Content deleted Content added

Inline

Revision as of 04:05, 9 October 2012

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

Shortcut

WP:RD/MA

Want a faster answer?

Main page: Help searching Wikipedia

How can I get my question answered?

Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
Post your question to only one section, providing a short header that gives the topic of your question.
Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
Don't post personal contact information – it will be removed. Any answers will be provided here.
Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
Note:
- We don't answer (and may remove) questions that require medical diagnosis or legal advice.
- We don't answer requests for opinions, predictions or debate.
- We don't do your homework for you, though we'll help you past the stuck point.
- We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

September 30

Bin packing problem

Hi all,

for my webshop I need some kind of algorithm, which given the size and weight limits of postal packages and the sizes/weights determines the best (cheapest) way to distribute the goods into postal packages. Does anyone know a solution for that problem? The stuff on google is filled with university-level math and allows only for one standard "bin" size... 85.181.101.47 (talk) 12:55, 30 September 2012 (UTC)[reply]

You're going to need to define the problem precisely. I assume more than one item is allowed per package, but is there a maximum number ? Do we have a table of package dimensions and shipping rates, and another of items dimensions ? Does weight matter ? Are there any rules about what must be in what packages (like refrigerated items in insulated packages) ? If the total number of items and packages is low, I suggest a brute-force method, where you try every combo to determine which is best. StuRat (talk) 16:23, 30 September 2012 (UTC)[reply]

Yes, there is more than 1 item per package, but no maximum limit (although there will be a practical one ;)); there is a table of package dimensions and the associated rates, and the list of items to be sent contains the size/weight data of each element. There's no rules about combinations of items and packages, and the orientation of goods doesn't matter in nearly all cases, so it can be left out. 85.181.101.47 (talk) 18:06, 30 September 2012 (UTC)[reply]

If they give weights for each item, that implies that there's a weight limit on each package to consider. Is there ? It would help if you could give us the exact wording of the problem, as slight variations in the setup can change what method is optimal. But, to outline a general method, you might want to sort items from largest to smallest, and place the largest item first, in the smallest package which will accommodate it, then place the second largest item in the smallest package in which it will fit, and continue through the list, trying to pack items along with other items as you go. This isn't guaranteed to give you the optimal solution, but should be a "good" one. If there's no maximum number, it may not be possible to always get the optimal solution, as calculating it would take an infinite amount of time. You might also want to create a hybrid program, which, when the number of items is low enough, does the brute-force calculations to find the optimal solution, but, when the number is too high for that, uses the method I just outlined. StuRat (talk) 04:00, 1 October 2012 (UTC)[reply]

Hard to give advice without an example of the table and typical items; I looked at http://pe.usps.com/text/dmm300/Notice123.htm but there are way too many options. If all your items come in boxes that should stay in pristine condition, and you want a way to stack them so the amount of empty space is minimised, I'm not sure if any algorithm can help you much with that. A computer controlled robot hand can stack things in order but filling a box by hand with say 40 rectangular items, based on a print-out of a 3-D arrangement seems quite challenging, and computing time might be excessive.

If you're sending items the way most Chinese webshops do it, everything wrapped in plastic foil, where the sum of the volumes is a good enough measure to know if they would fit a box, then it becomes more a knapsack problem (for volume or weight).

Usually price per unit of volume or weight is lower for larger parcels, so you'd want to use as few parcels as possible.

One possible approach would be: Start with the total volume plus (or minus) a certain percentage based on experience, find cheapest combination of parcels for the volume needed, check if largest items will fit in largest parcel (for long thin items, putting them diagonally might be an option), then use a knapsack algorithm based on the volume to fill the parcels, starting with smallest parcel (only using items whose length would fit). Verify if weight limits are respected.

In it's most general, it's a bounded multiple constraints (with m=2) multiple knapsack problem (and that doesn't include the item length), but the details would determine the best practical way to go about it. Sites like paypal offer solutions for calculating shipping costs but you'd have to integrate their system in your website I suppose. Ssscienccce (talk) 09:16, 1 October 2012 (UTC)[reply]

October 1

http://en.wikipedia.org/wiki/Ellipse#General_polar_form

In the section of this article "General polar form", the equaitons are given for expressing the polar form of the equation for an ellipse if we know a few parameters like the semi major and semi minor axes etc. To this end, the equation is given defining Q(theta). This equation contains the square root of the difference of two terms. Those terms are R(theta) and another term that varies independently of R(theta).

R(theta) is itself the sum of two terms, one of which includes (b**2 - a**2) and the other is (a**2 + b**2).

Since a and b can be arbitrarily close to one another, the first term involving (b**2 - a**2) can be arbitrarily small. also, a and b can be arbitrarily small, so that (a**2 + b**2) can be arbitrarily small. Consequently, R(theta) can be arbitrarily small.

Returning our attention to the square root term above, it seems that the argument under the square root can become negative since R(theta) can be arbitrarily small and the term subtracted from it can be positive and larger than R(theta). But then the square root is not defined, at least in terms of real numbers.

This leads me to wonder if there is some mistake in the given equations, since in physical reality the ellipse exists in the real plane, nd we should not expect to encounter negative square roots.

I am not expert in this area, so I might misunderstand and perhaps there is no problem, but I cannot seem to resolve this. I cannot seem to find other articles the cite the same equations (although I will keep looking). I would appreciate any clarification of any confusion on my part, or correction of a typo etc in the article. — Preceding unsigned comment added by 68.98.184.146 (talk) 00:17, 1 October 2012 (UTC)[reply]

I haven't verified the equations in detail, but, in all cases where the origin does not lie inside the ellipse, it is inevitable that r will have no real values for certain values of theta. 81.159.110.168 (talk) 01:42, 1 October 2012 (UTC)[reply]

That's right. You'll notice that if we put a plus minus sign before the radical, the number of real solutions for r(theta) equals the number of real solutions for the square root in the expression for Q(theta) -- namely either zero or two, or one in a razor's edge case. If the origin is outside the ellipse and you draw a ray from the origin, it either does not intersect the ellipse (no real solution; expression under the radical is negative), or it intersects the ellipse twice (expression under the radical is positive, giving two real solutions), or it has a single tangency point with the ellipse (expression under the radical is zero, so the number of distinct real roots is one). On the other hand, if the origin is inside the ellipse, and you allow the ray to have positive or negative length, then there are two intersections of the ray (actually line in both directions) with the ellipse; presumably the formula always gives real values for the square root when the origin is inside, and presumably the formula for r(theta) then gives one positive and one negative solution.

