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April 6

3d surface

Is a hollow sphere i.e a spherical shell with almost zero thickness a 3 d surface or a 2 d one??? Does there exist a 3 d surface by the way?? —Preceding unsigned comment added by 119.235.54.67 (talk) 06:53, 6 April 2010 (UTC)

The hollow sphere has two 2d surfaces, one inner and one outer. The boundary of a 4d sphere is a 3d surface. Bo Jacoby (talk) 07:25, 6 April 2010 (UTC).

Strictly speaking, I think a surface has to be locally two-dimensional. The generalisation of a surface to different numbers of dimensions is a manifold - however, I expect most mathematicians would understand that a "3d surface" informally means a 3-manifold. In your example, because you said the spherical shell has "almost zero thickness", each point in the shell needs three co-ordinates to determine its location. Therefore the shell is a 3 dimensional space. We can go further - because the space around each point in the interior of the shell is toplogically the same as 3d Euclidean space, we can say that the shell is a 3-manifold. However, you cannot travel as far as you like in any direction within the shell, because you may hit the inner surface or the outer surface. These surfaces are boundaries, so the shell is a "3-manifold with boundaries". The boundaries themselves are 2-manifolds, which we can also call surfaces. Note that if you had said that the shell had zero thickness, then it only has two dimensions (each point on the shell now only needs two co-ordinates to locate it) and so a zero-thickness shell would be a 2-manifold or surface. Gandalf61 (talk) 08:33, 6 April 2010 (UTC)

that all makes sense, except don't you always need three coordinates to locate a point, at least since its a sphere which can only exist in a 3 dimensional space? 68.171.233.51 (talk) 09:05, 6 April 2010 (UTC)

I think I answered my own question, since you could arbitrarily draw perpindicular axes within the sphere, identify one of the intersections as the origin and then go from there in the same way as if it were a standard 2d cartesian system; so it identifies the point internal to the sphere without referencing a 3rd dimension. Is that right?68.171.233.51 (talk) 09:09, 6 April 2010 (UTC)

To locate a point that is on the surface of a sphere you only need, at a minimum, two co-ordinates - to locate a point on the surface of the Earth, for example, you only need to know its longitude and its latitude. You could use a co-ordinate system with more dimensions - you could, for example, use Cartesian co-ordinates relative to the centre of the Earth - but you are then using more co-ordinates than you need. The fact that the surface happens to be embedded in a 3d space is irrelevant to toplogists, who focus on properties that are intrinsic to the surface itself. There are some surfaces, such as the Klein bottle, that cannot be embedded in 3d space without intersecting themselves - but they are still legitimate 2d surfaces. Gandalf61 (talk) 09:39, 6 April 2010 (UTC)

See also Spherical coordinate system - you can describe any point on a sphere with only $\theta$ and $\phi$. -- Meni Rosenfeld (talk) 11:39, 6 April 2010 (UTC)

I think the OP really had zero thickness in mind, but didn't know that was the right term for it. -- Meni Rosenfeld (talk) 11:39, 6 April 2010 (UTC)

April 7

Linear Combinations

Given a list V of positive numbers, what is a fast way to count how many lists U of nonnegative numbers give U.V<=N for a given N? I've found 'figurate numbers', but I can't think of a way to adjust the formula. Black Carrot (talk) 00:23, 7 April 2010 (UTC)

I don't understand, neither the problem, nor the connection to linear combination, nor the connection to figurate number. Can you give a simple example? Bo Jacoby (talk) 08:17, 7 April 2010 (UTC).

The OP seems quite clear to me. U and V denote vectors (of positive, resp. nonnegative integers) and U.V denote their scalar product. So, given V:=(v₁,...,v__k) and a number n, we want to compute the cardinality c(n) of the set

$\{u\in \mathbb {N} ^{k}\,:\,\sum _{i=1}^{k}u_{i}v_{i}\leq n\}$

It should be easy to write a generating function for the sequence c(n); actually it should turn out to be a rational function. From the GF one can derive asymptotics on the c(n)'s. Of course, a first approximation is the volume of the pyramid containing the integer points U. See also Schur's theorem#Combinatorics for a very close problem.--pma 09:01, 7 April 2010 (UTC)

re the clarity of Black Carrot's question, the word he missed is "integers". More confusion was caused by the fact that linear combinations usually refer to vector spaces over a field, which the integers are not. That said, I didn't find it difficult to understand what the question was about. -- Meni Rosenfeld (talk) 09:42, 7 April 2010 (UTC)

Even doing this for real numbers, finding the volume of a convex figure, isn't that easy. If the number N isn't too large I'd do it simply by counting using loops which stop when the number gets too large. My guess is that this is an NP problem in essence for integers. Dmcq (talk) 10:37, 7 April 2010 (UTC)

It's not an NP problem, as NP is a class of decision problems, and here we want a number as output, not a yes/no answer. The problem is in #P. Maybe you meant that it is NP-hard? That's quite possible, given its resemblance to the knapsack problem.—Emil J. 11:57, 7 April 2010 (UTC)

Actually in this case the volume is very elementary. --pma 16:33, 7 April 2010 (UTC)

Oh sorry yes you're right, it is just a n-dimensional pyramid. I had a picture of a load of faces but that's wrong. In that case calculating the number of integer points may be fairly straightforward. Dmcq (talk) 17:13, 7 April 2010 (UTC)

Thanks for the explanation.

Bo Jacoby (talk) 12:12, 7 April 2010 (UTC).

