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January 31

Harmonic mean problem

The sequence 20, 12, 6, 5, 2, 1 arose in a newspaper puzzle. It has the property that the cumulative HM from L to R is integral, with the specific values 20, 15, 10, 8, 5, 3. Is it possible to construct an increasing sequence without end with the same property?31.54.246.96 (talk) 23:30, 31 January 2015 (UTC)

According to my calculations, 1⋅2, 2⋅3, 3⋅4, 4⋅5, ... has cumulative harmonic means 2, 3, 4, 5 ... . --RDBury (talk) 03:44, 1 February 2015 (UTC)

True, analogous to 1, 3, 5, 7, ... for AM and 1, 4, 9, 16, ... for GM. I wonder if there is an HM sequence for all starting integers apart from 1.→31.54.246.96 (talk) 09:47, 1 February 2015 (UTC)

That GM sequence doesn't work, try 1, 4, 16, 64,... . -- Meni Rosenfeld (talk) 13:09, 1 February 2015 (UTC)

Actually there is one starting with 1: The sequence 1, 3, 5, ... are the cumulative HM's, of 1, -1⋅3, -3⋅5, -5⋅7, -7⋅9. If you're assuming positive integers in the starting sequence then 1 cannot be the first HM, since if the first harmonic mean is 1 then the second is at most 2, preventing it from being an integer. In fact, if the starting sequence is positive and the cumulative harmonic means are a₁, a₂, a₃, ... then a₂ < 2a₁, 2a₃ < 3a₂, 3a₄ < 4a₃, ... . The sequence a, a(2a-1), (2a-1)(3a-2), (3a-2)(4a-3) has cumulative HM's a, 2a-1, 3a-2, ... , so any starting point a > 1 is possible.

A more interesting question (meaning I don't know the answer) is what starting pairs a, b are possible. --RDBury (talk) 14:08, 1 February 2015 (UTC)

February 1

"For all vertices... but at most $\gamma n$ of them"

I am trying to decipher this:

Let $d(x)$ denote the degree of $x$ in $H$, and $d(x,y)$ the number of edges of $H$ containing both $x$ and $y$. Write $\pm \delta$ to mean a quantity between $-\delta$ and $\delta$.

For every integer $r\geq 2$ and reals $k\geq 1$ and $a>0$, there are $\gamma =\gamma (r,k,a)>0$ and $d_{0}=d_{0}(r,k,a)$ such that for every $n\geq D\geq d_{0}$ the following holds:

Every $r$-uniform hypergraph $H=(V,E)$ on a set $V$ of $n$ vertices in which all vertices have positive degrees and which satisfies the following conditions:

1. For all vertices $x\in V$ but at most $\gamma n$ of them, $d(x)=(1\pm \gamma )D$.
2. For all $x\in V$, $d(x)<kD$.
3. For any two distinct $x,y\in V$, $d(x,y)<\gamma D$.

contains a cover of at most $(1+a)(n/r)$ edges.

Specifically, I don't understand the first condition. "For all vertices" and "for at most $\gamma n$ vertices" seem like they contradict each other to me. --superioridad ^(discusión) 05:51, 1 February 2015 (UTC)

It is likely to mean "for all except at most γn of them". YohanN7 (talk) 09:54, 1 February 2015 (UTC)

Oh, as in "all but at most γn vertices". That makes complete sense, thank you.

Resolved

--superioridad ^(discusión) 10:47, 1 February 2015 (UTC)

Fitting cylinders inside a tube

How many cylinders with an external diameter of 34.2 mm can fit inside a cylinder with an internal diameter of 153.6 mm (no overlap allowed)? Roger (Dodger67) (talk) 19:31, 1 February 2015 (UTC)

At least 16, like so (the circles would actually touch each other, but I'm limited in what I can display here):

   O
  O O
 O O O
O O O O
 O O O
  O O
   O

Or maybe 20, like so:

    O
   O O
  O O O
 O O O O
O O O O
 O O O
  O O
   O

Or even 26, like so:

  O O
 O O O
O O O O
 O O O O
O O O O
 O O O O
  O O O
   O O

But I suggest you use MS Paint or some other simple drawing program to try to place them in different arrangements, to see if you can pack more in. StuRat (talk) 19:47, 1 February 2015 (UTC)

FWIW, the relevant articles are circle packing and Packing problem#Packing circles. -- ToE 20:19, 1 February 2015 (UTC)

And Circle packing in a circle. -- ToE 17:09, 2 February 2015 (UTC)

