Wikipedia:Reference desk/Archives/Mathematics/2008 July 18

= July 18 =

Exponential Exponents
Exponents or powers are the superscripts that in math mean to multiply the base the exponent's number of times. There are also numbers with exponents that have exponents. If you have an infinite number of exponents, does that make the number infinity or another infinite number?

i.e.

$$2^{2^{2^{2^{2^{2^{2^2...}}}}}}=\infin$$

or

$$2^{2^{2^{2^{2^{2^{2^2...}}}}}}\neq\infin$$

Earthan Philosopher (talk) 18:57, 18 July 2008 (UTC)
 * I don't think I've ever encountered an author who gave any meaning at all to such expressions. Algebraist 19:03, 18 July 2008 (UTC)


 * I think it would be like to asking if 1+1+1+1+1+1...=inf. Exponents are just a faster way to get there. I think.--Xtothe3rd (talk) 19:15, 18 July 2008 (UTC)


 * Is there any other way this be interpreted other than the limiting sense (you start with a finite tower, and exponentiate once for each additional term in the sequence)? RayAYang (talk) 19:49, 18 July 2008 (UTC)
 * Not that I can think of. Of course, you have to decide whether those twos are reals, complexes, ordinals, cardinals, 2-adics, or whatever, the answers being $$+ \infty$$, unsigned $$\infty$$, $$\omega$$, $$\aleph_0$$ and zero respectively. The only context I've seen infinite towers of exponentials is [[epsilon nought|ordinal

arithmetic]], which I doubt is what the OP cares about. Algebraist 20:09, 18 July 2008 (UTC)
 * The power tower $$a^{a^{a^{\vdots}}}$$ converges iff $$ e^{-e} \le a \le e^{1/e}$$ and is equal to $$\frac{W(-\ln(a))}{-\ln(a)}$$, where W is the Lambert W function. There are further explanations on the Mathworld article on "Power Tower". --XediTalk 20:12, 18 July 2008 (UTC)

Look at the Wikipedia article on tetration, which is the equivalent of the Mathworld article on "Power Tower", I would assume.  J kasd  20:19, 18 July 2008 (UTC)
 * Quite right, indeed, I should've thought of going on that article ! --XediTalk 21:28, 18 July 2008 (UTC)

The formal expression $$x=2^{2^{2^{2^{2^{2^{2^2...}}}}}}$$ satisfies the equation $$x=2^x.$$ So x has better be a solution to that equation. One of the solutions is the complex number x≈0.824679+1.56743i. Bo Jacoby (talk) 05:52, 19 July 2008 (UTC).