Ehrhart polynomial may be helpful to answer the question.--RDBury (talk) 13:45, 7 April 2010 (UTC)

Yea, that is the answer. Actually it includes Schur's theorem, which is essentially the case where the convex L is an "orthogonal simplex" as in the OP's question. The first approximation $c(n)={\frac {n^{k}}{k!\prod _{i=1}^{k}v_{i}}}(1+o(1))$ (as $k\to \infty$) may be deduced by an elementary rescaling argument. And that also holds if the v₁,..,v_k and n are just positive real numbers .( as the OP possibly wanted?). --pma 16:33, 7 April 2010 (UTC)

In what sense it is the answer? How is the Ehrhart polynomial easier to compute than brute force enumeration of the points? Come to think of it, how do you compute the polynomial other than evaluating it in k + 1 points by aforesaid brute force enumeration, and interpolating?—Emil J. 16:53, 7 April 2010 (UTC)

That solution does look seriously scary even if it must be better than counting if N is large. The closest I've come to looking at anything like that is the Bellows theorem for flexible polyhedra. Thinking about it, it looks an interesting subject to spend some time on. Dmcq (talk) 17:28, 7 April 2010 (UTC)

Well, yes, I just meant that the important and non-trivial information is that it is a polynomial of degree k. For a general polytope, of course, I agree that the most efficient way of computing the polynomial may be interpolating k+1 points (anyway, a finite task). Also, I don't know the details for a general polytope, but in the special case of the OP, it is easy to write down a generating series for the numbers c(n). It is a rational function with poles at certain roots of unity (below I also use v₀:=1 together with the other positive integers v₁,..,v_k); precisely

$\sum _{n=0}^{\infty }c(n)x^{n}={\frac {1}{\prod _{i=0}^{k}(1-x^{v_{i}})}}.$

(To check it, just expand the RHS into a product of geometric series). To get a closed formula for the c(n) one should write the partial fraction decomposition of this rational function; then expand the simple fractions into power series. It is at least clear that this way one gets a polynomial formula for c(n) (in form of some linear combination of binomial coefficients -with a linear combination depending on n periodically ). For a given, small list V this can even be computed explicitly by hand. I've never seen a general formula (i.e. for a general V). Note that the pole with higher order is 1, corresponding to $C(1-x)^{-(k+1)}$ in the decomposition in simple fractions, and it is responsible of the growth of the c(n) (this is the Schur's computation I mentioned above). Writing the next terms with higher order one gets more refined asymptotics on the c(n). --pma 19:32, 7 April 2010 (UTC)

It finally struck me, this is a variant of the Coin problem, in how many ways can you give change with coins of various denominations. Dmcq (talk) 21:18, 7 April 2010 (UTC)

Exact, that is exactly the subject of Schur's theorem. Note that the OP's variant, consisting in "≤N", can be written in the usual form "=N" adding a v₀:=1. This produces a further factor 1/(1-x) in the GF, which makes sense, if we recall what is the effect of multiplying a power series by 1/(1-x). pma --84.221.209.68 (talk) 08:04, 8 April 2010 (UTC)

Let us also not forget that the Ehrhart polynomial only works for polytopes whose all vertices are integer lattice points. In this particular case, this condition amounts to n being an integer multiple of every v_i, which is a fairly nontrivial restriction.—Emil J. 10:46, 8 April 2010 (UTC)

Right!! So that's not exactly a generalization of the Schur's situation. Actually, outside of the case you are saying, the c(n) are not necessarily a polynomial sequence: e.g. , we can change n euros with 1 and 2 euros coins in $\scriptstyle \lfloor n/2\rfloor +1$ ways: 1, 1, 2, 2, 3, 3,... : I liked so much the Ehrhart polynomial, that I forgot about it :-) .--pma 13:12, 8 April 2010 (UTC)

Elementarily $\{u\in \mathbb {N} ^{k}\,:\,\sum _{i=1}^{k}u_{i}v_{i}\leq n\}=\{u\in \mathbb {N} ^{k}\,:\,\sum _{i=1}^{k}{\frac {u_{i}}{a_{i}}}\leq 1\}.$ where the k variables $\ a_{i}={\frac {n}{v_{i}}}$ are edges of a k-dimensional pyramide having volume $\prod _{i=1}^{k}{\frac {a_{i}}{i}}={\frac {\prod a_{i}}{k!}}$, which is PMA's approximation. Bo Jacoby (talk) 14:33, 8 April 2010 (UTC).

As many lattice points lie on the boundary of the pyramide, a slightly larger pyramide include the same lattice points as inner points. So I suppose that the expression ${\frac {\prod (a_{i}+{\frac {1}{2}})}{k!}}$ gives a better approximation to the requested number of lattice points. Bo Jacoby (talk) 08:21, 9 April 2010 (UTC).

