# User talk:PMajer

Welcome!

Hello, PMajer, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful:

I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or place {{helpme}} on your talk page and someone will show up shortly to answer your questions. Again, welcome!

## Cat Theory RD Question

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)

## RD/S "expert"

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

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)

## FYI

I've made this edit :) hydnjo talk 23:12, 3 January 2009 (UTC)

Thank you! I don't completely understand but it sounds very nice :)--PMajer (talk) 00:50, 4 January 2009 (UTC)
Not of great importance but there is a list of folks who regularly contribute at one or more of the reference desks. I happened to notice your contributions at the Mathematics desk and so I added you to the list of people who are regular and positive contributors. I hope you don't mind  :) hydnjo talk 20:14, 4 January 2009 (UTC)
Thank you very much, I'm very glad to be both regular and positive!! ;) --PMajer (talk) 21:00, 4 January 2009 (UTC)
I've reformatted this section in order to help you understand some of the wiki syntax. You seem to be catching on just fine so I'll throw in a couple of links to illustrate disambiguation (one word has several meanings). Please let me know if you would like any further help. Also, you can click "" next to this section to see all of the wiki-syntax used :)
Regular as you/I intended = Protagonist so you can show "Regular" and have it link to "Protagonist" like this: [[Protagonist|Regular]].
Positive as you/I intended = Positive (linguistics) so you can show "Positive" and have it link to "Positive (linguistics)" like this: [[Positive (linguistics)|Positive]].
That vertical line "|" is called a pipe and using it is called "piping" a piped link. It lets you separate the name of the page you want to link to any name that you want : [[real pagename | any name]]. Another usage is called the pipe trick. My apologies if I've confused you, just let me know if there is something that I can try to explain better :)
hydnjo talk 23:50, 4 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)

## Your "sandbox" is at User:PMajer/sb

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)

Thank you again! I knew the pipe trick, anyway I will refer to you for help! --PMajer (talk) 14:30, 5 January 2009 (UTC)

## new WP:RDREG userbox

 This user is a Reference desk regular.

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. flaminglawyerc 03:08, 6 January 2009 (UTC)

## Why did you delete my reply on the WP:RD?

http://en.wikipedia.org/w/index.php?title=Wikipedia%3AReference_desk%2FScience&diff=267034236&oldid=267031174

SteveBaker (talk) 00:41, 29 January 2009 (UTC)

Steve, it was me?? ..Seems it is so; of course, it was completely unwillingly, for I am grateful to your response. Sorry. (By the way I noticed that you used the singular form "a dice" as a kind of support; it's a particular that reveals a very kind person). Thank you again and excuse me for the unwilled vandalism! --pma (talk) 11:45, 29 January 2009 (UTC)

## Parte presa

Molto grazie. Ecphora (talk) 02:28, 4 February 2009 (UTC)

You are welcome! Do not esitate to ask me directly here, if you have any unsolved question, for I usually check the RD/Maths, and seldom go around for the other RD (and remember I am definitely not an expert of your topics). --pma (talk) 09:50, 4 February 2009 (UTC)
I've done a little more work on this. A Venetian/Italian dictionary under "Parte" states:
"Parte chiamavasi a'tempi della Repubblica Veneta un Decreto o Legge o Risoluzione, ch' era presa a partito da un Consiglio tanto sovrano che suddito legalmente convocato."
In an Italian dictionary I have, "partito" means not only "party", but also "resolution" or "decision". I wonder therefor if "parte presa" should best be translated as "resolution taken" or "decision taken" (or "made"). This fits very well with the various Parte presa documents I have (which prompted by original question); they all involve some problem (thefts, tax avoidance, frauds, blasphemy, etc.) and do not involve any "parties" as such. They all contain the language "l'anderà parte ...", which as you recognized, could mean here "the resolution will be that ..." What do you think? Thanks again for all the help. Ecphora (talk) 00:44, 7 February 2009 (UTC)
Yes, I think your translation is correct, and notice that it's not in contradiction with the original meaning of "parte" as "piece". As far as I see, the typical situation, in an assembly where a decision has to be taken, by discussion and votation, is that the assembly divides in two or more groups according to their opinions, plus maybe a number of people still undecided and floating; they also gather in different parts of the room (otherwise everybody quarrels with each other). So the original phisical meaning of part as "piece" then also cover "group of persons with the same opinion" (party) and then also: the opinion itself shared by a given group; when one opinion wins (after votation or other) it became the decision, and we can well translate "parte presa" as "decision taken". Note that already the Latin word "pars" has a political acceptance of "party". --pma (talk) 13:07, 7 February 2009 (UTC)
Problem solved! Thanks once again. Ecphora (talk) 15:22, 7 February 2009 (UTC)

## Re: WP:RD/Math#Countable series in the rationals

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)

Of course, there's nothing special about 1/n: I could have used any sequence whose sum diverges but whose sum of squares converges. Also, the result still holds if the 2 in the question is replaced by any real strictly greater than 1. Algebraist 00:34, 6 February 2009 (UTC)
Thanks, yes, that's also what I thought, more or less. Clearly the point is to follow a first list of the rationals, to be sure not to miss any of them. --pma (talk) 13:45, 7 February 2009 (UTC)

