Wikipedia:Reference desk/Archives/Mathematics/2007 December 7

= December 7 =

Probability question
This site's business model has intrigued me, and prompted this question. Say every person has to place a bid on an item, between $0.01 and $4.00, with any amount (except fractions of cents) within that range being valid. Assume each bid amount is equally likely to be chosen. How many people must bid for the probability that every possible bid amount has been bid on at least twice (i.e. there are no unique bids) is greater than 50%? I have very little background in statistics, so I have no clue as to how to approach this. Thanks in advance to anyone who can help. -Elmer Clark (talk) 01:14, 7 December 2007 (UTC)
 * I'm not going to answer this (because I can't), but note that 'every possible bid amount has been bid on at least twice' is not the same as 'there are no unique bids'. If two people bid, both choosing $0.01, then there are no unique bids. Algebraist 02:04, 7 December 2007 (UTC)
 * I realize that, but I figure this is the factor that will in most cases determine whether there are any unique bids, so I'm considering just it for simplicity's sake. -Elmer Clark (talk) 02:16, 7 December 2007 (UTC)


 * Alright: Yours is a question obout multinomial distribution: I'm not sure this will be easy to compute. However, you could make some simulations to have a rough idea. With more time, I may be able to make some simulations. Not tonight. Too tired :( Pallida  Mors  03:47, 7 December 2007 (UTC)
 * Hem... I guess I'm not that tired to let the question sleep for some hours. I started to do some simulations. My first guess is that the number lies between 3000 and 4000 bids (bidders, I mean). Pallida  Mors  05:43, 7 December 2007 (UTC)


 * To get a biased guess at the answer, you can treat all people as indistinguishable. Then, it's a question of finding 1) the number of ways of putting X balls in 400 baskets, 2) the number of ways of putting X-800 balls in 400 baskets, then dividing the second by the first. The value of X (greater than 800) for which this is about 1/2 is probably a bit less than the right answer. I believe Figurate number describes counting of this type. Black Carrot (talk) 15:40, 9 December 2007 (UTC)


 * The fraction would be (X-401 choose 399)/(X+399 choose 399) = ((X-401)!/(X-800)!399!)/((X+399)!/X!399!) = (X-401)!(X)!/(X-800)!(X+399)!. This puts X at around 500000. Black Carrot (talk) 16:09, 9 December 2007 (UTC)


 * Sorry, Black Carrot, but I can't follow your analysis. Anyway, simulation gives a number between 3400 and 3500. Pallida  Mors  22:44, 9 December 2007 (UTC)

Going back to the unique bid auction site mentioned in the original post, and considering various bidding strategies from a game-theoretical point of view, it seems that the optimal strategy, if placing a bid cost nothing, would be to place bids on every sum below the maximum you were willing to pay. If everyone did this, it would effectively turn the system into a Vickrey auction, like the system employed e.g. by eBay. Of course, the fact that they do charge an "administrative fee" for each bid rather ruins this strategy, which is presumably the intent. (It's also how they make their money.)

One could, of course, attempt a more complex game-theoretic analysis of the system with the administrative fees included. However, if I wanted to get rich on such auction and wasn't above a bit of cheating, I'd forget the analysis and just try to find some way to gain insider access into the list of bids placed so far. Such abuse would be easy to detect if one kept doing it too often, but if one was cautious enough, it shouldn't be too hard to establish a reputation for late low "long-shot" bids, and then just once use the inside information to win, e.g., a house for $1.38. —Ilmari Karonen (talk) 04:32, 10 December 2007 (UTC)


 * Ok, well, I get the impression this is a much more complex question than I had realized. Thanks to everyone who gave input, especially to Pallida Mors with your estimate. -Elmer Clark (talk) 11:18, 12 December 2007 (UTC)

A question 8 years in the making
I'm a college student, and I was looking through my old school assignments, and I found an old math assignment from 6th grade. The only problem I had gotten wrong was as follows:

"Jessica had 12 paper fraction pieces, each of twelfths, sixes, fourths, and thirds. He [sic] did not have enough pieces to show 1/3 and 2/4 as equivalent fractions. How can he form enough by making one cut with a scissors?"

My answer had been: "Put them in a pile and cut them all in half."

