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May 27

Chess openings

May be it is a wrong place to ask, but I wondered if anyone studied chess openings in a case when one of the pieces is missing (an intentional handicap for a stronger player to play against a weaker one). (Igny (talk) 01:45, 27 May 2009 (UTC))[reply]

I believe the answer is simply, for the most part, no. The stronger side would rely on general principles to gradually change the odds. However, some games probably exist in some sources, and these may have some interest on the matter. Should I run across anything, as a chess enthusiast, I will let you know. But, you regard this as a math question. Perhaps it's not specific chess openings, but general principles of them that are your concern, and these would mostly apply to the situations you are considering. In short again, no.Julzes (talk) 04:33, 27 May 2009 (UTC)[reply]

The initial position of a game of chess is defined by the rules. If one of the pieces is missing then the game is not chess. Bo Jacoby (talk) 04:54, 27 May 2009 (UTC).[reply]

Handicap chess has a 200 year+ history, so I am pretty sure that someone (player or recreational mathematician) would have looked into the openings, even though I cannot point to anything specific. Do take a look at our articles and the references in it, and you may find some leads. Abecedare (talk) 05:01, 27 May 2009 (UTC)[reply]

"Looked into" is not the question I perceive, and I suspect finding a--by definition, detailed--source has probability 0.Julzes (talk) 06:27, 27 May 2009 (UTC)[reply]

I remember Garry Kasparov playing against a Brit, with a two-pawn handicap. If I recall correctly, a grandmaster on the side of the weaker player published some notes or ideas about the games' opening preparation. The missing pawns varied from game to game. I'll look in the Internet if there is something about it. Pallida Mors 14:56, 27 May 2009 (UTC)[reply]

This is the match. Unfortunately, I can't find any analysis online. Maybe there were on the now-defunct Kasparovchess.com. Pallida Mors 15:12, 27 May 2009 (UTC)[reply]

Here is a reference:

Staunton, Howard (1849). The chess-player's companion. H. G. Bohn.

It is a 510 pages book written by the famous Howard Staunton and devoted to studying openings in handicap chess games. You can download the full text from google books. Abecedare (talk) 16:24, 27 May 2009 (UTC)[reply]

I stand corrected.Julzes (talk) 21:22, 27 May 2009 (UTC)[reply]

Thank you, that was an interesting read. (Igny (talk) 22:21, 28 May 2009 (UTC))[reply]

Notation for derivatives ?

A question on the Science Refdesk made me wonder: In "English" we commonly use, ${\frac {dy}{dx}}$ to denote the derivative of y with respect to x; what is the equivalent notation in other languages ? In particular:

Do languages using an alphabetic writing system simply substitute the equivalent letter for the d's ?
What about languages with a logographic writing system ?

Abecedare (talk) 04:52, 27 May 2009 (UTC)[reply]

See Modern_Arabic_mathematical_notation#Calculus. Bo Jacoby (talk) 05:08, 27 May 2009 (UTC).[reply]

Also, Newton is spinning in his grave. --Stephan Schulz (talk) 06:18, 27 May 2009 (UTC)[reply]

Lol. To clarify: I used "English" to stand for English, French, German and other (all ?) Latin Alphabet based languages that I already knew used Leibniz's notation, which has largely displaced Newton's dot-notation (except in specific settings where there is only one obvious independent variable).

The Arabic usage seems to reflect (pun intended :) ) the Leinbiz notation too. What about Chinese, Japanese, Korean etc ? Is any really independent notation prevalent in any other language ? Abecedare (talk) 06:48, 27 May 2009 (UTC)[reply]

(ec) Generally letter symbols for common operators are not translated to languages. Compare notation in Derivative to respective articles in Arabic, Greek, Hebrew, Japan, Korean and Russian Wiki.
See also Derivative#Notations for differentiation and articles linked there. --CiaPan (talk) 06:58, 27 May 2009 (UTC)[reply]

Thanks for the interwiki links! With hindsight, that should have been the first place for me to look. Interesting to see that (unlike Arabi) Hebrew, Japanese etc use not only the notational scheme, but also Latin alphabets to denote functions and variable. Regards. Abecedare (talk) 07:59, 27 May 2009 (UTC)[reply]

The notation dy/dx is of course not characteristic of English nor of any other language. Mathematicians, in particular analysts, prefer more useful and less ambiguous notations for the derivative of a function y, like y' or Dy. The notation dy/dx is more useful in Physics and in applied sciences in general, where a quantity "y" may in principle be considered in dependence of any other quantity "x". --194.95.184.40 (talk) 16:51, 27 May 2009 (UTC)[reply]

Convergence of the series z^(n!)

