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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Where did the quaternions come from?

\mathbb {Q}

is the rationals... --Tango (talk) 21:58, 18 August 2008 (UTC)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]