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July 9

Ackermann function

The article gives A(1,n), A(2,n), A(3,n), and A(4,n), but what's A(m,n), without recursion? --138.110.206.101 (talk) 01:05, 9 July 2010 (UTC)

Didn't you notice the way it grew rapidly? It grows too rapidly to be representable using the functions and operations one normally comes across. Dmcq (talk) 01:16, 9 July 2010 (UTC)

The function is defined recursively. You can't change that. If there was a closed form expression for it, the article would give it (and the function wouldn't be very interesting, since it couldn't grow anywhere near as fast). --Tango (talk) 21:07, 9 July 2010 (UTC)

The fact that it's defined recursively doesn't mean that there isn't a closed form for it (e.g. Fibonacci numbers). --138.110.206.101 (talk) 21:32, 10 July 2010 (UTC)

Looking for a more intuitive way for expressing some property of some sets (and of a formula associated with them).

• Let $n$ be a natural number.
• For every $i\leq n$ let $S_{i}$ be a set.
• Let $\phi (x_{1},\ldots ,x_{n})$ be a well-formed formula (with the $n$ free variables as indicated).

Is there a more intuitive/common (or a shorter) way, for expressing the following property of $\{\phi ,S_{1},\ldots ,S_{n}\}$?

• • For every natural $j\leq n$ and for every $x_{j}\in S_{j}$, every $x_{1},\ldots ,x_{j-1},x_{j+1},\ldots ,x_{n}$ satisfying $\phi (x_{1},\ldots ,x_{j},\ldots ,x_{n})$ satisfy $x_{i}\in S_{i}$ for every natural $i\leq n$.

If you don't know of a more intuitive/common (or a shorter) way for expressing that property (n being general), then how about the simplest case of n=2 ? HOOTmag (talk) 10:43, 9 July 2010 (UTC)

Well, I would find it more intuitive stated as follows: if φ(x₁,...,x_n), then either x_i ∈ S_i for every i, or x_i  ∉ S_i for every i. YMMV.—Emil J. 11:48, 9 July 2010 (UTC)

Ok, thank you, but it's still a formal formulation, while I'm looking for a more intuitive expression.

Let's take the simplest case, where n=2. Do you know of an intuitive way for expressing the following property: every x,y satisfying φ(x,y) satisfy that x∈X is equivalent to y∈Y.

HOOTmag (talk) 13:03, 9 July 2010 (UTC)

n collectors bring samples of their coin collection to a bank. Each sample contains one each of some number of different types of coins. Some collectors may share coin types. The collectors each hand a clerk one coin from their collection. The clerk is new and mixes them all up. He needs to get the right type of coin back to the right collector, so he guesses which coins go where and computes phi, which tells him if he's entirely right or entirely wrong. Presumably he has another method of determining if he's entirely wrong--perhaps he remembered the second collector handed him a penny. Types of coins are used instead of coins themselves because coins are unique and not shareable among collectors, while types of coins are sharable. This is simpler if the $S_{i}$ are disjoint. Hopefully this is something like what you're after; I prefer Emil's formulation, personally. This would probably need to be reworked to fit whatever your application is, as well, since it's pretty far removed from the formal statements above. 67.158.43.41 (talk) 15:30, 9 July 2010 (UTC)

You've supplied an intuitive example for having to use the property I'm interested in, but I'm not looking for an intuitive example, but rather for a more intuitive - or a more common - description which may replace the formal description I've given for the property. HOOTmag (talk) 13:47, 11 July 2010 (UTC)

I do not know the answer but as a non-mathematician and following the Emil's formulation I would look for something like all-or-none... --CiaPan (talk) 13:24, 9 July 2010 (UTC)

Your suggestion is very similar to Emil's, except that yours uses no mathematical notations. HOOTmag (talk) 13:47, 11 July 2010 (UTC)

The solution set of $\phi$ is $\Pi S_{i}\cup \Pi {\overline {S}}_{i}$? —Preceding unsigned comment added by 203.97.79.114 (talk) 06:03, 11 July 2010 (UTC)

It's very similar to Emil's suggestion. Both of them are formal ways for describing the property I'm interested in, but I'm looking for a more intuitive - or a more common - way for describing that property. HOOTmag (talk) 13:47, 11 July 2010 (UTC)

Differential Forms

Let $\scriptstyle \Omega ^{n}(M)$ denote the space of differential $\scriptstyle n$-forms over a smooth manifold $\scriptstyle M.$ Furthermore, let $\scriptstyle E^{n}(M)$ denote the space of exact differential $\scriptstyle n$-forms over $\scriptstyle M.$ I would like to understand the quotient space:

$Q^{n}(M):=\Omega ^{n}(M)/E^{n}(M)\ .$

For example, is $\scriptstyle Q^{n}(M)$ connected? How could I calculate the homology groups of $\scriptstyle Q^{n}(M)?$ What about the homotopy groups of $\scriptstyle Q^{n}(M)?$ Even if you can't answer these question (which I can't either); any suggestions would be helpful. Thanks in advance. •• Fly by Night (talk) 19:13, 9 July 2010 (UTC)

speed cameras versus timestamped photos at 2 points.

The argument was made that speed cameras are fallible. So the city decided to put cameras at two ends of a road with timestamps (assume their clocks are properly synchronized).

If you were photographed at point A at 12:00:00 noon, and point B one minute later, at 12:01:00, where point B is 2 miles away, then it would seem like you were going two miles a minute, or 120 miles per hour. You would like to claim that you were going less than that. Is there any way, mathematically, that you could make the argument that being at A at noon and two miles down one minute later does not mean you had to go at 120 miles per hour or faster at any point along the trip? That you could go at various speeds, slowing down, accelerating, weaving and bobbing, and so end up at the specified point, but at no point in your stopping, going, accelerating, decelerating, backing into reverse, whatever else, would you have been going at 120 miles per hour or faster?

How would that mathematical argument look? Conversely what is the mathematical argument that you must have reached or surpassed 120 miles per hour at least at one point during the one minute interval? This is not homework. 84.153.230.67 (talk) 20:31, 9 July 2010 (UTC)

A distance-time graph makes it obvious to me that a constant speed between the points had to be 120 mph. Any other pattern will have to have had greater speeds somewhere. The UK approach is to put a limit on average speed between specified points, which is just distance over time regardless of any variation over the stretch.→86.160.105.64 (talk) 20:53, 9 July 2010 (UTC)

(edit conflict) To answer your first question: No. There is no way that you can be in one place at one time and then two miles away one minute later without travelling at at least two miles per minute. There's no need for a mathematical argument; it's just common sense. Let's assume you go in a straight line. Imagine your friend drives at a constant 120 mph and you decide to accelerate and decelerate. If you set off faster than your friend that you have already gone faster than 120 mph. If you travel slower than 120 mph at any point then your friend will go into the lead. If you want to catch him up then you will need to travel faster than he is, i.e. drive faster than 120 mph. If you bob and weave then you increase your distance and then you will have to travel further, so you will have to travel even faster to get to the same point at the same time as your friend. Remember:

${\text{speed}}={\frac {\text{distance}}{\text{time}}}$

As for the second question: it is possible for you to travel above 120 mph, but still be clocked as travelling at 120 mph. Assume the road has a width of w miles. You drive diagonally from (0,0) to (w,2), and it takes you one minute. Using Pythagoras' Theorem you have

${\text{speed}}={\sqrt {4+w^{2}}}\,{\text{mpm}}=60{\sqrt {4+w^{2}}}\,{\text{mph}}>120\,{\text{mph}}$ •• Fly by Night (talk) 21:06, 9 July 2010 (UTC)

The argument that proves you must have been going at 120mph at some point between A and B is the Mean value theorem. --Tango (talk) 21:12, 9 July 2010 (UTC)

I'll have to admit I wondered a while ago about having two similar cars and putting the same false numbers on both, or even just copying one, and arranging them to pass two cameras like that where it would be obviously impossible for them to go at speed. I suppose it would count as wasting time at best and they would take a dim view of it. :) Dmcq (talk) 21:32, 9 July 2010 (UTC)

Driving with false plates isn't exactly legal... --Tango (talk) 22:25, 9 July 2010 (UTC)

But who has not done it at least once? PST 10:35, 10 July 2010 (UTC)

Uncountable millions of people.→81.147.2.107 (talk) 14:03, 10 July 2010 (UTC)

You should get out more and meet some people. I recall the days when my grandpa used to take me out in his car, but I never quite understood why he had so many car plates with different numbers in his garage ... ;) PST 14:48, 10 July 2010 (UTC)

But now you're all grown up you do understand why your grandfather found it necessary? What was it - Resistance, gun-running, drugs, ...?→81.147.2.107 (talk) 23:35, 10 July 2010 (UTC)

None of these. My grandfather was never in "the business". He just wanted to impress his neighbor who was. But my grandmother was a different story ... PST 02:56, 11 July 2010 (UTC)

Disclaimer: "My grandfather" and "My grandmother" are entirely fictitious characters. Any resemblence they bear to real people, living or dead, is purely coincidental. PST 02:56, 11 July 2010 (UTC)

I should clarify: the MVT proves that you must have been going at exactly 120mph at some point. A straight line being the shortest distance between two points (Pythag. proves that) is all you need to prove that you must have been going at least 120mph at some point. --Tango (talk) 22:28, 9 July 2010 (UTC)

