1}{2}}+{\frac {C}{g(x)}}$, so one cannot pick any value of G valid for all x!" Since the sum cannot depend on x the only logical conclusion is that C=0 and G=1/2. The problem is essentially (G-1/2)e^xsin(x)+(-G+1/2)e^xcos(x)=C from which one must conclude G=1/2 and C=0 for that to be true for all x and the fact that G must be constant which is obvious from the fact the terms 1 -1 etc don't contain x.I think you might have been confused because I accidentally dropped an e^x in my reply but the logic is still the same and the C that I am using is not the same C that was used in the Proof by Integration its the subtraction of the constants which must as I have argued be 0.

There is no confusion over what it means for a function to be differentiable its just when dealing with sums for which you don't know the answer because they are divergent you can't differentiate them intially because of that reason but you want to assign a value to them and convergent series like power series are differentiable term by term within their radius of convergence and thus by differentiating them term by term you hope to be able to assign a value to them that would be consistent with convergent series? I probably should have just said regularisation rather than zeta function regularisation its just thats the most famous regularisation so its more commonly known. Dalek1099 (talk) 11:12, 7 July 2017 (UTC)

Anyway, if you want to place on a wiki your own thoughts, try Wikiversity, not Wikipedia. Boris Tsirelson (talk) 12:04, 7 July 2017 (UTC)

@Dalek1099: It ought not to depend on x, but one cannot painlessly dismiss that from the equation at hand, which, after all, is an equality of a set of functions and does not make any guarantee about the constant difference. See Mathematical fallacy#Calculus for an example of such fallicious reasoning. And one cannot use any regularization technique without explicitly mentioning it as such, and in the absence of such mention one must the standard definition of convergence. Also, one cannot generally differentiate a general series of functions term-by-term (see Weierstrass function for a dramatic counterexample) without the additional hypothesis of uniform convergence of the original series and of the series of derivatives. I do not doubt you have good intentions but I truly think you do not know the meaning of mathematical rigor.--Jasper Deng (talk) 15:13, 7 July 2017 (UTC)

There is no example of fallacious reasoning in the article you linked to the fact that there are undetermined integration constants has been dealt with and the fact that the integration constants must be equal for both antiderivatives as is derived from the result that G must be constant and the fact that two antiderivatives can only differ by a constant. If you want you could just differentiate the antiderivative and you would also reach the conclusion that G=1/2 . The whole point of the problem is for the sum 1-1+1-1+1-1+... to be assigned a constant value.When trying to prove any theorem rigorously there are always conditions the condition here being that G can be given a constant value.The whole point of differentiating term by term is for the summation properties to reflect those of series that are actually convergent(or more precisely uniformly convergent as you point out). Such techniques are common in regularisation and analytical continuation.This is perfectly fine to do after all the series isn't actually convergent so the series couldn't be differentiable in the usual sense but we could find a new way of assigning the series that makes what the series is now defined to sum to differentiable and one of the properties that we would want is that the term by term differentiation agrees with the differentation of the function that we have assigned as the sum of the series that condition would essentially be part of the definition of the summation. The main issue seems to be that you would have wanted all the constraints of the proof to be written down as well?

As for rigour that always depends on the branch of the Mathematics and as a whole unfortunately a completely rigorous theory of the summation of divergent series has not been developed. Heuristic approaches are very much part of how these sums are calculated by professional researchers partly as a result of this lack of a completely rigorous theory but on the basis of rigour anyway being arguably dependent anyway on the branch of Mathematics this essentially makes these approaches rigorous. Dalek1099 (talk) 09:25, 8 July 2017 (UTC)

Okay, you seem to be missing the fundamental point of mathematical rigor, which is that you must state and use correctly all assumptions and definitions you are making, no exceptions. The proofs as-is did not contain any method of regularizing the divergent series and one is not permitted to act like one was implicitly used. You can't say "differentiate both sides" because that is not a step performed by that proof, which directly proceeds to the conclusion without that step. Also, it is already a huge assumption to even assume G is a well-defined constant value. To differentiate both sides would require the assumption of the commuting of limiting operations on the left side which, as you know very very well, cannot be done in the usual sense.

