Wikipedia:Reference desk/Archives/Mathematics/2012 May 9

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May 9

Number Theory / Set Intersection

Suppose we have n sets:

${\begin{aligned}&{\text{Fix: }}q_{1},...,q_{n}\in \mathbb {Z} {\text{ and }}0\leq r_{1}<q_{1},...,0\leq r_{k}<q_{k}\\A_{1}&=\{m_{1}q_{1}-r_{1},...,m_{1}q_{1},...,m_{1}q_{1}+r_{1}\},\\A_{2}&=\{m_{2}q_{2}-r_{2},...,m_{2}q_{2},...,m_{2}q_{2}+r_{2}\},\\...,\\A_{n}&=\{m_{n}q_{n}-r_{n},...,m_{n}q_{n},...,m_{n}q_{n}+r_{n}\},\\&m_{1},...,m_{n}\in \mathbb {Z} \\&\forall x\in A_{i}\cdot x<{\text{lcm}}(q_{1},...,q_{n})\end{aligned}}$

Note that the last condition makes the sets finite and can be modified if it makes formulas simpler.

I'm interested in computing the cardinality of the intersection/union of these sets (i.e $|A_{1}\cup A_{2}\cup A_{3}|$) with a formula, with the ultimate goal of determining if $|A_{1}\cup A_{2}\cup ...\cup A_{n}|<{\text{lcm}}(q_{1},...,q_{n})$. Does anyone have any idea how I might go about doing this outside of doing an exhaustive check? Nkot (talk) 00:48, 9 May 2012 (UTC)

What use are the ⋛ and ⋚ symbols?

I came across the Unicode symbols LESS-THAN EQUAL TO OR GREATER-THAN (⋚) and GREATER-THAN EQUAL TO OR LESS-THAN (⋛). I was wondering what use these are in mathematics, and what the difference is. As a non-mathematician, A ⋚ B and A ⋛ B both seem to say "B could have any value irrespective of A. -- Q Chris (talk) 10:40, 9 May 2012 (UTC)

Exactly. Plasmic Physics (talk) 11:03, 9 May 2012 (UTC)

And that might mean A and B are comparable somehow. I guess this may be of use for example in partial orders, where some (pairs of) elements can be in 'less-than, equal, or greater-than' relations and some can not. --CiaPan (talk) 11:56, 9 May 2012 (UTC)

Those symbols could be used to answer the question "what is the relationship between A and B ?". Although I'm reminded of TV ads which say "We will loan you up to $1000 or more !". StuRat (talk) 16:37, 9 May 2012 (UTC)

Good question and I don't know. However in computing Not a number which is implemented in practically all floating point nowadays is not less than equal to nor greater than any number including NaN itself. In Perl <=> is an operator returning -1,0 or 1 depending on the the relation between the two sides or undef if NaN is used. Why the second symbol is needed as well is also a mystery to me but if I was doing these things I'd have instead put a diagonal stroke through the first symbol to produce NOT LESS-THAN AND NOT EQUAL TO AND NOT GREATER-THAN ;-) Dmcq (talk) 17:12, 9 May 2012 (UTC)

apples ⋚̸ oranges? I wouldn't put it past them, considering the surfeit of naïvely synonymous and possibly unused symbols of this type, e.g. {≠,≶,≷} or {=,≸,≹} or {<,≱,≨}, not to mention {⋛,⋚,⪑,⪒,⪓,⪔}. I bet the real reason was someone in the Unicode organization having a private joke at the expense of everyone else's puzzlement. — Quondum^☏ 18:12, 9 May 2012 (UTC)

Following up on apples ⋚̸ oranges, could we then say that McIntosh ⋚ Red Delicious?    → Michael J Ⓣ Ⓒ Ⓜ 15:55, 10 May 2012 (UTC)

Can I go for a quantum superposition of the two signs, ½ | ⋚ > + ½ | ⋚̸ > perhaps? Dmcq (talk) 16:44, 10 May 2012 (UTC)

Often these symbols are used in sentences like "a⋚b as c⋛d", which is shorthand for three "if and only if" statements: "a < b if and only if c>d" ; "a=b if and only if c=d"; and "a>b if and only if c<d". Duoduoduo (talk) 02:27, 12 May 2012 (UTC)

When are calculators allowed in math examinations and education?

