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Hope I'm editing this right, haven't added anything to a talk page before. Just to say I replied to your response on the maths RD question on http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics#Category_theory_-_Ob_and_Mor_functors, and thanks for the help so far! Spalton232 ( talk) 12:53, 16 October 2011 (UTC)
I crafted that link for everyone, not just "you":) DMacks ( talk) 18:26, 18 October 2008 (UTC)
Really brilliant! PMajer ( talk) 19:07, 18 October 2008 (UTC)
Thanks for providing a detailed answer to the problem involving number of progressions. Happy new year. Cheers-- Shahab ( talk) 06:50, 30 December 2008 (UTC)
I've made this edit :) hydnjo talk 23:12, 3 January 2009 (UTC)
I've changed your sig at the regs page to " pma ( talk)" to be consistent with your current sig. hydnjo talk 03:15, 3 February 2009 (UTC)
I've taken this opportunity to start your "try-out" or personal experimental page. This may be of help to you if you want to try something out without committing to an actual page just or see if it looks OK. Or, just play around - its your sandbox so do whatever you want there! ;) hydnjo talk 02:13, 5 January 2009 (UTC)
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The box to the right is the newly created userbox for all RefDesk regulars. Since you are an RD regular, you are receiving this notice to remind you to put this box on your userpage! (but when you do, don't include the |no. Just say {{WP:RD regulars/box}} ) This adds you to Category:RD regulars, which is a must. So please, add it. Don't worry, no more spam after this - just check WP:RDREG for updates, news, etc. flaming lawye r c 03:08, 6 January 2009 (UTC)
http://en.wikipedia.org/?title=Wikipedia%3AReference_desk%2FScience&diff=267034236&oldid=267031174
SteveBaker ( talk) 00:41, 29 January 2009 (UTC)
Molto grazie. Ecphora ( talk) 02:28, 4 February 2009 (UTC)
OT for the refdesk, but you did ask, so here goes: I find it vanishingly unlikely that a person using Cambridge university computers and asking three questions from a Cambridge problem sheet on a course being taught at the moment is anything other than a Cambridge maths undergraduate. All such persons have paid supervisors. Algebraist 21:36, 5 February 2009 (UTC)
Thanks for your helpful reply on the ref desk... The Bombing of Helsinki in World War II makes interesting reading on your point about a spread-out population and where the bombing went on. Cheers, Julia Rossi ( talk) 22:11, 5 February 2009 (UTC) Aaah! I just went to your article tip Aki Kaurismaki and discovered that I enjoyed the Total Balalaika Show and the Leningrad Cowboys without knowing about the Finnish connection. How coincidental and how clever, thanks again, Julia Rossi ( talk) 22:24, 5 February 2009 (UTC)
i went with this one! [1] -- Sonjaaa ( talk) 17:14, 5 February 2009 (UTC)
I can't believe it! That was from Teorema, one of my favorite movies, and all I associated were goats and cows :-)! Well spotted and thanks for your input. I hope you keep visiting the language desk. There has been a lack of native Italian speakers there, as far as I'm aware. --- Sluzzelin talk 08:52, 10 April 2009 (UTC)
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The Reference Desk Barnstar | |
Excellent contributions to Wikipedia:Reference desk/Mathematics, pma! I've not noticed you away from the reference desk for even a day, and a question on topology that you haven't answered. Good work! PS T 10:52, 16 April 2009 (UTC) |
I loved this quote:
"Personally I'm a bit sad about the fact that one day I must die, but the thought that everybody must die as well, greatly cheers me up."
Did you make it up yourself, or did you borrow it from somewhere. Not to accuse you or anything, it is just one of those perfect timeless quotes that sounds as if it should have being uttered by Dr. Johnson (or maybe Frankie Boyle!). I Google it, but got nothing. Frank Bruno's Laugh ( talk) 16:19, 14 June 2009 (UTC)
Dear User, in reference to your talk [2], can you tell me how to obtain the 'First' WP page referring to the Boubaker Polynom? thanks Rirunmot ( talk) 17:30, 16 June 2009 (UTC)
Hello. Thanks for all your help recently. I am very thankful. I will probably stop asking questions soon and spend time reviewing things I already know since my test is Thursday. But, before I do that, I have come to ask for help once more on a question from the past about a double integral. I never truly understood what you were saying here and I sort of forgot about it. But, this is a very important topic on the test and I am not good with this sort of thing. One thing you said was I could generalize some of the methods you mentioned on another question. So, I just used one of your methods to do the 1D case, I believe, which is at the bottom there. But, I'm not sure how that would generalize to the 2D case. So, if you are willing, will you please help me further in trying to understand this double integral problem? Thanks for all your help either way! I have also left a few additional comments at the double integral question to explain more. StatisticsMan ( talk) 20:34, 15 August 2009 (UTC)
Ok, I will resume and paste here the main facts, numbered, so you can ask me which point you are interested in and I'll expand it. The two dimensional statement was " an L1 function on whose integral over all rectangles [a,b]x[c,d] vanishes, is zero (a.e.)". You can prove it in several ways:
Okay, it would be helpful if you showed me more about #2. I guess I figured it out for the 1D case but I'm not sure about 2D.