How does a straight line being the shortest distance between two points prove anything here? At least the MVT requires differentiability. What if the car travels your straight line but the graph of its displacement versus time is continuous everywhere but differentiable only almost everywhere such that the car had velocity 0 at all points where its velocity is defined. What does Pythag. have to say about that? -- 124.157.197.248 (talk) 14:17, 10 July 2010 (UTC)

It means the ride must have had infinite accelleration at some points. Very jerky, it wouldn't do your neck any good. Dmcq (talk) 14:28, 10 July 2010 (UTC)

As with instantaneous velocity, acceleration (and all higher derivatives) is 0 almost everywhere and undefined elsewhere, but given small enough time intervals, arbitrarily large average velocities can be found arbitrarily close to 0 velocities, yielding you your points of "infinite [average] acceleration". My point was not to suggest a physically possible mode of travel while standing still, but to question Tango's assertion that the MVT is overkill. -- 124.157.197.248 (talk) 01:29, 11 July 2010 (UTC)

I don't understand what "a straight line being the shortest distance" has to do with this. What you really need is some integral inequalities applied to the car's velocity and speed. If the speed was bounded by $M<120\mathrm {MPH}$, then you would have

$2\mathrm {miles} =\left\|\int _{0}^{1\mathrm {min} }\gamma '(t)\,dt\right\|\leq \int _{0}^{1\mathrm {min} }\|\gamma '(t)\|\leq \int _{0}^{1\mathrm {min} }M\,dt=M\mathrm {min} <2\mathrm {miles} .$

-- Meni Rosenfeld (talk) 04:58, 11 July 2010 (UTC)

A caveat. If the road is curved, and the distance between the cameras is measured along the road, then the car can do this at less than 120MPH by going off-road. It will probably not be any more legal, though. -- Meni Rosenfeld (talk) 04:58, 11 July 2010 (UTC)

what about wide roads and keeping tight corners??

If you have a very, very, very long and curvy road, and it is also wide, then could you keep tight in taking the corners, and thereby, say, have two time-stamping cameras 30 minutes apart that "clock" you as going 90 MPH (as measured along the line going down the road, not as the bird flies) "be wrong", as you were only going 75, just keeping tight in corners? 92.230.71.228 (talk) 21:48, 11 July 2010 (UTC)

Not as big a difference as that (unless the road is almost as wide as it is long), but yes, it could make perhaps a fraction of one mph difference if the police measure the distance along the white line. I don't think you would be prosecuted for exceeding the limit by less than 1 mph, but you might be accused of dangerous driving if you go round corners at high speed on the wrong side of the road. Dbfirs 22:15, 11 July 2010 (UTC)

July 10

Online virtual Turing machine

I'm working on Turing machine programs, but after a while it gets tedious to go through all of the steps manually. Is there someplace online which will allow me to enter a Turing machine program and the input value, and will then simulate the Turing machine? --138.110.206.101 (talk) 21:24, 10 July 2010 (UTC)

Googling "Turing machine simulator" throws up a few candidates. Gandalf61 (talk) 17:33, 11 July 2010 (UTC)

Cosets

Hi. I am currently battling my way through a book on Group Theory and am having a bit of trouble with cosets. We have already stated that for a group G with a subgroup H and an index set I with the element $t_{i}$ being an element of the ith coset, $G=\bigcup _{i\in I}Ht_{i}$. We are now attempting to prove that $G=\bigcup _{i\in I}{t_{i}}^{-1}H$ is a decomposition of G into distinct left cosets. The first observation made is that if ${t_{i}}^{-1}H={t_{k}}^{-1}H$ then we must have ${t_{k}}{t_{i}}^{-1}\in H$ and hence $H{t_{k}}=H{t_{i}}$, which is impossible unless i=k. How exactly have we established $H{t_{k}}=H{t_{i}}$? Is it simply by saying that because we know ${t_{k}}{t_{i}}^{-1}\in H$, then $H{t_{i}}$ and $H{t_{k}}$ must be in H also and so we can form the respective left cosets and set them equal to each other? Or have we performed some inversion such as $(tH)^{-1}=Ht^{-1}$, where H is its own inverse, of sorts? That suggestion may be complete and utter rubbish; just remember, I'm new to this. Then it says i=k; is this just using that two cosets are either identical or have no element in common? I have more questions but I don't want to make this too long and scare off and potential readers so I'll save them until I have a response! Thanks asyndeton talk 21:53, 10 July 2010 (UTC)

You are right (informally) that you invert both sides of ${t_{i}}^{-1}H={t_{k}}^{-1}H$ to get $Ht_{i}=Ht_{k}$. To prove this, take $g\in Ht_{i}$, then $g=ht_{i}$ for $h\in H$. Then $g^{-1}=t_{i}^{-1}h^{-1}\in t_{i}^{-1}H=t_{k}^{-1}H$, so $g^{-1}\in t_{k}^{-1}H$... see where this is going? Finish that argument (to show $g\in Ht_{k}$), and you'll have shown $Ht_{i}\subseteq Ht_{k}$. The converse inclusion is exactly the same, with i and k swapped. Staecker (talk) 23:13, 10 July 2010 (UTC)

Thanks Staecker, I found that very helpful! It was also quite illuminating for the other questions I was planning to ask, so I shan't need to post them. Thanks again! asyndeton talk 13:11, 11 July 2010 (UTC)

1+1=1

I was surprisingly asked this question in an Arabic forum:

When can 1+1=2? I mean 1+1=1?

I answered: Impossible mathematically, but might be possible if it were a philosophical puzzle.

I knew then the one asked me had a Ph. degree in Maths and said it was possible in Mathematics and as an exceptional case related to Relativity theory. Does this make sense? Help me please! --Email4mobile (talk) 22:08, 10 July 2010 (UTC)

You probably mean: when can't 1+1=2 ? i.e. when can 1+1≠2, right?

Since you're looking for an answer in Relativity theory, so I think it's really rather simple and trivial (for anybody who has learnt that scientific discipline): let's assume that - when travelling at a speed of 1 mile per minute - you meet another vehicle travelling to the opposite direction at a speed of 1 mile per minute; then you won't see it travel at a speed of 2 miles per minute, although this is what should be expected intuitively.

The reasoning behind this phenonemon is that the mathematical operation required here is not a simple "addition" (+), so it's not really 1+1≠2...

HOOTmag (talk) 22:30, 10 July 2010 (UTC)

I think that if ZFC is inconsistent and you base your definitions of integers on sets, then "1+1≠2" is provable.

Other than that - if by 1, 2 and + you mean the integers and integer operation, then no, 1+1 can't be unequal to 2. But mathematical notation is overloaded, and you can use those symbols to mean different things. One way is to use + to mean some form of relativistic addition as explained by HOOTmag. This is usually defined for numbers less than 1, corresponding to fractions of the speed of light, as

$u\oplus v={\frac {u+v}{1+uv}}$

You can also have a probabilistic addition, corresponding to disjunction of independent events:

$u\oplus v=u+v-uv$

For general positive numbers you also have Pythagorean addition:

$u\oplus v={\sqrt {u^{2}+v^{2}}}$

All of these operations might be denoted by "+" itself if the context makes it more convenient.

Also, 1+1 can be -1, 0, 1, 10 or 11 depending on whether you use a field of characteristic 3, 2, boolean algebra, binary base or gray code. -- Meni Rosenfeld (talk) 05:14, 11 July 2010 (UTC)

I really am sorry for not revising what I wrote (probably because of that PhD). I meant 1+1=1? Can I/someone correct the title?

You can change the title when you edit the section - it's at the top of the edit window. I've done it for you.

So yes, 1+1=1 in relativistic and probabilistic addition, and in boolean algebra. Also in the real numbers modulo 1. -- Meni Rosenfeld (talk) 05:16, 11 July 2010 (UTC)

The far I know is that in Boolean operations + in this context refers to OR, not really addition. Addition in binary logic is still the same as any mathematical basic operation (i.e. 1 + 1 = 10 in binary system).

Regarding relativistic and probabilistic addition, I'm not well aware but I suppose there should be similar explanation or Mathematics would be explained in many wrong ways.--Email4mobile (talk) 05:42, 11 July 2010 (UTC)

Read my post again. "Addition" and "+" can refer to many things. When they refer to integer addition then yes, 1+1=2 and 1+1≠1. When they refer to something else the result can be different. You need to know what the "+" means whenever you encounter it (in the vast majority of cases it will be integer\real addition). And yes, it's the same addition in the binary\gray examples, just a different notation for the numbers. -- Meni Rosenfeld (talk) 06:33, 11 July 2010 (UTC)

Thanks Meni Rosenfeld. I was just afraid if there were really some exceptions in mathematics regarding basic operations. When it comes into notations then I will assume the man who asked me was absolutely right, although he should not have paid much attention for tht.--Email4mobile (talk) 08:29, 11 July 2010 (UTC)

Anyways, the man who (according to your testimony) is looking for a "relativistic" answer, has undoubtedly meant that - when you travel at the speed of light and you meet another vehicle travelling to the opposite direction at the same speed, then you won't see it travel at twice the speed of light, but rather at the speed of light...HOOTmag (talk) 11:46, 11 July 2010 (UTC)

Did I miss it, or did no one mention the additive trivial group?—msh210℠ 15:26, 13 July 2010 (UTC)

As an additive group, it doesn't have a 1 but rather a 0. 0+0=0 is not very unusual. -- Meni Rosenfeld (talk) 17:58, 13 July 2010 (UTC)

D'oh! You're right, of course, by convention (though naturally what symbol is used for the element of a trivial group is arbitrary, and '1' (or '5', or '}', or anything) can be used as well as '0').—msh210℠ 18:20, 13 July 2010 (UTC)

Girls, buses, cats AGAIN.