If the question in an exam is 0.345 * 0.435, I understand that calculators should not be allowed, since you are supposed to show that you know how to multiply decimal numbers. However, are non-programmable calculators always acceptable around the world when tackling other kind of problems? For example, take this questions: if you one of three coins three times, and one of them if unfair and tends to 0.69% of heads, the rest is fair. What is the possibility of obtaining 10 heads in a row? The teacher here could like to know if you get the concepts behind the calculation, but doesn't want to test your calculation abilities, so, would here normally be calculators allowed? — Preceding unsigned comment added by OsmanRF34 (talk • contribs) 13:08, 9 May 2012 (UTC)

Yes, in general, once you've proven your proficiency in basic math, they will allow you to use a basic calculator for that. However, not everyone follows this rule. StuRat (talk) 16:34, 9 May 2012 (UTC)

Usually it's possible for the teacher to construct problems that can be solved without having to do any difficult arithmetic. For example instead of using a probability 0.69, they could have a nice fraction like 2/3. They may also allow students to leave their answer unsimplified on questions where simplifying would require a lot of arithmetic. If there's one thing I hate to see as a math grader it's an answer to a problem written as an approximate decimal. Besides being ugly, it's harder to figure out what the person was doing, and if the answer is wrong, where they made a mistake. Rckrone (talk) 19:04, 9 May 2012 (UTC)

Its common to have two papers one which allows calculators and one which doesn't. This way you can test basic numerical skills as well as correct usage of a calculator. There are many parts of mathematics, say trig, where a calculator is essential except for specially chosen examples. --Salix (talk): 22:35, 9 May 2012 (UTC)

Question about the harmonic series

Okay, I have read this and basically it claims that, while the harmonic series is divergent when summed, if you remove certain terms from it, the sum converges. Their example is removing all the terms with the digit "9" in them. Basically, my question is: if the sum of the harmonic series is divergent, and then the sum of the harmonic series with all the terms with "9" in them removed becomes convergent, does that mean that the sum of the harmonic series with all terms except with "9" in them (i.e. the sum of all the terms we removed) is divergent?

I'm saying yes, because of the following reasoning: Let A be the harmonic series, let B be the harmonic series with all the digit "9" terms removed, and let C be the inverse series of B (AKA instead of removing the digit "9" terms, remove all the other terms instead). A = B + C. The sum of 2 convergent series is also convergent. If we assume that C's sum is convergent, then we have B and C both be convergent, therefore A is also convergent. But A is divergent because it's the harmonic series and we already know it's divergent. So therefore C is divergent.

Now, I want to take the concept further. Let's take the series C again, and this time, let's remove all the terms of C that contain the digit "8". Would this new series, D, be convergent? And would D's inverse, E, be divergent?

...

If E is divergent, can we do the same thing to E, but this time with all the digit "7" terms? And then take E, remove the digit "7" terms, get series F, invert it for series G.... can we do this an infinite number of times to get an infinite number of convergent series? Then, if we add all these convergent series, we'd get the harmonic series, a divergent one. So is it usually the case that a divergent series is the sum of an infinite number of convergent series?--24.130.151.188 (talk) 13:17, 9 May 2012 (UTC)

Summing an infinite number of series is different from decomposing a series into an infinite number of subseries (which prompts the question, is it even possible?). It's always the case that there exist an infinite number of convergent series equal to a divergent series, almost by definition. Consider your proposed convergent series of harmonic series terms with no 1s, 2s, ... 34s, 35s,.... and you will see it is not a decomposition of the harmonic series but has many repeated terms in the union, as it were. 208.38.20.91 (talk) 17:40, 9 May 2012 (UTC)

Yes, you can do this. Since the terms are all positive this simplifies things, as whether a series converges or not depends only on the entries in it and not their order. It's always easy to find some infinite subset $A\subseteq \mathbb {N}$ such that the sum of only the terms with indices in A converges, and there's not problem finding an infinite sequence of such subsets which are mutually exclusive and their union is $\mathbb {N}$.