And, for #6, I think I am actually understanding that. A previous result was if for every interval , then f = 0 a.e. Are you saying it is also true if we only have that integral is 0 for every interval with ? That makes sense I guess since if the endpoints are irrational, you can use the intervals inside or outside with rational endpoints and get arbitrarily close. Now, assuming that part is good, in the end you have for almost all y, for every . From the 1D case, this implies for all such y, f(x, y) = 0 for almost all x. So for almost all y, we have for almost all x that f(x, y) = 0. So, let E be the set of all (x, y) where the integral is not 0. Then, this says for almost all y, has measure 0. And, I already did another qual problem proving this implies mE = 0. Is that all right? Thanks for your help! StatisticsMan ( talk) 18:49, 16 August 2009 (UTC)
Hello there, I hope you're well. Your addition to the envelope article is a very nice piece of mathematics. But I think it might be a bit out of place in the article as it stands. The idea of an envelope is a simple one from differential geometry. When people talk about envelopes they are, by and large, talking about the envelopes of families of smooth submanifolds. Your addition seems very algebraic, and a little out of place. For example: Hölder's inequality? (and in turn Lp-spaces?) It seems to be a very algebraic and topological addition to what was, and ought to be, a differential geometric article. You obviously know a lot about the topic. Why don't you start another article or at least a new section? I think there's a lot of milage to be had in looking at envelopes from this point of view, but at the moment the example looks right out of place. It had me scratching my head, and I have a PhD in singularity thoery (e.g. discriminants, bifurcations sets, families of functions, etc) and differential geometry. If I was left scratching my head then the interested undergraduate (God help the laymen) would be totally lost. ~~ Dr Dec ( Talk) ~~ 12:50, 10 September 2009 (UTC)
Since I doubt that we'll get any feedback from the article's own talk page I've added a thread to the Wiki maths project asking for feedback. ~~ Dr Dec ( Talk) ~~ 15:29, 10 September 2009 (UTC)
Would you mind giving me a reference which discusses the convergence of the Binomial expansion for (1+x)^r where x and r are complex numbers? thanks a lot! HowiAuckland ( talk) 08:22, 3 October 2009 (UTC)
Hi, well I provided a self-contained proof in that section on binomial series because I did not have a better reference. I tried to make it as plain as possible. If you want a book reference, maybe a classic one on complex analysis is ok, like Titschmarsh's book. PS: Yes, sorry, it was mispelled; I answered in your TP.
Thanks for replying so fast. I am not familar with that book. Would you mind giving me the exact reference? i.e. title, full author name, year. Are you sure the spelling of his name is correct? I have searched in MathSciNet and found no articles by Titschmarch last name. HowiAuckland ( talk) 08:28, 3 October 2009 (UTC)
Given that you claim to be an Italian, I am surprised by how you mix up the spellings Davis and Davies. This is a common problem in the UK since one is an Irish surname and one is Welsh. Quite why an "Italian" might be subject to the problem is beyond me. ~~ Dr Dec ( Talk)~~ 23:45, 5 October 2009 (UTC)
I respect you decision to retire, but I hope to welcome you back again soon. Bo Jacoby ( talk) 09:20, 7 October 2009 (UTC).
I haven't really followed this fiasco but I do read the maths desk regularly. Just wanted to point out that the RD will quickly devolve into something like Yahoo! Answers if its good contributors retire every time they let their tempers get ahead of them. Zain Ebrahim ( talk) 13:22, 7 October 2009 (UTC)
Ok guys and girls, I retire the retirement; I just wanted to be flattered for a while ;-) -- pma ( talk) 18:50, 7 October 2009 (UTC)
Hi. I've only now noticed the barnstar you gave me. Thanks! -- Meni Rosenfeld ( talk) 07:37, 27 October 2009 (UTC)
Thanks! Sławomir Biały ( talk) 12:58, 15 November 2009 (UTC)
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Hi. Thank you for the detailed response.- Shahab ( talk) 14:37, 30 November 2009 (UTC)
Hi pma. I noticed that you reverted some vandalism recently [3]. Thanks for that. When you come across a user that has vandalised a page then it's a good idea to leave a warning on the user's talk page. If the user continues to vandalise they will build up a warning history and then can be reported to WP:AIAV and they will be blocked. For an example of a warning history leading to a block please see here. A proper warning history is necessary for a casual vandal to be blocked. (Of course there are many exceptions to this rule, e.g. WP:3RR.) You can cut-and-paste warning templates onto the user's talk page. A list can be found here. Alternatively, you could activate Twinkle in the gadget tab of your preferences section. This is an automated way of dealing with vandalism. It will revert the vandalism and, with some input from you, will leave a warning on the user's talk page. I've left a warning on the user's talk page, and I'll keep an eye on their future edits. ~~ Dr Dec ( Talk) ~~ 15:27, 15 December 2009 (UTC)
Thanks again for correcting my silly mistake at the reference desk. I knew that you had to be right, but I just could not see why (a bad day at the reference desk...). -- PS T 12:20, 19 December 2009 (UTC)
For no particular reason, except to say hello, and to say that your contributions to the project are appreciated:
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I was surprised at your comment on rckrone's comments on the circle problem. "A pretty good description"? That one person would be that confused is not surprising; that a mathematician would then endorse the comment is surprising. Michael Hardy ( talk) 04:59, 15 January 2010 (UTC)
pma, would you be willing to help me a bit more on a reference desk question I asked a couple weeks ago? If you are willing, just go to the link as I have asked my question there. Thanks. StatisticsMan ( talk) 15:47, 18 January 2010 (UTC)
Let's define, for and for
For any we have, by the mean value theorem (see Rckrone bound)
We wish to show that the family is locally normally summable in the uniform norm, that is, for any there exists a neighborhood of ) such that
This implies that the double sum
converges uniformly to a continuous function on
Consider an open covering of by open sets of the form
for real numbers and Let be one of these.
