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August 18

Sum of two disks

Does anyone know a proper reference to the following result:

Let $i\in \{1,2\}$ be an integer and $a_{i}$, $b_{i}$, $r_{i}$, $x_{i}$, $y_{i}$ be real numbers with $r_{i}>0$. Denote $B_{i}=B(r_{i},x_{i},y_{i})$ closed, convex disk in 2-dimensional euclidean space, with and radius $r_{i}$ and center $(x_{i},y_{i})$ and denote $(a_{i},b_{i})$ some point in $B_{i}$. Define $S$ to be the sum of two disks $B_{i}$ ie. $S=\{(a_{1}+a_{2},b_{1}+b_{2})|(a_{i},b_{i})\in B_{i},i=1,2\}$. Then the following holds:

Theorem: $S=B(r_{1}+r_{2},x_{1}+x_{2},y_{1}+y_{2})$.

88.72.195.199 (talk) 05:42, 18 August 2008 (UTC)

Does it really need a reference? It's pretty easy to prove. Let a = x+p, and b = y+q. Then a+b = x+p+y+q = (x+y)+(p+q). So, the x+y part commutes out in all cases. That means that, centered at the point x+y, the question is whether B(r1,0,0)+B(r2,0,0) = B(r1+r2,0,0). Every sum must land within that radius, by the triangle inequality, so all that's left is to show they cover the entire disc. Split it up into straight lines radiating out from the center. Along each line, each point is within r1+r2 from the center, and is therefore the sum of two vectors in the same direction, one of which is within r1 of the center, the other of which is within r2 of the center. So, every point is covered. Black Carrot (talk) 08:08, 18 August 2008 (UTC)

Thank you Black Carrot for the proof! But I really like to have a proper reference for quotation. It may be rather old. 88.72.215.10 (talk) 10:57, 20 August 2008 (UTC)

220 pages probability

Suppose that I have a book of 220 pages. I keep opening the book at random. How many times do I have to open the book until the probability that I'm looking at the only page I haven't seen yet becomes greater than 99%? How many times do I have to open the book until the probability that there are only ten pages I haven't seen yet becomes greater than 90%?

• Random here means random. Please ignore the physics of opening a book.
• This isn't the way opening a book usually works, but assume that when you open a book, you only see one page.

Thanks. —Preceding unsigned comment added by 24.7.54.224 (talk) 06:53, 18 August 2008 (UTC)

I don't think the first one will ever happen. For any number of openings, the odds that there is exactly 1 page left are pretty low, and the odds that you happened to land on it are even lower. The chances will be positive from 220 openings onwards, but I don't think they'll ever get that high. Black Carrot (talk) 08:10, 18 August 2008 (UTC)

Given that there is only one page that you haven't seen, the conditional probability that you see that page when you open the book at a random page is 1/220, and the conditional probability that you do not see that page is 219/220. The probability that you will see that one page at least once in the next n attempts is 1-(219/220)ⁿ, but probability of seeing that page on any specific attempt (given that you are only looking for one page) is always 1/220.

If you want to know how many times you expect to have to open the book until there is only 1 or 10 or whatever pages that you haven't seen, see our articles on the coupon collector's problem and coupon collector's problem (generating function approach). Gandalf61 (talk) 09:07, 18 August 2008 (UTC)

I agree. Two different openings of a book are mutually exclusive events and therefore, it does not matter if you have opened the book 1000 times prior to your next opening; the probability will be the same. The problem is analagous to the problem of tossing a coin: how many times would you need to toss a coin (call this number n) to obtain between n/2 - 10 and n/2 + 10 heads (not inclusive). There is no answer to this problem; however you would expect that the number of heads you would obtain is approx. half as many times as you have tossed the coin. In fact, there is a law in probability known as the Law of large numbers. It says (for this particular case), that the larger the number of coin tosses you make, the more likely you are to get approximately half as many heads as the number of times that you have tossed the coin. You should see the link, Law of large numbers if you want a more detailed discussion on this.

I hope this helps.

Topology Expert (talk) 07:03, 19 August 2008 (UTC)

I disagree with what you write on a few technical points. In particular, you write that as the number n of coin tosses increase, the probability to get n/2 heads also increases (let's suppose n is even, of course). This is not true: the probability of exactly n/2 heads is

${\frac {1}{2^{n}}}{\binom {n}{n/2}}\approx {\frac {\sqrt {2}}{\sqrt {\pi n}}}$

which decreases with increasing n (I used Stirling's approximation). This is elaborated upon a bit more in the fourth paragraph of the article on law of large numbers. Eric. 90.184.71.189 (talk) 11:17, 19 August 2008 (UTC)

I did not say this; I said that increasing the number of tosses of a coin, increases the probability that you will get a number of heads approximately equal to half as many times that you have tossed the coin. Your point is trivial; as you increase the number of times that you have tossed the coin, the binomial distribution tends to be very well aproximated by the normal distribution. Therefore, the probability of any exact value (no. of heads) tends to go to zero. Perhaps my wording was not clear enough.

Topology Expert (talk) 03:40, 20 August 2008 (UTC)

No worries, with the word "approximately" your statement is now correct. Eric. 90.184.71.189 (talk) 10:33, 20 August 2008 (UTC)

Clifford algebra

Is it possible to describe universal property of Clifford algebras using adjoint functors? 83.23.214.17 (talk) 14:33, 18 August 2008 (UTC)

a,b,c are three rational numbers..

a,b,c are three rational numbers and a2^1/2+b3^1/2+c5^1/2=0 , then prove that a=b=c=0 —Preceding unsigned comment added by 117.99.17.13 (talk) 15:35, 18 August 2008 (UTC)

As it says at the top of the page, do your own homework. The general case of this (with arbitrarily many primes and 1 if you want) can be solved quite neatly with a little Galois theory, but this instance can be done bare-handed with a bit of effort. Algebraist 15:39, 18 August 2008 (UTC)

I would start by re-arranging to get a2^1/2+b3^1/2 = -c5^1/2 and then squaring both sides. I'll let you take it from there. Gandalf61 (talk) 15:53, 18 August 2008 (UTC)