That section could use a drawing for clarity. The best I could find was in angular diameter

The angles the two tangent lines make with the horizontal are the minimum and maximum value for θ, outside of that, the value under the square root is negative. Ssscienccce (talk) 21:51, 1 October 2012 (UTC)[reply]

Like the previous respondent, I haven't verified the equations, but they appear to be consistent with the above analysis. I'll tag that sub-section as citation-needed. Duoduoduo (talk) 15:06, 1 October 2012 (UTC)[reply]

Complex functions

Can someone give an explicit 1:1 conformal map from the square to the half disk? Money is tight (talk) 08:06, 1 October 2012 (UTC)[reply]

I'd guess "no". A conformal map is supposed to preserve angles between crossing lines in the plane. The corners of the square can be thought of as the intersection point of two orthogonal straight lines (restricted to the square) crossing each other at an angle of 90 degrees. The images of these lines under the presumed conformal map (wich must be parts of the circle with a common point on the circle) can't possibly cross at 90 degrees. (It would be 180 degrees because its a circle.)

But, there is a homeomorphism (continuous and continuously invertible map) from the filled square in the real 2-dimensional plane to the filled circle. Divide each vector in the filled square with its length. YohanN7 (talk) 09:25, 1 October 2012 (UTC)[reply]

There is a map, it's guaranteed by the Riemann mapping theorem. But the proof is non constructive and I need an explicit function. Money is tight (talk) 11:20, 1 October 2012 (UTC)[reply]

I guess I was wrong then. But you must probably replace "square" by "interior of the square", and likewise for the circle. Better: use "open disk". Edit: "Open" is in the hypothesis of the Riemann mapping theorem.

I now recall something vaguely similar from a complex analysis text. [Churchill & Brown?]. I'll have a look later this evening. YohanN7 (talk) 11:46, 1 October 2012 (UTC)[reply]

The maps you need are discussed at http://www.encyclopediaofmath.org/index.php/Conformal_mapping (examples 5 and 9). Both are of the Schwarz–Christoffel mapping type (a digon and a tetragon), and the upper half-plane to square is given in our article here "explicitly". A source is Zeev Nehari's book "Conformal mapping". —Kusma (t·c) 12:07, 1 October 2012 (UTC)[reply]

You can get an idea of why something simple doesn't work from a little puzzle you might like at The Problem of the Right Wiggly Triangles. Dmcq (talk) 12:28, 1 October 2012 (UTC)[reply]

I don't agree with the solution for nr 2, that's not a wiggly triangle, that's a wiggly pentagon! Ssscienccce (talk) 22:00, 1 October 2012 (UTC)[reply]

I believe Paul Daniels catchphrase at the Bunko Booth about covers that "You'll like this...not a lot, but you'll like it." ;-) Dmcq (talk) 23:40, 1 October 2012 (UTC)[reply]

YohanN7, yes I do mean open square and open half disk, sorry for not pointing that out. Kusma, in Schwarz–Christoffel mapping which specific square does the half plane get mapped to (i.e. what are the coordinates of the vertices)? The integral is very difficult to evaluate. Money is tight (talk) 00:46, 2 October 2012 (UTC)[reply]

You might be able figure this out using the Jacobi elliptic functions, which map a rectangle into the upper half plane. But I have no experience doing this, I can just point you to the literature :( —Kusma (t·c) 10:12, 2 October 2012 (UTC)[reply]

October 2

Time derivative of the anomalies

Are there three types of angular velocity in motion on a ellipse, each corresponding to the time derivative of mean anomaly, true anomaly and eccentric anomaly?--188.26.22.131 (talk) 12:41, 2 October 2012 (UTC)[reply]

Yes. See Kepler's laws of planetary motion. Bo Jacoby (talk) 13:13, 2 October 2012 (UTC).[reply]

That shows how you use them to calculate the motion, but I'm not sure that's what the OP is looking for. Or maybe I just don't see it... Ssscienccce (talk) 17:35, 2 October 2012 (UTC)[reply]

The mean anomaly is

M=nt={\frac {2\pi }{P}}t.

So the mean motion is

{\frac {dM}{dt}}=n={\frac {2\pi }{P}}

where P is the period.

The eccentric anomaly E satisfies

M=E-\varepsilon \cdot \sin E.

So

{\frac {dE}{dt}}={\frac {n}{1-\varepsilon \cdot \cos E}}.

The true anomaly θ satisfies the equation

\tan({\frac {\theta }{2}})={\sqrt {\frac {1+\varepsilon }{1-\varepsilon }}}\cdot \tan({\frac {E}{2}})

Differentiate to obtain an equation for ${\frac {d\theta }{dt}}.$ Bo Jacoby (talk) 06:35, 4 October 2012 (UTC).[reply]

How much mathematics is essential?

My question comes down to what math should cover- is math the fundamental operations of addition, subtraction, division, and multiplication, coupled in applications such as equations etc.

And is calculus, differential equations etc. mostly experimentally derived workings that are useful in other subjects such as physics and finances?

So if i were to pick up a calculus book, most of the operations will be in the form of subtraction, division, miltiplication and division and just mostly operations of what we know? I know in linear algebra, there are directional shifts in matrices in regards to division, subtraction, addition, multiplication; but what is the real thing I should I know? — Preceding unsigned comment added by 108.50.249.125 (talk) 16:58, 2 October 2012 (UTC)[reply]

Know for what purposes ? If you get a job which doesn't require additional math skills, then the core set that everyone needs is addition, subtraction, multiplication, division, fractions, conversions, percentages, and interest rate calculations (using a formula). Some basic geometry, like calculating areas and volumes, would also be useful. I've never found the need to use algebra, trigonometry or calculus in "real life" (outside of work). StuRat (talk) 18:23, 2 October 2012 (UTC)[reply]

And I use all three routinely in my hobbies. —Tamfang (talk) 02:27, 3 October 2012 (UTC)[reply]

http://xkcd.com/1050/ . Just sayin'. --Trovatore (talk) 02:36, 3 October 2012 (UTC)[reply]

Calculus generally involves integration and differentiation which are more advanced mathematical operations than the arithmetic ones you have listed. Understanding those more advanced operations is essential to much of physics and most forms of engineering, though many people lead perfectly happy and successful lives while having no idea what calculus is about. Any decent introductory calculus text can teach you integration and differentiation, but these are not concepts that any student would be expected to already know before reaching calculus. So, in that sense, calculus is certainly not something that people already know simply because they have learned arithmetic. As others have said, what you "should" know can't really be defined without some idea of what you want to be able to do. Dragons flight (talk) 11:07, 3 October 2012 (UTC)[reply]

I agree with StuRat that his list is a good list of what everyone needs to know about math. And I agree with Tamfang that for many people, life can be immeasurably enriched by hobbies that use algebra, trig, and calculus. It's not clear to me what the context of your (the OP's) question is -- if you are a high school or college student wondering what areas of math to expose yourself to, one thing you should know is that the more math you learn, the higher-paying job you'll probably be able to get, the more your employment is likely to be shielded from economic recessions, and the more comfortable your life is likely to be. Duoduoduo (talk) 14:58, 3 October 2012 (UTC)[reply]

Another possibility for the list: I think that knowing some basic statistics is also extremely important for "real life", at least up to the point where you can comprehend that there are better methods of judging hypotheses than recounting a few anecdotes ("I took <insert quack cure> and my <insert malady> got better!"), but that these methods are not infallible ("Well, those 'scientists' were wrong about X, so why should I listen to what they say about Y?"). Though, of course, many would argue that statistics isn't a subset of maths. I think you could make the case that a bit of exposure to how "real" maths works (looking for patterns, proof, abstraction, etc.) is also important.