## Finnish humour

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)

No, pma, not at all. You brought another view to the events and the feelings behind them. I'm moved too by your experience in the alps. Conflict is always a tragedy, events getting out of hand and wiping out incandescent lives and people suffering without protection or choice in the matter. Then there's the humour, the joke against everything. The Finns were very clever to move the fires around to deceive the bombers that the city was elsewhere. The odds were amazing. The Leningrad Cowboys documentary made a big impression on me to see all those people enjoying themselves, regardless of history. You might be interested in this guy: Simo Häyhä. Julia Rossi (talk) 23:21, 5 February 2009 (UTC)

## quilt tessellation -- thanks so much!

i went with this one! [1] --Sonjaaa (talk) 17:14, 5 February 2009 (UTC)

## Nettles

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)

Thank you, my pleasure! I'm very glad to meet people who know Pasolini. --pma (talk) 09:03, 10 April 2009 (UTC)

## Reference desk Barnstar

 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! PST 10:52, 16 April 2009 (UTC)
Thank you, I'm honored to get a barnstar of topology from a topologist. I must confess that I answer questions at the RD/M also to be allowed to put sometimes silly jokes ;-) --pma (talk) 12:17, 16 April 2009 (UTC)

## RE: Death

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)

Hi Frank, I'm glad you appreciated. In fact, I meant to write something in style of philosophic consolatory, you know, appropriate for the circumstance, and the funny effect was not scheduled ;-) --pma (talk) 18:30, 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)

• Hi, actually, I don't know. Apparently, there is no more page on "Boubaker's polynomials"; I do not why, it has been removed. It was worth keeping at least the definition, I think. However there should be a way to read the removed pages, but I fear I don't know more than you; if you can't find it, try at the helpdesk... --pma (talk) 18:30, 16 June 2009 (UTC)
• Thanks, please, if you find this way, inform me in my talk page (or transmit to me the user name of he who can help), many thanks Rirunmot (talk) 21:20, 16 June 2009 (UTC)

## Double integral

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 ${\displaystyle \scriptstyle \mathbb {R} ^{2}}$ whose integral over all rectangles [a,b]x[c,d] vanishes, is zero (a.e.)". You can prove it in several ways:

1. Via the L1 density of step functions. The set ${\displaystyle \scriptstyle {\mathcal {S}}}$ of all "step functions" (that is, linear combinations of characteristic functions of rectangles) is a subspace of ${\displaystyle \scriptstyle L^{1}(\mathbb {R} ^{2})}$, dense in the L1 norm. This result implies in particular that there exists a sequence ${\displaystyle \scriptstyle (\phi _{k})}$ in ${\displaystyle \scriptstyle {\mathcal {S}}}$ converging almost everywhere to the function sgn(f) (note that it's only locally L1). Moreover, since ${\displaystyle \scriptstyle |\mathrm {sgn} (f(x))|\leq 1}$ for all x, we can take ${\displaystyle \scriptstyle \phi _{k}(x)\leq 1}$ for all x (because if needed we an replace it with the truncate sequence ${\displaystyle \scriptstyle \min(1,\max(-1,\phi _{k}(x)))}$, that still converges a.e. to sgn(f(x)) and of course is between -1 and 1). Thus the sequence ${\displaystyle \scriptstyle f(x)\phi _{k}(x)}$ converges a.e. to ${\displaystyle \scriptstyle f(x)\mathrm {sgn} (f(x))=|f(x)|}$, and it is dominated by ${\displaystyle \scriptstyle |f(x)|}$, so by the Lebesgue convergence theorem ${\displaystyle \scriptstyle \int _{\mathbb {R} ^{2}}f(x)\phi _{k}(x)dx}$ converges to ${\displaystyle \scriptstyle \|f\|_{1}}$. But ${\displaystyle \scriptstyle \int _{\mathbb {R} ^{2}}f(x)\phi _{k}(x)dx}$ is zero for all k because it is a linear combination of integrals of f over rectangles, and we conclude ${\displaystyle \scriptstyle \|f\|_{1}=0}$. RMK: the L1 density of step functions is a not difficult consequence of the L1 density of simple functions (that is, linear combinations of characteristic functions of measurable sets of finite measure), but still requires a little work.
2. Via convolution and approximation of the identity. Consider the characteristic function ${\displaystyle \scriptstyle \chi (x)}$ of the square [-1,1]x[-1,1], normalized by 4. Define as usual ${\displaystyle \scriptstyle \chi _{\varepsilon }(x):=\varepsilon ^{-2}\chi (x/\varepsilon )}$. Then ${\displaystyle \scriptstyle f*\chi _{\varepsilon }}$ converges to f in L1 as ${\displaystyle \scriptstyle \varepsilon \to 0}$: this is the most simple result about the approximation of the identity (warning: at the moment the wiki article about mollifiers is wrong in many points). By definition, for all x in ${\displaystyle \scriptstyle \mathbb {R} ^{2}}$, ${\displaystyle \scriptstyle f*\chi _{\varepsilon }(x)}$ is the integral mean of f on the square ${\displaystyle \scriptstyle [x-\varepsilon ,x+\varepsilon ]\times [x-\varepsilon ,x+\varepsilon ]}$, so it is identically zero, and we conclude as before. You may use as well the characteristic function of any other bounded set of positive measure ${\displaystyle \scriptstyle -\Omega }$ in place of the square [-1,1]x[-1,1], and the same conclusion follows assuming "the integral of f vanishes over every domain obtained from ${\displaystyle \scriptstyle \Omega }$ by means of homoteties and translations", since this ensures that the corresponding convolution ${\displaystyle \scriptstyle f*\chi _{\varepsilon }}$ vanishes identically.
3. Of course both proofs work in any dimension; actually "squares" or "cubes" in place of "rectangles" [a1,b1]x[a2,b2]..x[an,bn] are sufficient, and f in L1loc instead of L1 is also sufficient. In the second proof, you can use the segment [-1,0] for the one dimensional case: so define ${\displaystyle \scriptstyle \chi (x)}$ as the characteristic function of [-1,0], then ${\displaystyle \scriptstyle \chi _{\varepsilon }(x):=\varepsilon ^{-1}\chi (x/\varepsilon )}$. You can recognize that ${\displaystyle \scriptstyle f*\chi _{\varepsilon }(x)={\frac {F(x+\varepsilon )-F(x)}{\varepsilon }}}$ where ${\displaystyle \scriptstyle F(x):=\int _{-\infty }^{x}f(t)dt}$, and the proof can be rephrased in terms of properties of absolutely continuous functions, as you did.
4. It is interesting that in the one-dimensional case the same conclusion holds if you ask the condition only on the intervals of length 1. Indeed, this implies, by the sigma-additivity of integral, that the integral is 0 on all half-lines. By subtraction then the integral on intervals of any length also vanishes, and you are lead to the preceding case. This easily generalizes to more dimensions: if an integrable function on ${\displaystyle \scriptstyle \mathbb {R} ^{n}}$ has vanishing integrals over all unit cubes with edges parallel to the axis, it is zero a.e. You just have to play with the sigma-additivity and with translations, gaining the previous hypotheses. The reduced hypotheses are not sufficient in the case of ${\displaystyle \scriptstyle L^{\infty }}$ functions as shown by ${\displaystyle \scriptstyle \sin(x/2\pi )}$.
5. Still true, but less elementary, the following generalization to integrable functions on Rn : if the integral of${\displaystyle \scriptstyle f}$ over any translated ${\displaystyle \scriptstyle \Omega +x}$ of a given bounded set of positive measure ${\displaystyle \scriptstyle \Omega }$ vanishes, then again ${\displaystyle \scriptstyle f=0}$ a.e. Indeed, the assumption is equivalent to say that ${\displaystyle \scriptstyle f*g=0}$, where ${\displaystyle \scriptstyle g}$ is the characteristic function of ${\displaystyle \scriptstyle -\Omega }$. Applying the Fourier transform you have that the pointwise product of ${\displaystyle \scriptstyle {\hat {f}}}$ and ${\displaystyle \scriptstyle {\hat {g}}}$ is 0 a.e. But ${\displaystyle \scriptstyle {\hat {g}}}$ is a non-zero analytic function by the Paley-Wiener theorem, so it is a.e. different from 0, hence ${\displaystyle \scriptstyle {\hat {f}}}$ is 0 a.e., and since the Fourier transform is injective, ${\displaystyle \scriptstyle f}$ is 0 a.e.
6. You can also prove the result in the two-dimensional case (and in the n-dimensional too) as a consequence of the one dimensional case. Indeed you can reduce the problem to 1 variable using Fubini's theorem. For fixed a and b consider the function ${\displaystyle \scriptstyle f_{a,b}(y):=\int _{[a,b]}f(x,y)\,dx}$. It has vanishing integral over all intervals [b,c]. Therefore it is identically zero, according to the 1 dimensional case. This means that for a.e. y, the function ${\displaystyle \scriptstyle x\mapsto f(x,y)}$ has vanishing integral over the fixed interval [a,b]. But this is true for all a and b, so we can also say that for a.e. y the function f(x,y) has vanishing integral in x over all bounded intervals. To be precise, we sould argue in the usual way when dealing with "a.e. properties": let ${\displaystyle (I_{k})_{k}}$ be an enumeration of all finite intervals with rational end-points. The integral of f(x,y) in x over ${\displaystyle I_{k}}$ vanishes for almost all y, that is for all ${\displaystyle \scriptstyle y\in \mathbb {R} \setminus N_{k}}$, where ${\displaystyle \scriptstyle N_{k}\subset \mathbb {R} }$ has measure zero. So for all ${\displaystyle \scriptstyle y\in \mathbb {R} \setminus \cup _{k}N_{k}}$, hence for almost all y, it is true that ${\displaystyle \scriptstyle \int _{I_{k}}f(x,y)dx=0}$ for all k. But for any such y the equality ${\displaystyle \scriptstyle \int _{I}f(x,y)dx=0}$ immediately extends to any bounded interval I by continuity of the integral. By the one dimensional result you conclude.
7. You could also translate the two-dimensional (or n dimensional) hypotheses in terms of the integral function of f(x,y): that is, proving that ${\displaystyle \scriptstyle F(x,y):=\int _{-\infty }^{x}\int _{-\infty }^{y}f(s,t)dt\,ds}$ has vanishing distributional partial derivatives, ${\displaystyle \scriptstyle \partial _{x}F=0}$, ${\displaystyle \scriptstyle \partial _{y}F=0}$, or equivalently, it has a vanishing distributional gradient. Therefore F is constant, hence f is 0. But the proof of this fact would be essentially one of the above.