Hi all,

On a late-night gambling game show, the old chestnut was posed:

Five girls are traveling by tourist bus. Each girl has 5 baskets, each basket has 5 cats, and each cat has 5 kittens. The bus stops and 3 girls get off. How many legs are on the bus?

I figured, great, I'll use some of my free cellphone money that keeps piling up to send the R7.50 SMS to try my luck. It's obvious: Each basket has 5 cats +5*5 kittens = 30 cats = 120 legs, so each girl has 120*5 = 600 legs + 2 of her own = 602; 2 girls mean 1204 legs, duh.

Someone guesses that before they call me, and it's wrong. Oh, right, sneaky! The girls get off, leaving their cats behind, so 3004 legs left. Wash, rinse, repeat. Oh, right, the driver! 3006 then. Someone guesses that too. Also wrong. Same with 1206 (assuming a driver, but the girls take their livestock with them.)

The answer they finally gave was 2708. What? I can't figure out how that came to be. Any ideas? --Slashme (talk) 23:50, 10 July 2010 (UTC)

err... I hate to point it out, but there aren't any legs on the bus. Busses have wheels. --Ludwigs2 23:57, 10 July 2010 (UTC)

No idea. Well, a couple. Maybe there were 751 boys on the bus? Or, a shipment of prosthetic legs in the cargo hold? By the way, was this the 3:42 for St. Ives? I hate it when I have to take that bus.... --Trovatore (talk) 00:02, 11 July 2010 (UTC)

Trovatore alluded to the rhyme As I was going to St Ives. -- Wavelength (talk) 00:58, 11 July 2010 (UTC)

I'm pretty sure it's a scam. 74.15.137.192 (talk) 00:41, 11 July 2010 (UTC)

SlashMe, if this is that BrainBox game that comes on TV at night then it's an absolute scam and you've just wasted your airtime. A few months ago I read an article in the Cape Argus about how what they're doing is not technically illegal (they always award the prize if someone figures out the convoluted reasoning) but the puzzles themselves use so many unnamed assumptions that they are basically bogus and just a front to scam people of their money. Regards. Zunaid 06:56, 11 July 2010 (UTC)

OK, thanks everyone. I guess it was a school bus: I got schooled! --Slashme (talk) 08:08, 11 July 2010 (UTC)

There's a 'Running Car' from 'Origami to Astonish and Amuse' by Jeremy Shafer, at [1], it's the last model on the page. The car has four legs, but he's known for being truly demented and I don't think it would work full scale ;-) Dmcq (talk) 11:34, 11 July 2010 (UTC)

July 11

Simple Probability

Suppose that there have been 20 no-hitters in baseball history. What is the probability that there's been at least one on each day of the week?

The case for 7 no-hitters is simple enough, but I'm having trouble extending this to 8+ days. Help would be greatly appreciated. 74.15.137.192 (talk) 00:33, 11 July 2010 (UTC)

I only have time for this quick answer: I wonder if Schroedinger's method would help with this one? Michael Hardy (talk) 02:16, 11 July 2010 (UTC)

Isn't this what Stirling numbers of the second kind are for? $(S(20,11)*7!)/(7^{20})$should do it. —Preceding unsigned comment added by 203.97.79.114 (talk) 04:52, 11 July 2010 (UTC)

Shouldn't it be S(20,7)*7!/7^20? (There are seven days of the week, right?) 74.15.137.192 (talk) 09:22, 11 July 2010 (UTC)

This is the Coupon collector's problem. Robinh (talk) 19:27, 14 July 2010 (UTC)

ordinals

I'm interested mostly in countable ordinals in this question, in fact computable ones, I think. Is there a concept of a limit ordinal that represents a fixed point of the (informally) biggest operation seen so far? By that I mean ω is such an ordinal, and maybe 2ω and ω^ω are as well (not sure about further iterations of exponentiating), and the next thing after that is ε₀, but something like ω³+ω²·2+ω·3 would not be, even though it's a limit ordinal. The next thing after ε₀ would be ε₁ or maybe Γ₀. I don't mean a singular ordinal since there's only one of those for each cardinal. Am I making any sense? Maybe what I'm looking for is the places where various reasonable ordinal notations run out of gas. Thanks. 71.141.88.179 (talk) 01:16, 11 July 2010 (UTC)

The question does not make much sense without specifying rigorously which operations you take into consideration. It's fairly trivial to cook up an artificial ordinal notation system which "runs out of gas" at ω³ + ω²·2 + ω·3.—Emil J. 13:13, 12 July 2010 (UTC)

Thanks, I'm just asking if there is a way to formalize the usual informal way of explaining ordinals. You know, you've got 1,2,3...ω,ω+1,...,ω·2,...ω·3...ω²...ω³...ω^ω...ω^ωω...ε₀...

For each place that "..." appears in the above, what comes next? Obviously I've missed some intermediate ones, but the idea is that at each stage we've iterated some newly concocted ordinal operation and found a fixed point for it, as opposed to just finding the nearest limit ordinal by iterating the successor operation. Maybe each one corresponds to the order type of some natural combinatorial structure? 71.141.88.179 (talk) 19:28, 12 July 2010 (UTC)

Simplifying Radical Expressions

Hello. How would you simplify ${\sqrt {11+6{\sqrt {2}}}}$ to $3+{\sqrt {2}}$? If you let ${\sqrt {11+6{\sqrt {2}}}}=a+b$, how can you identify 11 as a sum of squares, $a^{2}+b^{2}$ and not anything else? Thanks in advance. --Mayfare (talk) 03:18, 11 July 2010 (UTC)

You can check the identity by letting $a={\sqrt {11+6{\sqrt {2}}}},b=3+{\sqrt {2}}$, computing a² and b², and checking whether they are equal. $11=3^{2}+({\sqrt {2}})^{2}$, a perfectly good sum of two squares in the extension field $\mathbf {Q} [{\sqrt {2}}]$.

71.141.88.179 (talk) 03:29, 11 July 2010 (UTC)

See Nested radical for the conditions where this can be done for nested square roots. Dmcq (talk) 11:25, 11 July 2010 (UTC)

In general, to find ${\sqrt {p+q{\sqrt {2}}}}$ where p and q are rational you proceed as follows:

${\sqrt {p+q{\sqrt {2}}}}=r+s{\sqrt {2}}$

$\Rightarrow p+q{\sqrt {2}}=r^{2}+2rs{\sqrt {2}}+2s^{2}$

$\Rightarrow p=r^{2}+2s^{2}{\text{ ; }}q=2rs$

$\Rightarrow 2r^{4}-2pr^{2}+q^{2}=0$

$\Rightarrow r^{2}={\frac {p\pm {\sqrt {p^{2}-2q^{2}}}}{2}}{\text{ ; }}s^{2}={\frac {p\mp {\sqrt {p^{2}-2q^{2}}}}{4}}$

The identification of coefficients in line 2 not only requies p and q to be rational, but also r² and s² must also be rational, and so ${\sqrt {p^{2}-2q^{2}}}$ must be rational too. Gandalf61 (talk) 17:26, 11 July 2010 (UTC)

Why does ${\sqrt {a^{2}-b^{2}c}}$ have to be an integer if some nested radical, ${\sqrt {a+b{\sqrt {c}}\ }}$, can be denested into a sum of surds? Thanks again in advance. --Mayfare (talk) 20:35, 12 July 2010 (UTC)

Angle

• Hi, How is the angle=arc/radius?Max Viwe | Wanna chat with me? 04:49, 11 July 2010 (UTC)

Radian should get you started on this. -- Meni Rosenfeld (talk) 05:22, 11 July 2010 (UTC)

Automorphism group of a Hilbert space

What is the largest subgroup of the diffeomorphism group of a 2n-1 sphere that is holomorphic when the sphere is taken to be the unit sphere of an n-dimensional complex space? 74.14.108.160 (talk) 06:58, 11 July 2010 (UTC)

What exactly are you looking for here? It doesn't make sense to say that a subgroup of maps that are also holomorphic? Do you mean a subgroup of holomorphic maps? Are you considering the diffeomorphism group of the sphere (as an abstract manifold) or the subgroup of the diffeomorphism group of the ambient space which preserves the sphere (as a submanifold of the ambient space)? •• Fly by Night (talk) 19:47, 14 July 2010 (UTC)

Yes, I meant to say the subgroup whose members are holomorphic maps rather than calling the subgroup itself holomorphic. The sphere was meant to be an abstract manifold and was only embedded in the complex space in order to give diffeomorphisms of the sphere a holomorphicity criterion. 76.67.74.129 (talk) 07:54, 16 July 2010 (UTC)

Mathematical induction

The article we have on mathematical induction seems to imply that assumption is an integral part of the process. This doesn't appear right to me. Statements made about an arbitrary k during the inductive step are only allowable because the existence of at least one suitable k is proven in the basis step. No speculation or assumption is done. Now, to the right person, there is no difference between "Let k be an integer for which the statement holds..." and "Assume the statement holds for some k..."; but to another reader, it's the difference between speculation and logical argument. The prevalence of the word assume in textbooks, at least over here, is part of the difficulty people seem to have with the legitimacy of induction. Am I being too picky or misunderstanding things, or is this something that should be looked at? Thanks for the help. —Anonymous Dissident^Talk 13:33, 11 July 2010 (UTC)

I always used to say "suppose it is true for some k" ... I also explained (once, not every time) that use of a logical contrapositive makes the argument clear and finite (rather than infinite induction). Dbfirs 22:00, 11 July 2010 (UTC)

The article explains the concept in several different ways so I think most readers will understand at least one version. My experience teaching this is that it's something of a pons asinorum in that there are always going to be a few people who have a very difficult time getting it. Keep in mind though that Wikipedia isn't intended to teach the subject from scratch, so a completely foolproof explanation isn't the goal.--RDBury (talk) 23:13, 11 July 2010 (UTC)

It's always a bit hard to find the right wording to explain this point when teaching induction, but once you see it it's not a difficult concept at all. The issue is that, yes, you make an assumption, but only temporarily. In logical jargon the assumption is discharged before the end of the proof, and doesn't count as an assumption for the proof as a whole.