Video Lectures on Topology and/or Differential Geometry

Does anyone know of any (good) video lectures on any of these two topics? Thanks. - DSachan (talk) 01:57, 7 April 2010 (UTC)

Open Courseware has some. I don't know if they are good. Youtube search also finds a bunch. You might also like The Catsters' youtube channel about category theory (I've heard positive reports about these videos). 66.127.52.47 (talk) 02:16, 7 April 2010 (UTC)

Thanks but OCW doesn't have any video lectures about these topics. What lecture did you see in youtube? I don't see any result. I want a good comprehensive kind of lecture series, not the bunch of 8 min. lectures given by highschoolers. The lectures by the lady seem nice, but as you said they are more on Category theory. - DSachan (talk) 02:28, 7 April 2010 (UTC)

Oh I see, there are several OCW courses on those topics but they only have written materials, not video. Sorry. On youtube I just typed "topology" and "differential geometry" into the search field and saw there were a bunch of hits. I didn't examine any of them. 66.127.52.47 (talk) 03:26, 7 April 2010 (UTC)

Additional: Terry Tao mentions this youtube (I haven't watched it) about Moebius transformations. Also, while they're not academic lectures, the topology videos from The Geometry Center are breathtaking. 66.127.52.47 (talk) 20:18, 7 April 2010 (UTC)

Lots of assorted math presentations at the MSRI website and their Internet Archive site 71.178.44.93 (talk) 20:43, 9 April 2010 (UTC)

Wow, those look great! I'm surprised I didn't know about them before. They seem to be very advanced though (research level), not that good for someone not already acquainted with a topic. 66.127.52.47 (talk) 06:57, 10 April 2010 (UTC)

Wow, that's unbelievably sexy. You made my day. Thanks. - DSachan (talk) 12:39, 10 April 2010 (UTC)

April 8

Biased dice

I have a probability question.
Firstly, imagine a biased dice is rolled 10 times and the results are as follows: 2, 4, 2, 1, 4, 4, 3, 4, 3, 3

One could estimate the probability of each particular value being rolled simply by dividing the number of occurences of that value by the total number of times that the dice was rolled. I believe this is known as a binomial distribution. However, my understanding is that a binomial distribution requires that the probability of a trial is independent of the trial's position.

I would like to know if there is a way of estimating the probability of each value being rolled in a particular trial if the probability is dependent on the position of the trial. Also, I would like to know if there is a way of estimating the probability of each value being rolled if the probability is dependent on the value rolled in the previous trial.--Alphador (talk) 09:21, 8 April 2010 (UTC)

The likelihood function of the six unknown probabilities is a Dirichlet distribution. Bo Jacoby (talk) 10:06, 8 April 2010 (UTC).

Huh??? A likelihood function is not a distribution at all. And the Dirichlet distribution has nothing to do with this likelihood function. The crude estimator given by the original poster is indeed the maximum likelihood estimate.

The original poster is confused as to the meaning of the term binomial distribution. The binomial distribution is the probability distribution of the number of successes in a fixed number of independent Bernoulli trials. Michael Hardy (talk) 19:54, 8 April 2010 (UTC)

The likelihood function is transformed into a Bayesian probability distribution by normalizing with trivially constant prior probabilities. Bo Jacoby (talk) 17:37, 9 April 2010 (UTC).

For the second question, the same technique works even without assumption of independence, but you may need more trials to get an accurate answer.

If you want more information about the system, you can build a 6x6 table, describing the number of times the roll was i when the previous roll was j. Then you can estimate the probability of every result given the previous result. If you then want to find the proportion of each result in a long run, this is basically the stationary distribution of a Markov chain.

The first question is a bit vague, and can range from virtually impossible to trivial. Can you clarify which kind of dependence you allow? -- Meni Rosenfeld (talk) 10:21, 8 April 2010 (UTC)

For example, suppose the value of the dice could be n, n+1, or n-1, where n is the position of the trial, and each of those three possibilities have an equal probability. However, you do not know this. Instead, you are only given a sample sequence, e.g. 2, 2, 3, 3, 6, 5, 7, 9. From that, determine the probability distribution, or an approximation thereof, for the nth term.--Alphador (talk) 11:17, 8 April 2010 (UTC)

Is it possible to conduct multiple experiments, so in the first experiment you get 2,2,3,3,6,5,7,9 and in the second experiment you get 1,3,2,5,4,7,6,8? If so, you can do many experiments and record the results for each position. Otherwise it becomes more difficult, and you'll have to enforce strong assumptions about the simplicity of the rule. For example, if you assume that each roll is evenly distributed between $an+b$ and $an+c$, you can estimate a, b and c. It can be more relaxed - you can assume any rule is possible, but more complicated rules are less likely - but it probably won't give good results if you don't have any additional information. -- Meni Rosenfeld (talk) 12:08, 8 April 2010 (UTC)

Maybe you are asking how to estimate the parameters of a Hidden Markov model (HMM). If you can describe the actual application you are thinking of, that might help get better answers. 66.127.52.47 (talk) 17:54, 8 April 2010 (UTC)

Vacuous truth

I attended a lecture on geometric probability by Gian-Carlo Rota at the Joint Mathematical Meetings in 1998. He read verbatim from prepared notes, later published here. On page 15, we read this:

I will engage for a minute in the kind of mathematical reasoning that physicists find unbearably pedantic just to show physicists that such reasoning does pay off. Let us ask ourselves the question: what is the value of the symmetric function of order zero of a set of n variables x₁, x₂, ... , x_n, say e₀(x₁, x₂, ... , x_n)? I will give you the answer and will leave it to you to justify this answer after the lecture is over. The answer is, e₀ = 1 if n > 0 (i.e., if the set of variables x₁, x₂, ... , x_n is nonempty), and e₀ = 0 if the set of variables is empty.

And then he shows how this leads us into the theory of the Euler characteristic!

I know one other sexy example, from applied statistics: the noncentral chi-square distribution with zero degrees of freedom is non-degenerate (IIRC it concentrates some probability at 0 and otherwise is continuous)—I should dig out the details; I haven't look at this is a while.

