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March 11

Suffix notation

Why are prefix and infix functions common, but suffix notation is very rarely used? The only example of suffix notation in pure math that I can think of is the factorial function. 76.67.75.94 (talk) 19:51, 11 March 2011 (UTC)

Derivative ($f'$)? Cycle notation?--203.97.79.114 (talk) 22:48, 11 March 2011 (UTC)

Well, (x)f is a little hard on the eyes... --COVIZAPIBETEFOKY (talk) 23:10, 11 March 2011 (UTC)

Though incidentally, I believe it is more common to see postfix (aka Reverse Polish Notation) than prefix on a calculator. --COVIZAPIBETEFOKY (talk) 01:14, 12 March 2011 (UTC)

The notation Xƒ is used in differential geometry to mean the derivative of the function ƒ with respect to the vector field X. Equivalent notations are d_Xƒ of (dƒ)(X), where the d is the exterior derivative. The reason being that you can think of a vector field as a differential operator. — Fly by Night (talk) 01:52, 12 March 2011 (UTC)

Would you count powers/exponentials as suffix notation? I have seen functions written using suffix notions (ie. f(x) written as xf) which makes composition of functions a little less confusing (applying f and then g is written xfg, which is the intuitive order, rather than g(f(x)) which seems backwards). However, I've only seen that in really old books. The notation we now consider conventional won out, but I don't know why. --Tango (talk) 02:20, 12 March 2011 (UTC)

I've seen postfix notation for group homomorphisms in recent papers (Eastern European authors, I think). It's also used in GAP: f(x) is x^f. Staecker (talk) 16:32, 12 March 2011 (UTC)

OP here. The derivative is a good example. The others, such as exponents, don't really count; 3² is the exponential with base 3 of two just as much as it is the square of three. Tango actually hit on the reason I asked the question: xfg is a more intuitive way to compose functions than g(f(x)), especially if one must deal with many functions being composed without being applied to a variable. I think prefix notation won just because it was invented first, but I was wondering if anyone the details of this. 76.67.74.178 (talk) 04:51, 14 March 2011 (UTC)

March 12

Random reverse Polish notation question

I was just wondering what something like (7x² + 5x + 1)/(3x − 6) would look like in reverse Polish notation. Although, now that I think about it, maybe it can't be done or just isn't done. It seems that this notation is mostly used for evaluating expressions, but I see no reason that it can't be used for more algebraic things. So, can it be done, and if so how? 10:42, 12 March 2011 (UTC) —Preceding unsigned comment added by 98.185.234.63 (talk)

7x2^*5x*+1+3x*6−/, or, if we decide − ought to be unary, 7x2^*5x*+1+3x*6−+/. Algebraist 11:02, 12 March 2011 (UTC)

Every expression can be represented in reverse polish notation without parentheses, assuming you've established how many arguments each operation requires. You can convert an expression of infix operators to postfix by converting the outermost operation first (ie, (some-expression) op (some-other-expression) to some-expression some-other-expression op), then converting the deeper expressions, eliminating the parentheses along the way.

BTW, the associative law written in postfix looks like "abc++ = ab+c+". --COVIZAPIBETEFOKY (talk) 15:39, 12 March 2011 (UTC)

Shouldn't it be "a b c + + a b + c + =" with the equality sign at the end? Bo Jacoby (talk) 12:27, 14 March 2011 (UTC).

Hm. Technically, yes. Of course I was indicating the two expressions which should be equal according to the associative law, written in postfix, and not thinking about applying postfix to the statement of equality itself. --COVIZAPIBETEFOKY (talk) 15:36, 14 March 2011 (UTC)

There are some other problems you need to deal with other than arities. For example, you need a way of distinguishing between 123+=1+23 and 123+=12+3. Algebraist 21:03, 12 March 2011 (UTC)

Yeah, so you also need spaces to separate the parameters. Is that all, or was there something else you were thinking of as well? --COVIZAPIBETEFOKY (talk) 04:22, 13 March 2011 (UTC)

That's all I was thinking of. I suppose another potential problem would be dealing with infinitary operations. I guess you'd either allow infinite strings in some way or replace the infinitary operation with an unary operation on sequences, or just decide not to use postfix notation at all for such things. Algebraist 04:29, 13 March 2011 (UTC)

See also: Forth (programming language)#Programmer's perspective. 93.95.251.162 (talk) 15:45, 15 March 2011 (UTC) Martin.

Flipping coins

Resolved

Me and my friend are flipping a weighted coin w a p probability of heads a number of times (p>0.5) and betting $1 even money on heads on each flip. We stop flipping the coin when we hit x dollars. What would be a fair over/under of number of games we end up playing? Also, what would be a fair over/under if my friend would lay me odds such that I only have to win the bet w a probability of q? I need this for a poker propbet. Thanks 93.141.6.170 (talk) 11:54, 12 March 2011 (UTC)

Do you have an infinite amount of money to bet ? If not, the possibility exists that you can run out, and this would enter into the calcs, depending on your stake. StuRat (talk) 21:30, 12 March 2011 (UTC)

In theory no, but in practice I'm rolled well enough to weather a bad luck run of 6-7 standard deviations below the mean so I think we can ignore that possibility. 93.136.39.209 (talk) 20:21, 13 March 2011 (UTC)

You might like to take a look at our spread betting article. Also, take a look at the Poisson distribution and Skellam distribution articles. — Fly by Night (talk) 00:01, 13 March 2011 (UTC)

Yes I think Skellam is what I was after although I'll have to read that article a few times to fully understand it; so far I've never needed anything beyond binomial and normal distribution so this is very new to me. Thanks to both of you for answering my question, I appreciate it. 93.136.39.209 (talk) 20:21, 13 March 2011 (UTC)

Your welcome. Note that another approach is to do a computer simulation of, say, a million trials, with different under/over values to determine which is the "fairest".