To prove $\forall nP(n)$, you have to prove two things: First, you have to prove $P(0)\,\!$; how you do that doesn't concern us here. More to the point, you need to prove $\forall k\left(P(k)\implies P(k+1)\right)$.

How do you do that? Well, how do you prove $A\implies B$ in general? You assume $A\,\!$ and conclude $B\,\!$. The assumption of $A\,\!$ is truly an assumption, but only for the length of time that you're in the little subproof of $B\,\!$. Once you're done with that, $A\implies B$ has been proved and the assumption is discharged (that is, it's no longer an assumption).

Exactly the same deal for the assumption of $P(k)\,\!$ when proving $P(k+1)\,\!$.

(Aside to Dbfirs: I don't think saying "suppose" rather than "assume" helps at all; they're synonyms.) --Trovatore (talk) 22:46, 11 July 2010 (UTC)

Not if you say "suppose we can find a value of k for which the statement is true". It conveys a different impression of the argument, avoiding the (unfounded) suspicion that we are assuming what we set out to prove. Dbfirs 07:26, 12 July 2010 (UTC)

Hmm, possibly. But they still have to deal with the proof-within-a-proof thing, and specifically that the proof-within-a-proof may have different assumptions than the whole proof, one way or another.

I think this is the same issue people have with proof by contradiction — there's a mental leap involved in "assuming" something that, come to find out, is not actually true. You see it over and over again in objections to, say, Cantor's diagonal argument; people will say "Cantor assumes you can list the real numbers! But you can't!", and not notice that that's actually the thing being proved. --Trovatore (talk) 07:46, 12 July 2010 (UTC)

"Suppose" and "Assume" are basically synonyms, so I don't see any real improvement. --Tango (talk) 09:24, 12 July 2010 (UTC)

This is exactly what the OP is talking about- to the uninitiated, these two words are not synonyms. In common language, "suppose" is somewhat of a hypothetical assertion of something which may or may not be true (this is the meaning mathematicians want), while "assume" means something is asserted without proof as being true in some wider or objective context (this is basically never what mathematicians mean). People beginning to read serious mathematics need to be taught that "assume" means "suppose", and it's natural for them to get confused otherwise. Staecker (talk) 11:58, 12 July 2010 (UTC)

Mathematical inductions consists of two steps: step 1: P(0), step 2: P(n)==>P(n+1). one of these is called the induction. A common misunderstanding is to call part 2 the induction. Logically it is part 1 which is the induction step. Induction is concluding from special cases. Bo Jacoby (talk) 07:44, 12 July 2010 (UTC).

I would say that the "induction" is putting the two steps together to reach a conclusion. Dbfirs 19:19, 12 July 2010 (UTC)

The word 'induction' is much older than the idea of 'mathematical induction'. So 'mathematical induction' got its name for a reason. That reason being that is contains an element of good oldfashioned philosophical induction. That element being P(0). Bo Jacoby (talk) 21:26, 12 July 2010 (UTC).

I'm skeptical of the claim that the name "induction" was applied because the method was thought of as philosophical induction from a single example. --Trovatore (talk) 23:40, 12 July 2010 (UTC)

Why else call it induction, if not because it contains an element of induction? (Although a single example is theoretically sufficient, in practice you also try P(1) and P(2) and P(3) to give a hint about the truth of the assertion before you work on the P(n)==>P(n+1) argument). Bo Jacoby (talk) 04:47, 13 July 2010 (UTC).

I too am skeptical of this. Do you have any evidence beyond your own inability to think of another explanation? Algebraist 09:33, 13 July 2010 (UTC)

Induction is concluding a general statement from observed special cases (and more generally, any form of reasoning that usually works but is not strictly guaranteed to be correct), but not from a single special case. One would check P(0), P(1), P(2), P(3), etc., and after reaching a sufficiently (depending on the application and desired level of confidence) high number and getting sufficiently bored, one would conclude that P(n) for every n by induction. Mathematical induction, which is indeed named after this "philosophical" induction (it does not actually have much to do with philosophy, it's the usual way of inferring general rules in natural sciences like physics), does the same in a sense, except that it involves a property (a proof of P(n) → P(n + 1)) which enables to generate an argument for all these special cases P(0), P(1), P(2), etc. uniformly (by chaining instances of the implication), and leads to the desired conclusion in a mathematically rigorous way. Claiming that the P(0) step is the induction is nonsense. As for calling P(n) → P(n + 1) the "induction step", this does not quite correspond to the origin of the term "induction" as described above, but that's irrelevant. In mathematics it is established terminology, with obvious etymology (it is the main step in what is called "(mathematical) induction", so we can as well call it the "induction step"), and there is nothing wrong with it.—Emil J. 13:21, 13 July 2010 (UTC)

Maybe it's language-dependent, but in Hebrew the two parts are often called "induction base" and "induction step" (בסיס האינדוקציה, צעד האינדוקציה), and I took the etymology to be: Both parts together constitute (mathematical) induction; the first is where we start, and the second is what allows us to step from each number to the next. -- Meni Rosenfeld (talk) 15:02, 13 July 2010 (UTC)

Gabriel's Horn

In Gabriel's Horn are the diagrams wrong? The text says x>=1 yet the diagrams seem to do x>0. -- SGBailey (talk) 14:16, 11 July 2010 (UTC)

Well the diagram does do some other shape than 1/x between 0 and 1, but it would be better to have a diagram that starts at x=1. I'd like one that showed the opening a bit so it looked more horn like rather than 2d. I'm surprised there's no reference to the problem of painting it. You only need a finite amount to fill the inside but would need an infinite amount to paint the outside. Dmcq (talk) 16:10, 11 July 2010 (UTC)

The outside may have infinite area, but a finite amount of paint of zero thickness would be required to cover it - just get two of them vertically. fill one with paint, fit the other inside then lift it out.→86.132.163.37 (talk) 15:46, 13 July 2010 (UTC)

July 12

order of growth

To show that log x grows more slowly then x^c (c > O) is it sufficient to show that log x/x^c goes to 0? How does that mathematically imply that log x is less then x^c for sufficiently large x. Thanks-Shahab (talk) 06:38, 12 July 2010 (UTC)

I can answer the second part. If $\lim _{x\rightarrow \infty }{\frac {\log x}{x^{c}}}=0$, then for every $\epsilon >0$ there exists $N\in \mathbb {R}$ such that for all $x>N$, $\left|{\frac {\log x}{x^{c}}}\right|<\epsilon$. Choosing $\epsilon =1$ gives us that for large enough $x$ (larger than a finite positive real number $N$), ${\frac {\log x}{x^{c}}}<1$, which implies that $\log x<x^{c}$ (since $x^{c}>0$ whenever $x$ is positive). RJaguar3 | u | t 06:44, 12 July 2010 (UTC)

Thanks for answering both parts.-Shahab (talk) 06:50, 12 July 2010 (UTC)

Calculus/Derivatives

Find the equations of both lines that are tangent to the curve y=1+x^2 and are parallel to the line 12x-y=1

—Preceding unsigned comment added by KRmiwaD93 (talk • contribs) 07:38, 12 July 2010 (UTC) I need some help finding the equation. Could I have an explanation please? Thanks.

What have you tried so far? 71.141.88.179 (talk) 08:08, 12 July 2010 (UTC)

The question didn't actually ask for both lines did it? Are you sure it didn't say the line that is both tangent to.. and parallel to ...? Dmcq (talk)

That's the way I read it. —Anonymous Dissident^Talk 08:45, 12 July 2010 (UTC)

If the line has to be parallel to $12x-y=1$ then it must have the same gradient. You then need to find out at which value of x the curve $y=1+x^{2}$ has that gradient. Readro (talk) 09:17, 12 July 2010 (UTC)

Well one can't have 'both lines' as no two points on the quadratic have the same tangent direction whereas a straight line only has the one tangent direction. Dmcq (talk) 11:32, 12 July 2010 (UTC)

In general, there is only one tangent to a given parabola that is parallel to a given line. The exception is that there are no tangents parallel to the axis of the parabola.--RDBury (talk) 13:36, 12 July 2010 (UTC)

Rewrite the linear part as $y=12x-1$, and then you mentioned derivatives in the title, what does differentiating with respect to x give and what use is that? Dmcq (talk) 14:33, 12 July 2010 (UTC)

Trig question

Resolved

I'm having a bit of trouble trying to solve the following identity verification:

[ (sec - tan)^2 + 1 ] divided by [(sec)(csc) - (tan)(csc)] entire fraction set equal to 2tan

I have gotten as far as converting the denominator in terms of sine and cosine, but I'm a bit stuck from there. Any help is appreciated! Thanks, WordyGirl90 12:07, 12 July 2010 (UTC)