It is convenient to bipartite the set of indices into the subsets:
Therefore, for any there are at most values of such that and in any case are among them.
Since for and we have
so
we can bound the sum on as follows:
On the other hand, for all either or In both cases, for any and
Note that for any one has so and the last inequality holds.
Thus
Remark. In the definition of there is no need of all the parameters a,b,c,d : you may better take just , and , so the open sets are a sequence and still cover the domain. But as I wrote it, it should be easier to check the inequalities. In fact you do not need precise bounds, and they are a bit annoying: it should have better made a more qualitative argument, but maybe this way it makes everything more concrete. -- pm a 10:04, 20 January 2010 (UTC)
Dear Sir, I found your reply to my question most unhelpful. I stated clearly that I wanted an explicit, worked example, and I gave a sample function to be worked with. Reading your reply makes me think that you have not read my original question. I wanted to see some εδ-proofs involving my sample function and its truncated Taylor polynomials. As I said: I know that my sample function is holomorphic and so equal to its power series, but I wanted it to be used as an example. Your reply is not dissimilar to other replies I have seen on that page. Many replies seem to attack the validity and rigour of the question instead of attacking the stated problem. This is most saddening. •• Fly by Night ( talk) 21:55, 12 March 2010 (UTC)
...at the reference desk. Since we are both Italian mathematicians may I ask you where do you work? I'm in Tor Vergata university.-- Pokipsy76 ( talk) 16:23, 13 March 2010 (UTC)
facile: pma=Pietro Majer a Pisa!
I've blocked the address for 3 months. Dougweller ( talk) 14:27, 11 May 2010 (UTC)
Hi Pmajer. I hope you are well. I am glad you read my question. I'll try to describe the whole problem to you, in fact I will be glad if you can give me any general advice regarding the problem. Consider the equation v+x+y-z = b (b is an even natural number). It is known that there exists a least natural number r(b) such that if {1,2...r(b)} is partitioned into 2 classes arbitrarily, at least one contains a solution to the given equation. My final goal is to find out a formula for r(b). To this end, I have written a computer program (using brute force) to estimate various values of r(b) for different b, and I use the bound r(b)<= b/2 as v=x=y=z=b/2 is a trivial solution. The problem is that the program starts taking an awful lot of time to estimate r(b) as b increases because the number of possible partitions increases as a power of 2. Is there any general advice on how to approach this problem, any techniques etc you can suggest. I will really appreciate it.-- Shahab ( talk) 02:33, 29 June 2010 (UTC)
Thanks for this example. It turns out to generalize nicely to uncountable dimensions if we assume the generalized continuum hypothesis. Details here. – Henning Makholm ( talk) 22:05, 16 October 2010 (UTC)
You are allusive, but elusive, in criticism of the article on Archimedes Palimpsest, particularly where it touches on Stomachion. It would surely help other users were you to present criticism in greater detail, preferably with references. —Preceding unsigned comment added by 133.31.18.68 ( talk) 05:45, 13 March 2011 (UTC)
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You're right, my additional observations did follow from what was already there, but they provided a perspective from the point of view of α as opposed to x. They also provided a link to the subject of conditional convergence. I believe they served a useful purpose. Would you reconsider including them? Rickhev1 ( talk) 14:27, 21 February 2012 (UTC)
Thanks very much. I think you have done an excellent job on this article. Rickhev1 ( talk) 15:09, 28 February 2012 (UTC)
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Hello,
I think you oversimplified Polya's proof. The inequality
is wrong: the best you can get in the right-hand side is e/n. This does not yield the sharp constant. To get the sharp constant, you need to apply the MA-MG inequality to akck, where ck are defined by
See the paper of Polya.
