Wikipedia:Reference desk/Mathematics

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June 17[edit]

Majority as logical quantifier[edit]

What kind of logical quantifier is that represented/expressed by the noun majority?-- (talk) 11:08, 17 June 2017 (UTC)

Normally it means more than half, so more than 50%. A recent example is the recent UK election. Before the election the Conservative party had a majority (of seats). But after the election they no longer have a majority, though they are still the largest party, so in a sense they won the election.--JohnBlackburnewordsdeeds 12:13, 17 June 2017 (UTC)
So, in math notation, that would be "> 0.5". To give an example, let's say the story problem is "Prove that the 5,6,7,8 rolls of a pair of dice represent the majority of dice rolls". So, if we used R2(5) to represent the probability of rolling a 5 with two dice, that statement becomes "Prove that R2(5) + R2(6) + R2(7) + R2(8) > 0.5". StuRat (talk) 17:27, 17 June 2017 (UTC)

June 18[edit]

What's the distance?[edit]

A boat take 23 min 35 secs to complete a course at average speed of 23.21 knots. How far has it travelled in miles, and how far in metres? Moriori (talk) 02:53, 18 June 2017 (UTC)

Pictogram voting delete.svg Please do your own homework.
Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. --Jasper Deng (talk) 03:02, 18 June 2017 (UTC)
(Yes, I know you've probably asked questions here before, but I firmly believe we shouldn't work out solutions without the asker first trying; knots = nautical mile/s).--Jasper Deng (talk) 03:03, 18 June 2017 (UTC)
Your apology is accepted. The figures I gave came from an actual America's Cup race this morning. I casually asked a couple of friends what the distance travelled would be, and got two different answers. Eureka, I know, I said, I'll ask at the wiki maths reference section which has all sorts of boffins involved. Unfriendly testy ones too I see. Moriori (talk) 03:19, 18 June 2017 (UTC)
Well, if you will use the same sort of wording that a homework problem would, it shouldn't be surprising if people think that's what you're doing!
A nautical mile is 1852 meters, but a knot is one nautical mile per hour (3600 seconds), not per second as Jasper said. So the distance in meters is 1852 × 23.21 × (23×60 + 35) / 3600 = 16,895 m, which should be rounded to 16,900 m since the inputs only have 4 significant digits. Now a mile is 1609.344 m; so divide by that number and you get 10.50 miles. -- (talk) 06:26, 18 June 2017 (UTC)
My bad, I'm so used to using seconds for time.--Jasper Deng (talk) 07:44, 18 June 2017 (UTC)
Reasonableness check - a nautical mile is a bit longer than a statute mile. So we have a boat travelling at a little more than 20 mph for a little less than half an hour. So ballpark answer is 10 miles. Gandalf61 (talk) 09:45, 19 June 2017 (UTC)

June 19[edit]

Solvability of the Rubik's cube group[edit]

I don't mean solvability of the puzzle, but rather, whether the group is a solvable group. To me, it seems like it is not solvable for similar reasons as the Galois groups involved in the Abel-Ruffini theorem. I've tried looking this up but couldn't find an answer.--Jasper Deng (talk) 05:45, 19 June 2017 (UTC)

All subgroups of a solvable group are solvable. But (even permutatiions of corner pieces) and (even permutations of edge pieces) are subgroups of the Rubik's Cube group, and and are not solvable (since they are simple and not Abelian). So we can conclude that the Rubik's Cube group is not solvable. Gandalf61 (talk) 09:17, 19 June 2017 (UTC)
Technically, the piece permutations form a homomorphic image of the Rubik group. But the homomorphism splits so there are isomorphic copies of these groups as subgroups. It should be mentioned that the Jordan–Hölder theorem comes into play here as well: the fact that if one non-solvable group appears as a quotient then there is no decomposition series with solvable factors. --RDBury (talk) 15:40, 20 June 2017 (UTC)

June 20[edit]

Random walk on a 2d rectangle graph[edit]

I have a 2d rectangle graph with 4 vertices: A,B, C, D. An agent start moving from an initial position p0∈{A,B,C,D}, either clockwise or counterclockwise. When encountering the vertex A, it can change its movement direction with probability of 0.5. How can I calculate the probability the agent will be at position P after t time steps? intuitively I guess after long enough time the probability to be in each of the 4 positions is approximately equal, but I am looking for an analytical solution. I am especially in a modeling that can be extended to other movement schems (i.e. various "direction flipping" points or other probabilities for changing the direction). Thanks! — Preceding unsigned comment added by (talk) 16:09, 20 June 2017 (UTC)