If you express the numerator in terms of cos and sin too it works out pretty simply - don't forget a famous identity involving the squares of cos and sin. AndrewWTaylor (talk) 12:57, 12 July 2010 (UTC)

If you divide said famous identity by the square of cos x then you get another identity in terms of tan and sec, which you can use if you fancy cancelling some terms first before converting to sin and cos. Readro (talk) 13:06, 12 July 2010 (UTC)

Thanks guys. A classmate solved it basically through factoring. I ended up using his method. Thanks anyway! WordyGirl90 19:58, 12 July 2010 (UTC)

integers solutions

Resolved

Hello. I have the following equation in three variables: 3x + 4y = 7z. I want to find out solutions to it which satisfy the following criteria: one they are all positive integers, secondly, they are all distinct, thirdly x,y,z are all less then 30. The only ways I know are to set it up as an integer programming problem, and use a software which I have, or to write a computer program which uses brute force. I want to then find out solutions for 2x + 5y = 7z, and in general I want to find out a way to solve ax + by = (a+b)z. What would be the appropriate way? Thanks-Shahab (talk) 14:25, 12 July 2010 (UTC)

$x\equiv y{\pmod {7}}$. For the general case, $x\equiv y{\pmod {a+b}}$. Gandalf61 (talk) 14:40, 12 July 2010 (UTC)

...assuming (without loss of generality) that a and b are coprime.—Emil J. 14:47, 12 July 2010 (UTC)

To write it more explicitly, integer solutions of the equation are exactly triples of the form (x, x + k(a + b), x + kb), where x and k are integers.—Emil J. 14:51, 12 July 2010 (UTC)

Gandalf61 how do I solve $x\equiv y{\pmod {7}}$ and from where did you get it. Thanks to you both for responding so fast.--Shahab (talk) 15:18, 12 July 2010 (UTC)

$x\equiv y{\pmod {7}}$ just means that x and y leave the same remainder when you divide them by 7 - see modular arithmetic. For a parametric solution, just pick two integers x and k and use Emil J.'s solution (x, x + 7k, x + 4k). Geometrically, these solutions all line on the plane spanned by the vectors (1,1,1) and (0,7,4). Gandalf61 (talk) 15:46, 12 July 2010 (UTC)

No, no...I mean how did you deduce that x,y such that $x\equiv y{\pmod {7}}$ give the solutions.-Shahab (talk) 16:37, 12 July 2010 (UTC)

On the one hand, if 3x + 4y = 7z, then 7 | 3x + 4y, hence also 7 | 4(y − x). As 7 and 4 are coprime, this implies 7 | y − x, i.e., $x\equiv y{\pmod {7}}$. On the other hand, if this conguence holds, then z = (3x + 4y)/7 = x + 4(y − x)/7 is an integer, and solves 3x + 4y = 7z.—Emil J. 16:43, 12 July 2010 (UTC)

Thanks, its all clear now-Shahab (talk) 16:51, 12 July 2010 (UTC)

July 13

the value of 0/0

i have a doubt for long days. what is the value for 0/0??? upto my knowledge i have guessed that this will have three values as shown below....could you please give me the right answer????

case (i):

 0/0=0

reason: zero divided by any number is equal to zero.

case(ii):

 0/0=1

reason: any number divided by the same number is equal to zero.

case(iii):

 0/0=infinite

reason: any number divided by zero is equal to infinite.

i have asked this question to many teachers. all said that the answer is infinite.....but they can't explain why we shouldnot consider the remaining cases.....please explain me —Preceding unsigned comment added by 122.183.208.130 (talk) 14:35, 13 July 2010 (UTC)

Division by zero is undefined. So, it does not have a value. You can argue that it is any value you like. So, in a sense, division by zero produces EVERY answer possible. -- kainaw™ 14:37, 13 July 2010 (UTC)

You mean division of zero by zero, I suppose. It does not make sense to make the value of x/0 a finite real number for nonzero x.—Emil J. 14:44, 13 July 2010 (UTC)

See indeterminate form. --Tango (talk) 14:40, 13 July 2010 (UTC)

In computing dividing one floating point 0 by another will give a special value called Not a Number, which basically is saying it is indeterminate. In reasoning about values in automatic proofs it is liable to be set to Bottom type which is yet another way of saying the same thing - but then again one might mean any value rather than no value. In some circumstances a value can be defined for it for instance x/x is indeterminate when x'=0 but it has a limit value of 1 at that point.You might as well ask how many angels can dance on the head of a pin? Dmcq (talk) 14:58, 13 July 2010 (UTC)

If someone with computer graphing skills can upload a graph showing together the three equations 0/x = y, x/x = y, and x/0 = y, then we might see where they lead for 0/0 = y.—Wavelength (talk) 15:11, 13 July 2010 (UTC)

There is nothing to graph here. For $x\neq 0$ we have $0/x=0,\ x/x=1$, and in the real projective line $x/0=\infty$. You also have $(2x)/x=2$ and $x/(7x)=1/7$. All of these approach the indeterminate form 0/0, and because of this, there is no natural definition for it. -- Meni Rosenfeld (talk) 15:23, 13 July 2010 (UTC)

Division by zero is also relevant. -- Meni Rosenfeld (talk) 15:24, 13 July 2010 (UTC)

Your teacher, even allowing them to be a bit sloppy, isn't right to say "infinite". More often the answer is something like "unknown". Imagine that you're trying to calculate the speed of an object (which is travelling at a steady speed) using the equation speed = distance / time. If you check how far its travelled after 2 seconds and find it has travelled half a metre, then its speed is 0.5 / 2 = 0.25 metres per second. If you had checked after just half a second you would find it had travelled an eighth of a metre, so its speed is still 0.125 / 0.5 = 0.25 metres per second. If you check how far it has travelled after zero seconds i.e. after no time at all, then you would find it hasn't travelled at all, so your calculation would look like 0/0. But this is what you would get no matter how fast or slow the object was travelling, so really "the object travels 0 metres in 0 seconds" gives no information. (What your teacher was thinking of was x/0 for any x other than 0 e.g. if an object travelled 2 metres in 0 seconds then it must have been going infinitely fast.) Quietbritishjim (talk) 00:36, 14 July 2010 (UTC)

A (slightly) more rigorous way of looking at it is to look at the graph. The page inverse function shows how to get from an equation like y=(something involving x) to x=(something involving y): by flipping about the diagonal line shown on that graph. In our case we want to go from y = 0 x to x = y / 0. But y = 0 x is a horizontal line, so x = y / 0 becomes a vertical line. This backs up the idea that if you try to take 0 / 0 you have any (or all) values, whereas y / 0 for any other y doesn't make sense (or is infinite, if you allow that). Quietbritishjim (talk) 00:43, 14 July 2010 (UTC)

The fraction a/b is 'the' solution to the equation bx=a. If b≠0 then this equation uniquely determines x, (because (b≠0 and bx=a and by=a) → (b≠0 and bx−by=a−a) → (b≠0 and b(x−y)=0) → (x−y=0) → (x=y)). If a=b=0 then the equation bx=a is 0x=0 which is true independent of the value of x, and so the equation does not define x, and so 0/0 is undefined. If a≠0 and b=0, then the equation bx=a is 0=a, which is false independent of the value of x. Bo Jacoby (talk) 03:20, 14 July 2010 (UTC).

time to break out truth

In case the above isn't convincing to you, let's break out actual reality - you know, truth. So, what is 0 - it is the absence of a single one or fraction of whatever you're counting, but not negative - you don't owe any either. For example, 0 apples is the complete absence of a single Apple. Now, I am about to do strictly integer division - no fractions. Let's say you drive to my house, pick up 2 apples, drive home, put them in your basket at home, and repeat until you can't make your next drive, since there aren't enough apples left. How many times can you make that drive (physically - we are talking about reality here, not math). You can physically do that half as many times (INTEGER DIVISION) as I have apples. If I have 8 Apples, you can drive to my house, pick up two, drive home, and repeat, 4 times (8/2 = 4). After the fourth time, you can't do that again, there not being enough apples at my house to make the trip. So, that is how many times you can do it. Now what if instead of coming to my house and picking up 2 apples, you come to my house and pick up 0 apples, drive home, and repeat. After how many trips will you have to stop, there not being enough apples at my house for you to make your next trip? There is no answer to "after how many do you stop" because you do not stop. It's not that you stop after infinite times - you do not stop; the division is not exhaustive: you can always do one more. So, if the normal definition of integer division would be subtracting the divisor one time until you can't do that any more* (because you run out of the dividend), then you will never meet that definition: you will never "run out" of the dividend when you're taking 0 from it. So, there is no solution to how many times you are going to be able to do that, if the process is never finished. This is like asking: if every time you say something, your wife says something, and every time your wife says something, you say something, then if you are the first to speak, how many times will the last person to speak have spoken? There is no "last person to speak" by that definition, so it is a meaningless question, it fails to refer. Asking what 5/0 is, is the same as asking what the value of the last prime is. There is no last prime, and there is no end result to the division. Saying 5/0 is "infinite" is just as wrong as saying the "last prime" is "infinite". The last prime is not infinite, since there is none.84.153.208.32 (talk) 10:13, 14 July 2010 (UTC)

• obviously this is the normal definition. When I ask you how many times you can remove 2 apples from 8, your answer isn't "Oh, you have a lot of choices. You can do it 0 times, 1 time, 2 times, 3 times, or 4 times - these are the physically possible alternatives facing you when you are looking at 8 apples and are to remove 2 a certain number of times, these are the possibilities of how many times you can do that.". Yes, "linguistically" the question "how many times can you remove 2 apples from 8" has the answer "0, 1, 2, 3 or 4", but that's not how we understand it. We understand "how many times until you can't do it again." That's the definition, and by that definition when you divide by 0 you do not get "any answer you want", any more than when you divide 8 by 2 you get "any answer you want, from the choices 0, 1, 2, 3, or 4" -- the definition is to do it until you can't do it one more time, and since there is no such point in the calculation, there is no answer when dividing by 0. It's like me telling you how to calculate Mango's number: you start with 2, then keep doubling (2, 4, 8, 16, 32) until you get to a negative number: that's Mango's number. What is the value of Mango's number? (Obviuosly it is not "infinite", or "any number you want", or architecture-dependent). No, there is no Mango's number. It's the same if we say Mango's number is defined as "4/0": there is no value, the definition does not admit of a point at which you can say, "Okay, now I have Mango's number", just as you can't reach a point where you can say "Okay, now I have the largest prime." You can't say "Okay, now I have the value of 4/0".