Best regards, Sasha ( talk) 15:37, 22 September 2012 (UTC)
Dear Sasha, actually I don't think that inequality is wrong (although it is a bit less obvious than the inequality with e/n). And I provided a simple derivation of it from a well-known inequality on the Stirling approximation. Therefore, can you please now
Thank you. I will in any case appreciate any contribution to improve that article. -- pm a 19:43, 22 September 2012 (UTC)
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Thanks for your update on "Function" , benifit a lot :) Xfcjscn ( talk) 08:00, 12 January 2018 (UTC) |
Thank you Xfcjscn! Receiving a wiki barnstar makes me feel proud but also a bit guilty, since I've not been here for quite a long time. Still I don't get what update you mean about Function :) -- pm a 11:40, 12 January 2018 (UTC)
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You have made a bold edit (introducing, for unclear reasons, yet another proof of the same result). It has been reverted. You can now seek to convince some other editors to agree with you, by using the talk page. -- JBL ( talk) 13:09, 25 July 2019 (UTC)
I see that you changed back to an older version on 2019-07-12, thereby undoing a number of changes I had made. I want to mention some possible problems in the current version.
1. In proving (iii) you state that if |x| = 1 and Re(alpha) <= -1, then \left|{\alpha \choose k}\; x^k \right| \geq 1. While this is true, I don't see that it is obvious. Instead of proving this, it would be simpler to observe that the series diverges since its terms are bounded away from zero by formula (5).
2. In your proof of (iv) it is true that formula (5) implies divergence if Re(alpha) < 0, but I don't see that it does if Re(alpha) = 0 but alpha neq 0.
3. In your proof of "Elementary bounds on the coefficients" you show that \left|{\alpha \choose k} \right|\leq\frac {M}{k^{1+\mathrm{Re}\,\alpha}},\qquad\forall k\geq1. I don't believe this establishes formula (5), since you also need that \left|{\alpha \choose k} \right| {k^{1+\mathrm{Re}\,\alpha}} is bounded away from zero; i.e. is >= m.
If you agree that these are indeed gaps in the proof, please let me know whether you want me to make the necessary changes or would prefer to do them yourself. Rickhev1 ( talk) 22:01, 14 September 2019 (UTC)
I'll defer to you on how to proceed. My earlier version did provide a self-contained proof of the whole shebang.
1. To keep things simple, I suggest in proving the part of (iii) for |x| = 1 & Re(alpha) <= -1 that you just use form. (5) to show that the terms |(alpha choose k) x^k| >= m. Thus, they can't converge to 0, whence the series must diverge.
2. Your new proof of (iv) is valid, but it depends on the gamma function. Your statement following form. (5) that it is sufficient for our needs is no longer true.
3. I don't see the point of establishing just the upper bound for |(alpha choose k)|. The way things are now, you need the Gamma function, not just form. (5). And even if you use my method or some other to prove (iv) using only form. (5), it's very unsatisfying to establish the upper bound but not the lower one for form. (5). So I would favor either using Gamma and eliminating the section on "Elementary bounds on the coefficients" or using a proof depending only on form. (5) and either giving a full proof of form. (5) in "Elementary bounds on the coefficients" or eliminating that section. Rickhev1 ( talk) 23:35, 15 September 2019 (UTC)
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Hello, can you add a reference to your paragraph or describe a bit more the proof of the inequality you use please (Forgive me if i do not use well the talk page). -- FRCNico ( talk) 15:21, 16 January 2022 (UTC)
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Hey, I noticed you added the power series of [W(x)]^r (the Lambert W function), with r an integer. It says there that it is obtained using the Lagrange inversion formula. I don't think i fully understand how, and i also didn't find any reference for that statement. The coefficients of the series seems to be just right, and I would love to understand how to obtain them myself. Thanks alot! BarakBud ( talk) 14:02, 30 November 2022 (UTC)
Dear BarakBud, the expansion of with real or complex exponent is just an extension of the Lagrange inversion formula to series with real or complex exponents, in the simplest case, namely series whose exponents are an arithmetic sequence of real or complex numbers (an instance of Hahn_series). At the time I also put a remark on this generalisation on the section on the Lagrange formula here Formal_power_series#The_Lagrange_inversion_formula: just follow the formal proof given there for the standard case, and adapt it to this case.
