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December 13

Meandric numbers

According to Meander (mathematics), and (sequence A005315 in OEIS), the third meandic number is eight. Ok, so I can find four distinct meanders, A1 to A4, with 23 crossings and along with their horizontal reflections, B1 to B4, make the required eight. But wait, some of them, C2 to C4, have a vertical reflection not homomorphic to any of the eight, and along with their horizontal reflections, D2 to D4, make a total of 14. What am I missing? SpinningSpark 11:29, 13 December 2014 (UTC)

The pdf linked in the article gives all 8. The problem is that yours are crossing the line 8 times, not 6. (It's 2n not 2^n, so you should have 6 crossings)Phoenixia1177 (talk) 12:59, 13 December 2014 (UTC)
Ah, thanks. I knew it had to be something simple. Just for the record, I found quite a few more with eight crossings after posting the diagram. SpinningSpark 13:17, 13 December 2014 (UTC)

December 14

Validity of (ab)^m = (a^m)(b^m)

(ab)m = ambm
I have questions regarding the validity of the above law of indices depending on what a,b,m are.

I'm trying to come up with the valid values that a,b,m can take so that the above law is always true for a given set of restrictions.
I'm trying to be as inclusive as possible, meaning that if something isn't wrong or results in something being wrong, then I won't exclude it. For example, if a complex number (not just a real number) also works in some situations of the above law, then I'll also try to include and describe those situations.

Note: For all cases, any situation that results in division by 0 (or 0 to a negative real number) and 00 are exceptions that have been noted. I'm not writing expressions that show I'm disallowing them so that I can reduce clutter in describing the different cases below.

Case 1:
If m ∈ {ℝ \ ℤ}, then a,b ∈ ℝ+
This is so we avoid the wrong results of 1 = -1 and i = -i (and other similar wrong results).

Case 2:
If m ∈ ℤ, then a,b ∈ ℂ
Apart from that note I made above, I don't see any problems with this case.

Case 3:
If m ∈ j2k+1 where j,k ∈ ℤ, then a,b ∈ ℂ
The power is rational, and its denominator cannot be even because taking even roots of negative real numbers results in the same problems mentioned in Case 1.

So, my questions are:
1. I paraphrased Case 1 and 2 from this wikipedia article (http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities). Is it correct for me to paraphrase it this way?
2. I'm not sure if Case 3 is correct though, and that is where I require help in checking its correctness. The reason I wanted another case (other than the 2 mentioned in that wikipedia article) is because (just from working a few examples out) there are some complex numbered bases that can be raised to a rational power (with an odd denominator) that satisfy this particular law of indices, so it feels too restrictive/un-inclusive if we don't cater to these cases.
Thanks. 175.156.52.140 (talk) 19:25, 14 December 2014 (UTC)

1. Cases 1 and 2 should be regarded as distinct functions. You cannot combine them sensibly so that the identity always holds, even though they will agree on every point for which they are both defined. If you regard the functions as distinct for case 1 (am ≝ exp(m⋅log a), where log is the real logarithm) and case 2 (am ≝ ∏m a, extended to all integer m through division where possible), these cases work perfectly. I would phrase case 1 without the exclusion of integers, i.e.
• Case 1: m ∈ ℂ and a,b ∈ ℝ+ is the domain of the first exponentiation function, defined in terms of log.
• Case 2: m ∈ ℤ and a,b ∈ ℂ is the domain of the second exponentiation function, defined in terms of repeated multiplication, with a ≠ 0 and b ≠ 0 for m < 0; interestingly we can define 00 for this function without problems if we wish to do so.
2. Case 3 does not work. There are problems like with even denominators whenever m ∈ ℚ ∖ ℤ for most complex bases. —Quondum 21:26, 14 December 2014 (UTC)

I've extended the case 1 function domain, since you're trying to to be maximally inclusive. The in the domain can be replaced with far more general objects in both cases (e.g. matrices, quaternions, etc.) if you wish to be maximally inclusive. —Quondum 01:49, 15 December 2014 (UTC)

zero knowledge comparison of two positive integers

Hi,

Can A and B run a Zero Knowledge Proof algorithm to learn whether A's or B's positive integer is larger (or if they are the same) without either of them learning (or giving away) any further information beyond this fact? They should not have to trust a third party, this is part of the field of mathematics called zero knowledge proofs. Thank you. 212.96.61.236 (talk) 20:28, 14 December 2014 (UTC)

Yes, see Secure multi-party computation for the general, highly inefficient methods, and Yao's Millionaires' Problem for a (presumably more efficient) method for this particular application. It is assumed that there is some known bound on the lengths of the numbers (which could be very loose), perhaps it can be generalized for a case that such a bound is not known a priori. -- Meni Rosenfeld (talk) 22:58, 14 December 2014 (UTC)
Meni or another: could you summarize the principle Yao's solution works on? (what is the general principle.) The math in that article is a bit hard for me to follow. 212.96.61.236 (talk) 23:14, 14 December 2014 (UTC)
As with your previous question, if A is allowed as many queries as they want, they can brute force B's integer. ("is it greater than 1?" yes. "is it greater than 2?" yes. "is it greater than 3?" no. "it's 3!"). If the range of the integers is limited, this can be done relatively quickly (see binary search) MChesterMC (talk) 09:34, 17 December 2014 (UTC)
Presumably, they will only run the process once, and they are both genuinely interested in the outcome, and thus will behave honestly. -- Meni Rosenfeld (talk) 22:17, 17 December 2014 (UTC)