July 14

Free stuff

What's the free ring over an abelian group? Is it this: given an abelian group A, take the free ring generated by A then quotient it by the ideal generated by all the i(a)+i(b)-i(a!b), where i is insertion of generators and ! is the addition in A?

Also I was reading about Freyd's adjoint functor theorem. If you run through the whole proof it seems to be somewhat constructive. My book (MacLane's 2nd edition) on page 125 proves the forgetful functor from compact hausdorff spaces to SET has a left adjoint using adjoint functor theorem; he finds a small solution set explicitly then applies the theorem. Did this suggest how to construct an explicit stone cech compactification? Similarly for Grp. Does the adjoint functor theorem suggest ways to construct free objects explicitly? Because free groups are so much more complicated than free abelian groups (you could put the abelian relation on the free group but why do that when you can use set of all cofinally 0 function from X to the integers) I wonder how they managed to get free groups in the first place. Money is tight (talk) 04:17, 14 July 2010 (UTC)

Regarding the first question, I don't specifically know that terminology, but I think it would be the group ring of Z over A. Using your phrasing I guess that would be the free ring generated by A modded out by the ideal generated by all the i(a)*i(b) - i(a!b). Rckrone (talk) 06:11, 14 July 2010 (UTC)

I think what you described is the tensor algebra over A with A considered as a Z-module. (Can anyone confirm if that is correct?) Rckrone (talk) 06:26, 14 July 2010 (UTC)

Sorry for the confusion I was in a rush. What I wanted was the left adjoint to the forgetful Rng->Ab, sending each ring to its underlying addition abelian group. I've never heard of this construction before. Also my 2nd question is more clearly stated as: How do people explicitly construct free objects for the first time? Is it raw ingenuity or is there some algorithm in doing so? Money is tight (talk) 09:20, 14 July 2010 (UTC)

For algebraic categories like the examples you mentioned there is a uniform "algorithm". If U and V are varieties such that the signature σ of U includes the signature τ of V and V includes all reducts of algebras from U, then the forgetful functor U → V has a left adjoint F: V → U defined as follows: F(A) is the quotient of the free U-algebra (which can itself be constructed as the quotient of an absolutely free σ-algebra, e.g. the term algebra, over the congruence generated by the defining identities of U) in generators A over the congruence ~ generated by the atomic diagram of A (i.e., f(a₁,...,a_k) ~ a for each k-ary operation f ∈ τ and each a₁,...,a_k,a ∈ A such that a = f^A(a₁,...,a_k)). (Incidentally, the coverage of universal algebra on Wikipedia is extremely poor.)—Emil J. 12:57, 14 July 2010 (UTC)

Oh, and the answer to your first question is yes, your construction is in fact a special case of the general construction above (more precisely, the general construction would also put i(−a) + i(a) into the ideal, but that's easily seen to be redundant).—Emil J. 13:22, 14 July 2010 (UTC)

Thanks! I've never had any exposure to universal algebra so I don't exactly know what you mean. I had a look at the online pdf version "A course in universal algebra" page 73, and I think this is how: Say we want the free group over a set X (with signiture multiplication binary inverse unary and 1 constant). Take the term algebra T(X), which is like the free magma except it's got terms with inverse and 1. Next take the intersection E of all congruences on T(X) such that the quotient is actually a group (since T(X) is the "absolutely free algebra" you described, it has no relations on it, not even associativity), and intersection of congrugence is a congruence. The quotient under E is a group too, the free group on X.

I really have no idea what's going on lol. I'll take a proper look some other time.

Another question: this "uniform algorithm" is not the standard way to construct free groups. Is it because this construction is so much more difficult to use practically than the standard construction using equivalence class of words? The free abelian group can be constructed from the free group by using further identifications, but why do that when we can use all confinally 0 functions to the integers. So sure, this algorithm proves every forgeful functor on some type of algebras has a left adjoint, but it might as well just be a pure existence because of the insane complexity in its construction. Money is tight (talk) 05:19, 15 July 2010 (UTC)

The congruence E can also be described "from below": t E s iff there exist n ≥ 0 and terms t₀ = t, t₁, ..., t_n = s such that for each i < n, t_i = u_i(v_i) and t_i+1 = u_i(w_i) for some terms u_i, v_i, w_i, where either v_i ≈ w_i or w_i ≈ v_i is an instance of one of the defining identities of groups (for example, we may have v_i = 1 and w_i = r⁻¹·r for some term r). The standard group-specific construction is actually equivalent to this: another description of E is that t E s iff (the flattened forms of) t and s are equivalent as words. In this way it gives more information, as it provides an explicit description of representants of the equivalence classes. This is useful for various group-theoretic considerations, but I don't think it is any easier than the general construction if you just need to prove that the free group is a free group. Of course, in the case of abelian groups the explicit description of free objects is so simple that it beats any generic construction.—Emil J. 12:28, 15 July 2010 (UTC)

Polish notation

What's the point of Polish notation? Not only is it harder to read, it's more difficult to work with algebraicly. For example, compare the normal version ab+cd to its Polish notation equivalent, + * a b * c d. --138.110.206.101 (talk) 16:45, 14 July 2010 (UTC)

It removes ambiguity. How do you KNOW that ab+cd=(ab)+(cd) and not ab+cd=a(b+c)d? You are using an assumption that multiplication precedes addition. Polish notation never requires assumptions or parenthesis. -- kainaw™ 16:59, 14 July 2010 (UTC)

It isn't really an assumption is it, when everyone is using previously agreed-upon rules? There isn't really any ambiguity to ab+cd is there? mislih 20:07, 14 July 2010 (UTC)

It's very easy to evaluate without ever rereading any of the input (which is helpful if the input is large). You just note each operation as you come to it and then evaluate it with the next one or two values (depending on the operator). One or both of those might itself be the result of an operator, so you proceed recursively. In your example you would say "Add... multiply a and b..." (at which point you have the two things + and ab to remember; you need not remember a and b separately or that you multiplied them) "...multiply c and d" (at which point you obtain cd and then immediately the overall result). In practice, RPN is used for this purpose because then only numbers (and never operators) have to be temporarily remembered (and the stack that remembers them can often then be smaller). --Tardis (talk) 18:59, 14 July 2010 (UTC)

So Polish or reverse Polish would be better for determining the value of a given expression, but isn't it worse for algebraic manipulation? --138.110.206.101 (talk) 19:01, 14 July 2010 (UTC)

Not really. You are just assuming it would be because you aren't used to seeing it. If you only learned Polish notation, you would look at all the precedence rules and parenthesis used in inline notation and think it was a confusing mess. -- kainaw™ 19:04, 14 July 2010 (UTC)

How would you, for example, solve x 2 ^ x 1 + 2 ^ + x - 1 - 0 = without converting to regular notation? It's a lot harder. --138.110.206.101 (talk) 19:09, 14 July 2010 (UTC)

You don't include = in the notation. Keep it as two separate expressions, one being x 2 ^ x 1 + 2 ^ + x - 1 - and the other being just 0. Anything you do to one you do to the other. ie: To get rid of "1 -", you use "1 +" and get x 2 ^ x 1 + 2 ^ + x - and 0 1 +. Since there are no parenthesis or order of operations, we had no issues with deciding what we could and could not do. The next thing to work on is the trailing x -. But, it is apparent that you will hit a point at which you want to have one side simply 0 and the other side will contain a squared x. You will have to simplify. How do you simplify? You memorize patterns. You've probably memorized them as inline patterns, but you could have memorized them as Polish notation patterns just as well. -- kainaw™ 19:25, 14 July 2010 (UTC)

English is like the standard arithmetic expressions if you consider verbs as operators. Latin is reverse polish as in Jim the ball hit and Irish is polish with the equivalent of hit jim the ball. I gues sin them polish or reverse polish might appear a bit more natural. Dmcq (talk) 20:54, 14 July 2010 (UTC)

In Hebrew, all three are correct. And the subject doesn't even have to precede the object, giving a total of 6 variations! (Though I'm unsure of the "Hit the ball Jim" variant). -- Meni Rosenfeld (talk) 09:50, 15 July 2010 (UTC)

Verb Object Subject doesn't seem to be all that common, and Object Verb Subject even less so - it says Klingon uses that order because it is so strange sounding :) Dmcq (talk) 11:15, 15 July 2010 (UTC)