I'm going to mark this question as resolved. Please remove the resolved tag if that article doesn't answer all your questions. StuRat (talk) 22:38, 14 March 2011 (UTC)

Linear Algebra help

Hello, I am stuck in a Numerical Linear Algebra problem, which is to solve the system in the form $A(X^{t=n+1})\cdot X^{t=n+1}=B(X^{t=n})$, where $A$ is a tridiagonal matrix. $A$ is a function of $X$ calculated at time $t=n+1$. I am assuming initial guess for $X^{t=n+1}=X_{0}^{t=n}$, then I calculate $A$, then I solve the system to get the new values of $X^{t=n+1}$. I repeat the procedure for convergence, but my system never seems to converge. Can anybody suggest some other approach via let's say splitting of $A$ or something cleverer. Thanks. - 14.139.128.12 (talk) 20:40, 12 March 2011 (UTC)

Your notation appears to be a bit nonstandard -- if you have a sequence of successive values of X, it is more common to name those with subscripts $X_{0}$, $X_{1}$, ..., $X_{n}$ instead of $X^{t=n}$. Or am I misunderstanding you.

Are you sure that the matrix A really depends on X? It would be a more standard linear algebra situation if A was a fixed matrix such that its product with X depends on X. If A does depend on X, you probably need to be more specific about what this behavior is before we can help much. Similarly for B, if $B(X^{t=n})$ does not mean just a fixed matrix multiplied by $X_{n}$ from the right.

If you can compute the matrix A without necessarily knowing X, your plan would be to invert A and multiply your equation with the inverse from the left. –Henning Makholm (talk) 00:40, 13 March 2011 (UTC)

First of all thanks a lot for showing interest to help me. In the current problem, $X$ is a column vector to be obtained as a solution of Linear system but the Coefficient matrix $A$ is the function of column vector $X$ and thus it's sort of implicit problem. The $R.H.S.$ contains a column vector $B$ which is again depending on $X$ at last time step. As,I know the value of $X$ at last time step, I know the $B$. In my first post, I mean the same by $B(X^{t=n})$.

Coefficient matrix $A$ is depending on column vector $X$ (which itself is the solution to be obtained), through some numerical schemes (Finite difference).Can u please suggest any standard algorithms to solve such kind of implicit Linear System? —Preceding unsigned comment added by 14.139.128.12 (talk) 10:36, 13 March 2011 (UTC)

Unfortunately, if your A is so secret that "some numerical schemes" is all you can disclose about it, I doubt anyone will be able to help you. –Henning Makholm (talk) 13:56, 13 March 2011 (UTC)

March 13

Matrices Question

Okay, suppose you have an n x n matrix such that there is only one '1' in every row and column and the rest of the matrix is filled with zeros. Take this 8 x 8 matrix for example, that I'll call R:
[0 0 0 0 0 0 1 0]
[0 0 0 0 0 0 0 1]
[0 0 0 0 0 1 0 0]
[1 0 0 0 0 0 0 0]
[0 0 0 0 1 0 0 0]
[0 0 1 0 0 0 0 0]
[0 0 0 1 0 0 0 0]
[0 1 0 0 0 0 0 0]

If you multiply R by itself six times (R ^ 6), you get the 8 x 8 identity matrix. Any multiple of 6, you also get the identity matrix. In this smaller 4 x 4 matrix, which I will call V

[0 1 0 0]
[0 0 0 1]
[0 0 1 0]
[1 0 0 0]

V ^ 6 yields the 4 x 4 identity matrix. Any multiple of 3 >= 6 you also get the 4 x 4 identity matrix.

Is there any reason for this? Any rule that can be used to describe this? — Preceding unsigned comment added by Aacehm (talk • contribs) 01:02, 13 March 2011 (UTC)

Not sure if this is what you're looking for, but if Aⁿ = I, then A^kn = I for any positive integer k. This is because you can write A^kn = Aⁿ ⋅ Aⁿ ⋅... ⋅ Aⁿ (multiplied k times), which is I ⋅ I ⋅ ... ⋅ I = I. In your examples you have R⁶ = I, and so you'll get I for any exponent that's a multiple of 6. You also have V³ = I, and so you'll get I for any exponent that's a multiple of 3. Staecker (talk) 01:36, 13 March 2011 (UTC)

The matrices you're speaking about are called permutation matrices. When one of them is used to represent a linear transformation on column vectors, it acts by permuting the components of the argument vector in a particular pattern. Any such permutation has a finite number such that repeated this many times its total effect is to do nothing, which is represented by an identity matrix. This is a consequence of the fact that a permutation is a product of independent cycles (which allows you to calculate the exponent needed to get the identity, as the least common multiple of the cycle lengths). –Henning Makholm (talk) 02:14, 13 March 2011 (UTC)

Calculation of s in Lucas–Lehmer primality test

Hi there, I haven't been to Wiki for a long time. Nice to be back! I have been wondering about the s in Lucas–Lehmer primality test. In the article, it was proven that $s_{i}=\omega ^{2^{i}}+{\bar {\omega }}^{2^{i}}$ where $\omega =2+{\sqrt {3}}$ and ${\bar {\omega }}=2-{\sqrt {3}}$. However, the calculation $s_{i}$ involes $\omega$ and ${\bar {\omega }}$, both not integers. Is there a formula for each and every value of $s_{i}$ starting with $i=0$ defined purely in terms of integers? Thanks. voidnature 03:11, 13 March 2011 (UTC)