Packing problem#External links points to The best known packings of equal circles in a circle (complete up to N = 2600) which gives densities which should help bound your answer. If you are able to pack n circles of diameter 34.2 inside a circle of diameter 153.6, you will have achieved a density of (34.2²/153.6²) n ≈ 0.0496 n, so your [Stu's] lowest guess of 16 would give a density of 0.793, while the optimal [best known] density for 16 circles is only 0.751. Likewise, fitting 15 would give you a density of 0.744, greater than the optimal [best known] density of 0.734. But you should be able to fit 14 because that gives you a density of only 0.694, far less than the optimal [best known] density of 0.747. So the answer is 14. (And they should rattle a bit.) -- ToE 21:10, 1 February 2015 (UTC)

Something seems wrong there, as the 16 circles in a diamond arrangement shown should be at most 4 small circle diameters across at any point, and 4×34.2 < 153.6. StuRat (talk) 21:18, 1 February 2015 (UTC)

How can four circles in a row touch going across horizontally and leave room in the middle for four in a column to touch going down vertically? -- ToE 21:24, 1 February 2015 (UTC)

Looks like I miscalculated by using the wrong type of symmetry for hexagonal packing. Specifically, I was thinking that a 90 degree rotation would produce the same 4 touching circles vertically, just like the 4 circles touching horizontally, but a hexagonal packing arrangement only allows for rotations in multiplies of 60 degrees, which excludes 90 degrees. StuRat (talk) 21:30, 1 February 2015 (UTC)

Yeah, hexagonal packing is optimal for the plane, giving a density of π/(2√3) ≈ 0.907. This density is surpassed when packing circles for n = 1 (with a density of unity), but I'm fairly sure (though I don't know how to prove it) that the optimal density for packing a circle will alway be less than this for n > 1, but will approach this value for large n. -- ToE 22:05, 1 February 2015 (UTC)

Note that the optimal solution is only exactly known for n≤13 and n=19, so if you do figure out a way to fit 15 you should write a paper on it. -- ToE 21:16, 1 February 2015 (UTC)

The site cited by ToE gives the ratio of the big radius to the small radius for each packing shown, which you want to be no more than 4.491228. The best known packing of 14 has ratio 4.328, and the best known for 15 has ratio 4.521; so your answer is 14, so far as is now known. —Tamfang (talk) 21:44, 1 February 2015 (UTC)

And I just noticed that the site even has a calculator which answers this sort of question with both numbers and a drawing, thanks to Eckard Specht at the Otto-von-Guericke University Magdeburg. (I wonder what the relation is to the German physicist Eckehard Specht at the same institution.) -- ToE 21:52, 1 February 2015 (UTC)

I find it amazing that so little is known with certainty. Makes me wonder if there is a grid computing project for finding better packings. -- Meni Rosenfeld (talk) 14:39, 2 February 2015 (UTC)

Good question. Two things I wonder are: A) Is there an easy proof that the density of packing n > 1 circles in a circle is always less than the hexagonal packing density in a plane? I though it seemed obvious at first, but I don't know where to start with a proof. B) For those n for which a proof is not know for the optimal density, but for which a strong lower bound is known for the density by way of a known (and perhaps conjectured optimal) packing (such as those listed in the external site), what techniques can be used to determine an upper bound for the density? My remark about Dodger67 publishing a paper if he manages to fit in a 15th cylinder was tongue-in-cheek, but I don't know how to find out if there is an upper bound on the density for n = 15 which, while being above the achieved 0.734, is below his required 0.744, in which case 15 cylinders are known to be impossible for the particular dimensions of his cylinders. -- ToE 17:09, 2 February 2015 (UTC)

One arrangement of 14, BTW, is like so:

  O O
 O O O
O O O O
 O O O
  O O

That's my original 16, but without the top and bottom circles. StuRat (talk) 21:59, 1 February 2015 (UTC)

February 2

Quiz Question

In a newspaper in the UK there was a quiz question that stumped me, I'd be grateful for any hints.

