User talk:Jitse Niesen/Archive13

Hello
I've seen your important contributions for the article Recurrence relation. I'm looking for the general (non-iterative) algebraic expression for the exact trigonometric constants of the form: $$\begin{align}\cos \frac{\pi}{2^n}\end{align}$$, when n is natural (and is not given in advance). Do you know of any such general (non-iterative) algebraic (non-trigonometric) expression? Eliko (talk) 08:26, 31 March 2008 (UTC)
 * Let me explain: if we choose n=1 then the term $$\begin{align}\cos \frac{\pi}{2^n}\end{align}$$ becomes "0", which is a simple (non-trigonometric) constant. If we choose n=2 then the term $$\begin{align}\cos \frac{\pi}{2^n}\end{align}$$ becomes $$\begin{align}\frac{1}{\sqrt{2}}\end{align}$$, which is again an algebraic (non-trigonometric) constant. etc. etc. Generally, for every natural n, the term $$\begin{align}\cos \frac{\pi}{2^n}\end{align}$$ becomes an algebraic (non-trigonometric) constant. However, when n is not given in advance, then the very expression $$\begin{align}\cos \frac{\pi}{2^n}\end{align}$$ per se - is not an algebraic expression but rather is a trigonometric (non-algebraic) expression. I'm looking for the general (non-iterative) algebraic (non-trigonometric) expression equivalent to $$\begin{align}\cos \frac{\pi}{2^n}\end{align}$$, when n is not given in advance. If not for the cosine - then for the sine or the tangent or the cotangent.


 * No, I don't know any such expression. -- Jitse Niesen (talk) 11:18, 31 March 2008 (UTC)


 * Thank you. Eliko (talk) 11:22, 31 March 2008 (UTC)


 * To Eliko: I think that for which you are asking is impossible. You want a closed polynomial with radicals, f (n), which satisfies: f (1) = 0; and 2&middot;(f (n+1))2 = 1 + f (n) for all n. I do not think that there can be any such thing. What degree (maximum power of n) would the formula have? JRSpriggs (talk) 11:32, 31 March 2008 (UTC)
 * The formula is not supposed to have a degree. Does Fibonnaci formula have a degree? Eliko (talk) 11:38, 31 March 2008 (UTC)

Well, if you are going to allow exponential formulas (like the Fibonnaci formula) rather than merely algebraic formulas, then why not simply replace the cosine by its exponential expression? $$ \cos\theta = \frac{e^{i \theta} + e^{-i \theta}}{2}$$ JRSpriggs (talk) 12:29, 31 March 2008 (UTC)


 * Your exponential expression is over the imaginary numbers, i.e. (according to Euler's formula) this is a trigonometric formula. I'm looking for a non-trigonometric formula (hence for a formula which is non-exponential-over-the-imaginaries). However, if it has not been sufficiently clear in my previous request - then you can add it now as an additional condition: non-trigonometric and non-exponential-over-the-imaginaries. Eliko (talk) 12:43, 31 March 2008 (UTC)


 * I hope you realize that it is easier to calculate trigonometric functions than real exponential functions. Instead of imposing a series of rather arbitrary and unclear conditions, why do you not explain for what you want to use this? JRSpriggs (talk) 02:46, 1 April 2008 (UTC)


 * This is an abstract question in pure mathematics. However, any solution may be helpful in answering other questions, like: finding a non-trigonometric proof for the following non-trigonometric claim: Every real interval includes a point x having a natural number n such that $$\begin{align}(x+i)^n\end{align}$$ is a real number. Note that this is an algebraic, non-trigonometric claim, so one may naturally expect that it may be proven by a non-trigonometric proof. Eliko (talk) 14:33, 1 April 2008 (UTC)


 * I think that your conjecture is probably true. However, to think that you can excise trigonometry from mathematics and expect what remains to hang together and function normally is unreasonable. JRSpriggs (talk) 07:53, 2 April 2008 (UTC)


 * What you call my "conjecture" - is actually a sentence (proven by trignometric means). However, there are many sentences which can be proven by both trigonometric means and non-trigonometric means (e.g, the sentence claiming that $$\begin{align}(-1)^n\end{align}$$ is either 1 or -1 for any natural n, etc.), so one may naturally expect that also the other (first) sentence may be proven by a non-trigonometric proof. So the abstract question (in pure mathematics) is now: whether any such non-trigonometric proof does exist (just as it exists in other fields). Eliko (talk) 11:20, 2 April 2008 (UTC)