You may also like to see a similar application of this Lagrange inversion formula for inverting a "polynomial" with non-integer exponents , which is a just a bit more challenging exercise (In case I'll add the details there; for the moment I won't spoil the pleasure of trying the computation :) but do not hesitate asking for details, because clarifying things would be useful for all users --me included).
https://mathoverflow.net/questions/249060/series-solution-of-the-trinomial-equation/249098#249098
Cheers! pm a 16:28, 30 November 2022 (UTC)
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Hope I'm editing this right, haven't added anything to a talk page before. Just to say I replied to your response on the maths RD question on http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics#Category_theory_-_Ob_and_Mor_functors, and thanks for the help so far! Spalton232 ( talk) 12:53, 16 October 2011 (UTC)
I crafted that link for everyone, not just "you":) DMacks ( talk) 18:26, 18 October 2008 (UTC)
Really brilliant! PMajer ( talk) 19:07, 18 October 2008 (UTC)
Thanks for providing a detailed answer to the problem involving number of progressions. Happy new year. Cheers-- Shahab ( talk) 06:50, 30 December 2008 (UTC)
I've made this edit :) hydnjo talk 23:12, 3 January 2009 (UTC)
I've changed your sig at the regs page to " pma ( talk)" to be consistent with your current sig. hydnjo talk 03:15, 3 February 2009 (UTC)
I've taken this opportunity to start your "try-out" or personal experimental page. This may be of help to you if you want to try something out without committing to an actual page just or see if it looks OK. Or, just play around - its your sandbox so do whatever you want there! ;) hydnjo talk 02:13, 5 January 2009 (UTC)
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The box to the right is the newly created userbox for all RefDesk regulars. Since you are an RD regular, you are receiving this notice to remind you to put this box on your userpage! (but when you do, don't include the |no. Just say {{WP:RD regulars/box}} ) This adds you to Category:RD regulars, which is a must. So please, add it. Don't worry, no more spam after this - just check WP:RDREG for updates, news, etc. flaming lawye r c 03:08, 6 January 2009 (UTC)
http://en.wikipedia.org/?title=Wikipedia%3AReference_desk%2FScience&diff=267034236&oldid=267031174
SteveBaker ( talk) 00:41, 29 January 2009 (UTC)
Molto grazie. Ecphora ( talk) 02:28, 4 February 2009 (UTC)
OT for the refdesk, but you did ask, so here goes: I find it vanishingly unlikely that a person using Cambridge university computers and asking three questions from a Cambridge problem sheet on a course being taught at the moment is anything other than a Cambridge maths undergraduate. All such persons have paid supervisors. Algebraist 21:36, 5 February 2009 (UTC)
Thanks for your helpful reply on the ref desk... The Bombing of Helsinki in World War II makes interesting reading on your point about a spread-out population and where the bombing went on. Cheers, Julia Rossi ( talk) 22:11, 5 February 2009 (UTC) Aaah! I just went to your article tip Aki Kaurismaki and discovered that I enjoyed the Total Balalaika Show and the Leningrad Cowboys without knowing about the Finnish connection. How coincidental and how clever, thanks again, Julia Rossi ( talk) 22:24, 5 February 2009 (UTC)
i went with this one! [1] -- Sonjaaa ( talk) 17:14, 5 February 2009 (UTC)
I can't believe it! That was from Teorema, one of my favorite movies, and all I associated were goats and cows :-)! Well spotted and thanks for your input. I hope you keep visiting the language desk. There has been a lack of native Italian speakers there, as far as I'm aware. --- Sluzzelin talk 08:52, 10 April 2009 (UTC)
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The Reference Desk Barnstar | |
Excellent contributions to Wikipedia:Reference desk/Mathematics, pma! I've not noticed you away from the reference desk for even a day, and a question on topology that you haven't answered. Good work! PS T 10:52, 16 April 2009 (UTC) |
I loved this quote:
"Personally I'm a bit sad about the fact that one day I must die, but the thought that everybody must die as well, greatly cheers me up."
Did you make it up yourself, or did you borrow it from somewhere. Not to accuse you or anything, it is just one of those perfect timeless quotes that sounds as if it should have being uttered by Dr. Johnson (or maybe Frankie Boyle!). I Google it, but got nothing. Frank Bruno's Laugh ( talk) 16:19, 14 June 2009 (UTC)
Dear User, in reference to your talk [2], can you tell me how to obtain the 'First' WP page referring to the Boubaker Polynom? thanks Rirunmot ( talk) 17:30, 16 June 2009 (UTC)
Hello. Thanks for all your help recently. I am very thankful. I will probably stop asking questions soon and spend time reviewing things I already know since my test is Thursday. But, before I do that, I have come to ask for help once more on a question from the past about a double integral. I never truly understood what you were saying here and I sort of forgot about it. But, this is a very important topic on the test and I am not good with this sort of thing. One thing you said was I could generalize some of the methods you mentioned on another question. So, I just used one of your methods to do the 1D case, I believe, which is at the bottom there. But, I'm not sure how that would generalize to the 2D case. So, if you are willing, will you please help me further in trying to understand this double integral problem? Thanks for all your help either way! I have also left a few additional comments at the double integral question to explain more. StatisticsMan ( talk) 20:34, 15 August 2009 (UTC)
Ok, I will resume and paste here the main facts, numbered, so you can ask me which point you are interested in and I'll expand it. The two dimensional statement was " an L1 function on whose integral over all rectangles [a,b]x[c,d] vanishes, is zero (a.e.)". You can prove it in several ways:
Okay, it would be helpful if you showed me more about #2. I guess I figured it out for the 1D case but I'm not sure about 2D.