Clearly you don't understand the meaning of term-by-term differentiation even for convergent series. The Weierstrass function is not differentiable, but it's also possible for a series to converge to a differentiable function (even an analytic one), even uniformly, and yet have the series of derivatives diverge, or worse, converge to the wrong value of the derivative. A good example is provided at Harmonic function#Remarks (not a series but a sequence, but one could easily turn it into a series by defining $a_{0}=f_{0},a_{n}=f_{n}-f_{n-1}$; it says x must be negative but the convergence is still uniform for $x=0$ as well) - in particular, the partial derivative with respect to x must be zero, but partially differentiating each term yields $e^{nx}\cos(ny)$ which has value 1 at the origin for all n. For term-by-term differentiation to be valid, the series of derivatives must converge to the derivative of the target of the series. Otherwise you are not actually differentiating the function represented on the left hand side. So one is not permitted to perform the rearrangement of terms necessary to factor out G in this proof (once again, no regularization allowed when not explicitly assumed).

@Dalek1099: But above all, the argument presented as-is is a howler in every sense of the word. It is very correct that many alternative summation methods sum the series to one half, but the argument itself has invalid steps as-is. I want to make it very clear that regularization methods are a red herring with respect to the argument's validity because you cannot use anything not assumed by the proof itself. Mathematical rigor does not depend on branch of mathematics: an argument is either valid or it is not, regardless of context. If you find a sum via a heuristic method, then you either have to prove that rigorously (which you're clearly failing to do here) or acknowledge that you have not actually proven anything. Heuristic methods even can disagree (see Fermat number#Heuristic arguments for density). Again, please stop writing in run-on sentences, it makes it a lot more difficult to understand your comments.--Jasper Deng (talk) 17:26, 8 July 2017 (UTC)

Here's an even better counterexample. The sequence of analytic functions $f_{n}(x,y)=e^{-nx^{2}}\sin(ny)/n$ is uniformly convergent to the zero function for all x and all y, and yet the sequence of partial derivatives with respect to y, $e^{-nx^{2}}\cos(ny)$, has the similar issue of taking value 1 at the origin for all n. The upshot? Even when the sequence of derivatives is itself convergent, one is not necessarily permitted termwise differentiation, and importantly, any regularization that agrees with the usual notion of (pointwise) convergence for convergent sequences will still have this problem.--Jasper Deng (talk) 11:02, 9 July 2017 (UTC)

Protection

Something happened on your user page? Kind regards MJ ^☕ 16:57, 17 July 2017 (UTC)

Long ago.--Jasper Deng (talk) 17:29, 17 July 2017 (UTC)

I noticed a Full protection And thought it must have been vandalised. Kind regards MJ ^☕ 17:37, 17 July 2017 (UTC)

Ransomware Page Edits

Jasper: Two questions if I may: (1) What is the process to resolve the dispute on ransomware (you reverted an edit on it today), and (2) Why are you defaulting to deletion of "from cryptovirology" as opposed to preserving it? Your expertise is welcome. 173.56.74.61 (talk) 23:06, 12 July 2017 (UTC)

@173.56.74.61: WP:BRD. You have not achieved the consensus necessary to insert it (also, argument from authority (alone) is not effective here because Wikipedia represents what reliable sources publish and generally, community consensus).--Jasper Deng (talk) 04:12, 13 July 2017 (UTC)

I will review this feedback and the wikipedia text on argument from authority. What steps did you take to help resolve the dispute? 173.56.74.61 (talk) 11:24, 13 July 2017 (UTC)

My latest post in the Ransomware Talk page includes multiple arguments in support of reinstating "from cryptovirology". You seemed to have noticed that I listed multiple arguments (since you said "alone" above). So far my latest post in the Ransomware Talk page has not been contested. I plan to reinstate the phrase "from cryptovirology" that was originally in the article. Please indicate if there are any process issues with this. I am appealing to your expertise. Again, it looks to be uncontested at this point. 173.56.74.61 (talk) 21:02, 18 July 2017 (UTC)

Nesat's ACE

Hi. So the JMA has upgraded 11W to TS Nesat during 03Z today, which is weird because normally the JMA would upgrade a system to a TS during the 6-hr intervals (00, 06, 12, 18Z). So when updating the ACE for Nesat, do we start at 03Z? Yes I know that there will be a BT coming out soon, but I'm talking about 'operationally'. Typhoon2013 (talk) 04:57, 26 July 2017 (UTC)

Do not do ACE for JMA data, see Talk:Accumulated cyclone energy#JMA West Pacific ACE questionable.--Jasper Deng (talk) 05:00, 26 July 2017 (UTC)

ACE by definition was designed for 1 min winds. But this probably warrant a broader much needed discussion. YE ^{Pacific Hurricane} 05:03, 26 July 2017 (UTC)