Are there other good examples of really substantial far-reaching consequences of such vacuities? Michael Hardy (talk) 20:09, 8 April 2010 (UTC)

The Dirac delta function was widely used in science and engineering despite not being a legitimate mathematical function at all. Trying to mathematically justify its use led to the development of distribution theory. Does that help? 66.127.52.47 (talk) 00:50, 13 April 2010 (UTC)

I don't see that that's an example of this sort of thing. Michael Hardy (talk) 03:35, 13 April 2010 (UTC)

April 9

Question about vacuous truth

Yes the question above me reminded me of a problem I've been having. If we say "For all x in A, there exist y in A such that x=y", would that be considered true? A here is empty. I know that if we have a "for all x in A, P(x)", where P(x) has no quantifiers, that would be considered true. But in here it has a "there exist" over an empty domain, although of course it comes after the "for all". Money is tight (talk) 04:07, 9 April 2010 (UTC)

Yes, the statement is true. -- Meni Rosenfeld (talk) 06:55, 9 April 2010 (UTC)

Let x be an element of A. OMG, there is no element of A, so this is a contradiction! Which means we can say absolutely anything we want to say about x. --COVIZAPIBETEFOKY (talk) 12:13, 9 April 2010 (UTC)

Informally, this property of the empty set becoms more intuitive if you phrase it as "there is no x in A for which P(x) is false". As there is no x in A - full stop - then this is always true. Gandalf61 (talk) 13:29, 9 April 2010 (UTC)

integer solutions

This question is about a statement in a book I am reading. I apologize if it is too trivial. There is an equation $c_{1}(x_{1}-x_{2})+sjdc_{1}=0$. Here j,d are natural numbers and the rest of the terms are integers, (c1>0). The book says that by letting |x2-x1|=jd|s| we have a solution. We need |s| for some other purpose and cannot simply make the solution in terms of s. My question is how is this the solution? Thanks-Shahab (talk) 05:03, 9 April 2010 (UTC)

Was there any additional context for this equation and the importance of its solutions? Since $c_{1}\neq 0$ this is equivalent to $x_{2}-x_{1}=sjd$. There's no need for absolute values to solve the equation. However, any solution also satisfies $|x2-x1|=jd|s|$, which perhaps is simply what the book needed for the rest of the argument. -- Meni Rosenfeld (talk) 07:19, 9 April 2010 (UTC)

x1 and x2 are supposed to belong to the set $\{a,a+d|s|,\cdots a+nd|s|\}$ and j is among {1,...n}. Sorry for missing that part before. The proof that I am reading is of Rado's theorem (Ramsey theory)-Shahab (talk) 09:17, 9 April 2010 (UTC)

Then it looks like |s| is more important for the argument than s itself, and maybe also that the roles of x1 and x2 are symmetrical. Then indeed it may be that $|x2-x1|=jd|s|$ is a more useful characterization of the solution. -- Meni Rosenfeld (talk) 09:49, 9 April 2010 (UTC)

LaTeX enumerate

\begin{enumerate} \item \end{enumerate}

I know I can label the parts whatever I want by doing \item [(b)] or whatever. But, what if I have several things that I want to be numbered in order and I have something to say in the middle. For example, the definition of a group sometimes is stated A group is a set and a binary operation satisfying the following properties: ... then the 3 properties are listed. Then, "G is said to be abelian if it also satisfies: and property 4 is listed. I have to end up doing two \begin{enumerate}'s to do this and on the second I have to manually input [4.]. That's not a big deal but what if it's more complicated like exercises in a textbook with many different sets of instructions to the different problems but all still need to be numbered in order. I don't want to have to manually type in [7.] through [56.] and what if I add a problem later, I'd have to renumber everything. What can I do? Thanks. StatisticsMan (talk) 20:11, 9 April 2010 (UTC)

I believe that you can use \setcounter{enumi}{4} at the beginning of a new enumerate. (Igny (talk) 22:00, 9 April 2010 (UTC))

Also you can look at \usepackage{mdwlist} to write

\begin{enumerate}
\item 1
\item 2
\suspend{enumerate}
text between enumerates
\resume{enumerate}
\item 3
\item 4
\end{enumerate}

(Igny (talk) 22:09, 9 April 2010 (UTC))

Cool, thanks a lot. I will try both of those out. They both seem like a great improvement over my current method. StatisticsMan (talk) 16:25, 10 April 2010 (UTC)

Eigenvector definition

Outside of the standard "a vector operation on a vector" type definition for an Eigenvector, consider an example in which a matrix is a collection of customer ratings for features of cars that they rented. A set of ratings is used to produce an Eigenvector. What does the Eigenvector represent? Is it proper to say that the Eigenvectors of the ratings simply normalize them so they can be compared? -- kainaw™ 21:09, 9 April 2010 (UTC)

I'm not sure eigenvectors would have much significance in that kind of situation. Say we had a square matrix A where the rows are various cars and the columns are various features. If we have a column vector v, we can think of it as a weighting of the various feature scores and then Av is a vector of composite scores for each type of car using that weighting. The property of v being an eigenvector is a relationship between v and Av, but these two are indexed over different sets: v over different features and Av over different cars, so the fact that they're proportional doesn't say anything useful. Eigenvectors are meaningful when the matrix represents an map from a vector space to itself. Rckrone (talk) 02:42, 10 April 2010 (UTC)

