Misplaced Pages

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

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

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

http://en.wikipedia.org/search/?title=Misplaced Pages%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)

Again, since you asked: I first saw this problem two years ago, and I liked it immediately. The key point is that the sum of the jumps needs to diverge (otherwise you can't possibly hit every rational) while the sum of their squares must converge. The obvious choice for such a sequence is 1/n, so I decided the nth jump would be (approximately) 1/n. The rest is easy. Start with your favourite enumeration of the rationals, starting with 0, 100, 1, -10 say. Then let 0 be the first term of your sequence, and jump towards 100, making the nth jump of length 1/n. Since the sum of 1/n diverges, you'll get to 100 in finitely many steps (the last jump will have to be a bit undersize, of course). As you go, cross off every number you hit from the list of rationals you started with. Now choose the first rational on your list you haven't crossed off (-10 in this case) and jump towards that, again making sure the nth jump overall is about 1/n. Of course, you might not be able to jump 1/n exactly, but if you stay close (between 1/n and 1/2n say), then nothing will go wrong (since the sum still diverges and the sum of squares converges), and since the rationals are dense, you'll always have somewhere to jump to. Continue in this fashion, and you're there.

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! --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

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)

Ask for information

Dear User, in reference to your talk, 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

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

1. Via the L 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 L 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 L). 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 L density of step functions is a not difficult consequence of the L 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 x, 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 L 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 ${\displaystyle \scriptstyle \times }$, 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 x, 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" x..x are sufficient, and f in L_loc instead of L is also sufficient. In the second proof, you can use the segment for the one dimensional case: so define ${\displaystyle \scriptstyle \chi (x)}$ as the characteristic function of , 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 R : 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 ${\displaystyle \scriptstyle f_{a,b}(y):=\int _{}f(x,y)\,dx}$. It has vanishing integral over all intervals . 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 . 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 ${\displaystyle }$, 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)

God gave me your bike

Unfortunately, it seems like the reference desk does not tolerate profound questions on religion -- I say this because it seems like all other topics are given license to spread and tangent off into topics that bear no resemblance to the original question, without incident, but if anything like that happens with a religious question, the question is terminated or asked to be so. But perhaps it's just not the forum...we can't just debate religion all day and all night forever.

That being said, I wanted to ask you about your response about "someone claiming that God gave them your bike." Why is that any different than the personal revelations of such individuals as Jesus, Mohammed, John Smith, etc.? DRosenbach 15:27, 24 August 2009 (UTC)

In a sense, it is not different. I have the greatest respect for the intimate religious feelings of people, but what many believers forget, incredibly, is that their faith is just an intimate feeling, and has no dialectic value. I'm saying nothing new, of course, and I hope you agree. There is an irremediable logic flaw in the usual argument "This is the word of God and God is omniscient, therefore I am right". Indeed, the whole sentence should be "I think that this is the word of God...", therefore, it is my word, not God's word. That being said with the due respect to Credonia Mwerinde, Moses, and all the other people you quoted, of course...

Why do you flare up? If the Jewish people claim that 3M people witnesses God speak to them, that's the best evidence one can hope for. I answer a question in proper order, and then defend against a comment made by a rabble-rouser, and you take aim at me for my annoying posts? And then you pepper in some "trolling," as though it justifies your position and weakens my argument -- you are the one who is being annoying. DRosenbach 21:56, 26 August 2009 (UTC)

Dear DRosenbach, I hope you didn't take it bad my recent comment here. Sorry for the shape form of my post, but the point is that the RefDesk is not the place for such claims of faith (of any faith). As to the claim about Revelation, the usual argument that "3M people witnessed God speak to them" , or that "it is reported in a text inspired by God", of course is not an evidence in support of the claim, but is part of the claim itself: a claim with no scientific evidence. On the other hand, from a religious point of view, one doesn't need any scientific evidence at all. Note that I've always aknowledged the right of everybody of having religious feelings. But having a faith, do not allow you to make other people aknowledge as true what you feel true. You certainly know how hideous the Chistian church has been through centuries under this respect; and of course they were convinced to be right, and that they were even following God's will: for instance, after Agostinus, they interpreted the parable of the Great Banquet (Lc 14:15-24) as God's justification to forcing people to convert. Wouldn't it be better if everybody recalled to precede any religious statement with the obvious "I personally believe, and as human being could be wrong, that &c.." (that God did this/ that he allowed me that/ that he wants us to do that/ that one must behave this way / that this is a sin, according to God's sensibility, or God's standards of decency, &c.).

Checking your preceeding posts, I see that you have a kind of mixed attitude in these topics: there is interesting doctrinal information from you, and here and there there's a kind of disposition to start religious quarrels (Jesus revelation is of lower quality than Moses's; the Ten Commandments are copyright material and gentiles should not be entitled to follow them,...&c). C'mon, don't you find it is a bit childish? Excuse me if you feel offended; as I told you I really do not mean to. I am sure that you will grow up in your faith and reach a mature and tolerant position, as several wonderful religious persons of your faith that I had the chance of meeting.

Maths refernce desk

I find your comments quite hurtful and unnecessary. I like to answer as many questions as I can in order to help people. Your attitude, like many other users on that page, is all wrong. I would recommend that you read wp:good faith and wp:civility when you have some free time. (p.s. I've been reading some of your other use talk page comments and it seems that I'm not imagining things: you could be a lot nicer) ~~ Dr Dec (Talk) ~~ 21:01, 7 September 2009 (UTC)

Okay, I take it back. In our last few correspondence you've been very civil, and most helpful. And for that, I would like to say thank you :o) ~~ Dr Dec (Talk) ~~ 22:36, 7 September 2009 (UTC)

Very good! I'm very glad. --pma (talk) 07:02, 8 September 2009 (UTC) In fact after reflecting I think I have to add my apologies: I did not realized that I was being so rude.

Envelope (mathematics)

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 L-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)

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:R →R , 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 R , that has nothing or little to do with general L spaces (no, indeed the link it was ok, it goes to Rn)

Well, I've posted the same message (more or less) on the talk page. I doubt that we'll get much feed back: the last new thread on that page, before mine, was more than a year and five months ago. Reading your last post, I get the feeling that you're over-complicating the issues I raised. The point is that your example doesn't fit into the page as it stands. If you add something to a page then you should modify the page so that it is capable of accepting the new information. E.g. you need to add sufficient new background and motivation so that whatever you add doesn't look out of place and doesn't just seem to appear out of the blue. It isn't just that your example doesn't use classical methods of differential geometry; it's that the example doesn't have any connection to anything else written on that page, before or after. For example, the article begins "In mathematics, an envelope of a family of manifolds (especially a family of curves) is a manifold that is tangent to each member of the family at some point." As far as I can tell, your T_s,t aren't even manifolds: they're manifolds with boundary. Anyone coming to look at the article is going to be looking for the envelope of families of smooth submanifolds, probably smooth curves in the plane. Adding your example will confuse people, and isn't really what anyone would be coming to the article to see. If the example is to stay then I think you need to add enough detail and motivation for it not to look so out of place. If you don't want to do that then I might suggest that you should think about removing it until a time that the article is at a stage so that it may be re-introduced. ~~ Dr Dec (Talk) ~~ 15:06, 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)

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 R. Is the Cauchy-Schwarz inequality in R 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.