As written, this problem is quite simple, just substitute in 0 for a, b, and c and solve. However, if the problem is to prove that there are no other solutions, this gets more complicated. StuRat (talk) 05:47, 19 August 2008 (UTC)

Proving that there are no other solutions is indeed required: the task as given is of the form "prove that if A, then B", where A in this case is "a2^1/2+b3^1/2+c5^1/2=0" and B is "a=b=c=0". This requires showing that there are no possible cases where A would be true but B false. I agree that the task would be absolutely trivial if it were the other way around. —Ilmari Karonen (talk) 06:47, 19 August 2008 (UTC)

Indeed. It says "given a2^1/2+b3^1/2+c5^1/2=0, prove that a=b=c=0", not "prove that a=b=c=0 is a solution to a2^1/2+b3^1/2+c5^1/2=0", they are very different problems. --Tango (talk) 22:21, 19 August 2008 (UTC)

Tensor products

Hello,

I'm trying to understand tensor products (of vector spaces) but I'm not sure how it works. On the construction given on that page, what additional equivalence relations should be made to recover $V\times W$ as the quotient ?

I'm especially trying to understand vector valued differential forms, as the construction $\Omega ^{p}(M,E)=\Gamma (E\otimes \Lambda ^{p}T^{*}M)$. I'm perfectly fine with ordinary real valued differential forms being $\Omega ^{p}(M)=\Gamma (\Lambda ^{p}T^{*}M)$, which implies $\mathbb {R} \otimes \Lambda ^{p}T^{*}M=\Lambda ^{p}T^{*}M$ which I'm not sure about (it makes some sense from the definition of tensor product with the quotient, but it doesn't fit in with my intuitive understanding of tensor products with identification of $L^{2}(U\times V,X)$ with $L(U\otimes V,X)$.)

So how does it work ? What is, for example, $\mathbb {Q} \otimes V$ compared to $V$ ? And really, why does $\Gamma (E\otimes \Lambda ^{p}T^{*}M)$ mean $E$-valued differential forms ?

Thanks. --Xedi^Talk 15:46, 18 August 2008 (UTC)

Well, to be terribly informal, if V is an m dimensional space and W is an n dimensional space then V⊗W is the space of m×n matrices. V×W would be an m+n dimensional space. You can't recover it as a quotient of V⊗W; it can even have a higher dimension than V⊗W. If V is a vector space over the reals then $\mathbb {R} \otimes V$ is naturally isomorphic to V in the same way that 1×n matrices are naturally isomorphic to row/column vectors. If V is a space of real-valued linear functionals then $\mathbb {Q} \otimes V$ could be seen as the corresponding space of quaternionic-valued linear functionals and $E\otimes V$ could be seen as the space of E-valued linear functionals. Does that answer your questions? -- BenRG (talk) 21:10, 18 August 2008 (UTC)

Where did the quaternions come from? $\mathbb {Q}$ is the rationals... --Tango (talk) 21:58, 18 August 2008 (UTC)

Guh, sorry, I had quaternions on the brain after participating in this marathon thread. $\mathbb {Q} \otimes V\cong V$ if V is over the rationals. -- BenRG (talk) 23:12, 18 August 2008 (UTC)

Ok, thanks a lot. Had to think about it for a while though. --Xedi^Talk 19:34, 20 August 2008 (UTC)

Fuel economy

From an old British Math book, is there adequate info to answer this?: Automobile A does 30 miles to a gallon of petrol and 500 miles to a gallon of oil. Automobile B does 40 miles to a gallon of petrol and 400 miles to a gallon of oil. If oil costs as much as petrol, which will be the cheaper automobile to run? Jonpol (talk) 15:55, 18 August 2008 (UTC)

Of course there is insufficient data to determine which is cheaper to run (what if one of them has an engine that requires constant expensive maintenance, and the other does not?), but there's certainly enough to determine which will leave you spending less on petrol and oil. Algebraist 15:59, 18 August 2008 (UTC)

If you let p be the price of a gallon of oil or petrol.

Then the cost per mile for A is p/30 + p/500, same for the other car, and you can then directly compare. As said above, this doesn't account for anything else, as the price of the car for example.--Xedi^Talk 16:04, 18 August 2008 (UTC)

For simplicity, I'd set p to 1 and use my calculator to determine the values of 1/30 + 1/500 and 1/400 + 1/40. --Bowlhover (talk) 19:45, 18 August 2008 (UTC)

The trick is to convert from mpg to gpm. Once you do that, it's quite easy to calculate the cost per mile. 202.147.44.80 (talk) 01:12, 19 August 2008 (UTC)

Hi Thanks, it was not so much the arithmetic but the wording of the question which I find ambiguous. When I first read it, it seemed that it could be interpreted to mean either you run your car on petrol or on oil (e.g. diesel). (There is an answer given at the back, which compares the cost for a 6000 mile journey (presumably using the LCM)). Jonpol (talk) 02:17, 19 August 2008 (UTC)

I don't think it means diesel, I think it means the oil you put in to serve as a lubricant (motor oil). I've not heard of diesel being referred to as "oil". --Tango (talk) 02:44, 19 August 2008 (UTC)

I agree with your interpretation, but the slang term "oil burner" is used to refer to diesel automobiles, and diesel fuel is essentially the same as heating oil. -- Coneslayer (talk) 17:13, 19 August 2008 (UTC)

The clue is "old British". It is only comparatively recently that diesel-engined cars have been common there, and whatever may be thought of the superior mpg such engines give, it is nothing like a factor of more than 10 compared with a petrol engine. Undoubtedly, lubricating oil is being referred to.…86.132.167.16 (talk) 22:13, 19 August 2008 (UTC)

August 19

Mathematicians do not use sliderule??