Indeed. An eigenvector is a vector which, when transformed, is proportional to itself. A matrix of ratings like that the OP describes isn't a transformation, so the concept doesn't make sense. --Tango (talk) 15:59, 10 April 2010 (UTC)

I would assume it was meaningless as the example you have given does not constrain the matrix to be square, where as having eigenvectors does. —Preceding unsigned comment added by 129.67.119.26 (talk) 15:45, 10 April 2010 (UTC)

Usually in these situations you compute eigenvectors not of the matrix itself, but of its Gramian. If the $(i,j)$ entry of X is the preference of customer i to product j (after centering and possibly scaling), then the eigenvectors of $X^{T}X$ are components from which you can compose customers, and the eigenvectors of $XX^{T}$ are components from which you can compose products. This is basically just PCA. -- Meni Rosenfeld (talk) 16:58, 10 April 2010 (UTC)

Symbol origins

What are the historical origins of the symbols for addition (+), subtraction (-), multiplication (x) and equal to (=)? I can understand the division one.--79.76.236.198 (talk) 22:25, 9 April 2010 (UTC)

Even though it's on a web site about English usage, this page has exactly the answers you want and appears to be well researched. By the way, if you want to read more about the history of mathematical symbols, Florian Cajori's two-volume book on the subject, published in the 1920s, is a good place to look. --Anonymous, 23:54 UTC, April 9, 2010.

April 11

Expressing an integer as a sum of powers of a prime

Hello all. This question is part of a proof that I am reading from a book. Essentially we have an integer i which has a prime p in its prime factorization. Now the book says let $\mu (i)$ be the largest integer such that ${\frac {i}{p^{\mu (i)}}}$ is an integer. I take it to mean that that $\mu (i)$ is the highest power of p in the prime factorization of i. Now the book states that $i=\sum _{j\geq \mu (i)}a_{j}p^{j}$ where $a_{j}$ 's are positive integers. I take it that it is implicitly assumed that this sum is finite, but I don't understand how to prove this statement formally. The next statement is that $1\leq a_{\mu (i)}<p$. I don't quite follow this one either. Can anyone please help me out? Thanks.-Shahab (talk) 06:34, 11 April 2010 (UTC)

The infinite sum might make sense if the book is describing p-adic numbers - the emphasis on a prime base suggests this is plausible. Alternatively, if $j\geq \mu (i)$ is a misprint for $j\leq \mu (i)$ then the book is just saying that i has an expansion in base p with $\mu (i)$ digits $a_{j}$, in which the most significant digit $a_{\mu (i)}$ is non-zero. Gandalf61 (talk) 08:31, 11 April 2010 (UTC)

The $a_{j}$'s are positive or nonnegative? If the latter then it makes perfect sense. All the large powers will have zero coefficient, and this is just the expansion in base p. The $j\geq \mu (i)$ just means that all the small powers are unnecessary because i is divisible by $p^{\mu (i)}$. All of them are $0\leq a_{j}<p$ and also, $a_{\mu (i)}$ can't be zero because then i would be divisible by $p^{\mu (i)+1}$. -- Meni Rosenfeld (talk) 10:45, 11 April 2010 (UTC)

Meni, what do you mean by saying that "The $j\geq \mu (i)$ just means that all the small powers are unnecessary because i is divisible by $p^{\mu (i)}$". Don't the small powers contribute to the sum? Isn't it better to say $j\leq \mu (i)$. Thanks-Shahab (talk) 11:04, 11 April 2010 (UTC)

There aren't any small powers. If a number is divisible by $10^{3}$ then its last 3 digits in base 10 are 0. For example, $98000=9\cdot 10^{4}+8\cdot 10^{3}$, no powers of 10 smaller than $10^{3}$ involved. Exactly this also happens for expansion to base p. Saying $j\leq \mu (i)$ would be incorrect. -- Meni Rosenfeld (talk) 11:15, 11 April 2010 (UTC)

Ok, thanks. But shouldn't the base be $p^{\mu (i)}$? And what does Gandalf61 mean when he says that it might be a misprint? Because the book does have a lot of those!-Shahab (talk) 11:20, 11 April 2010 (UTC)

Why should the base be $p^{\mu (i)}$? Of course we can expand the number in any base we want. For whatever reason, the book needs to use the base-p expansion. One of the things it uses to learn about the properties of this expansion (in particular, the power in which it starts) is our knowledge of $\mu (i)$.

I think Gandalf61 hasn't thought this through before he said it might be a misprint. -- Meni Rosenfeld (talk) 11:46, 11 April 2010 (UTC)

Indeed. Meni is correct - if all $a_{j}$ beyond some point are 0 then the sum is just describing the base p expansion of i in which least significant $\mu (i)$ digits are zero, and first non-zero digit is $a_{\mu (i)}$. Gandalf61 (talk) 10:45, 12 April 2010 (UTC)

how do you distribute poison?

Question:

A garage has 2 vans and 3 cars which can be hired out for day at a time. Requests for the hire of a van follow a Poison distribution with a mean of 1.5 requests per day and requests for the hire of a car fiollow an independent Poison distribution with a mean of 4 requests per day.

Find the probability that on a particular day there is at least one request for a van and at least two requests for a car, given that there are a total of four requests on that day.

I tried:

There are 2 possibilities: 2 vans and 2 cars or 1 van and 3 cars. So I found poisonpdf(1.5,2), poisonpdf(4,2), poisonpdf(1.5,1) and poisonpdf(4,3) but how do I combine them?