December 16

Forcing -vs- Boolean Valued Models

I have books on both - and my understanding is they are equivalent - but I wanted to ask which is better to study, in detail? I have a basic understanding of both, and intend to go into both in detail, but wanted a better idea of which is "more useful" in terms of seeing connections and understanding set theoretic things - as opposed to, say, actually proving some specific result. Our article on forcing indicates BV models, but I'd be curious to hear actual input to get a clearer idea of things. (Thank you for any answers, or help, the further I go, the harder it is to keep a clear sense of direction, all of you are immensely helpful and I am truly appreciative for all the help I've received in the past).Phoenixia1177 (talk) 15:07, 16 December 2014 (UTC)

I can't help but I'll point out that this is not about forcing in dynamical systems (e.g. [1]), but rather Forcing_(set theory). One might suspect that there are some analogies between the concepts based on the name but even that is unclear to me. SemanticMantis (talk) 19:17, 16 December 2014 (UTC)

This is a bit of a matter of opinion, but here's my take on it: Forcing is usually a more efficient way to actually make progress than Boolean-valued models are. A forcing partial order can be described in such a way that you can see "what it's trying to get at", what sort of generic object it's trying to add to the universe. The corresponding Boolean algebra is usually a lot less perspicacious. If you have to back off all the way to the general case (the algebra of regular open sets in the topology generated by cones from the partial order) then you've basically lost all useful intuition about what the elements of the Boolean algebra are.
However, it's very nice to know the Boolean-valued model approach for other reasons, to be able to have a more philosophically satisfying interpretation of certain results. For example, take the the result about making $2^{\aleph_0}$ equal to κ, for a given uncountable regular cardinal κ. What does that mean?
You can't capture it completely in proof-theoretic terms ("adding the assumption $2^{\aleph_0}=\kappa$ to ZFC cannot produce an inconsistency in ZFC unless there was one already") because κ is just some random cardinal, one that may or may not even be definable in the language of set theory, so there is no way to add $2^{\aleph_0}=\kappa$ as an axiom. It's a category error even to try.
Similarly, you can't get at it in any obvious way via countable transitive models (the usual fallback for talking about forcing in terms of concrete objects) because κ is not an element of any countable transitive model. There may be some way to make sense of it using countable transitive models that are Mostowski collapses of models containing κ; not sure on this point.
But with Boolean-valued models, there's a very clear interpretation. There's a Boolean algebra B such that the statement $2^{\aleph_0}=\kappa$ has probability 1 in $V^B$. --Trovatore (talk) 19:44, 16 December 2014 (UTC)

December 17

Coin flipping

If a coin has 50% chance of landing "tails" when flipped, and I flip it 25 times in a row, what are the chances that I never get 4 (or more) consecutive "tails"? What is the formula by which I can calculate this? --Theurgist (talk) 14:53, 17 December 2014 (UTC)

The number of sequences is a(29)=14564533 where a(n) is the nth Tetranacci number, so the probability is 14564533/225 or about 43%. See (sequence A000078 in OEIS) for formulas. --RDBury (talk) 00:10, 18 December 2014 (UTC)

December 18

Math & Value definition

If you receive 1 for 100, how much will you receive for 1,000,000,000,000,000,000.00?

How would you do the math?

How would you pronounce 1,000,000,000,000,000,000.00? - Is this 1 Zillion/Trillion?

What's the highest sum (word) known to human kind? Is it a 'Zillion'?

(Russell.mo (talk) 14:59, 18 December 2014 (UTC))

You might want to see names of large numbers. What you have here is a trillion (European naming) or a quintillion (US naming). YohanN7 (talk) 15:10, 18 December 2014 (UTC)
Just noting in passing here that so-called "US naming" is now almost completely standard across the English-speaking world. It may be in some sense less "logical", but it's more convenient, and that seems to have won out. See long and short scales. --Trovatore (talk) 20:15, 18 December 2014 (UTC)
Yes, more convenient, just like furlongs and pence, shillings and pound sterlings (Not)
Yes, it really is more convenient. To use the long scale, you have to use either locutions like "thousand million", "thousand billion", etc, or "milliard", "billiard", etc. The "thousand million" solution is longwinded; the other solution is virtually unknown in English. --Trovatore (talk) 21:47, 18 December 2014 (UTC)
You are right. But a universal naming with *-ion and *-iard would have been logical. At any rate, these are big numbers. Bigger numbers than this are called astronomical numbers. Yet bigger ones are called economical numbers (Feynman). YohanN7 (talk) 22:08, 18 December 2014 (UTC)
A Brazilian can get pretty high, especially around Carnival. Sławomir Biały (talk) 16:01, 18 December 2014 (UTC)
Nobody's answered the original question - you get one hundredth of the stated amount, i.e knock two zeros off before the decimal point.→86.157.199.240 (talk) 22:34, 18 December 2014 (UTC)
To answer another of the original questions, there is no such number as a zillion; it's just a slang word for a very, very big number. The -illion sequence is sometimes considered to end at a centillion (with 303 zeroes in what's called "US naming" or "short scale" above, or 600 zeroes in the other system), but as you will see at that article, others take the sequence still higher. In practice nobody uses these large names; they would express the numbers in other ways. As far as I know the googolplex is the largest number named with a single word; it ends with 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 zeroes and there is not nearly enough matter in the known universe to write it out in full. --65.94.50.4 (talk) 00:55, 19 December 2014 (UTC)
Thanks peeps, thanks to the ones, the most, who specified what was needed to know. I'm on two weeks holiday which started today, forced upon me, so I read the article stated in this post, otherwise you all would've heard me moaning for explanations. Thank you for stating the equation too.
Regards.
(Russell.mo (talk) 06:45, 19 December 2014 (UTC))
Resolved