Some thought should be given though how your particular proposed extraction continues. -- Meni Rosenfeld (talk) 09:27, 10 May 2012 (UTC)

geometry problem

Click to enlarge

We have square ABCD. We lop off identical areas DHG and EBF on the left and on the right and we do so at 135 degree angles. I wish to place 2 straight lines IJ and KL within resulting form HAEFCG such that the four resulting interior shapes will be equal in area. They are labeled as areas One, Two, Three, and Four. I know the amount of square units of area that I am aiming for. How do I determine the placement of lines IJ and KL? I can measure, with a ruler, lengths AE, FC, CG, and HA—which are of course all equal. And I can also measure, with a ruler, lengths EF and GH. Bus stop (talk) 16:20, 9 May 2012 (UTC)

The obvious solution, if you are free to measure with a ruler, is just to measure the location of those lines with a ruler, is it not ? StuRat (talk) 16:30, 9 May 2012 (UTC)

How? I need to know the placement of lines IJ and KL. I don't believe lines IJ and KL will intersect with one another at a 90 degree angle at the center. Nor will lines IJ and KL be at midpoints of any other lines. I am aiming for a certain number of square units that I know in advance. After lopping off the arbitrary but equal areas on the left and on the right, my sole concern is the resulting geometric shape HAEFCG and the four areas of which it's comprised. Bus stop (talk) 16:33, 9 May 2012 (UTC)

Oh, I thought lines IJ and KL already existed (it's customary to draw "proposed lines" with dashes, not solid lines). StuRat (talk) 17:39, 9 May 2012 (UTC)

You are right. Sorry about that. Bus stop (talk) 22:30, 9 May 2012 (UTC)

I presume you are free to calculate. Assuming you want to keep the result symmetrical (it does not have to be), each of the points I,K,L,J moves from the centre of the side of the square $EB\cdot FB \over 2CB$ (in the direction of points A and C as appropriate). — Quondum^☏ 17:27, 9 May 2012 (UTC)

Basically you start off with IJ KL perpendicular to each to each other when the area of the triangles DHG and EBF is zero. Say they cross at O and these original IJKL are ijkl. For non-zero triangles you then have to angle them away from the triangles so the area of AIOK for instance has one half of EBF taken from it. Then the area of the triangle OKk is one quarter that of EBF, or one half of that got by splitting the triangle by a line from O to B. You then have to move that triangle around to get sides along Ok and AB and halve the base. Dmcq (talk) 18:48, 9 May 2012 (UTC)

If you want the symmetric solution where AI = AK = CL = CJ then let DH = d, the side length of the square be s, and x = AI = AK = CL = CJ be the value you want to calculate. The total area you chopped off is d², and area one is (s/√2)(x/√2) = sx/2, so you want 4(sx/2) + d² = s². Then x = s/2 - d²/2s. Rckrone (talk) 19:25, 9 May 2012 (UTC)

If you only have lengths AE and EF, then you can use d = EF/√2 and s = AE + EF/√2, and you can substitute those values back into x = s/2 - d²/2s. Rckrone (talk) 19:36, 9 May 2012 (UTC)

These are the actual numbers:

• The four "Areas" have to be 67.5 square inches each. (That implies that the total area of HAEFCG is 270 square inches.)

• I know that the lengths of HG and EF are 6 inches each.

• For AH, AE, CG, and CF, I have measured with a ruler and I have determined that they are approximately 12.5 inches each.

I'm having trouble doing the math. Where along those 12.5 inch lines would lines IJ and KL terminate?