Um, the definition of the s_i's given first is purely in terms of integers. I suppose you don't think it is a "formula", but if we write it as $s_{i}=(\lambda x.x^{2}-2)^{i}(4)$ using Church's lambda notation, it looks perfectly like a formula to me. What exactly do you want that it doesn't have? –Henning Makholm (talk) 03:58, 13 March 2011 (UTC)

Thanks for such a quick reply, Henning, however I am quite unfamiliar with Church's Lambda Notation. I have read the Wikipedia article, but I still don't really get it. Can you please explain some of it please? By the way, by "formula", I meant like I didn't want a recursive relation but instead wanted something of a single equation such as $n_{i}=2^{n}$ so each and every value of the sequence can be computed immediately, without computing other values like $n_{i-1}$. Too, I am quite familiar with programming(specifically, mostly with C#.NET) and in the article about Church's Lambda Notation, it seems there is a deep relationship between programming languages and the notation, so don't mind explaining stuff in terms of programming stuff if required. Thanks!

voidnature 04:32, 13 March 2011 (UTC)

In practical programming terms, a lambda expression is just an anonymous function. The C# syntax for $\lambda x.x^{2}-2$ would be something like "x => x*x-2" or "delegate(BigInteger x) { return x*x-2;}", I think. I'm then taking the liberty of mixing in the conventional mathematical notation of using an exponent to indicate iterating the function i times.

Yes, there are deep connections to programming languages. The anonymous functions in modern programming languages trace their ancestry directly to Church's work in the 1930's on logic and number theory. In fact, the lambda calculus is held in higher esteem by computer scientists than by mathematicians in general. It is the favorite theory of programming language researchers, and everyone working with process-oriented languages, or parallel languages, or network-oriented or aspect-oriented or mixin-oriented languages or whatever wishes they could find a core calculus as general and well-fitting as the lambda calculus -- something that would make their favorite feature look like a fundamental primitive of computing rather than an ad-hoc graft-on.

If you define "formula" as something that avoids computing, specifically, the previous s_i's, may I suggest

$s_{i}=(\lambda x.x^{2}-4x-2)^{i}(6)-2$

? Less whimsically, my point is that there is no real principled distinction between a "formula for" something and an "algorithm for computing" it. Sometimes you may be familiar with notation that allows you to write one down one more succinctly than the other, but that's more a function of which notation you know, not an intrinsic property of the thing you're writing down. –Henning Makholm (talk) 05:14, 13 March 2011 (UTC)

No wonder the name "Lambda Expressions" for delegates! Thanks, but what would the argument "x" in $s_{i}=(\lambda x.x^{2}-4x-2)^{i}(6)-2$ be? Also, to be honest, I posed this question because I wanted to calculate the s' of a really large value without calculating the previous values. For such a huge value, it would probably be at least a few thousand times faster if the previous values were not required to be calculated. For the value, even if 100 iterations of i could always be done in 1 second, the final s would take slightly more than 1 whole year to compute. Thanks, Henning, you're great!

voidnature 05:26, 13 March 2011 (UTC)

Sorry, that should be $(\lambda x.x^{2}-4x-4)$, which my revised formula would apply to each $s_{i}$+2.

As for finding a fast algorithm, notice that the largest known Mersenne prime has an index of only several dozen million. If there were a known algorithm for testing a candidate as fast as you desire, surely much larger Mersenne primes would be known by now (or, alternatively, the unexpected absence of Mersene primes above i=43112609 would be all over the news). This strongly suggests that there isn't any known algorithm that will satisfy you. –Henning Makholm (talk) 14:09, 13 March 2011 (UTC)

If one simply put in a symbol like r instead of the square root of 3 and change r² to 3 wherever it occurs you'll never have to deal with non-integer values. The expressions will become quite big though. Far easier to just plug the non integer values into a calculator and if you see a .995 at the end just round it to the nearest integer because the result has to be an integer. Dmcq (talk) 10:57, 13 March 2011 (UTC)

It strikes me that using r like this might be quite an efficient method for computing very large values, see Exponentiation by squaring. I don't know what they actually do but I'd guess it would be something like this and doing a modulo operation at each step. Dmcq (talk) 11:11, 13 March 2011 (UTC)

Thanks for the answer, Dmcq. I have considered this method before I posted the question, actually. However, it requires at least ${\frac {i}{2}}$ addition operations, and considering the sheer size of "i" I am considering, and taking into account the requirement of calculating the binomial coefficients of ${\frac {i}{2}}$ values, many of them extremely large too due to the size of the considered "i" again, I have considered the method impractical. Thanks for your reply whatsoever.

voidnature 11:23, 13 March 2011 (UTC)

Yes I can see looking at that article that at best one would be replacing i big multiplies with 3i big multiplies so that doesn't save anything! Dmcq (talk) 12:17, 13 March 2011 (UTC)

Ah, and yes, I forgot to mention that if the type "double" could handle numbers of the size I am considering, I would have considered it too. The type "double" has to be used since nearly all methods in the System.Math class and virtually every other math library uses doubles.

voidnature 14:42, 13 March 2011 (UTC)

In a bit someone removed I pointed out it would require 3i big multiplies rather than i using exponentiation by squaring so yes it is no gain. The additions would be inconsequential in comparison in real life using exponentiation by squaring. I was answering your question as posed, which was not about producing an efficient algorithm but how to avoid the non-integers if using that formula. Dmcq (talk) 15:04, 13 March 2011 (UTC)

