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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

(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)[reply]

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)[reply]

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)[reply]

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

But who has not done it at least once? PS T 10:35, 10 July 2010 (UTC)[reply]

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

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 ... ;) PS T 14:48, 10 July 2010 (UTC)[reply]

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)[reply]

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 ... PS T 02:56, 11 July 2010 (UTC)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]