The teacher had circled it in red and wrote: "Nice try. Read the question again! :)"

What was the correct answer, what does the question even mean, and since when is 1/3 and 2/4 eqivalent fractions? --Ye Olde Luke (talk)  02:15, 7 December 2007 (UTC)


 * I... suspect the teacher intended to check whether you knew how to write 1/3 and 2/4 in common units (that is, as 4/12 and 6/12). I have to admit I find the question a little bizarre.  Does the question say that Jessica has 12 paper fraction pieces each of the four types (for a total of 48), or 12 paper fraction pieces distributed in some unspecified way over the four types? Michael Slone (talk) 02:32, 7 December 2007 (UTC)


 * I worded the question exactly the way it was on the paper. Unfortunately, I don't remember anything about this incident back from 6th grade, so I don't have any more information than what I've already posted. I'm guessing my 6th grade self didn't have a clue either, and merely put down something that might concievably work, even if it's not the best or most efficient answer. Apparently the teacher knew the right answer, so it can't be unsolvable. --Ye Olde Luke (talk) 02:43, 7 December 2007 (UTC)


 * The answer is clearly 42. Errm... what was the question again? --Lambiam 04:21, 7 December 2007 (UTC)


 * This is 6th grade, so the answer shouldn't be too hard to find. If we try solving the problem using both options (12 of each fraction, for 48 all-in-all; or 12 pieces distributed somehow over the four types), which one gives the most reasonable and succinct answer? --Ye Olde Luke (talk) 14:51, 7 December 2007 (UTC)


 * It's clearly the 12 in total. Otherwise, he would have enough pieces (10) to represent the given fractions as 12ths. Frankly, the question is too murky to be easily solved. Go find your teacher and ask :p mattbuck (talk) 15:31, 7 December 2007 (UTC)

I believe I am able to find a kind of solution, using a bit of imagination to complete the unclear indications of the problem. I suppose it means that we have at our disposal some fractions in twelfths, sixths etc., such that summing up one or more of them we cannot get 1/3 nor 1/2. So a priori we might have 2/3 (not 1/3), 1/4 and 3/4 (but not 2/4), 1/6, 4/6, 5/6 (but not 2/6 nor 3/6), and all fractions of the form k/12 for k=1, 2, ..., 11, excluding 4/12 and 6/12. Too many: we have fifteen of them.

Furthermore, 1/4 + 1/6 + 1/12 equals 1/2, and so does 1/12 + 2/12 + 3/12, so try removing 1/12; for similar reasons, remove 3/12 (1/4 + 3/12 = 1/2); in order to making it impossible to get 1/3, let us also remove 2/12 (for 1/6 + 2/12 = 1/3). So we are left with exactly twelve fractions. (I have not checked other possibilities for being left with exactly 12 fractions none of which sum up to 1/2 or 1/3.)

Cut in half "2/3", so you get 1/3, 1/3, 1/6 and the remaining ones, and the first three ones allow us to get both 1/2 and 1/3.

Sound too fanciful? Goochelaar (talk) 15:56, 7 December 2007 (UTC)


 * Sounds good. But how does that show 1/3 and 2/4 as equivalent fractions (i.e. 4/12 and 6/12)? --Ye Olde Luke (talk) 01:19, 8 December 2007 (UTC)
 * Oh, I read that as meaning "show 1/3 and 2/4 as such or in some equivalent form" i.e., not equivalent to each other, but equivalent to whatever you will get by cutting and playing around with Jessica's stuff: in my attempt, one is actually 1/3 and the other one 1/3 + 1/6 = 1/2 = 2/4. Goochelaar (talk) 16:53, 8 December 2007 (UTC)

Also, if you want to show 4/12 and 6/12, just use your one cut to divide the 1/6 piece into two 1/12ths. Then, combine a 1/12 + 3/12, and a 1/12 + 5/12, to make 4/12 and 6/12. Calforester (talk) 05:23, 9 December 2007 (UTC)
 * Ah, I understand! Thanks for your help, Goochelaar, and everybody else too! --Ye Olde Luke (talk) 01:02, 9 December 2007 (UTC)

Riemannian manifolds
questuion1 : One R^n can had two Riemannian manifolds (M,g) (N,h)?

and http://en.wikipedia.org/wiki/Conformal_map had said: --wiki-- A diffeomorphism between two Riemannian manifolds is called a conformal map

if the pulled back metric is conformally equivalent to the original one.

questuion2 : so conformal map can jest be a map not about angle?

questuion3 : conformal map can denote bijective map f:(M,g)->(N,h) then inverse f^-1(N,h)->(M,g)

between two Riemannian manifolds?