Your bit disappeared here, and I reinserted it later. Seems to have been an uncaught edit conflict; such glitches are not unheard-of with the MediaWiki software. –Henning Makholm (talk) 15:11, 13 March 2011 (UTC)

You can also write it as a series: $s_{i}=4s_{i-1}-s_{i-2}$ where $s_{0}=2$ and $s_{1}=4$ but that's probably not what you were looking for. 93.136.39.209 (talk) 20:17, 13 March 2011 (UTC)

No; I think you missed the 2-to-the in the OPs original formula. –Henning Makholm (talk) 20:33, 13 March 2011 (UTC)

Ah, Henning, as to your post above stating the impossibility of an algorithm as quick as I desire, I reject. This is because all I am asking is an algorithm to efficiently determine the value of s for a particular i, faster for large values but not to outrun the recursive algorithm for incrementing. A reasonable non recursive algorithm would be able to complete the calculation for huge values far faster than applying the recursive algorithm from the bottom up. For example, I want to start at the i value of 3321928097. If I use the recursive method, if 100 recursions could always be completed in 1 second, it would take 385 complete days to complete the calculation. What I am seeking is a faster method to initialize the s value. For incrementing, the square and minus 2 recursive method would be certainly quicker. I am just looking for a quick method to calculate any value, not attempting to be more efficient than the recursive algorithm while incrementing. Thankyou.

voidnature 10:26, 14 March 2011 (UTC)

As I said, if such a method were known, projects such as the Great Internet Mersenne Prime Search would already be using it to look for primes in the range of 2^3321928097-1, instead of mucking around in the forty-or-fifty-millions. They don't need the intermediate s_i's either (at least not modulo 2^p-1 for a different p than the one they're testing at the moment) -- if they are computing them it must be because they can't think of anything better. Note that computing the s_i without doing it modulo some M_p is a fool's errand; already s₁₀₀₀ is far too large to represent explicitly on any computer that would fit in the observable universe. –Henning Makholm (talk) 13:04, 14 March 2011 (UTC)

Thanks, Henning! Now I kind of get why you say my question as impossible. However there are two points I wish to state. Firstly, to represent M_3321928097, all you need are 3321928096 binary bits(The last bit must be false), about 3GB, and RAMs of such a size are perfectly practical. To leave space for the operating system, a 8GB RAM would be perfectly good for the purpose. Also, I really LOVE the .NET framework. In the System.Numerics namespace, there is the BigInteger class, allowing one to represent an integer of any size. Secondly, (I've discovered a mistake and take it back: It really is impossible to represent on a modern computer)I've thought I could maybe resort to the method of expanding the whole thing. In expanding, about half of the terms are cancelled. After that, I'll trying to put the terms back together in terms like (2 + 3)^a. I'll search for all possible methods to combine the terms such as adding and subtracting the same terms and greatly using the many identities. Hopefully, this method will turn out to be practical. Even if the calculation takes days, I'll save 100 times the original time, 385 complete days. Henning, can you please tell me about some of your thoughts on this method?

voidnature 14:28, 14 March 2011 (UTC)

Multinomial terms

Hi. Is there a "famous" example of the multinomial "choose" function? I want a numerical example that is known for some reason. Perhaps it appears in a proof, or perhaps some famous mathematician used it in a book, or maybe has some unusual property. I could use a random example such as ${10 \choose 1\ 2\ 3\ 4}=12600$ (which I just made up) but would much prefer a famous example. Anyone? Robinh (talk) 06:43, 13 March 2011 (UTC)

(OP) It doesn't have to just integers. ${\pi +e \choose \pi \ e}$ would be fine (if it was famous) Robinh (talk) 06:48, 13 March 2011 (UTC)

You mean something like how many ways can a bridge hand be dealt or something more complicated? Dmcq (talk) 15:10, 13 March 2011 (UTC)

That's what I thought at first, but 12600 does not match. It rather seems Robinh is referring to the multinomial theorem, where there is more than one number in the bottom half of the bracket. –Henning Makholm (talk) 15:14, 13 March 2011 (UTC)

How about ${\tbinom {52}{13,13,13,13}}$, though? –Henning Makholm (talk) 20:30, 13 March 2011 (UTC)

Thanks Henning and Dcmq: bridge is perfect. Everybody can interpret this. Sweet! Anyone got a non-integer famous example? Or an integer example which crops up in the combinatorics literature? Robinh (talk) 21:08, 13 March 2011 (UTC)

Form work of Plinth Beam or Shuttering of Plinth Beam

How to Calculate Form work of plinth beam in Running Feet —Preceding unsigned comment added by 182.178.71.47 (talk) 20:12, 13 March 2011 (UTC)

Note: this question was also asked on the science desk. I don't understand it well enough to figure out which desk is the proper one, but whomever can offer an answer, please add a pointer to the other copy.–Henning Makholm (talk) 20:47, 13 March 2011 (UTC)

I don't have an answer, but these links might help people understand the question better: Form work and shuttering: [1], plinth: [2]. StuRat (talk) 11:41, 14 March 2011 (UTC)

Quadratic Forms

I have a real quadratic form, say Q, in six variables. (These variables are actually coefficients in the Monge form expantion of a surface.) I wanted to think about the zeros of this quadratic form, so I calculated the Hessian matrix. It turns out that the Hessian is singular. Moreover, the Hessian has a kernel of dimension three. Let's write the quadratic form as Q(x) = xHx^T, where H is the Hessian matrix and x = (x₁, …, x₆). I'm interested in the solution set { x ∈ R⁶ : Q(x) = 0 }. So, I have a degenerate quadric that consists of a three dimensional linear space. The problem is that, or at least Maple claims, that Q(x) doesn't factor over R. Can anyone suggest a way of using the kernel to simplify and further understand Q(x)? The kernel is given by

$\ker(H)={\mathbf {R} }\langle (2,0,0,1,0,0),(0,1,0,-6,0,1),(0,0,-1,0,1,0)\rangle .$ — Fly by Night (talk) 20:45, 13 March 2011 (UTC)

The bit about not factoring over R sounds strange. A real quadratic form ought to diagonalize nicely, but perhaps that is not what it means? (One possibility is just that the diagonalization is not unique because there is a multiple eigenvalue).