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 ${\displaystyle \int _{a}^{b}f(x)\,dx=0}$ for every interval ${\displaystyle [a,b]}$, then f = 0 a.e. Are you saying it is also true if we only have that integral is 0 for every interval with ${\displaystyle a,b\in \mathbb {Q} }$? 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, ${\displaystyle \int _{I_{k}}f(x,y)\,dx=0}$ for every ${\displaystyle k}$. 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, ${\displaystyle E_{x}=\{x\in \mathbb {R} :(x,y)\in E\}}$ 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)

Yes, your conclusion of #6 is perfect. As to #2, it's not clear to me what's your doubt. If ${\displaystyle \scriptstyle \Omega }$ is a rectangle (or any measurable set of positive and finite measure) and if we denote ${\displaystyle \scriptstyle \chi (x)}$ the characteristic function of the set ${\displaystyle \scriptstyle \Omega }$ (normalized! I think I forgot to say it above), that is
${\displaystyle \textstyle \chi (x):={\frac {1}{\mathrm {m} (\Omega )}}\ }$ if ${\displaystyle \ \textstyle x\in \Omega }$ and ${\displaystyle \textstyle \chi (x):=0}$ otherwise,
then we have , for any ${\displaystyle \scriptstyle x}$
${\displaystyle \textstyle f*\chi (x)=\int _{\mathbb {R} ^{2}}f(y)\chi (x-y)dy={\frac {1}{\mathrm {m} (\Omega )}}\int _{x-\Omega }f(y)dy,}$
(because the support of ${\displaystyle \scriptstyle y\mapsto \chi (x-y)}$ is the set of all points ${\displaystyle \scriptstyle y}$ such that ${\displaystyle \scriptstyle x-y\in \Omega }$, that is ${\displaystyle \scriptstyle x-\Omega }$). Therefore ${\displaystyle \scriptstyle f*\chi (x)}$ is the integral mean of ${\displaystyle \scriptstyle f}$ on the set ${\displaystyle \scriptstyle x-\Omega }$; analogously ${\displaystyle \scriptstyle f*\chi _{\varepsilon }(x)}$ is the integral mean of ${\displaystyle \scriptstyle f}$ on the set ${\displaystyle \scriptstyle x-\varepsilon \Omega }$.
So in our hypotheses ${\displaystyle \scriptstyle f*\chi _{\varepsilon }(x)}$ is identically 0 for all ${\displaystyle \scriptstyle x}$ and ${\displaystyle \scriptstyle \varepsilon }$ if ${\displaystyle \scriptstyle \Omega }$ is a rectangle. As you see, everything follows from the general result:
If ${\displaystyle \scriptstyle f\in L^{1}(\mathbb {R} ^{n})}$, then ${\displaystyle \scriptstyle f*\chi _{\varepsilon }(x)}$ converges to ${\displaystyle \scriptstyle f(x)}$ in the ${\displaystyle \textstyle L^{1}}$ norm.
Do you know the proof? The usual way to show it is: first, prove it for the case when ${\displaystyle \scriptstyle f}$ is a continuous function with compact support. Indeed in this case the uniform continuity of ${\displaystyle \scriptstyle f}$ gives immediately the uniform convergence of the sequence, within the same bounded support, hence the ${\displaystyle L^{1}}$ convergence. Then one proves that the set of the ${\displaystyle \scriptstyle f\in L^{1}}$ for which the thesis hold true, is a closed linear subspace of ${\displaystyle \scriptstyle L^{1}}$. Therefore it is all ${\displaystyle \scriptstyle L^{1}}$ because of the density of the continuous functions with compact support. (It is true for any integrable function ${\displaystyle \scriptstyle \chi (x)}$ with unit integral, btw, and one also has the a.e. convergence: but that's not needed here ). Is it clear?
I think everything you say makes sense. I'm not sure I understand it well enough to repeat it though. But, I'm too the point where I think I need to just review everything I've done so far. I'm currently reading every qual problem I have written out, which is taking a long time, and I want to review theorems and definitions and all that. Thanks for all your help. You have definitely helped me out in a lot of ways and I am definitely more prepared to pass because of it. StatisticsMan (talk) 00:32, 19 August 2009 (UTC)
Good, I'm glad that I've been of help. It seems you are in a good shape for your qual. Do it, then I'll toast to it too, with the Ocean in between.
I passed! Thanks again. As I said on the Reference Desk, I believe I got the minimum correct to pass (I will see for sure in a couple days) and one of the ones I got was showing that if an integrable function has integral 0 over every interval, then it is 0 almost everywhere, a problem you helped me solve! StatisticsMan (talk) 14:14, 29 August 2009 (UTC)
Very good, great! And the next time you have to get the maximum! ;-)
Luckily, there is no next time. I just need to take classes and do research for the next 3 years. No more major tests! StatisticsMan (talk) 15:06, 29 August 2009 (UTC)