Thanks —Preceding unsigned comment added by 59.104.150.66 (talk) 05:01, 7 December 2007 (UTC)


 * On a Riemannian manifold the metric tensor can be used to define a Riemannian equivalent of angle - a concept that plays the same role as the normally understood angle does in Euclidean geometry. In the same way as conformal maps on the Euclidean plane preserve the "Euclidean" angle, so we can also define a class of functions between Riemannian manifolds that preserve the "Riemannian" angle - these are also called conformal maps. In fact, you could say that the "Euclidean" conformal maps are a special case of the more general "Riemannian" conformal maps. Conformal maps are invertible, and the inverse of a conformal map is also a conformal map. Gandalf61 (talk) 10:35, 7 December 2007 (UTC)

Solution to a nonlinear ODE
I want to solve:

y''' = C1*y + C2*cos(y) y(0) = y0 y'(1) = yp1 y''(1) = ypp1

If possible, I want to do it in closed form, but if not the evaluation should be as cheap as possible in terms of computational effort. Any ideas how best I can do it? Regards, deeptrivia (talk) 21:35, 7 December 2007 (UTC)


 * OK, I'm assuming you mean $$\dfrac{\partial^3y}{\partial x^3} = c_1y + c_2\cos y$$, with y(0) = y0 but I'm confused as to what yp and ypp might mean. mattbuck (talk) 00:06, 8 December 2007 (UTC)

They're just some known constants. deeptrivia (talk) 02:40, 8 December 2007 (UTC)

Since the function on the right is independent of x, why not just integrate with respect to y three times and the constant of integration can be solved for using the initial conditions. A Real Kaiser (talk) 05:16, 8 December 2007 (UTC)


 * It doesn't work that way, I'm afraid. Try $$y = y^2$$. By double integrating, you get $$-\log y = x^2 + C_1 x + C_2$$ or, with constants set to 0, $$y=e^{x^2}$$ and now $$y = (d/dx)(2x e^{x^2}) = (d/dx)(2xy) = 2y + 4x^2 y \ne y^2$$. The technique fails because you are integrating $$\int (1/y^2)d(dy/dx)$$, which doesn't work since there is no $$dy/dx$$ term in the integrand. SamuelRiv (talk) 06:34, 8 December 2007 (UTC)

Nonlinear ODEs do usually not have closed form solutions. Use numerical integration. Transform the third order differential equation in one variable into three first-order differential equations in three variables: dy1=y2&middot;dx, dy2=y3&middot;dx,  dy3=(C1&middot;y1 + C2&middot;cos(y1))&middot;dx. Set x=0. Now substitute the initial values for y1(0),y2(0),y3(0). Choose a small positive number for dx.
 * Compute dy1,dy2,dy3 from the ODEs. Compute y1(x+dx)=y1(x)+dy1, y2(x+dx)=y2(x)+dy2, y3(x+dx)=y3(x)+dy3. Set x=x+dx. Repeat.

This method is the simplest. I may not be the cheapest in terms of computational efford. Why do you want the cheapest method?

Bo Jacoby (talk) 10:09, 12 December 2007 (UTC).

Double Integral
Hello. Can I use a double integral when I add, subtract, multiply, divide, etc. the same number on both sides? For example, does $$(6x^2 + 2x)\times\iint_{}^{} \,\frac{1 - x}{2x} = \frac{5x}{3x + 1}$$ constitute $$(6x^2 + 2x)\times \frac{1 - x}{2x} = \frac{5x}{3x + 1} \times (6x^2 + 2x)$$? Thanks in advance. --Mayfare (talk) 23:59, 7 December 2007 (UTC)


 * I feel a bit confused about quite what you mean, but if you take any $$\int f(x)dx = F(x)$$ then $$g(x)\int f(x)dx = g(x)F(x)$$. Is that what you mean? If so, it works perfectly well for double integrals. mattbuck (talk) 00:10, 8 December 2007 (UTC)
 * Never forget the $$dx$$ and the like in integrals, they are very important. In this case in particular, it is not at all clear what your two variables are. It is also best to specify the domain over which you are integrating, in particular if it depends on some variables. Of course, you should check your formulae since something is obviously messed up. If you are asking if $$f(x)\int g(x)\ dx = \int f(x)g(x)\ dx$$, not only is this wrong for single integrals, it is not even meaningful for double integrals with standard notation. -- Meni Rosenfeld (talk) 17:02, 8 December 2007 (UTC)

In an integral, you use a dummy variable. In $$\int f(x) dx$$, the dummy variable is x. If, in the one expression, you use a variable as a dummy variable and as a normal variable, as in $$g(x)\int f(x) dx$$, confusion ensues. Something along the lines of $$g(x)\int^{x} f(x')dx' = g(x) F(x)\,$$ is a bit less ambiguous. Of course, you have to remember that, since this is an indefinite integral, there is a constant of integration, and where you put it is important - in this case, it's $$g(x) \left(F(x) + C\right)$$, NOT $$g(x) F(x) + C\,$$. Confusing Manifestation (Say hi!) 22:34, 9 December 2007 (UTC)