What do you know about the three nonzero eigenvalues of H? –Henning Makholm (talk) 21:32, 13 March 2011 (UTC)

Factoring over R and having a Hessian matrix which is diagonalisable over R is different. For example the quadratic for Q(x,y) = x² + y² doesn't factorise over R and gives the identity as its Hessian matrix. I think it might be because of the fact I have an eigenspace of dimension more than one. I'll post more details shortly. — Fly by Night (talk) 22:42, 13 March 2011 (UTC)

Ah, so by factoring you actually mean to write the form as a product of two linear forms? As far as I can see, that is impossible unless the positive-definite and the negative-definite subspaces both have dimension at most 1. –Henning Makholm (talk) 21:02, 14 March 2011 (UTC)

There are three other eigenvectors, two with negative eigenvalues and one with a positive eigenvalue. They are given below. The subscript denotes the eigenvalue:

$E_{154}=\langle (1,-6,0,-2,0,-6)\rangle \,,$

$E_{-36}=\langle (0,0,1,0,1,0)\rangle \,,$

$E_{-144}=\langle (0,-1,0,0,0,1)\rangle \,.$

I've managed to partially factorise the quadratic form. I get

$Q(\sigma ,a_{0},\ldots ,a_{4})=(\sigma -2a_{2}-12a_{4})(\sigma -2a_{2}-12a_{0})-9(a_{1}+a_{3})^{2}\,.$ — Fly by Night (talk) 18:47, 14 March 2011 (UTC)

Okay, so what I had in mind (which may or may not be helpful to you) would be to take $({\tfrac {1}{77}}E_{154},{\tfrac {1}{6}}E_{-36},{\tfrac {1}{12}}E_{-144})$, together with three spanning vectors for the kernel, and express the x vectors in that basis. Then

$Q(z_{1}{\tfrac {1}{77}}E_{154}+z_{2}{\tfrac {1}{6}}E_{-36}+z_{3}{\tfrac {1}{12}}E_{-144}+$anything from the kernel$)=2(z_{1}^{2}-z_{2}^{2}-z_{3}^{2})$

(bit of a shame about that factor of 2 -- it may be caused because I'm working from your formula Q(x) = xHx^T -- but if H is actually the Hessian as defined in our article, made of raw second derivatives, then the right formula is Q(x) = ½xHx^T and everything will look nicer).

Then the zeroes of Q are the vectors for which $z_{1}^{2}=z_{2}^{2}+z_{3}^{2}$ which form a cone in $(z_{1},z_{2},z_{3})$-space, direct-summed with the kernel of H. –Henning Makholm (talk) 20:55, 14 March 2011 (UTC)

recent old SAT I exams

Dear Wikipedians:

I am wondering, are there compilations of recent authentic old SAT I exams (after the recent SAT reform) for sale on Amazon? Also, I'm wondering if I could also download recent authentic old SAT I exams on line. I am especially interested in the quantitative sections.

Any help would be much appreciated!

174.88.32.181 (talk) 23:38, 13 March 2011 (UTC)

It wouldn't surprise me if some are circulating illicitly but I'm fairly sure they are never officially released to the public. There are plenty of study guides available whose questions are similar to the real ones though. IMO the main thing for it is get plenty of sleep the night before, and have an energizing breakfast. The questions are pretty easy but tedious and it's easy to make dumb errors if you're not well rested. 75.57.242.120 (talk) 04:43, 14 March 2011 (UTC)

March 14

complex analysis

(stupid question removed)...I take it this is correct wikiquette? Robinh (talk) 09:24, 14 March 2011 (UTC)

I think striking out your question and adding a retraction is better. -- Meni Rosenfeld (talk) 10:08, 14 March 2011 (UTC)

OK, I'll do that next time I ask a stupid question. best, Robinh (talk) 18:39, 14 March 2011 (UTC)

Quick identification of outliers

I know that this can be done using mean average and standard deviation, but I would like to know if there is any form of quick identification of outliers that I can study. For example, assume you have a person's yearly weight as 178, 191, 184, 81, 180. It is clear that the 81 is an outlier (it was measured in kg, not lb). What is the computationally fastest method for identifying 81 as an outlier? -- kainaw™ 14:53, 14 March 2011 (UTC)

Quickly removing outliers without thinking about them using an algorithm is what lead to the ozone hole being missed. What you said above is probably easiest, just using the average and some number of standard deviations you choose. Here's some real systolic/diastolic readings for a few days, you can see that just chopping out figures that seem out can be quite wrong here: 155/89 94/58 83/53 130/84 Dmcq (talk) 16:16, 14 March 2011 (UTC)