The question is: In one season of a football (soccer) league, one sixth of the goals were scored by defenders, two and a half times as many strikers, eighteen were own goals, and the remaining quarterby midfielders. If none were scored by goalkeepers, how many were scored by strikers? I added 1/6th to 5/12ths and thought that the total ( including the 18) would represent three quarters of the total and tried to solve for n, but got it wrong. What I can't understand is the own goals. As the could be scored by anyone, I can't see how you can write an equation that allocates them to any category. The answer is 45. Could anyone tell me how to solve it, many thanks.Widneymanor (talk) 10:21, 2 February 2015 (UTC)

I think you had the right method, but made an error in your arithmetic somewhere (or ended up looking for the wrong value). Since the soccer team contains only defenders, strikers, midfielders, and goalkeepers, own goals must not be counted in the other totals (possibly because they aren't thought of as the player "scoring"). 3n/4 - 1n/6 - (2.5)n/6 = 9n/12 - 2n/12 - 5n/12 = 2n/12 = 18, so n/12=9, so 5n/12=45 (the number scored *by strikers*) MChesterMC (talk) 10:42, 2 February 2015 (

Thank you very muchWidneymanor (talk) 13:34, 2 February 2015 (UTC)

Probability of cashing in

If I place a $1 bet on a six horse accumulator each day, and I always bet on the favourites, given that a favourite horse will win about 1/3 of the time, how many days will I have to bet before I win?

(I know statistics doesnt work like that but I'm not sure how to phrase the question - is it how many days do I need to bet before my chances of winning become 1? I know I could still bet forever and not win but I hope you understand the thrust of the question.) 5.148.151.18 (talk) 10:45, 2 February 2015 (UTC)

A good question to ask is what is the expected number of times to bet until you win. The answer is 3.

A different question is how many days to bet until your probability of winning at least one is greater than $p$. The answer is ${\frac {\log(1-p)}{\log(2/3)}}$, rounding as necessary. -- Meni Rosenfeld (talk) 12:34, 2 February 2015 (UTC)

Thanks - although I think if there was one horse with 3.1 odds I would have to bet 3 times, but this is a six-horse accumulator, meaning each horse needs to win, each with a 33% chance of winning the race. 5.148.151.18 (talk) 13:15, 2 February 2015 (UTC)

I forgot to mention that I have no idea what "six horse accumulator" means. I assumed that each day you bet on a horse, each day you have a probability of 1/3 to win, and the different bets are independent of each other. -- Meni Rosenfeld (talk) 14:38, 2 February 2015 (UTC)

Until we hear back on what an accumulator is in this sense, here's some WP articles that relate to the problem at hand: Martingale_(betting_system), Martingale_(probability_theory), Gambler's_fallacy, Gambler's ruin, Mathematics_of_bookmaking, Horse_racing#Types_of_bets. SemanticMantis (talk) 15:20, 2 February 2015 (UTC)

Apologies, you choose 6 horses and place a $1 bet. If the first one wins, the money won rolls to the second etc. If they all win, you get the payout. Assuming each horse has a 1/3 chance of winning, what is the expected number of days I need to bet in order to win? (assuming I have phrased that correctly, see above) 5.148.151.18 (talk) 16:43, 2 February 2015 (UTC)

If there is a six horse accumulator bet then the horses selected in each of six races must win for the bet to pay off. If each horse has a probability of winning of 1/3 then the probability that all six will win is (1/3)^6, which is 1/729. That is, this bet will pay off successfully, on average, once every 729 days. Further, if you start following this strategy today the "expected" number of days before your first win will be 729 days. What mathematicians mean by "expected" is "the average number of days you'll have to wait"; you might get lucky and get your first win before this time, or you might be unlucky and have to wait much longer, but "on average" you'll have to wait 729 days. RomanSpa (talk) 16:52, 2 February 2015 (UTC)

Also note that there is no number of bets that will guarantee a win. The more bets you make, the better your chance of having a win, but that chance never gets to 100%. And, by the time you do win, you are likely to have lost more money than you won. StuRat (talk) 16:57, 2 February 2015 (UTC)

Pólya Problem to Prove (Part I Chap 19)

Pólya's hypothetical teacher poses the following problem to prove to the hypothetical student: Two angles are in different planes but each side of one is parallel to the corresponding side of the other, and also has the same direction. Prove that such angles are equal. Through some work, he introduces the auxiliary problem to solve, adding an extra line segment, turning the angles into triangles, and going to prove that the triangles are congruent, but it gets to where the "teacher" says "Did you use the hypothesis? What is the hypothesis?" And the "student" says "We suppose that AB || A'B', AC || A'C'. Yes, of course, I must use that." But then the "teacher" says: "Did you use the whole hypothesis? You say that AB || A'B'. Is that all you know about these lines?" and the "student" says: "No; AB is also equal to A'B', by construction. They are parallel and equal to each other. And so are AC and A'C'."