To avoid potential misunderstanding, all six variations are possible in Polish as well, though the default word order is SVO as in English. Polish notation has nothing to do with Polish language, it's named after members of the Polish logical school who invented it.—Emil J. 11:41, 15 July 2010 (UTC)

Rubik's cube group

How many conjugacy classes has the Rubik's cube group? --84.61.131.18 (talk) 20:05, 14 July 2010 (UTC)

How large is the largest order of any element in the Rubik's cube group? --84.61.131.18 (talk) 20:41, 14 July 2010 (UTC)

We have some info on the group in Rubik's cube group. The answer to your questions is not there, but some of the stuff may be useful. I suppose it shouldn't be hard to compute conjugacy classes and maximal orders using some computer algebra system.—Emil J. 15:38, 15 July 2010 (UTC)

Off the top of my head there are elements of order 8.3.5.7=840. A complete listing of conjugacy classes would be rather lengthy, I'm guessing there are at least a thousand. Conjugacy class of Wreath products of the symmetric groups are fairly easy to work with so a computer algebra system may not be necessary.--RDBury (talk) 16:50, 15 July 2010 (UTC)

According to the Magma computer algebra system there are 81120 conjugacy classes and the largest element order is 1260. The group has elements of the following orders: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24, 28, 30, 33, 35, 36, 40, 42, 44, 45, 48, 55, 56, 60, 63, 66, 70, 72, 77, 80, 84, 90, 99, 105, 110, 112, 120, 126, 132, 140, 144, 154, 165, 168, 180, 198, 210, 231, 240, 252, 280, 315, 330, 336, 360, 420, 462, 495, 504, 630, 720, 840, 990, 1260. Most orders are made up of many classes of several sizes. Ignoring class sizes, and just sorting by order, one gets the following table:

Order table

Order	NrClasses	NrElems
1	1	1
2	74	170911549183
3	119	33894540622394
4	439	4346957030144256
5	5	133528172514624
6	7542	140621059298755526
7	3	153245517148800
8	316	294998638981939200
9	219	55333752398428896
10	136	34178553690432192
11	1	44590694400
12	20899	2330232827455554048
14	54	23298374383021440
15	190	14385471333209856
16	35	150731886270873600
18	4739	1371824848124089632
20	315	151839445189791744
21	66	39337151559333120
22	5	927085127270400
24	7590	3293932519796244480
28	114	97419760907673600
30	5155	1373347158867028224
33	23	15874019662233600
35	7	65526218912563200
36	8703	3768152294808760320
40	130	835897246403788800
42	1428	737199776831097600
44	4	100120377950208000
45	144	197329441659727104
48	739	911497647410380800
55	1	4854321355161600
56	47	671205306846412800
60	7371	4199961633799421952
63	57	264371433705308160
66	103	404051175250329600
70	37	210461722916290560
72	3289	1981453794190295040
77	2	187238109413376000
80	7	13349383726694400
84	1664	1697725818678067200
90	2177	1764876446897050368
99	27	104367909135974400
105	70	232824419423354880
110	1	4854321355161600
112	5	128726200221696000
120	1776	1947044011463147520
126	679	854783686296207360
132	36	637129677864960000
140	28	223125413717606400
144	330	714192029378150400
154	2	187238109413376000
165	12	213590139627110400
168	274	1050269239266508800
180	2015	2320395168471367680
198	55	759701292082790400
210	388	1053174509332070400
231	4	374476218826752000
240	84	407156203664179200
252	468	689877080447385600
280	6	68653973451571200
315	35	99309879652515840
330	12	213590139627110400
336	10	257452400443392000
360	460	571019888909352960
420	182	961155628321996800
462	4	374476218826752000
495	4	174755568785817600
504	70	238381852262400000
630	87	395380140162416640
720	10	120144453540249600
840	24	240288907080499200
990	4	174755568785817600
1260	8	51490480088678400

By hand calculations of the orders/sizes of all conjugacy classes are likely much too difficult due to the length, but David Joyner's Adventures in Group Theory (chapter 11, I believe) works out the elements of order 2 in quite some detail by hand. I believe Joyner also includes the by hand calculation of the maximum order, though it might be in one of his other books. JackSchmidt (talk) 17:25, 15 July 2010 (UTC)

Help with a definite integral

I am trying to solve this integral,

$f(x)={\frac {1}{2\pi }}\int _{0}^{2\pi }d\theta \,e^{ix\theta }\,\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\in \mathbb {Z} \!$

which, I believe, has the solution

$f(x)={\begin{cases}\pi ,&{\mbox{if }}x=0\\{\frac {1}{ix}},&{\mbox{if }}x\neq 0\end{cases}}\!$ .

However I am having a hard time getting this answer. Using integration by parts, I get

${\begin{aligned}f(x)&={\frac {\theta e^{ix\theta }}{2\pi \,i\,x}}{\Bigg |}_{\theta =0}^{\theta =2\pi }-{\frac {1}{2\pi }}\int _{0}^{2\pi }d\theta \,{\frac {e^{ix\theta }}{ix}}\\&={\frac {1}{ix}}\left(1-{\frac {1}{2\pi }}\int _{0}^{2\pi }d\theta \,e^{ix\theta }\right)\\&={\frac {1}{ix}}\left(1-\delta _{x,0}\right)\end{aligned}}$ .

This gives the correct answer when x is any integer other than zero, but is infinite when $x=0$. I could get the correct result for $x=0$ just by plugging that in at the beginning. Is there something wrong with my derivation above? Is there a more elegant way to arrive at the correct answer? Thanks mislih 20:03, 14 July 2010 (UTC)

$\forall x\;{\frac {de^{ix\theta }}{d\theta }}=ixe^{ix\theta }$, but $\int e^{ix\theta }\,{\text{d}}\theta \not \equiv {\frac {e^{ix\theta }}{ix}}+C$ because of the $x=0$ possibility, in which case you have $\int {\text{d}}\theta =\theta +C$, of course. If you write $\int e^{ix\theta }\,{\text{d}}\theta ={\frac {e^{ix\theta }-1}{ix}}+C$ (which is just using a different C), then the limit works out properly, so you might try that as an intuitive approach, but it's still technically undefined. So when you do integration by parts you have to split off the $x=0$ case just as if you were dividing by x on both sides of an equation. Then of course you're allowed to assume $x\neq 0$ later in the other branch, so you would avoid bringing in the Kronecker delta. --Tardis (talk) 23:36, 14 July 2010 (UTC)

Denesting Radicals

Hello. Why does ${\sqrt {a^{2}-b^{2}c}}$ have to be an integer if some nested radical, ${\sqrt {a+b{\sqrt {c}}\ }}$, can be denested into a sum of surds, ${\sqrt {d}}+{\sqrt {e}}$? If I rewrite ${\sqrt {a^{2}-b^{2}c}}$ in terms of $d$ and $e$, I get $d-e$. Thanks in advance. --Mayfare (talk) 22:17, 14 July 2010 (UTC)

If

${\sqrt {a+b{\sqrt {c}}}}={\sqrt {d}}+{\sqrt {e}},\,$

then squaring both sides yields

$a+b{\sqrt {c}}=d+e+2{\sqrt {de}}.\,$

Now if a, b, d, e are rational and √c is not, then we have

$a=d+e{\text{ and }}b{\sqrt {c}}=2{\sqrt {de}}.\,$

Therefore

$a-b{\sqrt {c}}=d+e-2{\sqrt {de}}=({\sqrt {d}}-{\sqrt {e}})^{2}\,$

and if everything in site is nonnegative, we can then say

${\sqrt {a-b{\sqrt {c}}}}={\sqrt {d}}-{\sqrt {e}}.\,$

Finally we have these two products:

${\begin{aligned}{\sqrt {a+b{\sqrt {c}}}}{\sqrt {a-b{\sqrt {c}}}}={\sqrt {a^{2}-b^{2}c}}\\[12pt]({\sqrt {d}}+{\sqrt {e}})({\sqrt {d}}-{\sqrt {e}})=d-e\end{aligned}}$

Therefore

${\sqrt {a^{2}-b^{2}c}}=d-e.\,$

So if d − e is an integer then we're done. Michael Hardy (talk) 23:50, 14 July 2010 (UTC)

July 15

Testing the bias on a coin

If a coin is flipped a number of times, and the results are n heads and m tails, what is the probability that the next flip will be heads?115.178.29.142 (talk) 03:22, 15 July 2010 (UTC)

There's no way to know. But the maximum likelihood estimate is n/(n+m). 67.119.12.184 (talk) 05:11, 15 July 2010 (UTC)

Assuming of course that you have no other information regarding the fairness of the coin, the probability that the next flip will be heads is P=(n+1)/(n+m+2). This formula also applies when n=m=0, unlike the approximate estimate P≈n/(n+m). Bo Jacoby (talk) 09:33, 15 July 2010 (UTC).