## Envelope (mathematics)

Dear Dec, of course, you are free to revert it or move it elsewhere, if you think it is out of place there; maybe also after hearing other people's opinion. I will consider in any case any solution of yours as aimed to improve the quality. However, for what concerns me, I wouldn't feel like making an article or a section out of it --but if somebody will, he's welcome of course. My idea was to provide a particular example of envelope that can be treated by elementary inequalities. In the same spirit, for instance, that it is didactically useful the elementary proof of the implicit function theorem in the case F(x,y(x))=0 with a function F:R2R , even if the general proof, at the level of Banach spaces, uses the local inverse mapping theorem in a definitely clarifying way, &c.
So, there are, if I understand, two objections: 1. it's not based on classical methods of differential geometry 2. it is difficult. I tend to have a unitary view of mathematics, so I wouldn't consider your first objection as an insurmountable difficulty, that is, we can work a little on it, showing the connections between different methods and languages of maths. In maths, connections are important: this is etymologically a truth. However, after all, what are the methods of a discipline is not a completely definite matter, for they vary with different schools, times, places.
As to the other point, maybe you can just show me the obscure points. I tried to make it short, because it was just intended to be an example. Also, you can certainly improve the language.
In any case, maybe we can stuff it in the astroid article, as a quick proof --generalized with no additional price, of the characterization of an astroid as envelope of unit segments with endpoints on the axes. What do you think?
PS: you are right on that Hölder inequality: I didn't check the link, I meant this Hölder inequality, in R2 , that has nothing or little to do with general Lp spaces (no, indeed the link it was ok, it goes to Rn)

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)

OK, good idea; and if we still don't get much feedback, we can decide something by ourselves. In the meantime, I'd be glad to have your impression of mathematical reader, about the readability of that example, regardless to its collocation. In particular, if in your opinion there is something to be explained in more detail. As to the laymen, I wouldn't worry for them more than enough -just because with no mathematical background not every mathematical article is accessible. I am more interested in the general mathematical readers (everybody: undergraduates, graduates, phd's, postdocs &c), that are the people that really look for this more technical information, as the RD/M experience shows.
So, as I mentioned in the article, the idea was making a classical example of an envelope of curves that in particular are boundaries of some sets. More precisely, an example of a case in which
boundary of a union of subdomains = envelope of the boundaries of the subdomains.
You can either see it as (i) an application of the notion of envelope to understanding a union of sets, or vice versa, (ii) the (classical) technique of treating an envelope of curves (more generally, of codimension one submanifolds) when they can be seen as boundary of some smooth subdomains.
1. How would you prefer to have the example explained, as (i) or (ii) ? I think (ii) is more natural here.
2. Is it clear to you what is the family of curves (the boundaries) in the example?
3. I understand that you are not familiar with the Hölder inequality in R2. Is the Cauchy-Schwarz inequality in R2 better known to you? In this case, do you think that doing the example in the particular case of α=1 would make it any easier?
4. Would you like to find a further explanation about why ${\displaystyle \partial \Delta _{\alpha }}$ is the envelope of the family of segments?
5. Other remarks?
Thank you very much for letting me know your impression! pma.
I've just got back to the house. I want to add three alternative definitions to the Enevelope article. I think your point about "boundary of a union of subdomains = envelope of the boundaries of the subdomains" will come under the third definition. I'm sorry, I don't have much time now. I'll address each of your points tomorrow evening. Sorry! ~~ Dr Dec (Talk) ~~ 22:24, 10 September 2009 (UTC)
Done! Let me know what you think. I'm off to bed now. ~~ Dr Dec (Talk) ~~ 23:03, 10 September 2009 (UTC)
Ok, good. I have few time till the next week, but I'll go and check it. The picture you added certainly improves the comprehension of the example 2. As I told you, one can keep it there, or move in the astroid page, or even in the Hoelder inequality page, as an example of a geometric application. My aim was, mainly, to have it as a quick reference.
As to the case of "boundary of a union=envelope of boundaries", I think that, of course in a rigorous form, it may provide a natural class of envelopes, but note that it is not as general as to cover all cases of envelopes even for curves in R2, so it couldn't be used as a definition.

## Complex Binomial Theorem

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)

## Davis and Davies

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)

And then? --pma (talk) 00:13, 6 October 2009 (UTC)

## Good bye

I respect you decision to retire, but I hope to welcome you back again soon. Bo Jacoby (talk) 09:20, 7 October 2009 (UTC).

## Quick comment

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 ok ok

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)

Welcome back! Bo Jacoby (talk) 20:19, 9 October 2009 (UTC).

## Thanks for the barnstar

Hi. I've only now noticed the barnstar you gave me. Thanks! -- Meni Rosenfeld (talk) 07:37, 27 October 2009 (UTC)

## Thanks

Thanks! Sławomir Biały (talk) 12:58, 15 November 2009 (UTC)

Hi. Thank you for the detailed response.-Shahab (talk) 14:37, 30 November 2009 (UTC)

## Vandalism

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)

Hi, thanks for the info. It will be useful (in the present case I guess it was unintentional, but you are right, it's a good idea to leave a warning)

Hello, PMajer. You have new messages at Hydnjo's talk page.
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## Thanks!