Is it true that Mathematicians do not use sliderule? I'm talking about back in the days where there is no such thing as pocket (electronic) calculators, and the only device for calculations are sliderule. Heard that Mathematicians do not use them and that most purchasers of sliderule are engineers. 211.28.51.233 (talk) 09:02, 19 August 2008 (UTC)

That is right. See Mathematics, it tends to be about general rules rather than particular numbers and you don't need a slide-rule for that. Engineers are more concerned about the concrete ;-) Dmcq (talk) 09:31, 19 August 2008 (UTC)

By the way you might like Human computer, they'd do the actual calculation very often when someone wanted to work some formulas or statistics out. Even engineers didn't use slide-rules all that much, many used mechanical calculators or could do just as good or better an approximation in their heads. They looked good though. Dmcq (talk) 09:49, 19 August 2008 (UTC)

It's not just a matter of sliderules, mathematicians don't use electronic calculators much either. As an example, my Uni Maths department offers 59 exams, only 23 of which allow the use of a calculator (and those are mostly statistics and probability), and it's not that they're trying to make the exams harder, you don't don't have any need for a calculator. --Tango (talk) 17:09, 19 August 2008 (UTC)

Right. In our department calculators are normally allowed for all exams, but (at least within pure mathematics) there's usually little point in bringing one; for some reason they don't tend to do topology at all well, and axiomatic set theory only slightly better. :-) —Ilmari Karonen (talk) 22:41, 19 August 2008 (UTC)

Try one of the new Möbius calculators. Wanderer57 (talk) 02:23, 20 August 2008 (UTC)

You need a better calculator. Just get one that's Turing-complete, program it to generate all the theorems of ZF, and wait for the result you want to appear on the screen. Algebraist 08:53, 20 August 2008 (UTC)

You may joke but the annual CADE ATP Systems competition which was on the 13th of this month says we'll soon be licking up to our robotic overlords and masters. Automated theorem proving says more about this competition. Dmcq (talk) 11:11, 20 August 2008 (UTC)

Pronumerals that satisfy an equation

I need to find all the combinations that satisfy this equation x² + y² + z² = 390. Is there a fast way to do it? 220.244.75.174 (talk) 10:14, 19 August 2008 (UTC)

You will probably notice that all such combinations are points that lie on the surface of a sphere with radius √390. I am not sure what kind of answer you are after but one nice way of characterizing the solutions to this is by using spherical co-ordinates:

x = √390*(cos(θ))*(sin(Φ))

y = √390*(sin(θ))*(sin(Φ))

z = √390*(cos(Φ))

for Φ belonging to [0,π] and θ belonging to [0,2π). Therefore, just choose any Φ and θ in the given range, and you can generate a triple that satisfy your solution. Of course, there may be other ways of characterizing the solutions to your equation but this is one way. For more information, you might like to have a look at Spherical co-ordinates.

I hope this helps.

Topology Expert (talk) 11:20, 19 August 2008 (UTC)

Firstly decide if the order of the numbers and signs matter. Then first simple method is to write a small program to find all the solutions and count them - or just list them all irrespective of order and then decide if the order matters. Second less simple method is to go through by hand taking the squares of 1 to 19 from 390 and check by hand if any of these is the sum of two squares. Third and most complex - learn up some maths Fermat's theorem on sums of two squares. The article doesn't say it but for two squares one can work out from the factorization how many ways of making up the number there are. There's also a theorem by Lagrange which says anything not of the form $4^{a}(8b+7)$ is okay. So with this one can be sure there is some such x,y,z and you just take the squares of 1 to 19 off and use Fermat's theorem. You don't have to work out the numbers but the problem is this last method only works if the order of the numbers matters. Dmcq (talk) 11:07, 19 August 2008 (UTC)

It is not specified whether the solutions to the equation have to be integers or not. However, now that I look at this, it seems more likely that only integer solutions are required. Otherwise, there would be uncountably many solutions.

My method still works though; just work when 390 multiplied with the square of cos (Φ) is a perfect square; this is an easy thing to do since the square of cos (Φ) has an absolute value less than 1. Then for each such angle Φ (there are only 19 such values), work out whether there exists an angle θ in the given range such that (sin (θ))*(sin(Φ)) squared times 390 is a perfect square; i.e work out all such numbers with absolute value less than 1 with the property of sin (θ); from there you can work out θ by using the inverse sine function. From here you reduce the possibilites of θ to ensure that x is also an integer.

I hope this helps.

Topology Expert (talk) 11:38, 19 August 2008 (UTC)

(There is a similar question at Wikipedia:Reference_desk/Computing#Excel_Question) Gandalf61 (talk) 12:59, 19 August 2008 (UTC)

Hyperbolic dynamics problem

Suppose x₀ is an hyperbolic fixed point of the map T.

Let T_ε be a map with the same hyperbolic fixed point such that for any x

$|T_{\epsilon }(x)-T(x)|<\epsilon$.

Question: can we say that for ε small the unstable manifold W^u(T,x₀) is close to the unstable manifold W^u(T_ε,x₀)? How close should they be? --Pokipsy76 (talk) 14:17, 19 August 2008 (UTC)

Pronumerals?

What are pronumerals? I don't think they were invented when I went to school. Wanderer57 (talk) 23:20, 19 August 2008 (UTC)

I've never heard the term before (perhaps it's commoner in some other language?) but Wikipedia and Google agree that it's another word for variable (presumably specifically a variable that represents a number). Algebraist 23:25, 19 August 2008 (UTC)

Thank you. I think mathematicians should be careful with this word invention business. Otherwise it could get as bad as with the physicists. Wanderer57 (talk) 23:57, 19 August 2008 (UTC)

Pronumeral is a fairly restrained example; it has a clear etymology which makes its meaning clear. Have you ever glanced at Morphism#See also? Algebraist 00:17, 20 August 2008 (UTC)

Thank you! That is jargoneering at its best. Idempotent endomorphism! Wanderer57 (talk) 02:09, 20 August 2008 (UTC)

Not quite the same thing. All the foomorphisms are (presumably) distinct concepts that someone (presumably) needed a name for. Pronumeral on the other hand looks like a silly name for something that already has a name. Just possibly, I suppose, it's intended to make a philosophical point (say, something that stands for an undetermined numeral, rather than an undetermined number, a numeral being a symbol for a number rather than the underlying noumenon). --Trovatore (talk) 07:35, 20 August 2008 (UTC)

I doubt any mathematician was involved in inventing the term pronumeral. It sounds to me like some maths teacher thing. <RANT>It seems that many of them are happier defining new words, saying how important their new words are and that parent's aren't capable of helping because they don't know them, and complaining about how badly the children spell the new words. I'm not a words person myself so I find it doubly annoying, I feel like saying why are you teaching maths if you're only interested in messing around with words?</RANT> Dmcq (talk) 08:49, 20 August 2008 (UTC)

*cough* improper fraction. Also check out "Lockhart's Lament" on K-12 math education in the US: [[1]]. Eric. 90.184.71.189 (talk) 10:46, 20 August 2008 (UTC)

Thanks for that. Can't fully agree - I think some rote learning is good, even I have to do it, but yeah I see here he's coming from. Dmcq (talk) 13:52, 20 August 2008 (UTC)

I haven't read any of the referenced articles, so will just ask a rhetorical question...