 * I suspect that you mean "theorem" rather than "sentence". I would think that the way to prove it would be to expand the power using the binomial theorem and then take the imaginary part and set it to zero. Then use the theorems about the number of roots of a polynomial as a function of the sign changes of its coefficients. JRSpriggs (talk) 05:15, 4 April 2008 (UTC)
 * You're definitely correct: I really meant "Theorem".
 * Unfortunately, I've never heard of "theorems about the number of roots of a polynomial as a function of the sign changes of its coefficients". Would you care to elaborate on those theorems (of which I've never heard), or to let me read about them on Wikipedia? Thank you in advance.
 * Anyways, "my" theorem states that the root belongs to a given interval, so I really don't know how "your" theorem (about the number of roots, rather than about the interval to which a root belongs) can help here. Eliko (talk) 08:08, 4 April 2008 (UTC)

(Unindenting) Unfortunately, I do not remember the name of the theorem or all the details, but there is a relationship between the number of sign changes in the coefficients and the number of positive roots. Since you can shift a root past zero by a simple substitution x = y+c for an appropriately chosen constant c, this gives a way to show that there must be a root in an interval (see also Viète's formulas). Or you could just increase n until you have both a positive and a negative value of the polynomial in the desired interval. Then there must be a root between them by the intermediate value theorem. JRSpriggs (talk) 11:11, 6 April 2008 (UTC)
 * As to your first proposal: Let's assume that our interval is (a,b). Now, for every constant c we can really substitute: x = y+c, and then x can belong to the interval (a,b) if and only if y can belong to the interval: (a-c, b-c). However, I really don't know how you can prove (without any trigonometric means) that y can really belong to the interval: (a-c, b-c).
 * By the way, Viète's formulas (for the sum of roots or for the product of roots etc.) - is well known, but it does not promise that (at least) one of the roots is real (as I want it to be), unless n is odd, but even when n is odd - some of the roots may still be unreal, so I really don't know how Viète's formulas are relevant for proving the existence of a (real) root which belongs to a given (real) interval.
 * As to your second proposal: I really don't know how you can prove (without any trigonometric means) that there really exists such an n "having both a positive and a negative value of the polynomial in the desired interval".
 * Eliko (talk) 12:29, 6 April 2008 (UTC)


 * JRSpriggs, you may be thinking of the Descartes' rule of signs or Sturm's theorem. -- Jitse Niesen (talk) 13:29, 6 April 2008 (UTC)
 * Cheers, Thank you; But I still don't know how these theorems may be helpful for proving the existence of roots within a given interval, as I've explained above. Eliko (talk) 13:37, 6 April 2008 (UTC)


 * To Jitse: Yes, thank you. I was thinking of Descartes' rule of signs, but Sturm's theorem (which I had not seen before) looks even better for this purpose. JRSpriggs (talk) 13:42, 6 April 2008 (UTC)


 * More precisely: Descartes' rule of signs is totally irrelevant here, because when we expand the power of $$\begin{align}(x+i)^n\end{align}$$ and set the imaginary part (of the received polynomial) to zero, then every two consecutive coefficients in the second polynomial include a zero, whereas Descartes' rule of signs is about consecutive nonzero coefficients. On the other hand, Sturm's theorem could have been relevant, but unfortunately it's about a (natural) n given in advance, so Sturm's theorem does not promise the very existence of such a (natural) n. My theorem states that such a (natural) n really exists. Eliko (talk) 14:03, 6 April 2008 (UTC)


 * I believe that Descartes' rule of signs does not require that all coefficients be non-zero. It works for sign changes between coefficients which are consecutive among the non-zero coefficients. For example, it implies that x2 - 1 = 0 has one positive real root because there is one sign change. Similarly, x5 - x + 1 = 0 must have either two or zero positive real roots. That is, skip over zero coefficients as if they were not there. JRSpriggs (talk) 07:07, 8 April 2008 (UTC)


 * If your conjecture is true, i.e. Descartes' rule of signs really allows zero coefficients to be skipped over as if they were not there, then Descartes' rule of signs could really have been relevant, but unfortunately it's about a (natural) n given in advance, so Descartes' rule of signs does not promise the very existence of such a (natural) n. My theorem states that such a (natural) n really exists. Eliko (talk) 09:17, 8 April 2008 (UTC)