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...). --PST 12:20, 19 December 2009 (UTC)

Don't worry, it happens to all of us... --pma (talk) 17:57, 19 December 2009 (UTC)

## Random act of kindness

For no particular reason, except to say hello, and to say that your contributions to the project are appreciated:

Hi Sławomir thank you I appreciate. Happy new year! --pma 19:28, 6 January 2010 (UTC)

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)

Well, we all agree that the original question was unclear and ill-posed, though not silly. As I understood it, Rckrone's approach was to start from the simplest and more direct interpretation of the question (which is a good rule for the RD, I think): precisely, the set of points is given, the radius R is given: how should we describe the admissible positions of the center of the surrounding circle of radius R and possibly locate it? I found his description clear, and satisfactory as a first answer. In fact it is the best answer one can give to the original question ("where is the center?"): the center is somewhere in a certain region, and that's it. I also suggested that simple variation of the question in order to save the wanted conclusion (namely, the unicity). One may further investigate on a distribution of probability for the center as a random variable; however this means changing the question, and possibly adding assumptions. Anyway, I'm glad you left a message here, and I hope you didn't take my comment on that comment as an indirect criticism, or underestimation of your mathematical comments, which I have the highest opinion of. --pma 14:28, 15 January 2010 (UTC)
Also, since you entered this matter, as to your non-mathematical comments, specifically, comments on other people's posts at the RD/M, my opinion its that they often reflect an excessive rigidity from your side. It is possible that somebody uses a term in a less common acceptance, or even improperly, even if the idea is correct. Also, it is possible that somebody has his own iterpretation of a previous unclear statement of somebody else, that differs from yours. I think maybe you should not necessarily decide that he's wrong, totally ignorant, &c and start biting. pma 12:34, 17 January 2010 (UTC)

## Ref Desk Question revisited

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)

Hi, sure, but would you precise better the question? --pma 18:02, 18 January 2010 (UTC)
Hi, I think I was precise. If you click on the link I put a few lines up, it goes to the ref desk archives of a question I asked a couple weeks ago. At the bottom of that, I added my understanding of what has been said so far and asked a question or two that I was not 100% on. Those are the questions I am talking about. Does that make sense? Thanks StatisticsMan (talk) 17:32, 19 January 2010 (UTC)

Let's define, for ${\displaystyle (z,\epsilon )\in \mathbb {H} \times (-1/2\,,+\infty )}$ and for ${\displaystyle (n,m)\in \mathbb {Z} \times \mathbb {Z} _{+}}$

${\displaystyle a_{n,m}(z,\epsilon ):={\frac {1}{(mz+n)^{2}|mz+n|^{2\epsilon }}}-\int _{n}^{n+1}{\frac {dt}{(mz+t)^{2}|mz+t|^{2\epsilon }}}.}$

For any ${\displaystyle (z,\epsilon )}$ we have, by the mean value theorem (see Rckrone bound)

${\displaystyle |a_{n,m}(z,\epsilon )|\leq (2+2|\epsilon |)\sup _{n\leq t\leq n+1}|mz+t|^{-3-2\epsilon }.}$

We wish to show that the family ${\displaystyle \{a_{n,m}\}_{(n,m)\in \mathbb {Z} \times \mathbb {Z} _{+}}}$ is locally normally summable in the uniform norm, that is, for any ${\displaystyle (z,\epsilon )\in \mathbb {H} \times (-1/2\,,+\infty )}$ there exists a neighborhood ${\displaystyle U}$ of ${\displaystyle (z,\epsilon }$) such that

${\displaystyle \sum _{(n,m)\in \mathbb {Z} \times \mathbb {Z} _{+}}\|a_{n,m}\|_{U,\infty }<+\infty .}$

This implies that the double sum

${\displaystyle \sum _{(n,m)\in \mathbb {Z} \times \mathbb {Z} _{+}}a_{n,m}(z,\epsilon )}$

converges uniformly to a continuous function on ${\displaystyle \mathbb {H} \times (-1/2\,,+\infty ).}$

Consider an open covering of ${\displaystyle \mathbb {H} \times (-1/2\,,+\infty )}$ by open sets of the form

${\displaystyle U_{a,b,c,d}:=\{(z,\epsilon )\in \mathbb {C} \times \mathbb {R} \;:\;|z|b,\quad c<\epsilon

for real numbers ${\displaystyle a>1>b>0,\;}$ and ${\displaystyle -1/2 Let ${\displaystyle \,\textstyle U:=U_{a,b,c,d}}$ be one of these.