Pronumerial is to numeral/number as pronoun is to noun -- true or false? At least, that's what it looked like to me. --Danh, 70.59.119.73 (talk) 23:53, 20 August 2008 (UTC)

Sounds like a plausible explanation to me. --Tango (talk) 00:38, 21 August 2008 (UTC)

There's a whole lot of other ones like that under Pro-form, I'd protest if I were tested on them though! :) Dmcq (talk) 09:53, 21 August 2008 (UTC)

Just struck me, if they had pronumerals then presumably they'll also have proangles, procircles, prosets.... Perhaps we should call any answer to a question a coquestion instead to get more like modern jargon? Dmcq (talk) 12:53, 21 August 2008 (UTC)

What do you call the device that you use to measure a proangle? -- Coneslayer (talk) 12:55, 21 August 2008 (UTC)

Well there are pro-objects in category theory (and by extension, within whatever domain of mathematics you care to formalise in a category), but they are something else again. -- Leland McInnes (talk) 13:31, 21 August 2008 (UTC)

August 20

Pi ^ Pi

When I calculate Pi ^ Pi , I noticed that a lot of digits repeat themselves. Why is that?

Pi ^ Pi = 36.46215960720791177099082602269212366636550840222881873870933592293407436888169990462007987570677485

211.28.51.233 (talk) 11:24, 20 August 2008 (UTC)

Is that 17 repeats in 10 digits? The Binomial distribution with p=1/10 and n=100 gives mean 10 and variance 9 - so t is about 2.3 standard deviations out. So it probably would only occur about say 1 in 100 times, or 1 in 50 if you'd also have written in about having very few repeats, or practically always if you've looked at quite a few of these numbers - surely you already tried out pi*pi and 2*pi and pi*e and e^pi and pie and e^(2*pi) and e+pi and pi^3 and 1/pi and pi/e and......? The human mind is marvelous in its ability to see patterns everywhere. Dmcq (talk) 12:37, 20 August 2008 (UTC)

100 digits! That's enough precision (way overkill, actually) to measure the "circumference" of the observable universe in Planck lengths! Saintrain (talk) 21:41, 20 August 2008 (UTC)

Yummmy, pie! -hydnjo talk 23:20, 20 August 2008 (UTC)

The first 10000 digits have only 1010 repeats, which is well within expectations. Anyway, why look at the digits base 10? Oded (talk) 17:42, 21 August 2008 (UTC)

Gosh that's really surprising. It means that there were only 3 excess repeats in the next 9900 places after the first 100 where there were 7. The chances of getting within only plus or minus 3 repeats would only happen about 1 in 20 times. together with the 1 in 50 for the extra 7 in the frst 100 that must come to a chance of only 1 in 1000. Surely one chance in a thousand is astonishing? :) Dmcq (talk) 22:35, 21 August 2008 (UTC)

My comment was an answer to the question "do the digits of $\pi ^{\pi }$ base ten repeat more often than expected". This was my interpretation of what 211.28.51.233 wanted to know. My answer was that there is no reason to expect this (though I guess we will never know for sure). Are you suggesting now that the repetitions occur closer to the expected average than independent trials would give? I hope not. Oded (talk) 00:11, 22 August 2008 (UTC)

I believe Dmcq is alluding to the fact that if you decide what you're interested in after viewing your data (as the OP appears to have done), it's possible to be surprized by anything. btw, we might know for sure some day, if anyone ever finds a good way of proving numbers to be normal. Algebraist 00:50, 22 August 2008 (UTC)

Yes sorry I should mark my jokes better. This sort of reasoning has been used in tests of telepathy, they find that the subject gets a high score and then a low score or one with the results displaced by one and say that's real proof of the effect, presumably caused by tiredness! There might(!) be such an effect but it would have to be separately tested, and one always has to worry about the nine out of ten cats prefer trick where you do a whole bunch of similar tests and choose the result you want Dmcq (talk) 16:19, 22 August 2008 (UTC)

The probability of this event (there are over 17 repeats of a particular digit) by Chebyshev's theorem (I could find the exact value but this is quicker), is at most 0.184 (approx.) so it is not that unlikely. In fact, you might find it interesting to know that in the decimal representation of pi, there is a time when the digit 1 is repeated 10000 times continually in a sequence(in the decimal expansion of pi, there is a time when any particular digit is repeated 10000 times continually in a sequence(or any number of times for that matter))

Topology Expert (talk) 13:52, 24 August 2008 (UTC)

Do you have a reference for this? Is this proven, or just conjectured? Oded (talk) 20:10, 24 August 2008 (UTC)

As far as I know there's no proof of such things, but the word "conjecture" doesn't really cover it; that suggests there's some serious or reasonable doubt that it's true. A lot of propositions fall into this category (no proof from any accepted set of enumerated axioms, yet no reasonable doubt about their truth) — Goldbach's conjecture is an obvious example. I think we need some third word between "conjecture" and "theorem" for these. --Trovatore (talk) 22:52, 24 August 2008 (UTC)

How about "fact"? Oded (talk) 23:22, 24 August 2008 (UTC)

Wouldn't that presuppose the normality of pi - the very thing that has yet to be proven? -- JackofOz (talk) 00:29, 27 August 2008 (UTC)

But that's the point; it has yet to be proved, but it is not really in doubt. Still, "fact" is possibly misleading, as in a mathematical context it could indeed lead people to think it had been proved. --Trovatore (talk) 00:39, 27 August 2008 (UTC)

And that was exactly my point. There's no way it's a fact - yet. It may be generally assumed to be true, and that may be a perfectly reasonable approach, but that doesn't make it a fact. I note that Fermat's Last Theorem is still called a theorem and not "Fermat's Law" or some such, despite having in fact been proven. -- JackofOz (talk) 00:47, 27 August 2008 (UTC)

Well, this is a context-dependent interpretation of the word "fact". It would be perfectly reasonable to describe it as a "fact" in the same sense that the word is used in, say, physics. --Trovatore (talk) 01:00, 27 August 2008 (UTC)

I have read this (about there being arbitrarily long strands of digits in the decimal expansion of pi), in a mathematics journal before. However, I do not remember the exact link but I think that it has already been proven.