Trouble with MathML
I'm sorry for coming out of the blue, but your name is referenced on the the relevant Mediawiki page as the person who is still working on this. I've just accepted the way the wiki generates formulae for quite some time now, but I've had it. I want my own system's better looking fonts to render them. I'm using FF 3.0 and I have both Matlab *and* Maple installed, so I know I have the high quality fonts installed in the system directory that are needed to support MathML. I've long since switched "on" the preference in my Wikipedia CP to enable MathML. Still, all I get are crummy rendered images. It is really annoying and the WP Formula authoring page is not helpful. What, oh what, must I do to get my MathML? Thanks in advance. --Dragon695 (talk) 23:44, 3 April 2008 (UTC)


 * The MediaWiki software that is running the website cannot do MathML. There is indeed an option in the preferences, but it doesn't do very much; only some extremely simple formulas are done in MathML, the rest uses rendered images as normal.
 * I did work on better support for MathML for a while, but I stopped this project. It's not as easy as I thought, support for MathML in the browsers is quite patchy, and I got almost no feedback from the programmers behind MediaWiki. So, I'm afraid it's not possible to get MathML on Wikipedia. -- Jitse Niesen (talk) 14:01, 6 April 2008 (UTC)

QR_Decomposition
Hi, thank you for your revision on article of QR_Decomposition. I am not sure which definition is ``more standard''. As far as I know, both of these two definitions exist. The first one (A is mxn, then Q is mxm, R is mxn) is used in many textbooks and also many numerical applications, such as LAPACK and Matlab. But the second one does exist too. Since I believe the first one may be used more frequently (partly due to the popularity of Matlab etc), I select it as the first candidate and the original one as an alternative one. It is just my personal opinion. Please correct me if necessary. —Preceding unsigned comment added by Realwhz (talk • contribs) 17:39, 7 April 2008 (UTC)

Question on Brent's method
Hi, could you maybe take a look at the question I asked on Brent's method's talk page? Thanks! laug (talk) 13:22, 10 May 2008 (UTC)

Advanced examples of mathematical induction
Hello there, I wonder if you would take the time to read the deletion discussion for Proofs involving the totient function (deletion discussion) The article on mathematical induction has the same format and cites an authoritative source, namely Mathworld. I don't understand why you would vote to keep one but not the other. The article on induction could be the start of a useful series of advanced examples, same as the totient function article. -Zahlentheorie (talk) 13:59, 10 May 2008 (UTC)


 * I didn't comment at the deletion discussion for Proofs involving the totient function, but if I had been forced to comment, I would have said that it doesn't belong on Wikipedia. The people that did comment were obviously of another opinion. I think the difference between Proofs involving the totient function and your article is that the former article is meant as a further elaboration of totient function in that it provides proofs of statements listed there, while your articles gives examples of proofs and is thus too much textbook material and not really encyclopaedia material. Another, less satisfactory, explanation is that we are not consistent in our deletion discussions; it is unfortunate but the outcome of a deletion discussion depends on which people participate in it.
 * Adding proofs to Wikipedia is a hazardous activity. Some people, including myself, think that detailed proofs do not belong here (with some exceptions, like Cantor's diagonal argument). Others disagree. If you do add a proof, especially an article solely about a proof, there is a decent chance it is deleted. I'm sorry you found this out only after all the work you put into it. -- Jitse Niesen (talk) 13:41, 12 May 2008 (UTC)

Winograd algorithm
Hi! Sorry, I managed not to see your comment on this talk page, so it took me almost a year to respond. Better late than never though, I guess :) -- Obradovi&#263; Goran ( t al k  20:10, 17 May 2008 (UTC)

New articles not showing up?
Hi, I noticed that new articles aren't showing up at the WikiProject_Mathematics/Current_activity. I don't know if it just takes them a little longer to make themselves known to User:Jitse's bot or if the bot is having problems. Anyway, I thought I should bring it to your attention. siℓℓy rabbit (  talk  ) 12:57, 4 June 2008 (UTC)
 * This actually goes back to mathbot, the machine on which it runs is down, and then nothing feeds Jitse's bot. I notified the system people about the machine, hopefully it will get fixed soon and all will come back to normal. Oleg Alexandrov (talk) 15:24, 4 June 2008 (UTC)