As to the question about calculus being "experimentally derived workings that are useful in other subjects", it is true that much of calculus was developed for the purpose of solving problems in other fields, and initially in a very informal way, but the advent of mathematical analysis put calculus on a firm theoretical footing and led to lots of beautiful mathematics. 130.88.99.231 (talk) 13:08, 4 October 2012 (UTC)[reply]

I agree that some basic knowledge of statistics is valuable. However, it's not the math that everyone needs to know (calculating standard deviations and such). What people need to know about statistics, they don't seem to teach in math class. There's how to ask an unbiased question (not "In the US Presidential election, do you intend to vote for the Muslim candidate or the Christian ?"), the importance of having a high level of participation among the people you poll (so no mailing out surveys, getting 1% back, and drawing conclusions from that 1%), the importance of polling a sample representative of the population (no surveys at country clubs), etc. StuRat (talk) 04:31, 6 October 2012 (UTC)[reply]

Calculus and differential equations

How do both relate? Are differential equations a part of Calculus? Are differential equations based on Calculus? Are differential equations related to, but independent of Calculus? — Preceding unsigned comment added by 80.58.205.34 (talk) 16:59, 2 October 2012 (UTC)[reply]

Differential equations are part of calculus, but they are ordinarily taught in a separate course, which students take after Calculus I and Calculus II. In other words, they are an advanced topic in calculus. Looie496 (talk) 19:53, 2 October 2012 (UTC)[reply]

Integral problems are a subset of differential equations, in particular those of the type

y'(x)=g(x)

, so that the solution is

y(x)=\int g(x)dx,

, and the main challenge is in solving the integral. In a generic differential equation, the left hand side is more complex. Sjakkalle (Check!) 20:13, 2 October 2012 (UTC)[reply]

"There is no such subject as calculus." — Paul Halmos Sławomir Biały (talk) 20:28, 2 October 2012 (UTC)[reply]

It's all very well to deny the existence of calculus, but there are limits! Marnanel (talk) 20:38, 2 October 2012 (UTC)[reply]

I find your humor to be quite derivative. Sławomir Biały (talk) 21:36, 2 October 2012 (UTC)[reply]

I found it partially derivative, but I am not so anti-derivative as you seem to be. 86.176.211.202 (talk) 02:41, 3 October 2012 (UTC)[reply]

Fast test for non-factorable polynomials of a special form

I have an arbitrary degree polynomial where every coefficient is one. Like x^117+x^87+x^30+x^23+x^13+x^4+x^2+x+1. Is there a fast, straightforward method for determining the factorability of such polynomials? Sebastian Garth (talk) 19:44, 2 October 2012 (UTC)[reply]

Just as an example, x^7+x^5+x^4+x^3+x^2+1 = (x^2-x+1) (x^5+x^4+x^3+x^2+x+1) and in general x^2-x+1 times (x^k+x^(k-1)+...+x+1) is (x^(k+2)+x^k+x^(k-1)...x^3+x^2+1) so there is at least one factorable polynomial with only unit coefficients for any positive integer degree greater than two.Naraht (talk) 19:57, 2 October 2012 (UTC)[reply]

Similarly (x^4-x^3+x^2-x+1) times (x^k+x^(k-1)+...+x+1) gives a similar result.Naraht (talk) 20:22, 2 October 2012 (UTC)[reply]

You may want to start at http://www.math.sc.edu/~filaseta/irreduc.html and some of the papers referenced.Naraht (talk) 20:40, 2 October 2012 (UTC)[reply]

Thanks, yes that looks somewhat promising. Or how about this: is there an efficient method for detecting whether or not there is a real root? Or is that just as difficult as factoring? Sebastian Garth (talk) 01:01, 3 October 2012 (UTC)[reply]

I can see some cases where the existence of a real root is clear:

If the constant term is 0 there is obviously a real root at x = 0. So from now on let's assume the constant term is 1, so f(0) = 1.
If the leading term (highest degree) has odd degree then there is a real root < 0.
If there are more terms with odd degree than with even degree then f(-1) < 0 so there is a real root between -1 and 0.
If there are the same number of terms with odd degree as with even degree then there is a real root at x = -1.
At the other extreme, if all terms have even degree, or if there is only one term with odd degree and it is not the leading term, then there are no real roots.

So the ambiguous case is when the leading term has even degree and there are more terms with even degree than odd degeer, but at least two terms with odd degree. I can't think of a simple method of handling that case (at least, not without using a root finding algorithm). Gandalf61 (talk) 09:36, 3 October 2012 (UTC)[reply]

If you sort the terms and find that every odd power is bracketed both above and below by an even power (i.e. the 5th power in x¹² + x⁵ + x⁴ is bracketed), then one can group terms and conclude there are no real roots. That reduces Gandalf's unresolved case to polynomials having two or more consecutive odd powers. Actually, one can use the same argument to say that if every set of N consecutive odd powers is bracketed both above and below by at least N consecutive even powers, then there are no real roots. Dragons flight (talk) 10:47, 3 October 2012 (UTC)[reply]

The number of real roots of the polynomial can be computed easily using Sturm's theorem.—Emil J. 12:44, 3 October 2012 (UTC)[reply]

That article looks like it could *really* use an example to be more clear. Say counting the number of real roots on a quintic equation..Naraht (talk) 13:09, 3 October 2012 (UTC)[reply]

I've added this to the references of Sturm's Theorem: Baumol, William. Economic Dynamics, section "Sturm's Theorem". Not a web resource, but I remember it has a very good example--maybe you can find it in your library. Duoduoduo (talk) 15:08, 3 October 2012 (UTC)[reply]

I've added an example of using Sturm's Theorem here. Thanks for the suggestion, Naraht. Duoduoduo (talk) 20:01, 4 October 2012 (UTC)[reply]

See also Descartes' rule of signs, which can sometimes prove that there exists at least one real root. Duoduoduo (talk) 15:12, 3 October 2012 (UTC)[reply]

Back to the original question, on factorization: see Factorization of polynomials and the various wikilinks therein. Duoduoduo (talk) 15:36, 4 October 2012 (UTC)[reply]

October 3

Log- Linear Plot given the coordinates?