Topology Expert (talk) 06:19, 27 August 2008 (UTC)

I very much doubt that. The only way they would know that is if the normality of pi has been established, which it hasn't; because otherwise you'd have to test every conceivable string of numbers to see if it actually turns up in the decimal expansion. Which is an absurd proposition. How many places has pi been calculated to now - billions? Well, that's just a drop in the bucket. How about I come with a string of 10^50 numbers and ask someone to tell me where it appears. Since we don't know that pi is normal, we can't say with absolute certainty that it would definitely appear, without actually checking. This is where the meaning of "fact" become important. It's one thing for mathematicians to bet their worldly goods on pi being normal and to perform calculations etc as if that were the case - that's one meaning of "fact"; it's another thing to prove it incontrovertibly - that's another. -- JackofOz (talk) 07:15, 27 August 2008 (UTC)

Yeah, but it's not as big a difference as it's been made out to be. The supposed "incontrovertible" or "apodeictic" certainty of mathematical proof is an idea that comes down to us from, oh, Euclid or something, but it doesn't really make sense when you stop to look at it closely. Even if the proof is correct, if the axioms are wrong, then the conclusion could be wrong, and how do we know the axioms are right? It's turtles all the way down -- this is the basic problem of foundationalism.

My view is that mathematics is an empirical science, not so different in kind from physics, though different in the type of objects it studies. The axioms of mathematics are not arbitrary; they are accepted for reasons and on the basis of evidence. The reasons for believing in the normality of pi are not as compelling as the reasons for believing in the Peano axioms, but I don't see that they're qualitatively in a different category. --Trovatore (talk) 07:27, 27 August 2008 (UTC)

Math grad student forum

Does anyone know of a good online forum/community for mathematics graduate students?Borisblue (talk) 13:24, 20 August 2008 (UTC)

It really depends what kind of forum you are looking for. For interesting problems and questions of the "problem-solving type", the Unsolved and proposed problems of Art of Problem Solving's College Playground are usually nontrivial, often difficult and sometimes even intractable. They are certainly of interest even to graduate students, and some of the people there are skilled, active researchers. I usually read the analysis-related boards there, but I suppose the same is true of the other sections. The sci.math.research newsgroup has more research-oritented questions. Apart from this, there are a number of blogs with a large readership aimed at people with at a graduate background in math, like Secret Blogging Seminar (written by graduate students), n-category cafe (a little too much emphasis on categories and abstract geometry for my taste), or Terry Tao's blog. ˜˜˜˜ —Preceding unsigned comment added by 70.82.45.229 (talk) 21:13, 22 August 2008 (UTC)

Compact

Just a dumb question? Is the set [1,3) compact in (0,3). It is closed in (0,3), but I am not sure about compactness. Given a homeomorphism to [a,inf) with the R topology I would think it is not compact. Is the case [1,3) in (0,3) analogous to not being bounded, like [a,inf) is in R? Thanks! Brusegadi (talk) 17:28, 20 August 2008 (UTC)

'compact in' doesn't mean very much. Compactness is a property of a topological space. We say a set A in a space X is compact if A is a compact space when equipped with the subspace topology. Whether a set is closed depends on the ambient space, but whether it is compact does not. Thus the question is 'is [1,3), with its natural topology, compact?'. It is not: a direct proof is given by the open cover [1, 3-1/n). In general, a subspace of Rⁿ, with the subspace topology, is compact iff it is closed and bounded (Heine–Borel theorem). Algebraist 17:33, 20 August 2008 (UTC)

Thanks! Brusegadi (talk) 04:47, 21 August 2008 (UTC)

Just a general rule for arbitrary topological spaces: If Y is a subspace of Z and if X is a subspace of Y, then X is compact relative to Y iff X is compact relative to Z. In other words, every open cover of X by sets open relative to Y have a finite subcover that covers X iff the same is true for open sets relative to Z. In fact, the subspace topology that X inherits from Y equals the subspace topology that X inherits from Z. A more interesting property is that if X is a subspace of Y, then X is compact iff every open cover of X by open sets of Y has a finite subcover that covers X. Try proving this yourself.

Also, closedness implies compactness in a topological space iff that topological space is compact itself. For your particular example, the space (0, 3) is not compact as a subspace of R and therefore the fact that [1, 3) is closed relative to (0,3) is not a criterion for compactness. Perhaps a useful thing to note also is that compact subspaces of Hausdorff spaces are closed and therefore, compactness is equivalent to closedness for compact Haudorff spaces. From this, the Heine-Borel theorem almost easily follows.

I agree with algebraist; compactness is a property of a space (X) itelf and it does not matter what space that X can be imbedded in. For instance, it only matters what topology of [1, 3) you are considering when determining compactness; the fact that [1,3) (with its usual topology) can be imbedded in (0,3) is irrelavent.

I hope this helps.

Topology Expert (talk) 13:12, 24 August 2008 (UTC)

August 21

Nothing?