It is convenient to bipartite the set of indices into the subsets:

${\displaystyle J_{1}:=\{(n,m)\in \mathbb {Z} \times \mathbb {Z} _{+}\;:\;-2(ma+1)/3

${\displaystyle J_{2}:=\mathbb {Z} \times \mathbb {Z} _{+}\setminus J_{1}.}$

Therefore, for any ${\displaystyle m\in \mathbb {Z} _{+}}$ there are at most ${\displaystyle \left\lfloor {\frac {8ma+2}{3}}\right\rfloor }$ values of ${\displaystyle n\in \mathbb {Z} }$ such that ${\displaystyle (n,m)\in J_{1},}$ and in any case ${\displaystyle n=-1,0,1}$ are among them.

Since for ${\displaystyle (z,\epsilon )\in U,}$ ${\displaystyle (n,m)\in J_{1}}$ and ${\displaystyle n\leq t\leq n+1}$ we have

${\displaystyle mb\leq |mz+t|\leq ma+|t|\leq m(2a+1),}$

so

${\displaystyle |mz+t|^{-3-2\epsilon }\leq m^{-3-2c}(3a+1)^{3+2d}}$

we can bound the sum on ${\displaystyle J_{1}}$ as follows:

${\displaystyle \sum _{(n,m)\in J_{1}}\|a_{n,m}\|_{U,\infty }\leq (2+2d)(3a+1)^{3+2d}\sum _{m=1}^{\infty }{\frac {8ma+2}{3}}m^{-3-2c}<+\infty .}$

On the other hand, for all ${\displaystyle (n,m)\in J_{2},\,}$ either ${\displaystyle n\geq 2ma}$ or ${\displaystyle n\leq -2(ma+1)/3.}$ In both cases, for any ${\displaystyle (z,\epsilon )\in U}$ and ${\displaystyle n\leq t\leq n+1}$

${\displaystyle |mz+t|^{2}=(m\mathrm {Re} \,z+t)^{2}+m^{2}b^{2}\geq (n/2)^{2}+m^{2}b^{2}\geq 1.}$

Note that for any ${\displaystyle (n,m)\in J_{2}}$ one has ${\displaystyle |n|\geq 2}$ so ${\displaystyle n^{2}/4\geq 1}$ and the last inequality holds.

Thus

${\displaystyle \sum _{(n,m)\in J_{2}}\|a_{n,m}\|_{U,\infty }\leq (2+2d)\sum _{(n,m)\in J_{2}}\left({\frac {n^{2}}{4}}+m^{2}b^{2}\right)^{-3-2c}<+\infty .}$

Remark. In the definition of ${\displaystyle U}$ there is no need of all the parameters a,b,c,d : you may better take just ${\displaystyle a=d\in \mathbb {Z} _{+}}$, ${\displaystyle b=1/a}$ and ${\displaystyle c=-1/(2+a)}$, so the open sets are a sequence ${\displaystyle U_{a}\subset U_{a+1},}$ 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. --pma 10:04, 20 January 2010 (UTC)

## My Reference Desk Question

Dear Fly by Night, I'm sorry that you were not satisfied with the answer. Note that I didn't, and don't have anything to object about the rigour of your question. I was just not sure about what you really wanted, because you presented two different situations. First, a generic function from C to C, etc. As you stated it, it was very natural to wonder whether you meant real differentiability. Second, the function exp(sin(z)). I tried to provide you with an answer covering all possible interpretations and stating the most relevant facts; this way you could then put a furthur question focusing better on what you really had in mind. Of course I'm not giving εδ-proofs before I'm sure of what a questioner want, otherwise I'm wasting my time and his time at the same time. The general idea is that after the first reply you may put a more precise question etc. After your reply, I understand that your interest was whether the Taylor polynomials of exp(sin(z)) converge uniformly on the complex plane, and whether they converge in some other senses. Is it this what you want? --pma 23:24, 12 March 2010 (UTC)
Dear Sir, I was clearly speaking about complex differentiability. My definition of ${\displaystyle f_{n}}$ used the notation ${\displaystyle f^{(k)}(z_{0})}$. This, I believe, is standard notation for ordinary differentiation, i.e. d^kf/dz^k (instead of partial differentiation). Since I used z as my variable, and used ordinary differentiation notation I would have thought that it were clear that I was differentiating with respect to z, with respect to the complex variable. •• Fly by Night (talk) 16:09, 13 March 2010 (UTC)
As I told you your question sounded a bit strange to me, since you were speaking of a function defined on C, but only differentiable on a neighborhood of 0, without any continuity assumptions outside the neighborhood. The point was not complex vs real differentiability, but since you required it to be infinitely differentiable on the neighborhood, incidentally, it could be really possible to me that you even meant real differentiability (for in the complex sense one usually just speak of "differentiability", since it implies C). But what is still not clear to you, is that mine is definitely not a criticism about your questions. My concern is only understanding what people want, and try to give an answer. I think I answerred your question under all possible interpretation of it and I hope it will of help. --pma 00:00, 14 March 2010 (UTC)

## Thanks a lot for your help...

...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!