I'm not looking to remove outliers. I'm looking to identify them for inspection. I have 10 years of weight/height information on over two million patients and I want to quickly mark which patients have data with outliers so the outliers can be verified as correct or as incorrect (and then corrected). I expect about 10% of the data to be outliers, so I don't want to go through all of it by hand. I am using standard deviation now, but it will take another hour or so to finish processing. That is why I'd like to study a different (faster) technique. -- kainaw™ 16:44, 14 March 2011 (UTC)

10% outliers! That's huge. The bit of statistics I do I normally just include all the data including outliers unless I'm very very certain about them. Are you counting data not supplied as outliers? I can't think of a better way of doing it than what you say. Dmcq (talk) 16:50, 14 March 2011 (UTC)

(edit conflict) You could look at the year-on-year delta and throw out any measurement that had a year-on-year change of more than, say, 50 pounds. But that 10% estimate is worrying. If I had a data set in which 10% of the observations were so far out of line that they were obviously incorrect, I wouldn't have much confidence in the quality of the remaining 90% either. Gandalf61 (talk) 16:55, 14 March 2011 (UTC)

10% is conservative. Most people in this position report around 50% error. I am assuming that I'm working with better data. The issue is units. Nurses take a weight measurement in pounds, but record it as kilograms. Height gets messed up as well because there are so many ways to write something like 5 feet 8 inches: 5'8", 5/8, 68, 1.7, etc... If you tell a computer that you just measured 5.8 meters, it will record the value and unit. What I'm doing is examining outliers to see if it is clearly an issue with units. -- kainaw™ 17:03, 14 March 2011 (UTC)

Instead of using the average and standard deviation, you might want to use the more robust median and inter-quartile range. In particular, if you have a large fraction of outliers (I doubt that with 50% it makes much sense to speak of outliers in a purely statistical sense), the estimate of standard deviation will be heavily influenced by the outliers, leading to an overestimation of standard deviation and a reduction of the efficiency of outlier detection. However, if I understand you problem correctly, you could maybe simply define a range of reasonable values (you wouldn't expect anybody to be 5.8 meters tall) and chop off everything outside this range. --Wrongfilter (talk) 19:42, 14 March 2011 (UTC)

That approach won't work in all cases. For example, some people weigh 100 pounds and others weigh 100 kilograms. However, if the same person changes from 100 lb to 100 kg between visits, that would be a red flag (unless they REALLY pigged out over the intervening holidays :-) ). StuRat (talk) 22:36, 14 March 2011 (UTC)

Correct. Height isn't much of an issue. Weight is. It is perfectly acceptable to have someone be 70lbs or 70kg. Similarly, 200lbs and 200kg is acceptable. The overlap between pounds and kilograms is huge. Similarly, the overlap between serum creatinine and urine creatinine is huge. The two labs are often mislabeled. So, if a person has a history of serum creatinine and suddenly the value jumps up beyond a standard deviation, I can flag it as a possible urine creatinine. There are many vitals/labs in which sloppy data entry is a major problem and I'm simply trying to flag possible errors as quickly as possible with a long-term goal of highlighting them during data entry to try to influence the humans to take a tiny bit of care in entering what should be accurate information. -- kainaw™ 05:23, 15 March 2011 (UTC)

You should take some logarithm of your data before doing anything else. Height and weight cannot have negative values, and so they cannot have a normal distribution. The logarithm of height or weight, however, may have a normal distribution. Taking the natural logarithm of 178 191 184 81 180 you get 5.18 5.25 5.21 4.39 5.19. Note that the log of the conversion factor between kilogram and pound is 0.79, and 4.39 + 0.79 = 5.18. So you are not looking for outliers in general, but rather for outliers that ceases to be outliers when increased by 0.79. Bo Jacoby (talk) 13:05, 16 March 2011 (UTC).

Power Series

It's been a couple years since I had complex analysis so I want to make sure I understand a certain aspect. So, my question is, have I put all this together correctly?

I'm looking at a certain function and I want to know a bit about the coefficients of its power series. I can prove that the function itself is bounded on any closed disc around any point in the complex plane. This means there are no poles or singularities for the function on the entire complex plane, i.e., it is holomorphic on the entire complex plane. Therefore, the function has a unique power series which converges to the function for all complex numbers, i.e., the radius of convergence is infinity. Therefore, we know $\limsup _{k\to \infty }|a_{k}|^{1/k}=0$, where $a_{k}$ are the coefficients in the power series. Since the terms are all nonnegative, this actually means the limit itself exists and is 0.

Is that all correct? Thanks StatisticsMan (talk) 17:07, 14 March 2011 (UTC)

Hmm, okay and the function is continuous. I thought of a counterexample to the above where I start with the same function but change its value at a few points and then it is no longer holomorphic as it is not even continuous. StatisticsMan (talk) 18:31, 14 March 2011 (UTC)

I'm not sure. Being continuous and bounded isn't enough to prove homolomorphisity, is it? Homolomorphisity is a differentiability condition. For example, take ƒ(z) = Re(z). This function is continuous and bounded on all of C, but it's not homolomorphic. — Fly by Night (talk) 19:13, 14 March 2011 (UTC)

You are correct. Okay, so I'm forgetting obvious things. I know the function is a composition of two entire functions and therefore is entire. How about everything after that? Thanks for the reply! StatisticsMan (talk) 19:17, 14 March 2011 (UTC)

It seems fine to me. I'm not sure I like the use of the phrase unique power series; but that's just a matter of taste I guess. You have a well defined and convergent power series expansion about each point of the complex plane. Take a look at function germ if you feel like more reading. These are equivalence classes of functions. A function and its power series look very different, but provided the function is holomorphic at the point of interest, they are essentially the same in a local way. That's what germs try to make concrete. — Fly by Night (talk) 19:35, 14 March 2011 (UTC)