You probably want to have a look at Markov chains. --Deacon Vorbis (talk) 16:28, 20 June 2017 (UTC)
If t = 0 mod 4 then the agent must be at vertex A, and if t = 2 mod 4 then it must be at position C. Gandalf61 (talk) 16:32, 20 June 2017 (UTC)
Please notice agents don't necessarily start at A, and change movement direction randomely when visiting point A. (talk) 16:44, 20 June 2017 (UTC)
I think this works: Label the positions with their numerical values 4, 3, 2, 1, such that the initial direction of movement is toward a lower number with wraparound. Let x_n be the position after n moves (n≥0). Then for n≥1 n≥x_0 we have:
If n = x_0 + 4k, Pr(x_n=4)=1;
If n = x_0 + 4k ± 2, Pr(x_n=2) = 1;
If n = x_0 + 4k ± 1, Pr(x_n=1) = 1/2 = Pr(x_n=3).
Loraof (talk) 19:29, 20 June 2017 (UTC)
can you please explain how did you derive this? — Preceding unsigned comment added by (talk) 20:42, 20 June 2017 (UTC)
I set it up so after n= x_0 iterations, we are at the value 4 with certainty. After that, every four more iterations in either direction brings us back to the value 4 with certainty. From value 4, two iterations more or less in either direction would leave us at 2 with certainty, whereas from 4 one iteration more or less moves us left or right, and hence to value 1 or 3, with the equal probablilties you gave as 0.5. (Note that I have changed the above slightly so that it applies for n ≥ x_0, since smaller numbers of iterations give a deterministic result.) Loraof (talk) 22:09, 20 June 2017 (UTC)
thanks! — Preceding unsigned comment added by (talk) 19:38, 21 June 2017 (UTC)

June 22[edit]

0^0 other values...[edit]

The standard explanation of the fact that 0^0 is undefined is the fact that lim x->0 of x^0 remains 1 and lim y->0 of 0^y remains 1. Considering this as part of the graph z=x^y, is there an easy way to figure out the line of the form y=kx that approaching (0,0), the limit is 1/2 (to pick a number).Naraht (talk) 13:12, 22 June 2017 (UTC)

The limit of x^(kx) is always 1. This can be found in the exercises of any calculus textbook in the chapter on limits. A good rule of thumb when dealing with limits of the form f(x)^g(x) is to begin by taking a logarithm. JBL (talk) 13:24, 22 June 2017 (UTC)

@Naraht: Take
for any real constant k and you'll get a curve approaching (0, 0) which maintains
Example graph at Wolfram Alpha: link. --CiaPan (talk) 13:40, 22 June 2017 (UTC)
Exponentiation#Zero_to_the_power_of_zero talks about this. In fact entering 0^0 gets there. Dmcq (talk) 14:21, 22 June 2017 (UTC)

Symmetric closure of transitive relation[edit]

Is there a transitive relation whose symmetric closure is not transitive? GeoffreyT2000 (talk) 14:50, 22 June 2017 (UTC)

Do you put any thought into your questions before you post them? If so, perhaps you could adapt your posting style to demonstrate it. If not, perhaps you should. --JBL (talk) 17:17, 22 June 2017 (UTC)
Please either help, or do not; being bitey is wasting everyone's time, and reflects poorly on our reference desks. SemanticMantis (talk) 18:10, 22 June 2017 (UTC)
I think JBL's comment is legitimate. IIRC the OP posts a lot of questions. I haven't followed this closely enough to recognize if there's an offending pattern, but if there is, it's justified to call them out on it. The reference desk does have rules. -- Meni Rosenfeld (talk) 18:23, 22 June 2017 (UTC)
Thank you, Meni. I do not think it is unreasonable to expect repeat users to include a sentence of the form "Here is what I have tried before posting this problem: ...." (In this case, such a sentence would not include "looking at relations on a 2 or 3 element set.") --JBL (talk) 18:41, 22 June 2017 (UTC)
I haven't thought about this stuff in a long time but symmetric closure would seem to preserve transitivity. If R is the transitive relation, and S is one symmetric closure, then you know (aRb & bRc) => aRc, and you know that bSa => (aRb or bRa). From here you need only to work out that (aSb & bSc) => aSc. Does that clear it up? SemanticMantis (talk) 18:10, 22 June 2017 (UTC)
I disagree, I think the answer to the OP's question is positive. I can think of at least three very nice examples. -- Meni Rosenfeld (talk) 18:23, 22 June 2017 (UTC)