Hi there guys - I was wondering about determining the convergence of the series $\sum ^{\infty }z^{n!}$ - particularly on the boundary of the radius of convergence. I compared it with $\sum z^{n}$ to show absolute convergence for $|z|<1$ , but what about $|z|\geq 1$ ? Just because that particular comparison shows $\sum ^{\infty }|z|^{n!}$ is convergent for -less- than 1, it obviously doesn't necessarily mean it won't also converge for larger |z|, just that that method of comparison doesn't work. I have a strong feeling the radius R will be 1, but even having shown it, what happens on the boundary? I can see that for $e^{i\theta }$ , any rational multiple of $\pi$ will diverge since the factorial will eventually make it an integer multiple of pi and then 1 for all terms after that - what about for general theta though?

I need this help quite urgently so any quick responses would be very kind, please help! Spamalert101 (talk) 17:34, 27 May 2009 (UTC)[reply]

Remember that

\sum _{n=0}^{\infty }a_{n}

cannot converge unless

\lim _{n\to \infty }a_{n}=0

. — Emil J. 17:46, 27 May 2009 (UTC)[reply]

Oh god, I was thinking about non-divergence for some reason, i.e. that it might not converge but it could still be bounded - how stupid of me! Okay, so it diverges on the boundary |z|=1, and for all R>1? Spamalert101 (talk) 18:00, 27 May 2009 (UTC)[reply]

It definitely diverges for |z|>=1 for the reason EmilJ gives. The individual terms tending to zero is necessary (but not sufficient) for a series to converge, although theoretically that doesn't rule out it being bounded. However, as you say, it diverges to infinity for any number with rational argument (you only said it for |z|=1, but the same reasoning works for |z|>1). Since the points with rational argument are dense and the partial sums are all continuous functions, I think that means it will diverge to infinity for all z. --Tango (talk) 18:59, 27 May 2009 (UTC)[reply]

It converges for |z|<1 by comparison test to geometric series, diverges for |z|>=1 because terms do not go to zero, diverges to infinity (is unbounded with monotone increasing norm) for |z|>1 by using the triangle inequality to establish that the ratio of the last term to the entirety of the preceding sum goes like |z|^n, diverges to infinity for rational-multiple-of-pi argument on the unit circle because it is eventually heading off from a partial sum with additions of the value 1, and should act like an ergodic random walk for all other values.Julzes (talk) 16:26, 28 May 2009 (UTC)[reply]

I'll get back to you with a proof of the fact that it is not analytically continuable.Julzes (talk) 21:27, 27 May 2009 (UTC)[reply]

Hold on now! This is homework. "Urgently" why? I take the last back.Julzes (talk) 21:31, 27 May 2009 (UTC)[reply]

True, this could well be homework. The OP has shown a clear attempt to do it themselves, though, so we are generally rather more tolerant. The OP doesn't seem to have asked about analytically continuing the function, though, so either way there is no harm in you going off on an interesting tangent. --Tango (talk) 21:40, 27 May 2009 (UTC)[reply]

it diverges to infinity for any number with rational argument etc. ???? EmilJ explication was clear enough. What on the moon is the need of adding this distinction between rational and not rational points?? Besides, divergence on a dense subset, for a sequence of continuous functions, does not imply anything on the other points of course. Bad! :-) --194.95.184.40 (talk) 07:31, 28 May 2009 (UTC)[reply]

EmilJ only proved it diverged, not that it diverged to infinity. The series 1-1+1-1+1-1... diverges, but the partial sums always stay within the set {0,1}. --Tango (talk) 07:43, 28 May 2009 (UTC)[reply]

The OP only asked about convergence, so it makes no difference whether it diverges to infinity or diverges in another way. 194.95 did not realize that you are answering something different than the original question, which is no wonder given that you did not formulate the new question explicitly. — Emil J. 13:08, 28 May 2009 (UTC)[reply]

The OP's follow up question mentioned boundedness, it was that that I was answering. --Tango (talk) 13:49, 28 May 2009 (UTC)[reply]

some detail added above, still planning something on continuability.Julzes (talk) 16:29, 28 May 2009 (UTC)[reply]

No. I'm not competent enough. Something about the degree of bad behavior as |z| approches 1 from below?Julzes (talk) 03:48, 29 May 2009 (UTC)[reply]

There is a good answer close to the top of May 29 below.Julzes (talk) 14:09, 29 May 2009 (UTC)[reply]