Well, the coefficients of a power series centered at 0 for an entire function are unique. That's all I care about. StatisticsMan (talk) 19:42, 14 March 2011 (UTC)

Oh, brilliant. Well you're fine then. — Fly by Night (talk) 19:48, 14 March 2011 (UTC)

Okay, thanks for the help. StatisticsMan (talk) 19:51, 14 March 2011 (UTC)

Problem with completing the square

I'm helping a friend with putting equations into vertex form, which requires completing the square.

y= - 2x^2+6x+1

This one is giving me problems in particular. The -2 in front of the X^2 in particular is throwing me off completely. If I divide through by 2, then I will get 3x eventually which doesn't mesh easily into a form that can be factored into a square. Need some help. ScienceApe (talk) 23:16, 14 March 2011 (UTC)

To start, you should divide both sides of y = 1 + 6x – 2x² by –2 to give

$-{\frac {y}{2}}=x^{2}-3x-{\frac {1}{2}}\,.$

Completing the square on the right hand side gives:

$-{\frac {y}{2}}=\left(x-{\frac {3}{2}}\right)^{2}-{\frac {11}{4}}\,.$

Finally, multiply back through by –2 to give

$y={\frac {11}{2}}-2\!\left(x-{\frac {3}{2}}\right)^{2}\,.$

Is that what you were looking for, or are you after something else? — Fly by Night (talk) 00:05, 15 March 2011 (UTC)

I would do it slightly differently:

$y=-2(x^{2}-3x-{\frac {1}{2}}\,)$

Completing the square yields:

$y=-2(\left(x-{\frac {3}{2}}\right)^{2}-{\frac {11}{4}}\,)$

Finally, multiply back through by –2 to get:

$y={\frac {11}{2}}-2\!\left(x-{\frac {3}{2}}\right)^{2}\,$ — Preceding unsigned comment added by StuRat (talk • contribs) 01:16, 15 March 2011 (UTC)

March 15

Martingales & Optional Sampling Theorem

Hello everyone,

I was hoping you could help me out with a question on martingales and the optional sampling theorem. The problem is as follows:

Suppose $a<x<b$, and let $\tau \equiv \inf\{t:B_{t}\not \in [a,\,b]\}$. For $\beta >0$, find an appropriate martingale and use the optional stopping theorem to prove

$\mathbb {E} \!\left[\exp \left({-{\frac {1}{2}}\beta ^{2}\tau }\right)|B_{0}=x\right]={\frac {\cosh {(\beta (x-(a+b)/2)}}{\cosh {(\beta (b-a)/2)}}}$.

Now first of all I'm having some trouble finding an appropriate martingale to apply to the question, could anyone offer a little help with this problem? Thankyou! :) Typeships17 (talk) 00:49, 15 March 2011 (UTC)

Could you explain your notation? For example, what is B_t? — Fly by Night (talk) 02:46, 15 March 2011 (UTC)

Have you tried the geometric Brownian motion $e^{\beta B_{t}-{\frac {1}{2}}\beta ^{2}t}$? This is a martingale. Some variant of this should work for the problem. Sławomir Biały (talk) 03:52, 15 March 2011 (UTC)

I think B_t is the position of a Wiener process at time t. -- Meni Rosenfeld (talk) 09:59, 15 March 2011 (UTC)

Yes, sorry, i forgot to say B_t is a Brownian motion (or Wiener process). Hopefully the rest is self explanatory. And thanks, I'll take a look at the geometric brownian motion, the left hand side would then arise pretty obviously, though where the right hand side comes from is currently beyond me, I'll see if I can figure anything out. Typeships17 (talk) 17:08, 15 March 2011 (UTC)

Let us know how it goes, either way. ;-) Sławomir Biały (talk) 12:36, 15 March 2011 (UTC)

Wait, if B_t is a Brownian motion, surely we can't set B₀=x? I was sure it was meant to be a BM, in keeping with the notation I've seen previously, but perhaps I'm just confused... Typeships17 (talk) 17:14, 15 March 2011 (UTC)

There's no significant loss in generality in taking x=0. Then it's the usual Brownian motion starting at the origin, if it helps. Sławomir Biały (talk) 17:55, 15 March 2011 (UTC)

Okay, so we want to say WLOG B_t is a Brownian motion, and prove $\mathbb {E} \!\left[\exp \left({-{\frac {1}{2}}\beta ^{2}\tau }\right)\right]={\frac {\cosh {(\beta (a+b)/2}}{\cosh {(\beta (b-a)/2)}}}$. The left hand side is clearly $\mathbb {E} \left[M_{0}\right]$, where $M_{t}=e^{\beta B_{t}-{\frac {1}{2}}\beta ^{2}t}$, and the optional sampling theorem says that equals $\mathbb {E} \left[M_{\tau }\right]$ - but how do I evaluate $\mathbb {E} \left[M_{\tau }\right]$? I'm not totally sure how to take the expectation when the step number is a stopping time rather than a 'fixed value' - do we use the iterated expectation formula? Advice would be much appreciated! Typeships17 (talk) 18:24, 15 March 2011 (UTC)

Consider the martingale

$X_{t}=e^{-{\frac {1}{2}}\beta ^{2}t}\cosh(\beta (B_{t}-(a+b)/2)).$

Evaluate $\mathbb {E} ^{x}[X_{\tau }]$ in two different ways. First, by optional sampling,