And, for #6, I think I am actually understanding that. A previous result was if for every interval , then f = 0 a.e. Are you saying it is also true if we only have that integral is 0 for every interval with ? That makes sense I guess since if the endpoints are irrational, you can use the intervals inside or outside with rational endpoints and get arbitrarily close. Now, assuming that part is good, in the end you have for almost all y, for every . From the 1D case, this implies for all such y, f(x, y) = 0 for almost all x. So for almost all y, we have for almost all x that f(x, y) = 0. So, let E be the set of all (x, y) where the integral is not 0. Then, this says for almost all y, has measure 0. And, I already did another qual problem proving this implies mE = 0. Is that all right? Thanks for your help! StatisticsMan ( talk) 18:49, 16 August 2009 (UTC)
Hello there, I hope you're well. Your addition to the envelope article is a very nice piece of mathematics. But I think it might be a bit out of place in the article as it stands. The idea of an envelope is a simple one from differential geometry. When people talk about envelopes they are, by and large, talking about the envelopes of families of smooth submanifolds. Your addition seems very algebraic, and a little out of place. For example: Hölder's inequality? (and in turn Lp-spaces?) It seems to be a very algebraic and topological addition to what was, and ought to be, a differential geometric article. You obviously know a lot about the topic. Why don't you start another article or at least a new section? I think there's a lot of milage to be had in looking at envelopes from this point of view, but at the moment the example looks right out of place. It had me scratching my head, and I have a PhD in singularity thoery (e.g. discriminants, bifurcations sets, families of functions, etc) and differential geometry. If I was left scratching my head then the interested undergraduate (God help the laymen) would be totally lost. ~~ Dr Dec ( Talk) ~~ 12:50, 10 September 2009 (UTC)
Since I doubt that we'll get any feedback from the article's own talk page I've added a thread to the Wiki maths project asking for feedback. ~~ Dr Dec ( Talk) ~~ 15:29, 10 September 2009 (UTC)
Would you mind giving me a reference which discusses the convergence of the Binomial expansion for (1+x)^r where x and r are complex numbers? thanks a lot! HowiAuckland ( talk) 08:22, 3 October 2009 (UTC)
Hi, well I provided a self-contained proof in that section on binomial series because I did not have a better reference. I tried to make it as plain as possible. If you want a book reference, maybe a classic one on complex analysis is ok, like Titschmarsh's book. PS: Yes, sorry, it was mispelled; I answered in your TP.
Thanks for replying so fast. I am not familar with that book. Would you mind giving me the exact reference? i.e. title, full author name, year. Are you sure the spelling of his name is correct? I have searched in MathSciNet and found no articles by Titschmarch last name. HowiAuckland ( talk) 08:28, 3 October 2009 (UTC)
Given that you claim to be an Italian, I am surprised by how you mix up the spellings Davis and Davies. This is a common problem in the UK since one is an Irish surname and one is Welsh. Quite why an "Italian" might be subject to the problem is beyond me. ~~ Dr Dec ( Talk)~~ 23:45, 5 October 2009 (UTC)
I respect you decision to retire, but I hope to welcome you back again soon. Bo Jacoby ( talk) 09:20, 7 October 2009 (UTC).
I haven't really followed this fiasco but I do read the maths desk regularly. Just wanted to point out that the RD will quickly devolve into something like Yahoo! Answers if its good contributors retire every time they let their tempers get ahead of them. Zain Ebrahim ( talk) 13:22, 7 October 2009 (UTC)
Ok guys and girls, I retire the retirement; I just wanted to be flattered for a while ;-) -- pma ( talk) 18:50, 7 October 2009 (UTC)
Hi. I've only now noticed the barnstar you gave me. Thanks! -- Meni Rosenfeld ( talk) 07:37, 27 October 2009 (UTC)
Thanks! Sławomir Biały ( talk) 12:58, 15 November 2009 (UTC)
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Hi. Thank you for the detailed response.- Shahab ( talk) 14:37, 30 November 2009 (UTC)
Hi pma. I noticed that you reverted some vandalism recently [3]. Thanks for that. When you come across a user that has vandalised a page then it's a good idea to leave a warning on the user's talk page. If the user continues to vandalise they will build up a warning history and then can be reported to WP:AIAV and they will be blocked. For an example of a warning history leading to a block please see here. A proper warning history is necessary for a casual vandal to be blocked. (Of course there are many exceptions to this rule, e.g. WP:3RR.) You can cut-and-paste warning templates onto the user's talk page. A list can be found here. Alternatively, you could activate Twinkle in the gadget tab of your preferences section. This is an automated way of dealing with vandalism. It will revert the vandalism and, with some input from you, will leave a warning on the user's talk page. I've left a warning on the user's talk page, and I'll keep an eye on their future edits. ~~ Dr Dec ( Talk) ~~ 15:27, 15 December 2009 (UTC)
Thanks again for correcting my silly mistake at the reference desk. I knew that you had to be right, but I just could not see why (a bad day at the reference desk...). -- PS T 12:20, 19 December 2009 (UTC)
For no particular reason, except to say hello, and to say that your contributions to the project are appreciated:
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I was surprised at your comment on rckrone's comments on the circle problem. "A pretty good description"? That one person would be that confused is not surprising; that a mathematician would then endorse the comment is surprising. Michael Hardy ( talk) 04:59, 15 January 2010 (UTC)
pma, would you be willing to help me a bit more on a reference desk question I asked a couple weeks ago? If you are willing, just go to the link as I have asked my question there. Thanks. StatisticsMan ( talk) 15:47, 18 January 2010 (UTC)
Let's define, for and for
For any we have, by the mean value theorem (see Rckrone bound)
We wish to show that the family is locally normally summable in the uniform norm, that is, for any there exists a neighborhood of ) such that
This implies that the double sum
converges uniformly to a continuous function on
Consider an open covering of by open sets of the form
for real numbers and Let be one of these.