When logy is graphed as a function of x, a straight line results. Graph straight lines, each given by two points, on a log linear plot, and determine the functional relationship (The original x-y coordinates are given)

(x1,y1) = (0,5) and (x2,y2) = (3,1)

The correct answer is y= 5 * (0.58)^x Could someone tell me how to arrive at this answer? thank you! — Preceding unsigned comment added by 24.86.170.175 (talk) 00:41, 3 October 2012 (UTC)[reply]

Let's see. You start with:

y = a(b)^x

You plug in x_₁ for x and y_₁ for y:

5 = a(b)⁰ = a(1) = a

So, a = 5:

y = 5(b)^x

Now plug in x_₂ for x and y_₂ for y:

1 = 5(b)³

Divide both sides by 5:

0.2 = b³

Take the cube root of both sides:

0.5848 = b

Also, since you are dealing with logarithms, I suppose I should put that last "finding the cube root" step in terms of logs. Here you find the antilog of log(0.2)/3, or the antilog of -.233, to get 0.5848. StuRat (talk) 01:16, 3 October 2012 (UTC)[reply]

This process worked with this question but when i tried applying it to other questions it does not work any longer. For example, the same question, except with coordinates (-2,3) and (1,1). I first substituted the first set of coordinates into the equation to get 3b^2= a Then i substituted the second set of coordinates to get a = b. After a bit of algebra my answer was y = 1/3 * (1/3)^x. However, this is incorrect. I have spent some time trying to figure this out but I still do not understand. — Preceding unsigned comment added by 24.86.170.175 (talk) 01:40, 3 October 2012 (UTC)[reply]

The second given point tells you ab=1, not a=b. —Tamfang (talk) 02:26, 3 October 2012 (UTC)[reply]

Oh! that was a very silly mistake. Thank you to the both of you for helping — Preceding unsigned comment added by 24.86.170.175 (talk) 02:40, 3 October 2012 (UTC)[reply]

You're quite welcome. StuRat (talk) 02:57, 3 October 2012 (UTC)[reply]

Resolved

Generating function of a particular sequence

I have a sequence in n which is $\sum _{g=0}^{n}{\binom {g+m-1}{m-1}}z^{n-g}$ where m is a natural number and z is some indeterminate, and I'm trying to find a rational generating function for this sequence. In other words, a rational function f(x), such that when expressed as a power series,

$f(x)=\sum _{n=0}^{\infty }\sum _{g=0}^{n}{\binom {g+m-1}{m-1}}z^{n-g}x^{n}.$

With z = 1, this is just $f(x)=1/(1-x)^{m+1}$ , but the z term complicates things.

If it's helpful, one way to think about the sequence is that the nth element is equal to the sum of all monomials of degree n in m+1 variables with one variable evaluating to z and the rest to 1. Another way is that it counts the number of monomials of degree ≤ n in m variables but weighting the sum based on their degree. I'm not totally sure that the sequence does have a rational generating function, but I have some reasons to believe it should.

Anyone have any ideas? Rckrone (talk) 04:51, 3 October 2012 (UTC)[reply]

Ok now I'm feeling a bit silly. I spent a while on this earlier and got nowhere, and then 30 min after I post it here the answer occurs to me. It should just be

f(x)={\frac {1}{(1-x)^{m}(1-zx)}}

, right? Rckrone (talk) 05:24, 3 October 2012 (UTC)[reply]

Model theory

Does every (first order) proposition having a model, constitute a part - of a finite class of (first/second order) propositions that has a unique model (up to isomorphism) for the individuals? HOOTmag (talk) 16:23, 3 October 2012 (UTC)[reply]

Why trig doesn't work?

let say I have this triangle. Why I didn't get the correct answer when I used inversed of cosine or inversed of tangent or inversed of sine? I saw the answer key used cosine law. Why only cosine law gives the correct answer?Pendragon5 (talk) 22:42, 3 October 2012 (UTC)[reply]

This is not a right triangle, so none of the side-length ratios will be equal to the sine, cosine or tangent of the angle. However the law of cosines can be applied to any triangle (not just right triangles) so you can use it to compute cos θ. Rckrone (talk) 23:34, 3 October 2012 (UTC)[reply]

October 4

Finitely generated abelian groups

I'm stumped on what should be an easy problem. The book defines a free finitely generated abelian group as one with a basis, i.e., a linearly independent generating set. Then an exercise defines a torsion-free group in the usual way, and asks me to prove that a finitely generated abelian group that is torsion-free must be free. I know this follows from the fundamental theorem of finitely generated abelian groups, but I get the impression I'm not supposed to import a big gun to solve this problem.

Can I just start from the fact that I've got a finite generating set, add the fact that there's no torsion, and show that there must be a basis? I know I can't count on the basis being a subset of my generating set, because I've got a simple counter-example: the set {2,3} generates the integers, but neither {2} nor {3} does it alone. Help! Thanks in advance. -GTBacchus^(talk) 20:07, 4 October 2012 (UTC)[reply]

Suppose

X=\{x_{1},\ldots ,x_{n}\}

is a generating set for a torsion-free abelian grop. If

i\neq j

then we can replace

x_{i}

by either

-x_{i}

or

x_{i}+x_{j}

and we will still have a generating set with

n

elements.

Suppose that

X

is not linearly independent, so there is a linear dependence

a_{1}x_{1}+\cdots +a_{n}x_{n}=0

. First, we can assume that each coefficient is non-negative since if

a_{i}<0

then we can replace

x_{i}

by

-x_{i}

and

a_{i}

by

-a_{i}

.

Now, if at least two of the coefficients are non-zero, say

0<a_{i}\leq a_{j}

then we can replace

x_{i}

by

x_{i}+x_{j}

and since

a_{i}x_{i}+a_{j}x_{j}=a_{i}(x_{i}+x_{j})+(a_{j}-a_{i})x_{j}

, we have a linear dependence for this new generating set with non-negative coefficients and the sum of the coefficents is smaller than before. Hence after doing this finitely many times we must obtain a generating set which has a linear dependence with only one non-zero coefficient, and so the corresponding generator has finite order. But the group is torsion-free so this generator must be trivial, and so the group has a generating set with

n-1

elements.

It follows that if

X

is a generating set with the smallest possible cardinality, then

X

is a basis. 60.234.242.206 (talk) 11:40, 6 October 2012 (UTC)[reply]

October 5

Mathematics degrees

Is there any stand alone degree of Algebra, Analysis, Combinatoric, Geometry? In the same way that you can study computational science independently. — Preceding unsigned comment added by 83.41.157.164 (talk) 09:29, 5 October 2012 (UTC)[reply]

Well, there are certainly lots of postgraduate degrees covering specific branches of mathematics. Googling gave me this course for a "Bachelor of Arts in Mathematical Sciences - Combinatorics and Optimization" at the University of Montana (Googling for "degree geometry" turns out not to be very helpful). You might need to be a bit more specific about what kind of degrees you are interested in, and which countries. 130.88.99.231 (talk) 17:54, 5 October 2012 (UTC)[reply]

Combinatorics and Optimization is an emphasis at Montana, not a stand-alone degree. From their website: The department offers a Bachelor's Degree with emphases in Pure Math, Applied Mathematics, Mathematics Education, Combinatorics and Optimization, Combined Mathematical Sciences-Computer Science and Statistics.' [my bolding]. I strongly doubt that any American undergraduate department offers such a specialized degree as Combinatorics and Optimization, Algebra, etc.--to fill up the required in-major credit hours would require courses that get so deep that they would not be undergraduate level. But some do offer separate degrees in Pure Mathematics and in Applied Mathematics and in Statistics. Duoduoduo (talk) 18:14, 5 October 2012 (UTC)[reply]

The University of Waterloo offers several specialized undergraduate majors in mathematics, including Combinatorics and Optimization. See [1]. —Bkell (talk) 21:02, 5 October 2012 (UTC)[reply]

Incompleteness theorem

Once when I was feeling brave and tried to understand Godel's incompleteness theorem, I thought the essential result was that it is possible to formulate syntactically valid statements (in some applicable system) that can be neither proved nor disproved. However, in an article I'm currently reading, it says that the theorem "revealed that mathematics must contain true but unprovable propositions", and again in the intro to the Wikipedia article "there will always be statements about the natural numbers that are true, but that are unprovable within the system".