Similar question: In the morning, the number of people entering a bank per minute has a Poison distribution with mean 3, and independently, the number of people leaving the bank per minute has a Poison distribution with mean 2.

Find the probability that during a particular minute in the morning, exactly one person will enter the bank given that the total number of people entering and leaving the bank is exactly 4.

Does that mean 3 people left the bank? poisonpdf(3,1) and poisonpdf(2,3)? —Preceding unsigned comment added by 59.189.218.247 (talk) 10:38, 11 April 2010 (UTC)

Distributing poison is illegal. I can help you distribute Poisson though :)

Let's start with the first question. Are there any additional sections to this question? Because the number of vehicles in the garage is irrelevant to the question asked - perhaps it needs to be interpreted in some special way.

I'll answer the question as written. This is a conditional probability problem - you need to divide the probability that $v\geq 1,\ c\geq 2,\ v+c=4$ by the probability that $v+c=4$.

To calculate the first probability, note that the probability of the union of two mutually exclusive events is the sum of probabilities, and the probability of the intersection of two independent events is the product of their probabilities. -- Meni Rosenfeld (talk) 11:04, 11 April 2010 (UTC)

LOL OMG MY SPELLING FAIL! For the first question, that is the third part. The first two parts are:

i) Find the probability that not all requests for the hire of a van can be met on any particular day. [SOLVED: 1 - poissoncdf(1.5,2)=0.191]

ii) Find the least number of vans that the garage should have so that, on any particular day, the probability that a request for the hire of a van for that day has to be refused is less than 0.1. [SOLVED using graphic calculator, table of poissoncdf(1.5,X)>0.9, answer is 3.]

I think that the probability that v + c = 4 would be poissonpdf(5,4), correct? Which events are mutually exclusive? I know the requests for vans and requests for cars are independent. THANKS! —Preceding unsigned comment added by 59.189.218.247 (talk) 11:32, 11 April 2010 (UTC)

The events that are mutually exclusive (and together exhaust the requirement $v\geq 1,\ c\geq 2,\ v+c=4$) are: The first is v=2, c=2. The second is v=1, c=3. So you need to sum the probabilities of the first and the second. To find the first, for example, you need to multiply $P(v=2)P(c=2)$. -- Meni Rosenfeld (talk) 11:51, 11 April 2010 (UTC)

Solved! THANKS! May come back with more questions soon, hehe.

April 12

Simple way to weight average ratings by total ratings?

I've got a website where users can rate items from 1 to 5. The site won't get a lot of traffic, and I expect the "total ratings" range for the items to be as high as 100 and as low as 4 or 5. Each item can be rated on an integer scale of 1 to 5. I'd like to be able to show a top 10 list of best-rated items, but because of the wide variance in total ratings I feel this needs to be weighted in some fashion to favor items with more votes. Unfortunately, I've never taken a stats class and really don't know what to do. Could some of you kind folks please suggest some relatively simple approaches to solving this problem? Thank you! (Calculations will be done in PHP/MySQL if that's relevant, though I doubt it'd matter) 59.46.38.107 (talk) 00:52, 12 April 2010 (UTC)

You should just use a simple cut-off. For example, list the top rated products of those with 10 or more ratings. StuRat (talk) 00:54, 12 April 2010 (UTC)

I agree, a simple cut-off is best. The number of ratings doesn't say anything about the quality of the product, just the confidence you can have in the average weighting, so you don't want to take the number of ratings into account for a top 10 list. The products with only a handful of ratings are just as likely (given no additional information) to be underrated as overrated. If you want to do something more complicated, you could do a variable cut-off and choose the cut-off for each product based on the variance of the ratings for that product. If you have had 5 ratings and they have all rated a product 3 then you can probably have more confidence in that rating than the rating of a product with 20 ratings of 1 and 20 ratings of 5. I'm not sure what function of variance would be best. To be honest, I think a simple cut-off is better, a variable one would be being complicated for the sake of it rather than to get significantly better results. --Tango (talk) 01:17, 12 April 2010 (UTC)

To me, a small number of identical ratings would indicate the likely use of sock-puppets, by one individual. StuRat (talk) 02:10, 12 April 2010 (UTC)

A product with only a handful of ratings, given what those ratings are, is not as likely to be underrated as overrated. A product with a few high ratings is more likely to be overrated than underrated. This is basically regression towards the mean. In other words, if your prior is that the product is average, the more evidence you have that the product is good, the higher your assessment of its expected quality. Thus a product with many good ratings should be placed higher than a product with a few excellent ratings (depending on the exact numbers, of course).

Personally I would use a parametric model with Bayesian updating for each product, but I guess that falls outside what the OP can easily work with. -- Meni Rosenfeld (talk) 08:34, 12 April 2010 (UTC)

Thank you so far. If I go with a simple cutoff should that limiter be a function of the total userbase so that as the userbase grows more votes are required? Or should it be a fixed value? 59.46.38.107 (talk) 01:27, 12 April 2010 (UTC)

I think you might want to increase the cutoff number once you have more reviews. You don't necessarily need to code in a formula for this, though, just have a fixed number which you edit and change from time to time. Using a formula is more likely to introduce problems. If you're determined to use a formula, though, you could only compare those products with more than the median number of reviews. StuRat (talk) 02:08, 12 April 2010 (UTC)

As a rule of thumb, if you have about 30 values (from a normal distribution), average and standard deviation become somewhat reliable measures. So you can probably stop increasing the cut-off once it reaches 30. This will also give great new products a chance to show up eventually. --Stephan Schulz (talk) 08:40, 12 April 2010 (UTC)

You might want to look into the IMDB system since they have some of the same issues. As I recall, they use the straight average when the number of votes is high enough and otherwise they use a weighted average. The exact formula is on their website somewhere.--RDBury (talk) 10:12, 12 April 2010 (UTC)

Equations on cartesian planes with interesting properties

Hi. The following functions can be graphed using a graphing calculator (or is there some kind of online/HTML program that can do that?), but have interesting properties, such as having Y values with no X values, looking like a seismograph, etc. My question is what are some equations with similar properties, and what causes them? This is not homework.