It is convenient to bipartite the set of indices into the subsets:
Therefore, for any there are at most values of such that and in any case are among them.
Since for and we have
so
we can bound the sum on as follows:
On the other hand, for all either or In both cases, for any and
Note that for any one has so and the last inequality holds.
Thus
Remark. In the definition of there is no need of all the parameters a,b,c,d : you may better take just , and , so the open sets are a sequence and still cover the domain. But as I wrote it, it should be easier to check the inequalities. In fact you do not need precise bounds, and they are a bit annoying: it should have better made a more qualitative argument, but maybe this way it makes everything more concrete. -- pm a 10:04, 20 January 2010 (UTC)
Dear Sir, I found your reply to my question most unhelpful. I stated clearly that I wanted an explicit, worked example, and I gave a sample function to be worked with. Reading your reply makes me think that you have not read my original question. I wanted to see some εδ-proofs involving my sample function and its truncated Taylor polynomials. As I said: I know that my sample function is holomorphic and so equal to its power series, but I wanted it to be used as an example. Your reply is not dissimilar to other replies I have seen on that page. Many replies seem to attack the validity and rigour of the question instead of attacking the stated problem. This is most saddening. •• Fly by Night ( talk) 21:55, 12 March 2010 (UTC)
...at the reference desk. Since we are both Italian mathematicians may I ask you where do you work? I'm in Tor Vergata university.-- Pokipsy76 ( talk) 16:23, 13 March 2010 (UTC)
facile: pma=Pietro Majer a Pisa!
I've blocked the address for 3 months. Dougweller ( talk) 14:27, 11 May 2010 (UTC)
Hi Pmajer. I hope you are well. I am glad you read my question. I'll try to describe the whole problem to you, in fact I will be glad if you can give me any general advice regarding the problem. Consider the equation v+x+y-z = b (b is an even natural number). It is known that there exists a least natural number r(b) such that if {1,2...r(b)} is partitioned into 2 classes arbitrarily, at least one contains a solution to the given equation. My final goal is to find out a formula for r(b). To this end, I have written a computer program (using brute force) to estimate various values of r(b) for different b, and I use the bound r(b)<= b/2 as v=x=y=z=b/2 is a trivial solution. The problem is that the program starts taking an awful lot of time to estimate r(b) as b increases because the number of possible partitions increases as a power of 2. Is there any general advice on how to approach this problem, any techniques etc you can suggest. I will really appreciate it.-- Shahab ( talk) 02:33, 29 June 2010 (UTC)
Thanks for this example. It turns out to generalize nicely to uncountable dimensions if we assume the generalized continuum hypothesis. Details here. – Henning Makholm ( talk) 22:05, 16 October 2010 (UTC)
You are allusive, but elusive, in criticism of the article on Archimedes Palimpsest, particularly where it touches on Stomachion. It would surely help other users were you to present criticism in greater detail, preferably with references. —Preceding unsigned comment added by 133.31.18.68 ( talk) 05:45, 13 March 2011 (UTC)
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You're right, my additional observations did follow from what was already there, but they provided a perspective from the point of view of α as opposed to x. They also provided a link to the subject of conditional convergence. I believe they served a useful purpose. Would you reconsider including them? Rickhev1 ( talk) 14:27, 21 February 2012 (UTC)
Thanks very much. I think you have done an excellent job on this article. Rickhev1 ( talk) 15:09, 28 February 2012 (UTC)
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Hello,
I think you oversimplified Polya's proof. The inequality
is wrong: the best you can get in the right-hand side is e/n. This does not yield the sharp constant. To get the sharp constant, you need to apply the MA-MG inequality to akck, where ck are defined by
See the paper of Polya.
