Wikipedia:Reference desk/Archives/Mathematics/2007 December 18

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December 18

s = cos(a).sin(a)

What is the correct mathematical way to prove that the max value of s is when a= 1/4pi or 3/4pi radians? I went to take derivative but that makes it go around in circles :)--Dacium (talk) 03:59, 18 December 2007 (UTC)

Use trigonometric identities :

$\cos(a)\sin(a)={\frac {\sin(2a)}{2}}$

Then you just have to get the maximum of $\sin(2a)$

--Xedi (talk) 04:20, 18 December 2007 (UTC)

(Edit conflict/previous older edit follows) Or use them in the derivative:

${\frac {d\cos(x)\sin(x)}{dx}}=\cos ^{2}(x)-\sin ^{2}(x)=2\cos ^{2}(x)-1$

Pallida  Mors 06:08, 18 December 2007 (UTC)

Go on... $\dots =2\cos ^{2}(x)-1=\cos ^{2}(x)+\cos ^{2}(x)-1=\cos ^{2}(x)-\sin ^{2}(x)=\cos(2x)$
--CiaPan (talk) 09:38, 18 December 2007 (UTC)

Go on?? Why on Earth?

$2\cos ^{2}(x)-1=0$

...is all we need to find the maxima. Pallida  Mors 13:57, 18 December 2007 (UTC)

It's not all – you'll have to solve an equation, which in turn requires you to recall (or find in trig. tables) such $x$ value(s) which give $\cos x={\tfrac {1}{\sqrt {2}}}$, which may be (a little) harder than recalling cosine's zeros. :) --CiaPan (talk) 09:33, 19 December 2007 (UTC)

They used to teach us cos(45º). Seriously. Pallida  Mors 21:47, 19 December 2007 (UTC)

You can do this with calculus, but it's easier to use geometry. Your function gives the area of a rectangle inscribed in a circle of radius 1/2 such that the angle between the base and the diagonal is a. The area is maximized by an inscribed square.

This is a form of the arithmetic mean-geometric mean inequality. For any angle a, the inequality

cos(a) sin(a) ≤ (1/2) (cos(a)² + sin(a)²) = 1/2

holds. It holds with equality when a = π/4, that is, an angle of a = π/4 maximizes the function. Michael Slone (talk) 06:01, 18 December 2007 (UTC)

5 Cats and 3 mice

in a 5 x 5 grid. [1] Is this possible please? - CarbonLifeForm (talk) 09:42, 18 December 2007 (UTC)

Yes it is. It's not easy, though - After several failed attempts, I have written a program to find a solution. Would you like me to present a solution, or do you want to try again now that you know it is possible? -- Meni Rosenfeld (talk) 10:24, 18 December 2007 (UTC)

I give up! Thank you. - CarbonLifeForm (talk) 10:27, 18 December 2007 (UTC)

That is annoying. mattbuck (talk) 10:55, 18 December 2007 (UTC)

Please tell me the solution. - CarbonLifeForm (talk) 10:56, 18 December 2007 (UTC)

* C C * *
* C * * C
* * * * C
M * * * *
M * * M *

This is the only solution, up to symmetry (rotations and reflections). -- Meni Rosenfeld (talk) 12:28, 18 December 2007 (UTC)

boo

Exact, I got the same but reflected and rotated. It wasn't that hard, actually, took me 10 minutes of attempts. I got it by trying to understand how to minimize the impact of 4 cats on the board, getting three mice correctly placed and finally adding the fifth cat and moving one of the three mice. Before attempting it, speeding up your trial-and-error process with "tricks" is also useful. --Taraborn (talk) 13:49, 18 December 2007 (UTC)

I gave it a try, too. However, I misplaced one cat and it ate one of the mice. And I had no spare rodent, I'm afraid. :( Pallida  Mors 14:04, 18 December 2007 (UTC)

Sophomore's dream

Pretty simple question, yet one I haven't been able to find an answer to. Why is the mathematical identity sophomore's dream called that? --Dr. Skullthumper (talk) 21:32, 18 December 2007 (UTC)

As nobody else answers, I'll give a try. I suppose that it is called so because it is "too good to be true": the argument of the integral and the summands of the series are very similar, almost as if you could substitute sums for integrals. For similar reasons, somebody calls "freshman's dream" the congruence $(x+y)^{p}\equiv x^{p}+y^{p}\mod p$. It makes true what normally is not: being able to distribute exponents with respect to a sum. Goochelaar (talk) 15:40, 19 December 2007 (UTC)