If you were pretty certain there should be little or no bias either way then (n+10)/(n+m+20) might be better. You can see it depends on your prior probability. Dmcq (talk) 09:48, 15 July 2010 (UTC)

Dmcq's result assumes that the coin has been flipped n+m+18 times given n+9 heads and m+9 tails. The prior probability can not be chosen freely. No information about the fairness of the coin means that the prior probability of heads, (H|), is 1/2, and the prior probability of tails, (T|), is also 1/2. The wanted conditional probability of heads, giving the facts, F, (that there were already n heads and m tails), is computed by Bayes' formula together with the formula for the hypergeometric distribution

$(H|F)={\frac {(F|H)(H|)}{(F|H)(H|)+(F|T)(T|)}}$ $={\frac {{\frac {{\binom {n+1}{n}}{\binom {m}{m}}}{\binom {n+m+1}{n+m}}}{\frac {1}{2}}}{{\frac {{\binom {n+1}{n}}{\binom {m}{m}}}{\binom {n+m+1}{n+m}}}{\frac {1}{2}}+{\frac {{\binom {n}{n}}{\binom {m+1}{m}}}{\binom {n+m+1}{n+m}}}{\frac {1}{2}}}}$ $={\frac {n+1}{n+m+2}}$

Bo Jacoby (talk) 11:11, 15 July 2010 (UTC).

That is a pretty good prior if you'd never come across a coin before but most people recognize that coins are usually pretty unbiased. Now supposed we tossed the coin once and it came up heads. That would give the chance of the coming coming up heads as 2/3. That means if I bet 100 quatloos on tails and the other person 200 on heads then we should break even over a number of throws - in three goes on average I'd lose twice and they'd lose once. Now would you really be willing to bid against me if I put 150 quatloos on tails to your 200 on heads? Dmcq (talk) 12:00, 15 July 2010 (UTC)

Indeed. Bo is assuming we have no information about the fairness of the coin, but that's not the case. We know it's a coin, which means it is very likely to be very close to fair. --Tango (talk) 12:42, 15 July 2010 (UTC)

If you already know that the coin is fair, then don't test it, just tell the OP that the probability is 1/2 because you know that it is a fair coin. But the OP is not sure, and that's why he tests it. If you have noticed (or have been told) that coins are usually pretty unbiased, it must be because coins have been flipped before, and then the values of n and m are high. It doesn't need to have been the same coin every time. Now the OP clearly assumes that "a coin is flipped a number of times, and the results are n heads and m tails", so we are justified in assuming that n and m are the correct counts. Or perhaps the OP is using the word coin figuratively meaning some device providing random values called heads and tails. In any case (n+1)/(n+m+2) is the correct formula for the probability that the next flip show head. Bo Jacoby (talk) 14:46, 15 July 2010 (UTC).

Previous coin tosses aren't the only kind of evidence you could have. You could have the word of the person that gave it to you that is isn't a fixed coin, for example. Or simply the knowledge that most coins in existence are fair. Knowing that a different coin was tossed and got certain results isn't the same as knowing that the coin in your hand was tossed and get certain results. --Tango (talk) 14:53, 15 July 2010 (UTC)

I feel this is rapidly heading towards at least one side of at least one sheep seems black territory. :) Dmcq (talk) 15:33, 15 July 2010 (UTC)

That's nonsense. There is no dichotomy between "know for certain the coin is fair" and "have no idea what a coin is, and hence use a maximum-entropy prior". If you know that 99% of coins are fair, then when given a random coin (which you have no reason to suspect was chosen based on its fairness or lack thereof), you'll say there's 99% that this coin is fair. If you then toss it 3 times and it's all heads, you'll still have a high credence to it being fair (the exact amount depends on the continuous prior distribution for the bias of biased coins), and thus your credence for the next toss being heads will be close to 50%. As more evidence is collected where this coin lands heads more then it should, your posterior will shift towards the coin having the bias implied by the Heads:Tails ratio.

Even if past coin tosses is the only knowledge of coins you have (as opposed to, say, proficiency in metallurgy and rigid body mechanics), you can't compress this knowledge to 2 numbers m and n. If you sample 2 coins and toss each many times, then "one coin always lands heads, one always tails" is very different, wrt your conclusions about the bias of coins, from "each coin lands a roughly equal number of heads or tails". Your probability distribution for the bias of coins can take any form depending on what observations you have made. -- Meni Rosenfeld (talk) 15:51, 15 July 2010 (UTC)

The arcsin for numbers >1

How do you find the arcsin for numbers greater than 1 (which I believe is a complex value) —Preceding unsigned comment added by 115.178.29.142 (talk) 04:44, 15 July 2010 (UTC)

By the way, I know the real part is pi/2 . but what is the imaginary part?115.178.29.142 (talk) 05:15, 15 July 2010 (UTC)

Sin(z) is defined in terms of the exponential function exp(z) as sin(z)=(exp(iz)-exp(-iz))/2i, and exp(z) is in turn defined in terms of a power series (see article Exponential_function). So when you want to find arcsin of x, you solve for z in x=(exp(iz)-exp(-iz))/2i Money is tight (talk) 05:24, 15 July 2010 (UTC)

Which will give you the expression in Inverse trigonometric functions#Logarithmic forms.—Emil J. 11:51, 15 July 2010 (UTC)

$\arcsin(x)={\frac {\pi }{2}}\pm i\,\operatorname {arccosh} (x){\text{ for }}x>1$ Gandalf61 (talk) 13:00, 15 July 2010 (UTC)

Comparing two averages

If you are comparing two samples from completely separate populations, you can use a two sample t-test. Is there anyway to compare a population's mean to a subset's mean? For instance, you look at 2000 students with an average weight of 150 pounds. Then you choose 50 of those students and get their average weight. Is there anyway to check that their average weight is not significantly different from the population's average of 150 pounds? Like: an average of 149 might not be significant, but an average of 220 is significant. Remember, you don't have the average and standard deviation of the whole population without those students.

Basically you're trying to show that the group of students you chose as a whole is not significantly heavier than the other students at the school. 160.10.98.106 (talk) 15:21, 15 July 2010 (UTC)

Theoretical computer science

Suppose F is a boolean function with n variables, and I want to determine whether there exists an assignment of true or false to those variables such that F is true. With a nondeterministic Turing machine, this can clearly be solved in time n, so this is in NP. However, a deterministic Turing machine requires time 2ⁿ to check all 2ⁿ possible assignments of variables, so this is not in P. Therefore, P!=NP. What's the problem with this proof? --138.110.206.101 (talk) 16:25, 15 July 2010 (UTC)

You're assuming there's no better method for a deterministic Turing machine to use than straightforward brute force search. You're also taking "polynomial-time" to mean polynomial in the number of variables, while I think the only sensible meaning for it here is polynomial in the length of the proposition. Algebraist 16:31, 15 July 2010 (UTC)

[ec] The problem is that you have assumed that brute-forcing all possibilities is the most efficient algorithm to solve this. To prove that a problem is not in P you have to show that each and every algorithm that solves it is superpolynomial, not that one particular algorithm is superpolynomial. -- Meni Rosenfeld (talk) 16:33, 15 July 2010 (UTC)

Are there any polynomial-time deterministic algorithms for determining whether a boolean equation has a solution? --138.110.206.101 (talk) 16:40, 15 July 2010 (UTC)

We don't know. That's exactly what the P =? NP problem asks for. However, there are SAT solvers which do better than the brute force assignment search (they achieve time complexity $2^{\epsilon n}$ for some $0<\epsilon <1$), which should at least persuade you that the problem is nontrivial.—Emil J. 16:43, 15 July 2010 (UTC)

This is a very common mistake for those beginning to study P/NP. You are not using a standard nondeterministic Turing machine. You are using an infinite-length nondeterministic Turing machine. What if the maximum number of values that your machine can store is m. You have n values to check. If m>n, no problem. If m<n, you have a problem. You cannot do this in simply polynomial time. A basic rule: If your "polynomial" time solution begins with "get a tape long enough to store all the values" or "get enough memory to store all the variables" or "get a rope long enough to stretch around all the posts", it is not a polynomial solution. -- kainaw™ 17:24, 15 July 2010 (UTC)

What on earth are you talking about? The standard Turing machine, as used in the definition of NP, has (at least potentially) an infinite tape. It can get away with a polynomially bounded tape as it is a poly-time machine here, but it certainly cannot be bounded by a constant. That would be equivalent to a nondeterministic finite automaton, which can only recognize regular languages, a tiny subset of P (let alone NP). The OP's NP algorithm is correct, and it is in fact a textbook example (SAT being among the most popular NP-complete problems).—Emil J. 18:04, 15 July 2010 (UTC)

I was attempting to use language to make it understood to the questioner. When you say "infinite tape" and you know what you are talking about, those just learning about P/NP hear "the tape starts out with an infinite amount of data already written on it and all I have to do is read it quickly from one end to the other". I have argued for many years that the language used in describing P/NP leads to more confusion than anything else. -- kainaw™ 18:07, 15 July 2010 (UTC)

Whereas telling the OP that he or she made a mistake when they actually did not (in this part of the argument, that is) is supposed to reduce confusion, right?—Emil J. 18:15, 15 July 2010 (UTC)

Implicit function theoem

My text book writes "...note that the cain rule applied to F(x,g(x)) = 0 gives

$\mathbf {D_{x}} F(\mathbf {x} ,g(\mathbf {x} ))=[{\frac {\partial F}{\partial x}}(\mathbf {x} ,g(\mathbf {x} ))][\mathbf {D} g(\mathbf {x} )]$

which is equivalent to

${\frac {\partial g}{\partial x_{i}}}=-{\frac {\frac {\partial F}{\partial x_{i}}}{\frac {\partial F}{\partial z}}}.$

I was wondering the first equation can be arrived at. Oh, and F:ℝⁿ⁺¹ → ℝ. PS Do you have to bold the D everytime you write the derivative matrix? Because it's hard to bold on paper...74.15.137.192 (talk) 19:31, 15 July 2010 (UTC)