How does the notion of "true" get involved in this? How do we know that the statements that cannot be proved or disproved are "true"? Why is the theorem described in those terms rather than in my original terms? 86.176.210.107 (talk) 20:28, 5 October 2012 (UTC)[reply]

The crucial point is "... within the system". Thus, we may be able to prove the truth of a statement that is expressible in a given formal system (under some semantic model), but is not a theorem of the system. Meaning that using the axioms and rules of the system, the statement is not provable, but reasoning outside the rules of the system is not constrained in the same way. Any formal system of symbols and rules has no inherent meaning; it requires a semantic model to match an interpretation (and hence what a string expresses as we would understand it). — Quondum 21:08, 5 October 2012 (UTC)[reply]

I see, thanks. Is there any reason why it is important to emphasise the existence of true statements that are not provably true (within some system), rather than any other combination, such as false statements not provably false, or statements that we can never know the truth of (in any way) not provable one way or the other (in some particular system)? 86.176.210.107 (talk) 21:17, 5 October 2012 (UTC)[reply]

I would phrase things somewhat differently from the way Quondam did. The point, which is often obscured in popular treatments, is that the Goedel sentence of a formal theory is an assertion about a very clear collection of objects, the natural numbers. If you agree that (i) there are such things as natural numbers and (ii) it is well-specified what they are, then you must agree that the Goedel sentence of (say) Peano arithmetic is either really true or really false.

The Goedel sentence of a consistent theory T is necessarily true. This is because what the sentence asserts is of the form "for every natural number n, n is not the Goedel number of a proof of (something complicated that we don't need to worry about the details of here) within T ", and if that were false, then there would be such a natural number n, and it would follow, by the arguments of the proof of the incompleteness theorem, that T was inconsistent.

The reason it is important to emphasize the truth of the Goedel sentence of a consistent theory is that, otherwise, it is easy to make the mistake that many popularizers make, and suppose that the "choice" between adding the Goedel sentence as an axiom, and adding its negation as an axiom, is a neutral one. It is not neutral; it is very clear — the Goedel sentence of a consistent theory is true. --Trovatore (talk) 21:27, 5 October 2012 (UTC)[reply]

(i) When you say "the Goedel sentence", you mean a constructed statement that is not provable or disprovable within T, right? (ii) How do we know that things are really "true" about the natural numbers without invoking some other system that itself suffers from the flaw (or incompleteness) that Godel exposed? 81.159.104.36 (talk) 22:55, 5 October 2012 (UTC)[reply]

For your (i), yes, but something a bit more specific than that. I mean specifically a sentence that encodes "I am not provable in T".

For your (ii), it is easy to show, assuming very little, that if T is consistent, then the Goedel sentence of T (call it G_T) is true. Now, T itself does not prove that G_T is true, but that is only because T also does not prove that T is consistent (this is the second incompleteness theorem, and in fact that's basically how it's proved).

So as to how we know that G_T is true — well, we know that only to the extent that we know that T is consistent, which is a whole 'nother discussion. Depending on what T is, perhaps we don't know that T is consistent. But that doesn't change the fact that the Goedel sentence of any consistent theory is true. --Trovatore (talk) 01:22, 6 October 2012 (UTC)[reply]

To the OP: I agree with the other people here, that it's possible to prove - outside the system - that the Goedel sentence is true. However, I think it's redundant to indicate that the Goedel sentence is true, i.e. Goedel's full discovery is not fully reflected by the fact that the Goedel sentence - not provable within the system - is true; Goedel's discovery (or rather Goedel-Rosser discovery) is fully reflected by the fact that the Goedel's sentence - not provable within the system - is also not disprovable within the system. It's really a stronger claim, because: on one hand, this version of Goedel's discovery entails also that there is a sentence (whether the Goedel sentence or its negation) - which is not provable within the system - and is true (assuming every sentence about natural numbers must be either true or false); On the other hand, the other version of Goedel's discovery - which only claims that the Geodel sentence (not provable within the system) is true - is a weaker version (of Goedel's discovery), because it may make us suspect that the (true) Goedel sentence may still be disproved within the consistent system, that may disprove true sentences and still stay consistent. HOOTmag (talk) 19:07, 6 October 2012 (UTC)[reply]

You're right, there is a small extra piece that is not captured by saying that the sentence is true. It is nevertheless important to emphasize that the sentence is true (given that the theory being studied is consistent). This is at least arguably really the main point, because the pre-Goedel incarnation of formalism hoped to be able to capture truth by formal provability, and this is the hope that was dashed. Moreover, without saying that loudly and explicitly, readers tend to come away with the impression that the sentence is "neither true nor false" or some such, an impression that many popularizers are guilty of spreading.

Certainly the point that the Goedel–Rosser sentence of a consistent theory can neither be proved nor refuted by the theory even if the theory proves false things, is also worth saying. But to me it's a little bit of a secondary point. --Trovatore (talk) 19:19, 6 October 2012 (UTC)[reply]

So we almost agree, and may disagree - a little bit. I agree that - from a historical point of view (as well as from an "anti-populistic" point of view) - Goedel's version for his discovery is more important than Rosser's one, because Goedel's version directly-refutes what was hoped (and what is still being claimed by many popularizers). However, in my opinion, from a purely mathematical point of view - Rosser's version is more important, because it's mathematically stronger - and not less aesthetical !!! it's not easy at all to conclude - Rosser's aesthetical version - from Goedel's version, whereas it's quite easy to conclude Goedel's version from Rosser's version - once we assume an (almost) obviuos (or definitely-obvious) assumption: "every sentence about natural numbers must be either true or false" (as opposed to what those popularizers think). Btw, also you indicated this assumption, by your stating: "If you agree that (i) there are such things as natural numbers and (ii) it is well-specified what they are, then you must agree that the Goedel sentence of (say) Peano arithmetic is either really true or really false". HOOTmag (talk) 20:27, 6 October 2012 (UTC)[reply]

Hmm — I guess I think the most important mathematical consequence of the theorems is the hierarchy of consistency strength, for the most part of theories that have wellfounded models (for example, various large cardinal assumptions). From this perspective, the non-arithmetically-sound theories for which Rosser closes a loophole are not all that interesting in the first place. From the perspective of a pure proof theorist, I can see your position might make sense. --Trovatore (talk) 20:34, 6 October 2012 (UTC)[reply]

