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March 31

Strange result in the quotient of a number and the logarithm of it's powers?

Define f(N, X) = N / log(N^X). Why is it that f(2, X) = f(4, X) for any given X, always? This seems like a rather unusual result. Is there a simple explanation for it?Earl of Arundel (talk) 03:29, 31 March 2017 (UTC)[reply]

4 / (log 4^X) = 4 / (log 2^(2X)) = 4 / (2 log 2^X) = 2 / (log 2^X). Double sharp (talk) 03:30, 31 March 2017 (UTC)[reply]

Right, thanks, but in terms of the graph produced when plotting out possible values, it just seems strange that both 2 and 4 have the same intersection point, doesn't it? Earl of Arundel (talk) 03:48, 31 March 2017 (UTC)[reply]

If you hold X constant and take the partial derivative with respect to N, you get (log N - 1) / (X log^2 N). This equals zero when log N = 1, or N = e, so you have a turning point there. Since your function is "nice" in the sense that it is continuous as N changes in the region around e, it is then not surprising that you get the same values for a value of N below e and a value of N above e. {2, 4} just happens to be the only such pair both of whose members are integers. Double sharp (talk) 04:28, 31 March 2017 (UTC)[reply]

Yes, but since 2 and 4 are not equidistant from e, isn't it kind of odd that the inflection would result in a crossing at the same point for both of those values? I don't know, maybe I'm just overthinking this! Anyway, thanks for the insight. Earl of Arundel (talk) 04:39, 31 March 2017 (UTC)[reply]

Your function is equal to (N / log N) / X. The point is that 2 / log 2=4 / log 4.Burzuchius (talk) 08:29, 31 March 2017 (UTC)[reply]

A very simple explanation for this common intersection just occurred to me: observing that N^K = NK only when N = K = 2 or N = 4, K = 1/2 (excluding the trivial case where K = 1, of course), since log(N^K) = log(N) * K then obviously (2 / (log(2) * 1)) = (4 / (log(2) * 2). Earl of Arundel (talk) 16:37, 1 April 2017 (UTC)[reply]

For any k > 0 take n = exp( log(k) / ( k - 1 )) and it's a solution to n^k = nk. --‎Joel B. Lewis 11:22, 2 April 2017 (UTC)[reply]

Interesting! Thanks, Joel. Earl of Arundel (talk) 14:38, 2 April 2017 (UTC)[reply]

Just found a neat set of equivalences there: n = exp(log(k)/(k-1)) = (((k^k)^(1/(k-1)))/k) = ((1/k)^(-(1/(k-1)))) = k^(1/(k-1)). Earl of Arundel (talk) 16:16, 2 April 2017 (UTC)[reply]

@Earl of Arundel, Burzuchius, and Joel B. Lewis: Hmm.

@Double sharp: hi. Remember me? We met on Einsteinium. —usernamekiran[talk] 22:05, 2 April 2017 (UTC)[reply]

The Existence of a Soltuion in Tower of Hanoi

Is that true that the problem of Tower of Hanoi has a solution for any "legal" starting position: that is, as long the disks are ordered in increasing order in each of the 3 towers, there is a solution? עברית (talk) 10:33, 31 March 2017 (UTC)[reply]

The Wikipedia article titled Tower of Hanoi has extensive discussion of the solution. --Jayron 32 11:05, 31 March 2017 (UTC)[reply]

Yes: you can transform any legal position into any other legal position. Double sharp (talk) 11:17, 31 March 2017 (UTC)[reply]

One can solve the puzzle from any initial position without violating rules of move, not just from "legal". --CiaPan (talk) 12:56, 31 March 2017 (UTC)[reply]

That isn't altogether obvious except after one stares into space for a while and then declares it obvious :) Dmcq (talk) 14:31, 31 March 2017 (UTC)[reply]

@Dmcq: I never said it's obvious. :) --CiaPan (talk) 22:56, 31 March 2017 (UTC)[reply]

I guess a legal move is one that doesn't make the position any more illegal? —Tamfang (talk) 07:54, 3 April 2017 (UTC)[reply]

I won't dare to define 'more illegal' ;) Anyway, a move is legal if it obeys rules: only one disk is moved at a time and that disk lands either on a greater disk or on an empty stack. --CiaPan (talk) 20:41, 4 April 2017 (UTC)[reply]

Thank you! עברית (talk) 06:09, 1 April 2017 (UTC)[reply]

Could someone fix an error?

I took a class in college where we did what is done in 99 Bottles of Beer#References in computer science. There's no close parenthesis and I'm pretty sure one is needed, but I want to keep what was used in the first place.— Vchimpanzee • talk • contributions • 22:02, 31 March 2017 (UTC)[reply]

Where exactly are you seeing a parenthesis missing? There's only one pair of parentheses in the section you cited, and both are present. --76.71.6.254 (talk) 00:33, 1 April 2017 (UTC)[reply]

For some reason the section heading has changed, but I still only see a left parenthesis. Maybe this is one for the WP:VPT.— Vchimpanzee • talk • contributions • 15:38, 1 April 2017 (UTC)[reply]

It changed from "References in science" to "References in computer science" in [1] and that's what I see. If you refer to

O(\log N)

then I see an image displaying "O(log N)" for both the first anmd third setting of Math at Special:Preferences#mw-prefsection-rendering. Have you enabled the second setting "LaTeX source (for text browsers)"? There I see "$ O(\log N) $" Some browsers with LaTex features may try to render it as a formula. PrimeHunter (talk) 15:52, 1 April 2017 (UTC)[reply]

I did not do that. I don't use it enough.— Vchimpanzee • talk • contributions • 16:19, 1 April 2017 (UTC)[reply]