$y=x\sin(0.5\pi )^{2}-2x^{-1}/9{\sqrt {x-\tan(180/\pi )}}$

$y=\sin(\cos(\tan(89x+3)))+\sin(x+\pi )^{-1}$

$y=\tan(60x)(\arctan 30x)$

$y=\tan(\tan(x^{2}))$

$y=\sin(\cos(x^{2})x^{-1})-\tan(x^{2})/\arcsin(x)$

$y={\sqrt {(x-\sin(x^{2})+x^{-1}-3x)^{3}}}$

Thanks. ~AH1^(TCU) 01:14, 12 April 2010 (UTC)

Y-values without x-values is very common. (The graph of the constant function y=0 has no x-values when y is not zero.) X-values without Y-values means that the function is not defined for some x-values. For example is y=1/x not defined when x=0. This function has a Pole for x=0. The functions in your examples have poles in the x-values where y is not defined. Bo Jacoby (talk) 13:08, 12 April 2010 (UTC).

For plotting online, try FooPlot and Wolfram Alpha. (I prefer the former for most straightforward plotting, although Alpha will do, e.g., implicit functions.) 94.168.184.16 (talk) 23:28, 12 April 2010 (UTC)

Set theory

if S = {i | 0<i<=50}, |S| = 10, A and B are subsets of S, |A|=|B|=5, sum of all integers in A is equal to that of all integers in B, then prove that there exist atleast one such pair A and B.Rajeev 26987 (talk) 07:22, 12 April 2010 (UTC)

If $S=\{i|0<i\leq 50\}$ then $|S|\neq 10$. Did you mean $S\subseteq \{i|0<i\leq 50\}$? -- Meni Rosenfeld (talk) 08:09, 12 April 2010 (UTC)

I guess A and B have to be disjoint? If not, let A be any subset of S, with B=A. Staecker (talk) 10:56, 12 April 2010 (UTC)

Think of the elements of A and B as pairs. For each pair A_j and B_j for j=1 to 10, if j is even, ensure that A_j=B_j+1. If j is odd, ensure that A_j=B_j-1. So, there are 5 cases where B_j is one more than A_j and 5 where it one less. They cancel each other out, making A and B sum up to the same value. -- kainaw™ 12:29, 12 April 2010 (UTC)

Subconical shapes?

What exactly is a subconical shape? I've read many descriptions of individual Native American mounds that speak of the mound as being subconical; however, as you can see from the picture of the subconical Dunns Pond Mound, this apparently can mean "vaguely round and slightly raised in the middle", and definitely not very much like a Cone (geometry). Nyttend (talk) 11:33, 12 April 2010 (UTC)

Let me rephrase the question — is there some official definition of "subconical" or "subcone"? Thanks. Nyttend (talk) 11:36, 12 April 2010 (UTC)

Well, Wiktionary has a definition, but it just says "somewhat cone shaped", so not very helpful; Webster's says "slightly conical", which is just as useless. Googling random examples, "subconical" seems to often used to describe the shape of a flattened or truncated cone - possibly what we maths types would call a conical frustum. Gandalf61 (talk) 13:27, 12 April 2010 (UTC)

I was thinking of the shape a powder takes when dumped at a central point onto a plane, like the sand in the bottom of an hour-glass. This yields a slightly bowed-out cone. Salt storage facilities shape their containers to match this form. Here's an example: [1]. StuRat (talk) 01:55, 13 April 2010 (UTC)

Lowering Summed Indices for Tensors

Let's say we have a tensor $A_{\alpha \beta }$. The tensor is defined as $A_{\alpha \beta }=g^{\lambda \gamma }S_{\alpha \gamma }T_{\lambda \beta }$ where $S_{\alpha \gamma }$ and $T_{\lambda \beta }$ are tensors with $g^{\lambda \gamma }$ the inverse metric, and I want to lower indices of the inverse metric. How do I achieve that, since the inverse metric is summed over.The Successor of Physics 13:29, 12 April 2010 (UTC)

You can move those indices down and move the other indices they are contracted with up. Count Iblis (talk) 17:01, 12 April 2010 (UTC)

$A_{\alpha \beta }=g^{\lambda \gamma }S_{\alpha \gamma }T_{\lambda \beta }=g_{\lambda \gamma }S_{\alpha }^{\gamma }T_{\beta }^{\lambda }=S_{\alpha \lambda }T_{\beta }^{\lambda }=S_{\alpha }^{\lambda }T_{\lambda \beta }$. Bo Jacoby (talk) 20:56, 12 April 2010 (UTC).