Best regards, Sasha ( talk) 15:37, 22 September 2012 (UTC)
Dear Sasha, actually I don't think that inequality is wrong (although it is a bit less obvious than the inequality with e/n). And I provided a simple derivation of it from a well-known inequality on the Stirling approximation. Therefore, can you please now
Thank you. I will in any case appreciate any contribution to improve that article. -- pm a 19:43, 22 September 2012 (UTC)
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Thanks for your update on "Function" , benifit a lot :) Xfcjscn ( talk) 08:00, 12 January 2018 (UTC) |
Thank you Xfcjscn! Receiving a wiki barnstar makes me feel proud but also a bit guilty, since I've not been here for quite a long time. Still I don't get what update you mean about Function :) -- pm a 11:40, 12 January 2018 (UTC)
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You have made a bold edit (introducing, for unclear reasons, yet another proof of the same result). It has been reverted. You can now seek to convince some other editors to agree with you, by using the talk page. -- JBL ( talk) 13:09, 25 July 2019 (UTC)
I see that you changed back to an older version on 2019-07-12, thereby undoing a number of changes I had made. I want to mention some possible problems in the current version.
1. In proving (iii) you state that if |x| = 1 and Re(alpha) <= -1, then \left|{\alpha \choose k}\; x^k \right| \geq 1. While this is true, I don't see that it is obvious. Instead of proving this, it would be simpler to observe that the series diverges since its terms are bounded away from zero by formula (5).
2. In your proof of (iv) it is true that formula (5) implies divergence if Re(alpha) < 0, but I don't see that it does if Re(alpha) = 0 but alpha neq 0.
3. In your proof of "Elementary bounds on the coefficients" you show that \left|{\alpha \choose k} \right|\leq\frac {M}{k^{1+\mathrm{Re}\,\alpha}},\qquad\forall k\geq1. I don't believe this establishes formula (5), since you also need that \left|{\alpha \choose k} \right| {k^{1+\mathrm{Re}\,\alpha}} is bounded away from zero; i.e. is >= m.
If you agree that these are indeed gaps in the proof, please let me know whether you want me to make the necessary changes or would prefer to do them yourself. Rickhev1 ( talk) 22:01, 14 September 2019 (UTC)
I'll defer to you on how to proceed. My earlier version did provide a self-contained proof of the whole shebang.
1. To keep things simple, I suggest in proving the part of (iii) for |x| = 1 & Re(alpha) <= -1 that you just use form. (5) to show that the terms |(alpha choose k) x^k| >= m. Thus, they can't converge to 0, whence the series must diverge.
2. Your new proof of (iv) is valid, but it depends on the gamma function. Your statement following form. (5) that it is sufficient for our needs is no longer true.
3. I don't see the point of establishing just the upper bound for |(alpha choose k)|. The way things are now, you need the Gamma function, not just form. (5). And even if you use my method or some other to prove (iv) using only form. (5), it's very unsatisfying to establish the upper bound but not the lower one for form. (5). So I would favor either using Gamma and eliminating the section on "Elementary bounds on the coefficients" or using a proof depending only on form. (5) and either giving a full proof of form. (5) in "Elementary bounds on the coefficients" or eliminating that section. Rickhev1 ( talk) 23:35, 15 September 2019 (UTC)
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Hello, can you add a reference to your paragraph or describe a bit more the proof of the inequality you use please (Forgive me if i do not use well the talk page). -- FRCNico ( talk) 15:21, 16 January 2022 (UTC)
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Hey, I noticed you added the power series of [W(x)]^r (the Lambert W function), with r an integer. It says there that it is obtained using the Lagrange inversion formula. I don't think i fully understand how, and i also didn't find any reference for that statement. The coefficients of the series seems to be just right, and I would love to understand how to obtain them myself. Thanks alot! BarakBud ( talk) 14:02, 30 November 2022 (UTC)
Dear BarakBud, the expansion of with real or complex exponent is just an extension of the Lagrange inversion formula to series with real or complex exponents, in the simplest case, namely series whose exponents are an arithmetic sequence of real or complex numbers (an instance of Hahn_series). At the time I also put a remark on this generalisation on the section on the Lagrange formula here Formal_power_series#The_Lagrange_inversion_formula: just follow the formal proof given there for the standard case, and adapt it to this case.
You may also like to see a similar application of this Lagrange inversion formula for inverting a "polynomial" with non-integer exponents , which is a just a bit more challenging exercise (In case I'll add the details there; for the moment I won't spoil the pleasure of trying the computation :) but do not hesitate asking for details, because clarifying things would be useful for all users --me included).
https://mathoverflow.net/questions/249060/series-solution-of-the-trinomial-equation/249098#249098
Cheers! pm a 16:28, 30 November 2022 (UTC)
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