That last statement by the "student" is what I don't get from the original problem statement. I copied the original problem statement verbatim in full. From that first italicized bit of text at the beginning of my question, where is it to be gotten that any line from one angle is also equal to any other by construction? It seems to me only to say the angles are in different planes, and that corresponding sides are parallel and in the same direction. I don't see anything about construction or how I can assume lengths are equal from that statement. Can anyone explain? 20.137.2.50 (talk) 18:34, 2 February 2015 (UTC)

Here's my attempt to reconstruct (in outline) the proof Pólya has in mind; not sure that it explains the statement but it can't hurt. First prove a lemma (*) that if AB = A'B' and AB || A'B' and in the same direction, then AA'B'B is a parallelogram. This should be easy, even if you don't assume the points are in the same plane. Note the lemma is false if AB and A'B' are in opposite directions; you get a self intersecting quadrilateral.

We are given two angles A, A' with corresponding sides parallel and in the same direction. Add points B, B', on corresponding sides so AB = A'B' and C, C' on the other sides so AC = A'C'. AB || A'B' and AC || A'C' so AA'B'B and AA'C'C are parallelograms by (*). Then BB' = AA' = CC' and BB' || AA' || CC'. Apply (*) again to get BB'C'C a parallelogram. This gives you BC = B'C' and, with AB = A'B' and AC = A'C', ABC ≝ A'B'C'. --RDBury (talk) 17:19, 3 February 2015 (UTC)

percentage

Is 0.3408 % 0.0408 percent than 0.3 % or 13.6 % greater? — Preceding unsigned comment added by 199.119.235.181 (talk) 18:56, 2 February 2015 (UTC)

It depends on the base:

0.3408% is 0.0408% of 100 more than 0.3%

0.3408% is 13.6% of 0.3 more than 0.3%

Now, where would you use each ? Let's say the interest rate on your checking account just went up from 0.3% to 0.3408%. The bank might claim that's a 13.6% increase, but you will only make .0408% more on your money, so I'd use that figure. In other words, you're still not getting much. StuRat (talk)

So if 0.3 % of population group A has a disease, and 0.3408 % of group b has a disease, group b has 0.0408 percent more people with disease than group a. — Preceding unsigned comment added by 199.7.157.123 (talk) 23:01, 2 February 2015 (UTC)

February 3

MATHEMATICS

03/02/2015. 1.1 Use the substitution method to solve the following simultaneous equation. 1.2 b=2a 6a-b=8. 1.3 q=p-5 p+2p=20. 1.4 2x+3y=2 x-y=1. 1.5 3x-y=1 x-2y=-8. — Preceding unsigned comment added by 2A03:2880:3010:6FF6:FACE:B00C:0:1 (talk) 05:16, 3 February 2015 (UTC)

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. --Kinu ^t/_c 05:21, 3 February 2015 (UTC)

Our article System of linear equations may be helpful. Bo Jacoby (talk) 12:23, 3 February 2015 (UTC).

Work the problem even if you're not sure what to do. Keep trying. When you get an answer, you can check to see if you're correct by plugging your solution values back in to the original equations, and seeing if they evaluate to true expressions. If not, start over. It's easier to do something correct from scratch than to catch a mistake you made in a page full of calculations. Also consider asking a classmate for help, or your instructor. If this is a college level class, your professor and TA should both have office hours. Math is much more fun and rewarding if you can work in a group, so do engage with your classmates too. 14:39, 3 February 2015 (UTC)

I have done all three questions. You can have my solutions for $40 dollars per hour. I spend 1 hour solving it. Alternatively you can solve it yourself. 175.45.116.65 (talk) 00:45, 4 February 2015 (UTC)

I have done all four questions. You are welcome to buy my solutions. I charge $300 dollars per hour for this work, but since it only took me 6 minutes the cost to you is $30 dollars. This is cheaper than the previous solver. In addition, you can be more confident in my work for the following reasons: first, I counted the number of questions you posted correctly (four, not three); and, second, the previous solver does not appear to have spotted the typing error in question 1.3. I have three different solutions for question 1.3, depending on whether you wish to stick with the question as you originally typed it, or which of the two possible obvious corrections to the typo you wish to use. As further proof that I am a better and more reliable solver than the other guy (as well as cheaper!), I will tell you that I expect the correct version of problem 1.3 to be: q=p-5 p+2q=20. As further evidence of my efficiency and reliability in diverse fields, I note that I have made more than 10,000 edits to Wikipedia, while my rival has made only 3. I am also prepared to furnish you with notarised copies of my degree certificates, which confirm that I am (inter alia) a mathematics graduate from one of the world's greatest and most well-known universities, and studied at a college within that university noted for its mathematical prowess. I am prepared to compensate you for losses incurred in the use of my solutions, to a maximum of $250,000, should my solutions be found to be incorrect. Finally, as a special bonus, each purchase of solutions from me will include not just the solutions, but also your choice of an elegant six-piece set of wine glasses or a small cuddly teddy bear suitable for children between the ages of 7 and 15. RomanSpa (talk) 03:28, 4 February 2015 (UTC)