I thank you all in advance and I'm sorry I can't figure this out for myself. Bus stop (talk) 22:26, 9 May 2012 (UTC)

If the area of all four regions is 270, and HG and EF are 6, then the square has area 288, so it has side s = 12√2. Then AH is 9√2 ≈ 12.73, but that's not really relevant since we already have the side length. Anyway, x = s/2 - d²/2s = 6√2 - 18/24√2 = 45√2/8 ≈ 7.95, so the ends of IJ and KL are about 7.95 inches from A and C. Rckrone (talk) 00:57, 10 May 2012 (UTC)

Thanks a lot. That is great. I really appreciate it. It is actually for a painting I am working on. I couldn't proceed without this information. Thanks again. Bus stop (talk) 03:26, 10 May 2012 (UTC)

p-adic integration

I don't know whether integral calculus can be carried on a field of p-adic numbers. --84.61.181.19 (talk) 21:14, 9 May 2012 (UTC)

The short answer is "yes". A slightly longer answer is that, when it is equipped with its usual topology induced by the p-metric, the p-adic field is a locally compact topological field. So there is a unique Haar measure normalized so that the ring of p-adic integers (which is compact and open in the p-adic field) has measure 1. As a result, the Lebesgue integral is well-defined. A good chunk of modern number theory requires integrals over p-adic fields to be well-defined. This goes back at least to the thesis of John Tate from the 1940's. Sławomir Biały (talk) 23:20, 9 May 2012 (UTC)

I don't know whether p-adic valued functions can be integrated. --84.61.181.19 (talk) 07:55, 10 May 2012 (UTC)

Yes, they can (as Sławomir Biały said above). If you want the technical details, googling "p-adic integration" throws up texbook and lecture notes links like this and this. Gandalf61 (talk) 11:13, 10 May 2012 (UTC)

Actually, it is namely integration over the Haar measure which is the principal operation in p-adic analysis, because there is no differential operators. All usable operators on functional spaces are pseudodifferential (i.e. require an integration), like Vladimirov's operator. Incnis Mrsi (talk) 11:33, 10 May 2012 (UTC)

Someone should mention p-adic measure (apparently it is redirected to p-adic distribution (not the other way around).) -- Taku (talk) 02:57, 11 May 2012 (UTC)

Time before ruin

A player starts with x chips and plays a game where in each move he tosses a weighted coin and depending on the results of coin toss wins 1 chip with probability p (p<0.5) or loses 1 chip with probability 1-p. He continues the game until he loses all his chips. Since p<0.5 the risk of ruin after infinite amount of moves is obviously 100%. But how do we calculate the average amount of moves the player will make before he busts? 93.136.193.231 (talk) 22:55, 9 May 2012 (UTC)

Let f(x) be the expected number of moves to go bust if the player has x chips. After 1 move, with probability p the player will gain a chip and then have expected bust time of f(x+1), and otherwise will lose a chip and have an expected bust time of f(x-1). Therefore f(x) = 1 + pf(x+1) + (1-p)f(x-1), which gives us a recurrence relation we can use to solve for f. We also have that f(0) = 0 as an initial condition. Solving for f(x) gives f(x) = x/(1-2p) + c(((1-p)/p)^x - 1) for some constant c. I'm pretty sure the right answer is where c = 0 so f(x) = x/(1-2p), but I seem to be missing some additional constraint. Maybe someone can help me out here. Rckrone (talk) 01:41, 10 May 2012 (UTC)

Starting with x₀ chips, after n plays the player has x_n = x₀ − n(1−2p) ± √n(1−2p)2p chips. He busts when x_n = 0. Solving for n gives n = x₀/(1−2p) ± √n2p/(1−2p), and the average amount of moves the player will make before he busts is x₀/(1−2p) . Bo Jacoby (talk) 05:56, 10 May 2012 (UTC).

Thanks for the help guys. I'd also like to know how to calculate the probability of player starting with x chips busting after making at least n moves. 64.94.77.175 (talk) 11:17, 10 May 2012 (UTC)

The distribution of x_n has mean value x₀ − n(1−2p) and standard deviation √n(1−2p)2p. The distribution is approximately normal when n is not very small. The inequality x_n<0 means that x₀ − n(1−2p) ± √n(1−2p)2p < 0 or ±1 < −(x₀ − n(1−2p))/√n(1−2p)2p. The probability for this to happen is Φ(−(x₀ − n(1−2p))/√n(1−2p)2p) where Φ is the Normal distribution#Cumulative distribution function. Bo Jacoby (talk) 13:41, 10 May 2012 (UTC).

A slight error in the above; the standard deviation is √n(1−p)p. — Arthur Rubin (talk) 20:00, 16 May 2012 (UTC)