If "the non-arithmetically-sound theories for which Rosser closes a loophole" hadn't been "all that interesting in the first place" (as you state), then also the inconsistent systems wouldn't have been all that interesting in the first place; However, Goedel did find it important to emphasize that the theories he was talking about - were the consistent ones - in order to exclude the inconsisent ones, although the inconsistent theories are not more interesting than the theories you claim to be "uninteresting", i.e. the consistent theories which prove false sentences too. Hence: since Goedel did find it important - to indicate that the theories he was talking about - were consistent, then he must also have found it important - to indicate that the consistent theories he was talking about - proved true sentences only. All of this shows, that Rosser's version is also important for "practical" purposes, because Goedel's version must assume that the Peano system - if consistent - proves true sentences only, whereas Rosser's version does not have to assume that. Note that this assumption is not more trivial/obvious than the assumption that the Peano system is consistent. HOOTmag (talk) 21:42, 6 October 2012 (UTC)[reply]

(OP) Thanks. I cannot pretend to understand all of that, but I appreciate your help. Moving away from the exact literal meaning of what Godel originally said, I thought I had read in some places that more complicated statements, not directly concerning the natural numbers, have been shown to be undecidable in Godel's sense, and that this implies that it might (or will) never be known whether those statements are true or false. Is that correct? 86.160.216.227 (talk) 20:36, 6 October 2012 (UTC)[reply]

Ah, this is much more subtle. All that has so far been shown is that certain such statements are undecidable in certain formal theories, or at best certain classes of formal theories. There has never been a mathematical proof that, for a given statement, we can never know in any way at all whether it is true or false, and it is unclear what a such a proof could even look like. There is an excellent paper on this question by Peter Koellner called On the question of absolute undecidability — you might google for it. Be warned that it's not light bedtime reading. --Trovatore (talk) 20:39, 6 October 2012 (UTC)[reply]

October 6

Root problem

Can someone explain for me how to do problem 3? I have the answer key but unable to understand what is it saying. I don't get how (p^2 - 1)/4 is the product of the roots. And the part where it say "which checks vs. the coefficient of the x term" --> What is this even mean? 65.128.190.136 (talk) 03:05, 6 October 2012 (UTC)[reply]

They are talking about simplifying this expression:

x² - px + (p²-1)/4

into these roots:

[x + (p+1)/2] [x + (p-1)/2]  (See correction below.)

The first step is to resolve:

(p²-1)/4

which can be rewritten as:

p²/4-1/4

into it's roots:

[(p/2)+(1/2)] [(p/2)-(1/2)]

or:

[(p+1)/2] [(p-1)/2]

You might recognize this as a difference of squares problem, where (a+b)(a-b) = a² - b². StuRat (talk) 03:35, 6 October 2012 (UTC)[reply]

Why when I multiplied this out. I got everything the same except x^2 + px instead of - px. How did they get -px? Can you show me? Thanks!65.128.190.136 (talk) 03:57, 6 October 2012 (UTC)[reply]

Oops, I had the signs wrong on the roots. Here's the corrected version:

[x - (p+1)/2] [x - (p-1)/2]

StuRat (talk) 04:11, 6 October 2012 (UTC)[reply]

Okie! I think everything is clear! Thanks!65.128.190.136 (talk) 04:20, 6 October 2012 (UTC)[reply]

Your quite welcome. I'll mark this Q resolved. StuRat (talk) 04:33, 6 October 2012 (UTC)[reply]

Resolved

Relations: any usual way to denote/describe the following relation?

1. Let R be a relation. Is there a better way, to denote (mathematically by R), and/or describe (verbally by R), the following relations?

{ (a,(b,c)) | ((a,b),c)∈R }
{ ((a,c),(b,d)) | ((a,b),(c,d))∈R }

.

2. If R and S are relations, then the relation:

{(a,c) | there exists b, such that (a,b) ∈ R and (b,c) ∈ S}

is usually-denoted as S∘R, and is usually-described (verbally) as the composition of S and R. Is there a better way, to denote (mathematically by R,S), and/or describe (verbally by R,S), the following relation?

{ ((a,c),b) | (a,b)∈R and (b,c)∈S }

84.229.139.122 (talk) 18:01, 6 October 2012 (UTC)[reply]

.

No. Looie496 (talk) 23:44, 6 October 2012 (UTC)[reply]

Not sure what would be considered "better", but writing the relation as a function seems a bit clearer to me. if S is f(x) and R is g(x) then S∘R is f(g(x)). Ssscienccce (talk) 00:16, 9 October 2012 (UTC)[reply]

Statistically, is this a valid argument?

A common argument for the existence of life on other worlds in our universe is that there are just so many planets out there that there that it would extraordinarily unlikely for life to not evolve on at least some of them. Basically, this argument is making the claim that it is very unlikely that the probability of life emerging on a planet, p, is very low. But is this reasoning valid? Is it really possible to justify the claim that it's more unlikely that p < 10^-20 than p > 10^-20? 74.15.136.9 (talk) 20:42, 6 October 2012 (UTC)[reply]

No, that's a weak argument, especially when accompanied by "it evolved here, so it can't be that unlikely". The problem is that only on planets with intelligent life can they know that life has evolved there, so, the observed rate on your own planet will always be 100%, no matter how low the overall rate is. Now, if we had evidence that life had evolved independently several times on Earth, that would be different. Life does seem to have evolved quite early on Earth, and that would support the idea that it's not all that rare. However, again, had it evolved a billion years later, there would be nobody around to observe it yet, so our observation is again biased. StuRat (talk) 21:00, 6 October 2012 (UTC)[reply]

Also note that, according to the many worlds hypothesis, there may be an infinite number of planets. If so, then, no matter how rare the existence of life is, there must not only be more than one, but an infinite number of worlds with life. This doesn't mean that any are accessible from Earth, however. StuRat (talk) 21:10, 6 October 2012 (UTC)[reply]

The OP specified that (s)he was only talking about our universe. Rojomoke (talk) 22:28, 6 October 2012 (UTC)[reply]

Well, that depends on your definition of "universe". I'm using def 1 under Wikt:universe, and you are using def 2. Note that the etymology supports def 1. StuRat (talk) 22:39, 6 October 2012 (UTC)[reply]

This issue is often discussed in terms of the Drake equation, which helps to make the numerics more specific. I think you might find our article on it interesting. Looie496 (talk) 23:42, 6 October 2012 (UTC)[reply]

The really unlikely scenario is that life evolved on three planets, or ten, or some other small number. Most probably, either life is unique to Earth, or it has evolved independently billions of times in the Universe. 86.160.216.227 (talk) 00:29, 7 October 2012 (UTC)[reply]

On a related point Nick Lane, in his bookPower, Sex, Suicide: Mitochondria and the Meaning of Life argues that evolution of multicellular lifeforms is exceedingly unlikely (this is not the question of any life evolving, though), should this be of interest. The point of life evolving being either very likely or very unlikely is a good one, and supports the point that the originally questioned reasoning has no merit. — Quondum 12:53, 7 October 2012 (UTC)[reply]