Thus does capitalism work, and thus is a new service industry born... RomanSpa (talk) 03:31, 4 February 2015 (UTC)

Alternatively, and cheaper still, you might solve the equations yourself. I promise you that they're not difficult once you try; just carefully follow the instructions given to you by your teacher, and you'll be fine. RomanSpa (talk) 03:34, 4 February 2015 (UTC)

Will you also charge for the time it took you to write down this offer? It probably took more than 6 minutes. -- Meni Rosenfeld (talk) 14:49, 4 February 2015 (UTC)

February 4

Cubes of the form 3 x² ± x y + 5 y² with x, y coprime

Are there any cubes of the form $3x^{2}\pm xy+5y^{2}$ with x and y coprime ? I've found no solutions for $|x|,|y|\leq 10^{3}.$ — 79.113.247.229 (talk) 13:26, 4 February 2015 (UTC)

Work through the method here- I think you'll find that x must be 0 for any 3x^2 +- x*y + 5*y = k^3- so no. DTLHS (talk) 20:14, 4 February 2015 (UTC)

Characteristic polynomial without determinant

The usual definition of the characteristic polynomial of an endomorphism A of a finite dimensional vector space uses the determinant. I faintly remember that, way back in time at university, there was an approach that didn't need neither determinants nor eigenvalues. I can't find anything on the net, and I can't figure it out by myself.

As far as I remember, the trick was to decompose the vector space into subspaces that were stable under the endomorphism (corresponding to the Jordan blocks, but those were not yet defined at this stage of deduction).

It worked roughly by taking a basis b1, ..., bn, then computing b1 A^0, b1 A^1, ... until the set of vectors was no longer linearly independent. Then construct a polynomial from the coefficients of b1 A^k= sum ai b1 A^i, then find a new basis containing b1 A^0 to b1 A^(k-1) for the first subspace. Continue iteratively for the subspace consisting of the remaining vectors of the basis. Finally, multiply all the polynomials together.

Anyone out there who has seen such a thing? 93.132.26.165 (talk) 13:51, 4 February 2015 (UTC)

You could get the matrix into say, upper triangular form, read off the eigenvalues, and construct the characteristic polynomial from the eigenvalues. Down with Determinants! is a fun paper going into the math behind a determinant-free approach. --Mark viking (talk) 14:12, 4 February 2015 (UTC)

I had seen that before. That guy uses the concept of eigenvalues and, even worse, restricts himself to vector spaces over the complex numbers. The above approach would work for any field. Eigenvalues, a volume function (p(0)(-1)^n = det A) and Cayley-Hamilton would be for free, based solely on the concept of vector spaces and linear independence. 93.132.26.165 (talk) 14:52, 4 February 2015 (UTC)

What you're doing is constructing the characteristic polynomial from the cyclic subspaces of the matrix. Off the top of my head, I would check Hoffman and Kunze "Linear algebra". They emphasize the cyclic subspace perspective fairly heavily. Sławomir Biały (talk) 14:17, 4 February 2015 (UTC)

That's in the line of the thing ... but I'm fairly decadent ... . I was hoping for something on the internet. Well, probably I will step out into the cold and get the book, in physical form, from the library ;-) . 93.132.26.165 (talk) 14:52, 4 February 2015 (UTC)

You might want to check out Advanced Linear Algebra by Steven Roman (GTM 135) as well. The characteristic polynomial is defined as the product of the elementary divisors, and cyclic subspaces are clearly involved. (Disclaimer: I haven't read the chapter, but at least determinants are nowhere in sight.) YohanN7 (talk) 16:15, 4 February 2015 (UTC)

Thanks, this is even available online. 488 pages. This will take some time. 93.132.26.165 (talk) 16:43, 4 February 2015 (UTC)

Ouch, this starts with theorem 0.1 stating det A^t=det A. .... So determinants are a prerequisite, not something to be derived. Standard approach, not what I am looking for. 93.132.26.165 (talk) 17:28, 4 February 2015 (UTC)

Identifying a vertex that is in every maximal matching of a bipartite graph

A bipartite graph is never factor-critical, and so must have a vertex v that is in every maximal matching. Is there any characterisation of such vertices/way to identify them? --2404:2000:2000:5:0:0:0:C2 (talk) 02:02, 5 February 2015 (UTC)

Actually simply identifying whether one of the partitions contains such a vertex would be sufficient, if that's possible.