'Cmon, something? --hydnjo talk 23:17, 22 August 2008 (UTC)

There was something, but it was immediately reverted by its poster. Algebraist 23:19, 22 August 2008 (UTC)

But if you really want something, here goes: Is there any place in the world where today, the anniversary of one of the founding fathers of modern mathematics, is celebrated as it deserves? Algebraist 23:24, 22 August 2008 (UTC)

Lots of people have big celebrations on Newton's birthday, does that count? ;) --Tango (talk) 17:58, 23 August 2008 (UTC)

Newton's birthday is on the 4th of January. I was speaking of the 21st of August. Where are these celebrations, anyway? I never noticed them, and I spent the last four years at his college. Algebraist 22:07, 23 August 2008 (UTC)

Walked into that one. I take it you're using old style? Algebraist 22:09, 23 August 2008 (UTC)

I use whatever calendar is necessary for my jokes to work! --Tango (talk) 02:27, 24 August 2008 (UTC)

Ah, I misread you - I switched the words "where" and "today" and thought you were using "today" to means modern times and were just asking a general question. I doubt anywhere celebrates Cauchy's birth - he doesn't get all that much recognition within the mathematical community that I've seen, certainly not in the wider population. I suppose his home town might raise a toast to him, but that's probably it. --Tango (talk) 02:34, 24 August 2008 (UTC)

August 22

Semplifying my problem above

Since nobody seems to be going to answer to my question above I'll try with a semplified version of the problem which only requires to know standard calculus to be understood.

Let $T$ be a map (from $\mathbb {R} ^{2}$ to itself) and $x_{0}$ be the unique fixed point of $T$.

Let $x_{1}$ be a point such that the sequence $x_{n+1}=T(x_{n})$ converges to $x_{0}$.

Let $T_{\epsilon }$ be a map which also has $x_{0}$ as its unique fixed point and such that

$|T(x)-T_{\epsilon }(x)|<\epsilon$ for any $x$.

Question: if we fix a neighbourhood $B$ of $x_{1}$ is it true that for $\epsilon$ small enough there exists $y_{1}$ inside $B$ such that $y_{n+1}=T_{\epsilon }(y_{n})$ converges to $x_{0}$?

--Pokipsy76 (talk) 09:03, 22 August 2008 (UTC)

At the very least, you're going to need some continuity assumptions on T. What are you assuming? Algebraist 11:42, 22 August 2008 (UTC)

You are obviously right. Let's assume $T$ and $T_{\epsilon }$ are continuous on $\mathbb {R} ^{2}$.--Pokipsy76 (talk) 18:32, 22 August 2008 (UTC)

Smooth injective counterexample (I think): $T$ and $T_{\epsilon }$ both map the plane to $B_{\epsilon /2}(x_{0})\!$; T is contractive while $T_{\epsilon }$ is a rotation around $x_{0}$ composed with a monotonic radial map which expands some neighborhood of $x_{0}$. -- BenRG (talk) 13:28, 22 August 2008 (UTC)

But T is fixed: it cannot have a behavoiur which depends on epsilon.--Pokipsy76 (talk) 18:32, 22 August 2008 (UTC)

Oops. Then what if $T(x)=x_{0}$? No longer injective but still continuous. -- BenRG (talk) 21:28, 22 August 2008 (UTC)

For T bijective, how about $T(r,\theta )=({\text{sinh}}^{-1}\,r,\theta +\Delta \theta )$ and $T_{\epsilon }(r,\theta )=({\text{sinh}}^{-1}\,r+\epsilon \,{\text{tanh}}\,r,\theta +\Delta \theta )$, for any nonzero angle $\Delta \theta$? -- BenRG (talk) 21:47, 22 August 2008 (UTC)

Yes I'd been wondering what he meant, perhaps he meant maps satisfying the Banach fixed point theorem? Dmcq (talk) 16:43, 22 August 2008 (UTC)

No, the maps need not be contractions.--Pokipsy76 (talk) 18:32, 22 August 2008 (UTC)

The page you refer to in the previous post deals only with hyperbolic fixed points for flows, not for maps. Should this be fixed? Oded (talk) 19:19, 22 August 2008 (UTC)

Not sure what the previous question was but if the map isn't a contraction then for instance $T(x,y)->(x-(|x|+|y|)/(1+|x|),y)$ when iterated moves points on the line $(x,0)$ for positive x towards $(0,0)$ but has no other fixed points and no other points go to $(0,0)$. It is also continuous. Points with non-zero y will never converge to $(0,0)$ no matter how close to a point on the line. So $T_{\epsilon }$ can be the same as $T$. Does this do the job as a counterexample? Dmcq (talk) 19:54, 22 August 2008 (UTC)

Sorry I see you want there to be no point within a short distance of $x_{0}$ which goes to the fixed point. Yes I see that previous counterexample now. Just have $x_{0}$ be the fixed point at the centre of a rotation and that is a counterexample. It always maps to itself and nothing else does no matter how close. Dmcq (talk) —Preceding undated comment was added at 20:12, 22 August 2008 (UTC)

The the distance between the points $T_{\epsilon }(y_{n})$ and $T(x_{n})$ would have to decrease with increasing n. In fact, if z_n is the distance between $T_{\epsilon }(y_{n})$ and $T(x_{n})$, then z_n would have to converge to 0. —Preceding unsigned comment added by Topology Expert (talk • contribs) 13:37, 24 August 2008 (UTC)

Category of monoids

Is the category of monoids (in Set) Cartesian closed? 83.23.209.193 (talk) 15:25, 22 August 2008 (UTC)

About the Möbius strip

I'm reading about topology, and I can't figure this one out.