Let p = the probability of life forming on a given planet, and let the planets be independent as to whether or not they support life. (Or if you don't like that independence assumption for two planets in the same star system, let p be the probability that a star system supports life, with each star system independent.) N is the number of planets. Before observing any planets (even Earth), the number of planets with life is binomially distributed with mean pN. If we hypothesize that p is much smaller than 1/N, then the probability of there being even one planet with life is very small, and hence not consistent with the observation that at least one does in fact exist. So a maximum likelihood argument for getting an estimate of p, conditional on knowing #planets with life >0, will not choose an estimate for p that is very much below 1/N. So let's make the conservative assumption that p = 1/N rather than p>1/N. Now, what is the probability that with N-1 planets that have not been monitored for life, none of them support life? The probability that any one of those planets does not support life is 1-p = 1-(1/N). The probability that all N-1 of them do not support life is [1-(1/N)]^(N-1). We could set N = 10^11 (a conservative estimate of the number of stars in our galaxy and assuming an average of one planet per star -- use your own estimate for the number of planets if you prefer), but my calculator can't deal with that. But if I assume one billion planets, N=10^9, I get a probability of .3679 that no other planets have life. I think if you let N go up from there by a couple orders of magnitude you'd get a much lower probability. Duoduoduo (talk) 14:00, 7 October 2012 (UTC)[reply]

The observation that we exist that you want to condition on has to be taken into account in the correct way. This is a non-trivial issue, see here. See also the sleeping Beauty problem. Count Iblis (talk) 15:44, 7 October 2012 (UTC)[reply]

The way I'd look at it is to get intelligent life we have to get a suitable planet, then life has to arise, then complex multicellular life, then intelligence. If we are the only intelligent life in the universe then the ratios of the probabilities of each of these happing per unit time by maximum likelihood is I think proportional to about 1/(time it took to happen). They seem to be in the proportion about 8 billion to perhaps a fraction of a billion to 2 billion to 2 billion. It seems that for a suitable planet lie will arise pretty quickly. The interesting thing is that the other factors are of about the same order of magnitude and the least likely to happen per unit time is a suitable planet for life. If there was one stage which was very difficult I would expect it to dominate the others so for instance if intelligence was the hard thing we'd have had multicellular life in less than a billion years and have spent the next 4 billion waiting for intelligent life. This argument does unfortunately give the chancs of there being life elsewhere, only ratios of probabilities, but having no relatively small probability means to me that there i no particular reason forus to be the only intelligent life - it is quite possible there are others just they are in other galaxies and if we wait around a few more billion very possibly we'll see some in our own galaxy - and there is very likely to be life already in this galaxy. Dmcq (talk) 00:30, 8 October 2012 (UTC)[reply]

I think this thread is diverging from the original question, which seems to be asking whether it is valid to argue that life has at least some significant probability as with the Drake equation within the field of statistics. And I think that the correct answer is that it has no validity as a statistical argument; it is merely a plausibility argument given a number of (statistically) untested conjectures. For example, if you start with the given that we have one life-supporting planet, it would be false to conclude on statistical grounds that the probability of this arising is non-vanishing. Thus, simply put, the argument is invalid from a statistical perspective. Additionally, what observational evidence we do have tends to support (also using the plausibility approach) the idea that treating the one known instance of a life-bearing planet as a representative sample leads to a wildly high estimate of the probability of life arising. — Quondum 04:44, 8 October 2012 (UTC)[reply]

All statistical results are based on Bayesian priors, it is probability theory that doesn't need assumptions as it is the mathematical idealization. I think there is quite enough evidence to derive reasonable statistics about life with plausible assumptions, as I pointed out above the ratios of the various timescales indicates there is no particular sticking point. The major problem is the number of planets that are suitable for life in a galaxy like ours and how often life is totally wiped out by something like a huge meteor or local supernova. I don't think we can say anything much about what has happened in other galaxies though I guess in the future we could have a galactic scale SETI program. Dmcq (talk) 00:21, 9 October 2012 (UTC)[reply]

October 7

silly question

Hello, when I try solving the equation a/(b+c)=d/c for b using pencil and paper, I get b=ac/d-c (like this:
a/(b+c)=d/c
(multiply by b+c) a=d(b+c)/c
(multiply by c) ac=d(b+c)
(divide by d) ac/d=b+c
(substract c) ac/d-c=b.)
(One could also flip both fractions and save a step: a/(b+c)=d/c; (b+c)/a=c/d; b+c=ac/d; b=ac/d-c.)
However, when I input the formula into WolframAlpha, telling it to solve for b, it says that b=c(a-d)/d. Is it doing this for no reason (the way the equation rearranging algorithm works) or is it because the latter form is preferred in refined circles (is it?) and WA maybe even takes extra steps to arrive at it? Thanks in advance. Asmrulz (talk) 08:47, 7 October 2012 (UTC)[reply]

You have the right answer if you, as I hope, put the "-c" outside the fraction. A few algebraic manipulations of your answer is all that is needed to arrive at the answer your program gave

You started with

b={\frac {ac}{d}}-c

convert the "

-c

" into a fraction which has a common denominator with the first fraction

b={\frac {ac}{d}}-{\frac {cd}{d}}

Now you arrive at the program's answer

b={\frac {ac-cd}{d}}={\frac {c(a-d)}{d}}

Sjakkalle (Check!) 09:38, 7 October 2012 (UTC)[reply]

Also, you asked if there was any reason for this. Well, there is no "best" standard, but in many cases a single fraction (one term) is a tidier expression than a single fraction with something subtracted from it (two terms). Sjakkalle (Check!) 10:00, 7 October 2012 (UTC)[reply]

simple math wikicode

I just wasted a 45 minutes trying to find the code for a simple division.
I saw a chart in an article that has the wrong division results and I want to show the correct math.
I don't just want to change the results in the chart, I want to show the math with the math code in the article code.
1,152,000,000/365 $\neq$ 1,152,000,000/365.26,
so if I change the results to the later I don't just want to replace the numbers,
I want to show that how it was calculated in the article code. I know it's possible,
I have seen it with the automatic conversion of Celsius to Fahrenheit, and I did find the addition code: {{Addition|1|1|1}}
Somebody please help.
24.79.38.15 (talk) 11:03, 7 October 2012 (UTC)[reply]

So in the mean time I find that this works {{addition|360/365.26}} = 0.98559929912939
But it has too many sig difs. I can`t believe this is so hard to find.
24.79.38.15 (talk) 11:26, 7 October 2012 (UTC)[reply]

Maybe

{{round|360/365.26|3}}, giving 0.986? — Quondum 12:32, 7 October 2012 (UTC)[reply]
Thank you very much, I will use that. And thanks for ``Roaming the foam.`` 24.79.38.15 (talk) 14:33, 7 October 2012 (UTC)[reply]

October 8

How do you decompose this into orbits?

