Wikipedia:Reference desk/Archives/Mathematics/2013 June 6

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June 6

Shortcut methods or tricks for solving mathematical and numerical problems

It would be very helpful for me if I got shortcut methods (or tricks) for solving mathematical and numerical problems in physics. Someone, please, give me links to some websites and books related to my question. Britannica User (talk) 01:16, 6 June 2013 (UTC)

Do you talk about rules of thumb or more complex approaches? Listening to ASTR 160: Frontiers and Controversies in Astrophysics (probably an earlier edition, though). I learned a lot of interesting math, e.g. pi=3 and pi²=10. Listening to Charles Bailyn doing maths on the blackboard is a real learning experience ;-). --Stephan Schulz (talk) 06:48, 6 June 2013 (UTC)

One last question, I promise!

The last thing I was having trouble with for my exam preparation was proving |cos x - cos y| < |x-y| for any x, y in the reals. I get that you'll probably need to use the triangle inequality here? Or maybe the definition of cos? I'm unsure how to approach this. — Preceding unsigned comment added by 211.31.22.140 (talk) 08:32, 6 June 2013 (UTC)

(disclaimer: this is definitely a physicist's proof, and may not stand up to full mathematical rigour) You get equality iff x=y. both sides of the equation are continuous (not smooth, but that doesn't matter), so if the statement is false you would have a counterexample at small x-y.

cos(x)-cos(y) = 2sin(x+y/2)sin(x-y/2) = (for small angle) x-y sin (x+y/2)

Subsititute this into the inequality, and you get |x-y||sin(x+y/2| < |x-y| => |sin(x+y/2)|<1, which is true.

There's probably a considerably more rigorous way of doing this, however. MChesterMC (talk) 10:25, 6 June 2013 (UTC)

You can prove this quite easily by using the Mean value theorem - the basic idea (or "physicist's proof") is that as the derivative of cos is -sin, the slope of the line joining two points on the graph is always < 1 (with a little care to prove strict inequality where x != y). AndrewWTaylor (talk) 10:59, 6 June 2013 (UTC)

I got the term "physicist's proof" from a lecturer. Rougly defined as a proof that doesn't fully prove anything but looks good enough and will annoy the mathematicians. Contrasted with an engineer's proof, which is "eh, looks about right" MChesterMC (talk) 13:31, 6 June 2013 (UTC)

First of all, the inequality isn't true. If x = y, then |cos x - cos y| = |x-y| and you've got a poor exam writer. To prove |cos x - cos y| <= |x-y|, set x = A-B and y = A+B. then cos x - cos y = 2 sin A sin B and x-y = -2B. Then we get |2 sin A sin B| <= |-2B|. Since |sin A| <= 1, then |2 sin A sin B| <= |2 sin B| and the inequality becomes |sin B| <= |B|. That is a pretty well-known inequality, but if you wanted to prove it, note that f(B) = B - sin B has f(0)=0 and the derivative f'(B) = 1-cos(B) >= 0 for all B so f is increasing and thus f >= 0 for B >= 0. Since f(-B) = -f(B), it holds true for negative B, too, and you are done. --Mark viking (talk) 12:03, 6 June 2013 (UTC)

The statement is false: |cos(x) – cos(y)| = |x – y| if x=y. For other cases, however, the inequality is true.
A derivative of sine (or cosine) is between −1 and +1 (inclusive), and it is −1 or +1 only in separate points, so each chord of the sinusoid has slope between −1 and +1 (exclusive). That gives an answer: for two different arguments the difference between sine (or cosine) values is always less (with respect to absolute values) than the difference between arguments. --CiaPan (talk) 12:15, 6 June 2013 (UTC)

What are "complex curves"?

In Snyder's An Album of Map Projections, the term "complex curve" is used a lot. What does the term refer to and what are they (the curves) specifically named in the discussion of map projections? Czech is Cyrillized (talk) 08:54, 6 June 2013 (UTC)

It's defined in the Glossary near the start of the document: "Complex curves Curves that are not elementary forms such as circles, ellipses, hyperbolas, parabolas, and sine curves." AndrewWTaylor (talk) 10:22, 6 June 2013 (UTC)

I guess a straight line would be considered elementary as well...? --CiaPan (talk) 06:08, 7 June 2013 (UTC)

The term "complex curves" sound very generic for me. Are they algebraic curves on map projections? Czech is Cyrillized (talk) 15:10, 6 June 2013 (UTC)

I think they're almost never algebraic. One of the simplest projections whose graticules are not circles or straight lines, for instance, is the transverse Mercator projection. The graticules are given as transcendental functions involving the natural logarithm and trigonometric functions. Sławomir Biały (talk) 00:18, 9 June 2013 (UTC)

Some aren’t even analytic, and can only be plotted using numerical methods or recursion.—Odysseus1479 08:13, 9 June 2013 (UTC)

Chess Knight move math problem.

I was thinking about chess and had this idea. 1-Rook, move 0, 90, 180, and 270 degrees . 2-Bishop move between those lines (45, 135.....). 3-Queen have rook and bishop moves

Now, i have a question about the knight. It is true that the knight moves between the queen moves (but by jumping because, because the way the chess board is) or my "math" is wrong? If no, how a knight that moves like that would move?201.78.152.67 (talk) 18:36, 6 June 2013 (UTC)

Yes, you could think of the knight moving like that but for a specific distance and being able to hop over other pieces. So it would move at 30, 60, 120, 150 ... degrees. 105.236.239.190 (talk) 18:50, 6 June 2013 (UTC)

First you say yes and then you say it move at 30, 60, 120....? The middle point between 0 and 45 as some example is 22.5 degrees, If knight moves as I was talking before he would move 22,5 degrees, 67.5, 112.5, 157.5, 202.5, 247.5, 292.5 and 337.5 degrees.

What I am asking is, if the knight move at those degrees (transposed to a board made of squares) or not? And if not tell me what are the degrees that the knight move and then how a knight would move if he moved at (22.5, 67.5.....)? 201.78.152.67 (talk) 19:01, 6 June 2013 (UTC)

If you work out the trigonometry, it is approx 18, 72, 108, 162 ... degrees. AndyTheGrump (talk) 19:02, 6 June 2013 (UTC)

I think I found my answer here ( http://www.chessvariants.org/index/listcomments.php?itemid=PcKnight ) a guy said "I have noticed some pages in this site (I forget which ones) described the 'hippagonals' as 'the lines halfway between diagonals and orthogonals'. They are not. The hippagonals are at inverse tangents of -2, -.5, .5 and 2 from forwards, which are at about 26.6 degrees from the orthogonals (18.4 degrees from the diagonals). The lines halfway between the diagonals and orthogonals are 22.5 degrees from each, and will not cross the centres of any squares no matter how big a board they are on."

So if this is right, knight is 26.6 from orhogonals and 18.4 from diagonal. and he say it would be impossible to have a knight that move at 22.5 and touch the middle of the squares.201.78.152.67 (talk) 19:20, 6 June 2013 (UTC)

Yup. 26.6 degrees is right - I wasn't thinking straight earlier. Just as well I don't play chess any more... AndyTheGrump (talk) 21:08, 6 June 2013 (UTC)