Molto onorato :) I'm just a PhD student instead... --Pokipsy76 (talk) 19:27, 17 March 2010 (UTC)

## IP you just noted for vandalism

I've blocked the address for 3 months. Dougweller (talk) 14:27, 11 May 2010 (UTC)

well done! :-)

## Hi...My RefDesk problem

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)

## Dual spaces

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)

Thank you Henning, nice construction and very clear explanation. It also leaves interesting questions (e.g. what about not using GCH...). It seems worth add some information to the articles on vector spaces and duals. --pma 10:17, 18 October 2010 (UTC)
I managed to find a brute-force approach that works without GCH. More elementary, but less elegant. –Henning Makholm (talk) 05:38, 19 October 2010 (UTC)
Very good. So summarizing, if ${\displaystyle \mathrm {dim} (V)}$ is infinite, ${\displaystyle \mathrm {dim} (V)<\mathrm {dim} (V^{*})\leq 2^{\mathrm {dim} (V)},}$ and at least in the case ${\displaystyle \mathrm {dim} (V^{*})=\aleph _{0},}$ the right inequality is an equality; all without using CH.--pma 08:47, 20 October 2010 (UTC)
Well, ${\displaystyle \mathrm {dim} (V^{*})=\aleph _{0}}$ never happens. I'm not sure about the right inequality at all. A slight modification of my latter argument will prove ${\displaystyle \mathrm {dim} (V^{*})\geq 2^{\mathrm {dim} (V)}}$, with equality when the scalar field is small enough. But I don't think we've excluded the possibility of ${\displaystyle \mathrm {dim} (V^{*})>2^{\mathrm {dim} (V)}}$ for huge scalar fields. –Henning Makholm (talk) 09:59, 20 October 2010 (UTC)
In fact, for any set ${\displaystyle X}$ and any field ${\displaystyle F}$ there is an extension field ${\displaystyle K\supset F}$ such that ${\displaystyle \mathrm {dim} _{K}(K^{\aleph _{0}})\geq |X|}$. Namely, construct ${\displaystyle K}$ by adjoining ${\displaystyle |X|}$ independent (formal) transcendentals ${\displaystyle (\alpha _{x})_{x\in X}}$ to ${\displaystyle F}$. Then the set
${\displaystyle \{(1,\alpha _{x},\alpha _{x}^{2},\ldots ,\alpha _{x}^{n},\ldots )\mid x\in X\}\subseteq K^{\aleph _{0}}}$
is linearly independent. –Henning Makholm (talk) 05:02, 21 October 2010 (UTC)
Thanks, very interesting! I made a typo above, of course I meant ${\displaystyle \mathrm {dim} (V)=\aleph _{0}}$ and, yes, the right inequality should not be true for large fields. I will think about it as soon as I have moer free time. --pma 17:16, 21 October 2010 (UTC)

## Stomachion criticism

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)

I think you mean this. You are right that I gave no details and references; my post is exactly an invitation/request to people in the field in order to fill the gap, and present the criticism in detail and with complete references, as you are saying. --pma 15:08, 14 March 2011 (UTC)

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Thank you for the notification (I didn't realize it was linking to a disambiguation page). Fixed now. --pma 00:13, 17 February 2012 (UTC)

## Binomial series conditions for convergence

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)

Yes, your observation is indeed meaningful, so I have re-included. My concern was about not expanding the article too much, as happened to some other ones. Cheers! --pma 19:34, 26 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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## Carleman's inequality

Hello,

I think you oversimplified Polya's proof. The inequality

${\displaystyle (n!)^{-1/n}\leq {\frac {e}{n+1}}}$ for all ${\displaystyle n\geq 1}$

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

${\displaystyle c_{1}\cdots c_{n}=(n+1)^{n}~,\quad n\geq 1~.}$

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

• provide a counterexample to that inequality (i.e. a particular value of n for which it does not hold); or
• show where is the flaw in the argument I mentioned; or
• just explain what is unclear to you, or what makes you suspicious about the validity of that inequality (note that I was not referring to any Polya's proof).

Thank you. I will in any case appreciate any contribution to improve that article. --pma 19:43, 22 September 2012 (UTC)

Dear P.,
I am sorry, it seems that the inequality is correct (your explanation works for n=2,3,..., and for n=1 it is directly verified). I was in a hurry, my apologies.
Best regards,
Sasha (talk) 20:30, 22 September 2012 (UTC)
Dear P.,
sorry for bothering you again. Could you please clarify the phrase "...optimal -otherwise it would be attained by some non-negative and non-identically zero convergent series."? I agree that the constant is optimal, and that it follows from the proof, but I do not understand the argument that you propose.
Best regards, Sasha (talk) 20:35, 22 September 2012 (UTC)
Dear Sasha,
Yes, I think you are right, that sentence is obscure. The fact that the constant e is optimal should be just a consequence of the fact that it can be approached arbitrarily by a sequence of the form (1,1/2,1/3,...,1/N,0,0,0,..) taking N large enough, right? Maybe I had something in mind, but at the moment I think it is better to remove it. --pma 22:36, 22 September 2012 (UTC)
Dear P.,
thanks again! I agree with both things you wrote.
I have looked again at Polya's paper (which is a didactic exposition of his paper from the 20's), it seems that he chose the coefficients in a more complicated way in order not to use very precise asymptotics for n!. Asymptotically his c_k behave like k, so the difference between the two choices is not very dramatic. Which argument is better is of course a matter of taste.
Best regards,
Sasha (talk) 01:01, 23 September 2012 (UTC)

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