Wikipedia:Reference desk/Archives/Mathematics/2010 July 20

= July 20 =

Chess books
Hi, I'm not sure if this is the best place to ask this question, but here it goes. Would anyone be able to recommend some good chess books for self-study? I consider myself a competent player, but everything I've learnt was from playing and I figure a more formal education would both improve my game and be an enjoyable way to spend some of my free time over the summer. 74.15.137.192 (talk) 06:37, 20 July 2010 (UTC)


 * If no one answers here, you can ask at WT:CHESS, -- Wavelength (talk) 18:55, 20 July 2010 (UTC)


 * I guess, "My System" by Aaron Nimzowitsch is one of the best options... And (to keep neutrality) - something by Siegbert Tarrasch - for example, "Die moderne Schachpartie" or "Three hundred chess games". --Martynas Patasius (talk) 18:13, 23 July 2010 (UTC)


 * Those may be fine books but they're both a bit dated, and I personally think both authors tended to go on a bit about their Theories of Everything. There are so many books, and even so many from more modern authors, that my recommendation would be to go to a bricks-and-mortar bookstore and browse their Chess section, and pick something that interests you based on reading it a bit.


 * I played some tournament chess but never studied enough; like a lot of amateurs in games, I preferred playing it to working on it. I could have been better, and still could be better, if I had done more study.


 * Something you should keep in mind: you can spend a lot of time studying quite different things in chess. The endgame deserves study all on its own, because most of the time in the opening and middle games, you are making decisions based on what the endgame would look like if you got that far.  In that case, you have to know what things are advantages and what things only look like advantages; there are cases where having as much as a two-piece advantage will still not produce a win (though that's extreme); even if you have a winning advantage, it isn't a win if you cannot complete it!  As far as I know, the definitive book on the endgame is still by Rueben Fine, and its age is a lot less of an issue than books on opening and middle game issues, because so much of the endgame technique, etc., is still the same.  So it is reasonable advice to study the endgame first.


 * I always enjoyed Bobby Fischer's My 60 Memorable Games; he includes extensive notes, often entire other games, and tells a lot of what's going on behind the scenes. I don't recommend him because he was such a likable person or anything, but I think the book was a landmark and shows some wonderful games and moves.  Be sure to check out his game with Tigran Petrosian where they both queened pawns with their original queens still on the board, and the one with Robert Byrne where, at the moment Byrne resigned, the two grandmasters in the spectator room were explaining that Byrne had a won game.


 * rc (talk) 05:44, 27 July 2010 (UTC)

Dot product
Why does a · b = x1x2+ y1y2 (where a = (x1, y1), b = (x2, y2))? I have read the proof in Dot product, but I didn't understand. Thank you. 60.7.71.152 (talk) 08:46, 20 July 2010 (UTC)
 * Use the notation a = (a1, a2), b = (b1, b2).
 * (a+b)2&minus;(a&minus;b)2 = (a2+b2+2a&middot;b) &minus;(a2+b2&minus;2a&middot;b) = 4a&middot;b
 * According to Pytagoras
 * (a+b)2&minus;(a&minus;b)2 = ((a1+b1)2+(a2+b2)2)&minus;((a1&minus;b1)2+(a2&minus;b2)2) = 4(a1b1+a2b2)
 * so a&middot;b = a1b1+a2b2. Bo Jacoby (talk) 09:18, 20 July 2010 (UTC).


 * The definition of a · b ensures that the identity
 * a + b|2 = (a + b) · (a + b)
 * holds for all vectors a and b. Gandalf61 (talk) 10:04, 20 July 2010 (UTC)


 * Usually that's the way the dot product is defined, so the answer you're looking for depends on what interpretation or use of the dot product you are trying to reconcile that definition with. Could you be more specific about your question? 24.99.214.240 (talk) 18:13, 20 July 2010 (UTC)

y = dy/dx + x
Is there an analytic way to solve equations of the type like $$y = \frac{dy}{dx} + x$$ for y?

The reason I ask is because I created my own problem; $$e^{ix}=\frac{dy}{dx}+iy$$ and tried to solve it, but found that I was unable to do it (the answer, or at least one of the answers, is y=sin(x)).––220.253.172.214 (talk) 08:51, 20 July 2010 (UTC)
 * As an aside, I love chucking RefDesk questions into Wolfram Alpha for kicks; it was able to solve both your equations, although I can't help you with the actual analysis: http://www.wolframalpha.com/input/?i=y+%3D+dy%2Fdx+%2B+x and http://www.wolframalpha.com/input/?i=exp%28i*x%29+%3D+dy%2Fdx+%2B+i*y If I hazarded a guess I'd say there was an integrating factor involved. Zunaid 09:24, 20 July 2010 (UTC)
 * Wolfram Alpha also classifies both your equations as first-order linear ordinary differential equations so I would presume you need to learn the methods that attack those types of problems. Haven't check our articles but hopefully they start you off in the right direction. Zunaid 09:34, 20 July 2010 (UTC)
 * Substitute y=z+x+1 into y = dy/dx + x. Then dz/dx=z. Then z=Aex. Then y=Aex+x+1. Bo Jacoby (talk) 09:40, 20 July 2010 (UTC).
 * It's just a standard integrating factor question. Multiply by $$e^{ix}$$ and you get $$e^{2ix}=e^{ix}\frac{dy}{dx}+ie^{ix}y$$, which simplifies to $$\frac{d(ye^{ix})}{dx}=e^{2ix}$$.
 * Integrating and rearranging, you get the family of solutions $$y=ce^{-ix}+\frac{1}{2i}e^{ix}$$, where c is the constant of integration. If you set $$c=\frac{-1}{2i}$$, you get the solution $$y=\frac{e^{ix}-e^{-ix}}{2i}=\sin(x)$$, which you had already encountered. Readro (talk) 10:59, 20 July 2010 (UTC)