Let  $M$  be the set of  $m\times n$  matrices, and let  $G=GL(\mathbb {R} ^{m})\times GL(\mathbb {R} ^{n})$ , where an element  $(P,Q)\in G$  acts on an element  $A\in M$  by  $(P,Q)A=PAQ^{-1}$ . How do you decompose  $M$  into  $G$ -orbits? I see no general way of doing it, since you can have general invertible P and Q, and then general A. I can only get special cases, for example that the zero matrix in  $M$  is in an orbit by itself. Widener (talk) 13:48, 8 October 2012 (UTC)[reply]

The orbit of a matrix A consists of matrices that can be obtained from A by a sequence of elementary row and column operations. We can get a unique representative of each orbit as follows. Start with A. If there is a nonzero entry in the matrix, we can switch rows and columns to move this element to the upper left corner, and row multiplication to make this element 1. Then we use row and column addition to make all other entries in the first row and column zero. We repeat the same process with the  $(m-1)\times (n-1)$  submatrix omitting the first row and column. In this way, we show that each orbit contains a matrix whose main diagonal starts with r 1s (where  $r\leq \min\{m,n\}$ ), and the rest of the matrix is zero. This r is unique: the action of G preserves the rank of the matrix, hence r equals the rank of A. In summary, there are exactly  $1+\min\{m,n\}$  orbits, and each orbit consists of matrices of rank r for some  $r\leq \min\{m,n\}$ . This works over any field, not just  $\mathbb {R}$ .—Emil J. 15:09, 8 October 2012 (UTC)[reply]
Thank sWidener (talk) 04:05, 9 October 2012 (UTC)[reply]

4 of a kind probabilities

In Texas Hold'em, what are the odds of hitting all 13 individual 4 of a kind hands?  — Preceding unsigned comment added by 66.215.136.120 (talk) 17:15, 8 October 2012 (UTC)[reply]

Zero.  You only get seven cards, so you can only hit at most one 4 of a kind hand.  Of course that's not the answer you are looking for, but if you want a better answer, you'll have to ask the question more clearly. Looie496 (talk) 20:30, 8 October 2012 (UTC)[reply]

If you are asking about the odds of doing so in 13 out of 13 sequential hands, then we can do that math, given the probability P, of getting 4 of kind in each individual hand, and assuming you return the cards to the deck and reshuffle after each hand, with no jokers or other wilds:

Hand  Odds of getting 4 of a kind you haven't already had
----  ---------------------------------------------------
  1      P
  2      P(12/13)
  3      P(11/13)
  4      P(10/13)
  5      P(9/13)
  6      P(8/13)
  7      P(7/13)
  8      P(6/13)
  9      P(5/13)
 10      P(4/13)
 11      P(3/13)
 12      P(2/13)
 13      P(1/13)

This multiplies out as P¹³(12!/13¹²) = (0.00002056)P¹³.  See Poker probability (Texas hold 'em) for some possible values to use for P (depends on several things).  If we assume you don't put the cards back in the deck after each hand, then the odds go up (the very last hand is guaranteed to be a 4 of a kind in the one remaining rank). StuRat (talk) 20:50, 8 October 2012 (UTC)[reply]

Number theoretic function -- growth, properties

Define  $g(n)$  to be the maximum value of k in Z such that k is divisible by every positive number less than or equal to its own nth root. I've calculated that  $g(1)=2,g(2)=24$ , but I'm not sure about good ways to bound g in order to calculate higher values. Does anyone have insight about the properties and growth of this function? Cheers. —Anonymous Dissident^Talk 00:56, 9 October 2012 (UTC)[reply]

2, 24, 420, 27720, 720720, 36756720, 5354228880, 481880599200, 25619985190800, 10685862914126400, 876240758958364800, 113035057905629059200, 24792356033967973651200, 9690712164777231700912800, 2364533768205644535022723200, 396059406174445459616306136000
oeis series A003102, the definition given is Largest number divisible by all numbers < its n-th root, but there's obviously no difference between the series for less than and the one for less than or equal to since a number is always divisible by it's root. See the page for references. Ssscienccce (talk) 01:39, 9 October 2012 (UTC)[reply]

Probability question

Hi,
If I am given a random number, r, that has been chosen uniformly between 1 and n, though n is unknown to me, it seems I ought to be able to say something about n (in a probabilistic sense). For example, if r = 2, then it seems intuitively more likely that n = 10 than n = 10^100000000000. However, I cannot seem to grasp a way to fully explain or quantify this. Is it just because there are "more big numbers than small ones"? E.g. I am unsure, in the same scenario, whether it is more likely that n = 10 than n > 10^100000000000. Can anyone shed any light on this problem? 86.160.83.247 (talk) 03:02, 9 October 2012 (UTC)[reply]

Stabilizer of cosets

What is the stabilizer of the coset  $[aH]$  for the operation  $G$  on  $G/H$ ? My answer is simply the set of all  $g\in G$  such that  $gaH=aH\implies a^{-1}gaH=H$ , so if g conjugated with a is in H then it is part of the stabilizer. Is there a more elementary way of expressing this? Widener (talk) 04:04, 9 October 2012 (UTC)[reply]

@@ Line 266: / Line 266: @@
 Let <math>M</math> be the set of <math>m \times n</math> matrices, and let <math>G = GL(\mathbb{R}^m)\times GL(\mathbb{R}^n)</math>, where an element <math>(P,Q) \in G</math> acts on an element <math>A \in M</math> by <math>(P,Q)A = PAQ^{-1}</math>. How do you decompose <math>M</math> into <math>G</math>-orbits? I see no general way of doing it, since you can have general invertible P and Q, and then general A. I can only get special cases, for example that the zero matrix in <math>M</math> is in an orbit by itself. [[User:Widener|Widener]] ([[User talk:Widener|talk]]) 13:48, 8 October 2012 (UTC)
 :The orbit of a matrix ''A'' consists of matrices that can be obtained from ''A'' by a sequence of [[elementary matrix|elementary row and column operations]]. We can get a unique representative of each orbit as follows. Start with ''A''. If there is a nonzero entry in the matrix, we can switch rows and columns to move this element to the upper left corner, and row multiplication to make this element 1. Then we use row and column addition to make all other entries in the first row and column zero. We repeat the same process with the <math>(m-1)\times(n-1)</math> submatrix omitting the first row and column. In this way, we show that each orbit contains a matrix whose main diagonal starts with ''r'' 1s (where <math>r\le\min\{m,n\}</math>), and the rest of the matrix is zero. This ''r'' is unique: the action of ''G'' preserves the [[rank (linear algebra)|rank]] of the matrix, hence ''r'' equals the rank of ''A''. In summary, there are exactly <math>1+\min\{m,n\}</math> orbits, and each orbit consists of matrices of rank ''r'' for some <math>r\le\min\{m,n\}</math>. This works over any field, not just <math>\mathbb R</math>.—[[User:EmilJ|Emil]]&nbsp;[[User talk:EmilJ|J.]] 15:09, 8 October 2012 (UTC)
+::Thank s[[User:Widener|Widener]] ([[User talk:Widener|talk]]) 04:05, 9 October 2012 (UTC)
 == 4 of a kind probabilities ==