Pinging @Yellow Evan: too. Well Keith Edkins is the one who updates the ACE for WPac from time to time and I wanted to help. Though I did a JTWC version at the bottom of Talk:2017 Pacific typhoon season/ACE calcs. Should we now switch to the JTWC version or keep both? I really personally think that we follow the JTWC but I knew that we always follow RSMC. Also, JD, I wanted to know where to get the JTWC BT data so I could do the same for previous seasons, especially how I did not finish the 2015 ACE for two years. Typhoon2013 (talk) 06:08, 26 July 2017 (UTC)

Well, @Keith Edkins: you shouldn't - ACE is not even defined for anything other than 1-minute sustained winds.--Jasper Deng (talk) 06:21, 26 July 2017 (UTC)

Just to note that I have created a discussion over in the talk page of the ACE article. One more thing, what source do you use for the JTWC BT? Well I know that it is from the JTWC but what's the link? Thanks so much. Typhoon2013 (talk) 07:00, 26 July 2017 (UTC)

For this storm, this and this (you will have to change the URL once JTWC updates it for having Nesat's name).--Jasper Deng (talk) 07:12, 26 July 2017 (UTC)

Oh no I am talking about like the one you recently did to the 2016 PTS where you made changes to many storms' intensity (eg Megi classified as a C4 instead of a C3, Mirinae as a C1 instead of a TS etc) like this edit. Thanks and sorry if I made my previous message unclear.Typhoon2013 (talk) 08:53, 26 July 2017 (UTC)

Was it so hard to navigate to "Best Track Data" on the JTWC site yourself? I thought "per best track" would be clear about this.--Jasper Deng (talk) 08:55, 26 July 2017 (UTC)

ACE classifications

Ok so I've recently just downloaded the BT for the 2015 and 2016 seasons, and I have decided to work on the WPac ACE by switching the JMA BT to the JTWC BT, as you have stated in the other talk page. So just to clarify 1) subtropical storms are included in the ACE, just like what Lionrock did? And 2) Also how about Monsoon depressions? Omais' BT stated that the JTWC had 35kt winds though it was a Monsoon Depression. Thanks. Typhoon2013 (talk) 01:18, 28 July 2017 (UTC)

Consensus

Hello Jasper Deng -- You wrote, "I should also add that Wikipedia processes are not courts of law. No one (with the exception of the Wikimedia Foundation itself) is above the consensus of the community."

Aside from all the happy talk on the pages you read when you start editing, it seems to me that WP doesn't work that way. An admin can blow a page away speedily and more importantly, can delete all the talk that went with it. Out of sight, out of mind, is how the saying goes. If a district attorney were free to destroy all the evidence and court transcripts after he got a conviction, that would make his reversed on appeal record look pretty good, wouldn't it?

These guys are sharp. They are happy to post your name on a list of suspects (a permanent list), but they won't engage you in discussion. They have all the secret tools, so they must be right. It's the same with the FISA court. "Of course we are doing the right and fair thing. We just won't tell you what it is." If I had more confidence, I would be silent. My example is a weak case, with not so much as a post, positive or negative in response. The part that irks me is that the automated message for all SPIs requests a response. If no one reads it, what's the point? Same with articles for deletion. The tag asks for editing help to make the article better. One person wrote me to observe that because I had responded to a request to work on a tagged article, that I was suspect. I am therefore a duck.

Nah. This consensus thing is happy talk. But thanks for stalking the conversation. Rhadow (talk) —Preceding undated comment added 22:17, 4 August 2017 (UTC)

@Rhadow: Speedy deletion is one of the actions for which there is pre-existing consensus. Whenever someone disputes an admin's actions and asks that admin about it, the burden is on that admin to reply and explain their actions. All admin actions can be undone if consensus is for it. Administrators who repeatedly flaunt consensus are typically removed from the role. This is not analogous to the law, which must be followed to the letter. You can view all administrator actions at Special:Log - as far as I know, only CheckUser and oversight logs are private, and that due to the need to protect user privacy.--Jasper Deng (talk) 03:18, 5 August 2017 (UTC)

Dreaded 45 day list at ACC

Hello , just a poke to remind you to login to the ACC interface to avoid account suspension. Thanx, - FlightTime (open channel) 21:12, 15 August 2017 (UTC)

Thank you, - FlightTime (open channel) 01:41, 16 August 2017 (UTC)

Hurricane wind speeds

Thank you for correcting that. I was not aware it was rounded to the nearest increment of 5. --Figfires (talk) 20:24, 2 September 2017 (UTC)

WP:AN

Hello. I'm not sure if this is necessary. I am notifying you of a message I left at the Administrator's noticeboard about closing the AfD in the way consensus seems to be breaking - and that is "merge". Here is the diff. The purpose of this post is to simply notify you only because you created the AfD. Regards. Steve Quinn (talk) 19:48, 11 September 2017 (UTC)