 * The W function is analytic, right? Can that expression be extended to give an answer for 2? Black Carrot (talk) 02:35, 19 July 2008 (UTC)
 * Yes, it does give an answer, albeit not a very sensible one (approximately 0.8246785461-1.567432124i), sort of in the same way you could say $$\sum_{n=1}^\infty n = \sum_{n=1}^\infty \frac{1}{n^{-1}} = \zeta(-1) = -\frac{1}{12}$$ --XediTalk 03:00, 19 July 2008 (UTC)
 * $$\sum_{n=1}^\infty n = -\frac{1}{12}$$? Does that have any deeper meaning? In similar cases (1+2+4+8+...=-1, for example), the strange answer is actually because you're working in a strange metric without knowing it (2-adic in the case of that example), but I can't think of a metric in which 1+2+3+4+... would converge. --Tango (talk) 04:01, 19 July 2008 (UTC)
 * I'm not really aware of how to put this into context, but, in this particular example, -1/12 is the Ramanujan sum of $$\sum_{n=1}^{\infty}n$$, which might give a few clues. I would, too, like to know what mathematical theory is best suited to these kind of tricks, which seem to mainly involve a sum convergent on only part of its domain, extending it analytically and getting a "sum" for where the sum doesn't initially converge. --XediTalk 04:39, 19 July 2008 (UTC)
 * See the article on Analytic continuation. The sum 1+2+4+8+... is the value f(2) = &minus;1 of the function f(x) = 1+x+x2+... = (1&minus;x)&minus;1. The sum 1+2+3+4+... is the value f '(1) of the function f '(x) = 1+2x+3x2+... = (1&minus;x)&minus;2, so f '(1) is not the Ramanujan sum (which I do not understand). Bo Jacoby (talk) 06:00, 20 July 2008 (UTC).
 * $$\left ( \frac{1}{(1-x)^2} \right )\!\! \Bigg | _{x=1}$$ is not going to help as a value for $$\sum_{n=1}^{\infty} n$$ seeing as it isn't defined... --XediTalk 15:58, 20 July 2008 (UTC)
 * Right! Is Ramanujan's &minus;1/12 consistent with analytic continuation? Bo Jacoby (talk) 08:17, 21 July 2008 (UTC).
 * Well, using another method, you define the Riemann Zeta function for Re(s) > 1 as $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$. This doesn't converge for Re(s) < 1 but you can use that to give values for the sum, hence giving $$\sum_{n=1}^{\infty}\frac{1}{n^{-1}}=\sum_{n=1}^{\infty}n=\zeta(-1) =\frac{-1}{12}$$ which is what the Ramanujan summation gives, not totally sure why though, there must be some argument to explain that if we define that sum to have a value, -1/12 is the only sensible one to give. --XediTalk 01:32, 22 July 2008 (UTC)
 * OK. The sum 1+2+3+4+5+... appears in two ways: either as the infinite value for x=1 of the differential quotient f '(x) = 1+2x+3x2+... = (1&minus;x)&minus;2 of the geometric series f(x) = 1+x+x2... = (1&minus;x)&minus;1, or as the value &minus;1/12 of the zeta function ζ(s) for s = &minus;1. Does a sum like 1&minus;2+3&minus;4+5&minus;6+... = f '(&minus;1) = (1&minus;(&minus;1))&minus;2 = 1/4 have a different Ramanujan sum? Bo Jacoby (talk) 11:28, 22 July 2008 (UTC).

Well now that it is explained to me I suppose that it doesn't have to be just $$2^{2^{2^{\vdots}}}$$ or any exponent, it could be like $$1+1+1+1...$$, and thus I have my answer, because $$2^{2^{2^{2^{2^{2^{2^2...}}}}}}=1+1+1+1+1+1...=1\infin=\infin$$ and so there for $$2^{2^{2^{2^{2^{2^{2^2...}}}}}}$$ must equal $$\infin$$

Earthan Philosopher (Talk) 03:20, 19 July 2008 (UTC)
 * See also Knuth's up-arrow notation. $$2\uparrow\uparrow \infty$$. Tlepp (talk) 05:41, 19 July 2008 (UTC)


 * In reply to the side topic, above, there is an article on 1 + 2 + 3 + 4 + · · ·, along with several other related series in Category:Mathematical series. Confusing Manifestation (Say hi!) 22:45, 20 July 2008 (UTC)

Reversi
This question was originall posted at wp:rd/m --Shaggorama (talk) 20:39, 18 July 2008 (UTC) Hello. What is the largest number of possible moves on anyone's turn at anytime for Reversi? For example, dark has four legal moves on its first turn. Thanks in advance. --Mayfare (talk) 18:19, 14 July 2008 (UTC)
 * One challenge is working this out is that you would need to know if a given situation is actually possible during a game. Situations could be constructed to yield a high number of possible moves (16 is easy, more than that shouldn't been too hard), but verifying that they are achievable under the rules of the game would be much harder. --Tango (talk) 21:14, 18 July 2008 (UTC)
 * Out of interest, what's your 16? My trivialist idea yields 28. Algebraist 21:19, 18 July 2008 (UTC)
 * Columns of one colour in the sequence .BWB.BWB or similar - white can go in any of the 16 blank spaces. I don't doubt that there are far better arrangements, that's just the first one I thought of. --Tango (talk) 23:03, 18 July 2008 (UTC)