Why do a paper strip and a Möbius strip have different intrinsic topologies? Or as they ask it, How does a Two-Dimensional creature (Flatlander) living on a Möbius strip logically implies that it's not a paper strip?... —Preceding unsigned comment added by DarkLaguna (talk • contribs) 15:43, 22 August 2008 (UTC)

I'm not sure what you mean by a paper strip; I shall assume you mean a cylinder, i.e. a Möbius strip without the half twist. In this case, a difference is that the paper strip is orientable while the Möbius strip is not. Concretely, if a flatlander set off on a trip round the Möbius strip, he would end up back where he started, but would find that he had become the mirror-reflection of his former self. Algebraist 15:48, 22 August 2008 (UTC)

Why would he be a mirror image? If the "circumference" of the mobius strip is the length of the original piece of paper, then after 2 laps he is back where he started as he was. After one lap, he is still "himself" but is just on the other side of the piece of paper - which stops him knowing he is the other way round etc. Are you assuming that the mobius flatland allows interaction between the two "sides" of the strip? If so why is that assumption justified / necessary? -- SGBailey (talk) 22:13, 23 August 2008 (UTC)

A truly two-dimensional strip doesn't have 'sides', just as our three-dimensional space doesn't, and the flatlanders live in it rather than on it. Of course, the OP used 'on' in the question, so my interpretation may be wrong. If they are living on the surface of the strip, then there's no topological difference between their home and a cylinder. Algebraist 22:16, 23 August 2008 (UTC)

Unless you consider the sides to be joined by the boundary, allowing you to cross over. Then you would know you were on a Moebius band, not a cylinder, if you could go half way round, cross over the boundary and get back to where you started. I'm not sure how to describe that topologically, but it works in the intuitive sense of an ant walking on the surface. (Actually, if you consider the two sides to be distinct, and joined by the boundary, then I think you have the Klein bottle.) --Tango (talk) 22:44, 23 August 2008 (UTC)

No, that's not the Klein bottle. Tango, I suggest you read Algebraist's entry again. Oded (talk) 22:51, 23 August 2008 (UTC)

If you consider a paper Möbius strip as a three-dimensional object, and then consider its boundary (smoothing out that pesky edge), then you just get a torus, not a Klein bottle. The same applies if you start with a non-Möbius strip. Tango's method doesn't work since going 'half way round' doesn't make sense topologically. Algebraist 23:01, 23 August 2008 (UTC)

Oh yeah, it is a torus - the ends are identified with a half-rotation, not a reflection, that doesn't help then. My method depends on the boundary being recognisable in some way. If it is, then the Moebius band is recognisable because there exists a closed curve which crosses the boundary precisely once. If the boundary isn't recognisable, then it's just a torus and there's no difference. --Tango (talk) 00:20, 24 August 2008 (UTC)

Ah, I see. Just for fun, I wonder if there's some sensible way to topologize the 'surface' of a Möbius strip so that the boundary is topologically recognizable. Any ideas? Algebraist 00:26, 24 August 2008 (UTC)

I am not 100% if I understand your question the way you meant it, but the natural thing is to think of it as a laminated 2-manifold. In this case this just means that on the surface there is a curve which you require the transition maps of the manifold atlas to respect. Perhaps it can also be thought of as an orbifold; I'm not sure. Oded (talk) 00:57, 24 August 2008 (UTC)

I assumed you could do something like that. I was just wondering if there's a sensible answer to the question 'what topological space is the surface of a Möbius strip' (which is, as it stands, not quite well-posed enough to necessarily have a unique answer) which allowed you to retain the boundary. Algebraist 01:18, 24 August 2008 (UTC)

I guess it comes down to definition. What is required for it to count as the surface of a Moebius strip? I suppose everywhere other than the boundary should be locally Euclidean, otherwise imagining flatlanders living there becomes rather difficult, and that's really all you need since you only need to consider the surface near a section of the boundary, so basically what you need is a way to identify the boundaries of two half-planes in such a way as to make it recognisable. I'll need to think about what that actually means... Obviously, if we consider it as a differentiable manifold, then it's easy - that's probably the best way to do it! By the way, aren't you British? What's with all the z's? --Tango (talk) 01:02, 24 August 2008 (UTC)

I am a citizen of the Kingdom of Kent, owing no allegiance to anything north of the Thames. Remember Otford! At present, however, I'm preparing to commence my studies at Oxford. Algebraist 01:18, 24 August 2008 (UTC)

Ok, how about this: Take two closed half-planes, identify the boundary and then, instead of giving it the usual quotient topology, give it the topology formed by taking the union of the topologies of each half-plane as a subbasis. Away from the boundary, everything is as it should be and the boundary is clearly identifiable since it's the only place you can have line segments that are open. Exactly how you do the same thing with a Moebius band, I don't know, but the principle should be exactly the same locally, it's only globally that things are different. --Tango (talk) 02:25, 24 August 2008 (UTC)

I agree with Algebraist (as usual). Here are a few other ways to think about it. First, the boundary of the cylinder consists of two disjoint circles, while the boundary of the Mobius strip is just one circle. You could object to this by saying that you are interested in the open Mobius trip (without the boundary) and the open cylinder. Another observation is that if you remove from the cylinder a simple closed curve, you disconnect it. However, if you remove the "equator" from the Mobius strip, what you get is a cylindar. Oded (talk) 16:28, 22 August 2008 (UTC)

On beyond expm1

The expm1 function effectively removes the first term of the Taylor series of $e^{x}$ without the cancellation caused by doing it manually. Does anyone know of an extant function (a CS question) or an expression (a numerical math question) that would remove the first two terms and accurately calculate $e^{x}-1-x$? (Example: an expression of expm1 in terms of exp and sinh.) --Tardis (talk) 16:07, 22 August 2008 (UTC)

Ummm, why not just compute the Taylor series minus whichever terms you want to exclude? Dragons flight (talk) 16:56, 22 August 2008 (UTC)

I would like to avoid switching between multiple algorithms (as the series would converge slowly for large x) and to avoid rounding error in summing the series explicitly. I would trust a simple combination of standard functions (with known, small errors) more than a long sequence of elementary operations. --Tardis (talk) 17:19, 22 August 2008 (UTC)

The version on the python list expm1(x) = 2*exp(x/2)*sinh(x/2) looks incredibly inefficient and does not improve accuracy. I would not take this as a base for any numerical algorithm. --Salix alba (talk) 09:02, 23 August 2008 (UTC)

You must have misread the formula. The only area where expm1 is problematic is around 0, and the sinh formula neatly avoids the cancellation there. It should be at worst about 2x or 3x more expensive than exp. It's obviously more efficient to use a custom Taylor series or interpolating polynomial with a fancy argument reduction scheme, but it's up to you whether you want to spend an evening implementing and debugging that or if you're happier typing in a simple formula that just works^TM. Fredrik Johansson 10:38, 23 August 2008 (UTC)

It might be worth having a look at the source code of some real implementations, for example[2]. It seems that to calculate exp(x) use x=k ln(2) + r, with |r|< ln(2)/2, k integer. So exp(x) = exp(r) * exp(k ln(2)) = exp(r) * 2^k. The use a Reme algorithm to find a 5 degree polynomial in r² which can efficiently and accurately calculate exp(r)-1. --Salix alba (talk) 09:54, 23 August 2008 (UTC)

Sine function

It seems that $Im(x^{x})\times |x|^{|x|}=sin(x\times \pi )$ for negative numbers. Why is that? Is there a name for this equality? Thanks --helohe (talk) 22:37, 22 August 2008 (UTC)

I don't know of a name for it, but I have a proof of it. It's easier to change all the xs to -xs and assume x is positive.