Probability

I have b buckets that each can hold n items, and I have i kinds of items. I fill every bucket with unique (different kinds) of items. Items need not be unique across buckets. What's the probability that there are at least l of each kind of item in the buckets? I want to get an idea of the plausibility of untrusted distributed backups but it appears I fail at mathematics. --194.197.235.240 (talk) 19:17, 12 April 2010 (UTC)

You mean you're going to scatter these items around the buckets randomly? If yes, how do you know a given kind of item doesn't occur in a bucket more than once? Really aren't you better off populating the buckets deterministically so you can control where and how often each item is backed up? Maybe you also want to read about erasure codes rather than relying on pure replication. 66.127.52.47 (talk) 23:42, 12 April 2010 (UTC)

Taylor-like Series

I was thinking about smooth functions $f:\mathbb {R} \to \mathbb {R}$, and I defined an operator ${\mathfrak {F}}$; given by

${\mathfrak {F}}(f):=\lim _{n\to \infty }\sum _{k=0}^{n}{\frac {d^{k}\!f}{dx^{k}}}{\frac {x^{k}}{k!}}.$

This operator has a nice property. It can be shown that

${\frac {d}{dx}}{\mathfrak {F}}(f)=2{\mathfrak {F}}\left({\frac {d}{dx}}f\right).$

Working out some explicit examples, one the sine, cosine and exponential functions gives

${\mathfrak {F}}(\sin x)=\sin(2x),\ {\mathfrak {F}}(\cos x)=\cos(2x),\ {\mathfrak {F}}(e^{x})=e^{2x}.$

I used Matlab to get these last identities. It doesn't seem obvious to me, for example:

${\mathfrak {F}}(\sin x)=\sin x+x\cos x-{\frac {x^{2}}{2}}\sin x-{\frac {x^{3}}{6}}\cos x+\cdots .$

I have a couple of questions:

1. How do we prove the expression for, say, the sine function?
2. What is the domain and range of this operator? (I think that if $f$ has a convergent power series in a n'hood of $x_{0}\in x$ then ${\mathfrak {F}}(f)$ converges in a n'hood of $x_{0}$).
3. Is this operator already known and studied? If so, does it have any uses?

•• Fly by Night (talk) 19:39, 12 April 2010 (UTC)

First question: if you plug in f(x)=e^(ax) to your formula, you see that ${\mathfrak {F}}(f)(x)=e^{2ax}$. Now note that your operator is linear in f, and sin x is a linear combination of exponentials. 129.67.37.143 (talk) 21:48, 12 April 2010 (UTC)

OK, so it's quite straightforward for the exponential case:

${\mathfrak {F}}(e^{\lambda x})=e^{\lambda x}+\lambda xe^{\lambda x}+{\frac {(\lambda x)^{2}}{2!}}e^{\lambda x}+{\frac {(\lambda x)^{3}}{3!}}e^{\lambda x}+\cdots =\left(\sum _{k=0}^{\infty }{\frac {(\lambda x)^{k}}{k!}}\right)e^{\lambda x}=e^{\lambda x}\cdot e^{\lambda x}=e^{2\lambda x}.$

but since I'm only considering real functions and you need complex function to make sine from exponentials: $2sin(x)=i(e^{-ix}-e^{ix})$, I don't see this can prove it. •• Fly by Night (talk) 22:11, 12 April 2010 (UTC)

For analytic functions you can sum the Taylor expansion and show that

${\mathfrak {F}}f(x)=f(2x)$

Count Iblis (talk) 23:18, 12 April 2010 (UTC)

just to give a bit more of a hint, show that ${\mathfrak {F}}f(x)=f(2x)$ when f is x^n, then extend to power series by linearity 129.67.37.143 (talk) 08:29, 13 April 2010 (UTC)

While you may be working with real functions, it is often possible (and in many cases easier) to work in the realm of complex functions, and as long as your functions are analytic in a region including the real axis, your proofs are valid for the real function as well. The identities between exponential and trigonometric functions is a good example. (Analogously, you might have a function acting on the integers, but to prove some property it's easier to extend the function into the reals.) Confusing Manifestation(Say hi!) 00:58, 13 April 2010 (UTC)

Adomian decomposition method

Hello Math Reference Desk. I have recently come across the Adomian decomposition method in a paper I am reading. I have never heard of this technique, and our article is very vague and stubby. Can anyone enlighten me on the purpose, utility, and general procedure for Adomian decomposition? I understand it is a "semi-analytic" method to solve PDEs, but I'm not clear whether that means it's "analytic" or "numerical." Probably the most important part of my question: is this method common/widespread, or is it an esoteric technique? Thanks, Nimur (talk) 23:22, 12 April 2010 (UTC)

The Wikipedia article doesn't say much, but that other site has some links about it. 66.127.52.47 (talk) 23:50, 12 April 2010 (UTC)

April 13

Three Dimension

If P (a,b,c) and Q (x,y,z) are two points in three dimensional space, how can we find equation of line PQ.

There are many methods. See system of linear equations. -- kainaw™ 03:41, 13 April 2010 (UTC)

I think the simplest method is the vector equation: r = OP + λPQ where r is the position vector of any point on the line, OP is the vector from the origin to P, and PQ is the vector from P to Q. The parameter λ indicates how far along the line we have travelled. λ = 0 at point P, λ = 1 at point Q etc. Other people will have different preferences, and the various methods are all equivalent. Dbfirs 08:04, 13 April 2010 (UTC)

If you just need a Cartesian equation, you can eliminate λ to give: (x - x₁)/a = (y - y₁)/b =(z - z₁)/c where the co-ordinates of point Q are (x₁, y₁, z₁) Dbfirs 08:13, 13 April 2010 (UTC)