Nevada-tan
any reason why there should not be a redirect from XXXX to Nevada Tan?

oh and I've cited the name on the Nevada Tan article, with photographic proof, that was shown by Fuji TV, it should be considered to be a highly reliable source Sennen goroshi (talk) 20:22, 16 June 2008 (UTC)

Sorry, I only just saw the talk message on my page. I agree with you in part, that to give someones name as a murderer without proof is a really shitty thing to do, however Fuji TV themselves gave the name, due to their inability to correctly remove it from the images they displayed. Sennen goroshi (talk) 20:26, 16 June 2008 (UTC)


 * Okay, let's discuss it on Talk:Nevada-tan. -- Jitse Niesen (talk) 20:33, 16 June 2008 (UTC)


 * OK, let's do that. Actually I am going to sleep, it's 5.40am in Japan now. In the meantime I have removed the name, I still feel the name should stay, but there is no harm in hiding it, while we are sorting this out. Sennen goroshi (talk) 20:43, 16 June 2008 (UTC)

Why Euclid's Axioms Are Unacceptable (Taxicab Geometry)
Hello, Jitse Niesen! I have received your message on my Talk Page. I will abstain from posting "Euclid axioms" into the Taxicab geometry page, but I would like to take you up on your offer. Please let me know why Euclid is not a reliable source and why three editors have opposed my motion. Thank you for your time and effort.

(Rallybrendan2006 (talk) 15:31, 18 June 2008 (UTC))

General relativity FAC review
Thanks once more for your helpful comments at the FAC review. It would be great if you could hide your resolved comments in a cap box template, to make it easier for the FAC directors to see which issues have been resolved and which haven't. Thanks! Markus Poessel (talk) 18:15, 26 June 2008 (UTC)


 * Well, the article was promoted before I had time to do this, and find out what this cap box template is (I thought they used strike-through? ah well, I'm surely getting old here). Anyway, congratulations are due for surviving the process and getting a shiny star! -- Jitse Niesen (talk) 16:16, 27 June 2008 (UTC)


 * No problem. The template I was talking about is the one that automatically hides all resolved comments (you can see examples at numerous FAC discussions). When I use it, I simply copy it from the first other example I can find. Thanks for the congratulations, Markus Poessel (talk) 18:02, 27 June 2008 (UTC)

Matrix congruence refimprove
Hello. The reason I placed a refimprove tag on this article that it has only one reference, and that is relevant only to the material on Sylvester's law of inertia. There is nothing to support the main thrust of the article on matrix congruence. I happen to know that it is correct from my own personal knowledge, but that's not what WP:V wants us to do. Richard Pinch (talk) 20:46, 27 June 2008 (UTC)


 * Thanks for the explanation and even more for adding the additional references. I didn't know which part of the article was supported by the one reference that was already there, so I'd appreciate it if you could be a bit more explicit in the future (I agree that a reference for the definition of matrix congruence is required). On the other hand, I should have asked you about the refimprove tag instead of summarily removing it; and I apologize for that. -- Jitse Niesen (talk) 12:26, 29 June 2008 (UTC)

Why you removed Membrane analogy from Differential equation?
Hi,

I hope there is a very good reason to remove it. Wouldn't it better to improve it instead of removing it? Anyway, it is one of good examples of using similarity of equations.

User: Mossaiby —Preceding comment was added at 09:56, 29 June 2008 (UTC)


 * At the present, Membrane analogy is a very poorly written article. If you are familiar with the subject, why do you not fix it before attempting to link to it from other articles? JRSpriggs (talk) 10:54, 29 June 2008 (UTC)


 * Indeed, that was one reason. The other reason is that there are hundreds of analogies, and that it makes no sense to list them all. There are already two examples listed and I think that is enough. The membrane example did not strike me as better than the examples that are already there. -- Jitse Niesen (talk) 12:35, 29 June 2008 (UTC)

polar
Thanks a lot, and I ask you to take a final look at my reanswere. Mozó (talk) 14:55, 29 June 2008 (UTC)

Thanks, thanks, thanks. Now, I see everything. ( claim 2 is a bit different that you mean on it, but the counterexample still works. After all, I will try to figure out how to formulate prop. 1 in terms of limits of functions (without sequential limits)) Regards, Mozó (talk) 17:54, 29 June 2008 (UTC)