Doing that, we have:

$Im((-x)^{-x})|-x|^{|-x|}=Im((-x)^{-x})x^{x}=Im({\frac {1}{(-x)^{x}}})x^{x}$

Then, since x is positive, x^x is real, so can be taken inside the imaginary part:

$=Im({\frac {x^{x}}{(-x)^{x}}})=Im({\frac {x^{x}}{(-1)^{x}x^{x}}})=Im((-1)^{-x})$

We then use $e^{i\pi }=-1$ and get:

$=Im(\left(e^{i\pi }\right)^{-x})=Im(e^{-ix\pi })$

Then we use $e^{i\theta }=\cos(\theta )+i\sin(\theta )$ and the result follows immeadiately. --Tango (talk) 00:02, 23 August 2008 (UTC)

Oh, yes thats it. Thanks --helohe (talk) 10:49, 23 August 2008 (UTC)

August 23

Pyramid Volume section contains wrong derivation of volume: a negative sign in the integral...

Hi,

Just noticing in the Pyramid_(geometry) following "The volume is given by the integral"

There's a negative sign (-A / 3h^2) and there shouldn't be! Right? link title

How would I fix this? I don't know how to generate the math images... Oh wait, it's

${\frac {A}{h^{2}}}\int _{0}^{h}(h-y)^{2}\,dy={\frac {A}{3h^{2}}}(h-y)^{3}{\bigg |}_{0}^{h}={\frac {1}{3}}Ah.$

Right? Thanks! InverseSubstance (talk) 18:12, 23 August 2008 (UTC)

No, the sign is correct. Algebraist 18:16, 23 August 2008 (UTC)

More explicitly, the sign gets flipped once by the (implied) substitution for $-y$, and it gets flipped again because you evaluate at $y=h$ where the antiderivative is 0 and subtract the positive evaluation at $y=0$. Hope this helps. --Tardis (talk) 18:26, 23 August 2008 (UTC)

Wait. I understand the evaluation part. But not the integration part. So it seems to me what your saying is, fex:

$\int (1-x)dx=-{\frac {1}{2}}(1-x)^{2}$ Right? What rule of integration does this fall under? Thanks... InverseSubstance (talk) 19:25, 23 August 2008 (UTC)

There is no rule for that, but it's useful to note that $\int (a+bx)^{n}\,dx={\frac {(a+bx)^{n+1}}{(n+1)b}}+c$ for complex a and b. x42bn6 Talk Mess 20:15, 23 August 2008 (UTC)

It's called the substitution rule. You write $u=1-x\Rightarrow du=-dx,dx=-du$ so $\int (1-x)dx=\int u(-du)=-\int u\,du=-{\frac {u^{2}}{2}}+C=-{\frac {(1-x)^{2}}{2}}+C$. For definite integrals, you have to transform the limits too, unless you immediately (as I did here, and as the original example does) switch back to the original variable before evaluating. --Tardis (talk) 04:34, 24 August 2008 (UTC)

You can see it's true by differentiating, the key rule there being the chain rule, the derivative of -x (and, therefore, 1-x) is -1. --Tango (talk) 20:50, 23 August 2008 (UTC)

How to find the shorter path?

What algorithms can you use to find the shorter path in a net (like a very complex subway net with thousands of nodes)? —Preceding unsigned comment added by Mr.K. (talk • contribs) 19:16, 23 August 2008 (UTC)

Is Shortest path problem#Algorithms what you're looking for? -- BenRG (talk) 19:44, 23 August 2008 (UTC)

Geometric figure for xⁿ

If x is a segment, x² is a square, x³ a cube, what geometric figure is x⁴, and is there a figure for every xⁿ? —Preceding unsigned comment added by Mr.K. (talk • contribs) 19:17, 23 August 2008 (UTC)

x⁴ is the hypervolume of a tesseract, and x⁵ a five-dimensional hypercube, and so on. There can't be a three-dimensional figure with a volume of x⁴, because the units come out wrong. For example if x is 2 meters, then x⁴ is 16 meters⁴, but a volume has to have units of meters³. -- BenRG (talk) 19:49, 23 August 2008 (UTC)

It is interesting to note that the first solution to the quartic equation was published together with the first solution to the cubic equation (by Cardano, I think), but Cardano downplayed the solution of the quartic equation, because at that time raising things to the power of 4 was commonly regarded as meaningless, as there is no corresponding geometrical object in three dimensions. Eric. 213.173.233.179 (talk) 09:48, 24 August 2008 (UTC)

August 24

Math question

There's a quarter-circle ABC with centre O and radius unit 1. There is a point D which is the midpoint of OC and DB is parallel to OA. Ive tried to draw it roughly using characters cos i wasnt bothered to draw and upload a proper picture but it should give you a good idea of it.

 A |\\\
   |    \\  B
   |      / \\
   |    /     |
   |   /      | \\
   |  /       |    \\
   | /        |      \\
   |____________________
   O          D       C

I have been asked to "BY considering separately the triangle OBD and the sector OAB, show that the area of the shaded region OABD is 1/24(2π + 3 sqrt(3) )" I get sqrt(3)/8 for OBD and 1/12π for OAB. And cant get to the required answer. What am I doing wrong? By the way, this is not homework, Im just doing questions as part of my revision for my resits. --212.120.246.239 (talk) 17:19, 24 August 2008 (UTC)