$\mathbb {E} ^{x}[X_{\tau }]=X_{0}=\cosh(\beta (x-(a+b)/2))$

Second, since at $t=\tau$, $\cosh(\beta (B_{\tau }-(a+b)/2))=\cosh(\beta (a-b)/2)$ is deterministic,

$\mathbb {E} ^{x}[X_{\tau }]=\mathbb {E} ^{x}\left[e^{-{\frac {1}{2}}\beta ^{2}\tau }\right]\,\cosh(\beta (a-b)/2)$

as desired. Sławomir Biały (talk) 00:16, 16 March 2011 (UTC)

That's brilliant, thank you Sławomir! You know I don't think I've ever really seen an argument like that before, it's very nice. Thanks again :-) Typeships17 (talk) 10:25, 16 March 2011 (UTC)

Singular value decomposition

Just checking that the article singular value decomposition is correct. Given M is an mxn matrix and M=UΣV^T, it claims U is an mxm matrix. However, examples I've seen make U an mxn matrix, Σ an nxn matrix, and V an nxn matrix. It is possible that the article has switched the meaning of U and Σ compared to other articles I've read. -- kainaw™ 04:14, 15 March 2011 (UTC)

The article makes sense to me. In particular, U is supposed to be an unitary matrix, and for that it has to be square. –Henning Makholm (talk) 04:42, 15 March 2011 (UTC)

See Singular_value_decomposition#Reduced_SVDs. The "vanilla" SVD indeed has square U and V matrices, but to save space you can drop unnecessary rows and columns. The "Thin SVD" is the form you've seen. -- Meni Rosenfeld (talk) 09:43, 15 March 2011 (UTC)

Thanks. That explains it perfectly. -- kainaw™ 12:27, 15 March 2011 (UTC)

Latex Pictures

I want to put some large images into my latex file.

However because of the margins of my page, the pictures get pushed to the right of the page. I cannot make them centered unless I make them very small.

I am using the graphicx package, as follows:

\begin{figure}[H] \begin{center} \includegraphics[scale=0.4]{picture.png} \caption{etc} \end{center} \end{figure}

However the "center" isn't actually centering it, due to the fatness of the margin on the left.

How do I get the picture to be actually centered, overriding the margin? Like choosing "in front of text" when you're putting a picture into microsoft word? 130.102.158.15 (talk) 05:31, 15 March 2011 (UTC)

Hmm, in plain TeX I would just add \hss on either side of the over-wide box, but that doesn't seem to work in LaTeX. Perhaps one of the suggestions in this TeXblog post will work for you. —Bkell (talk) 10:28, 15 March 2011 (UTC)

Expression for partial sums of sequence

Is there a closed-form expression for the nth term of the sequence

1, 3, 5, 8, 11, 14, 18, 22, 26, 30, ... (sequence A060432 in the OEIS)

which are the partial sums of the sequence

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...

containing one 1, two 2s, three 3s etc. ? The asymptotic behaviour of the nth term (of the first sequence) is

${\frac {(2n)^{\frac {3}{2}}}{3}}$

but I am wondering if an exact expression is known. Gandalf61 (talk) 14:41, 15 March 2011 (UTC)

The k(k+1)/2 term is k(k+1)(2k+1)/6 from the formula for the sum of squares, so you need to find the next k down that gives your n and add up from there

k(k+1)/2 = n would give k = (-1+sqrt(1+8k))/2 so we choose the next integer below this as k. That will give a sum for all the terms up to the k(k+1)/2 one and we just add (n-k/k(+1)/2)k to the sum. Dmcq (talk) 15:11, 15 March 2011 (UTC)

Okay, so putting that all together I get:

$n(k+1)-{\frac {k(k+1)(k+2)}{6}}{\text{ where }}k=\left\lfloor {\frac {{\sqrt {8n+1}}-1}{2}}\right\rfloor$

Not pretty, but it works. Thank you. Gandalf61 (talk) 16:00, 15 March 2011 (UTC)

You can also write in a slightly less not-pretty way as

$nk-{\frac {k^{3}-k}{6}}{\text{ where }}k=\left\lfloor {\frac {{\sqrt {8n+1}}+1}{2}}\right\rfloor$

-- Meni Rosenfeld (talk) 16:13, 15 March 2011 (UTC)

That's positively beautiful compared to what I expected :) Dmcq (talk) 18:54, 15 March 2011 (UTC)

Calculator

How do you get the "exact" result on your calculator? (for example: square root 2/2 instead of 0.707106781) Thanks. — Preceding unsigned comment added by Atacamadesert12 (talk • contribs) 16:46, 15 March 2011 (UTC)

What kind of calulator do you have? Most fancy (graphing) calculators can convert rational decimals to fractions pretty well, but most can't do the kind of thing you suggest with irrational numbers. (I don't know of any ordinary handheld calculator that can do that.) Staecker (talk) 17:35, 15 March 2011 (UTC)

By the way, you can do this sort of thing at Plouffe's Inverter, which for your example gives various different forms, some more useful than others. Staecker (talk) 17:43, 15 March 2011 (UTC)

I have Sharp EL-520W — Preceding unsigned comment added by Atacamadesert12 (talk • contribs) 18:17, 15 March 2011 (UTC)

I have no knowledge about this calculator, but if you want to do computations and get an exact result, you can use Wolfram|Alpha (example). -- Meni Rosenfeld (talk) 19:43, 15 March 2011 (UTC)

You are asking about doing symbolic algebra on your calculator. There have been some calculators that can do it, but these days it probably makes more sense to use a computer, either with software like Sage or (if you must) with a web service like Wolfram Alpha. 75.57.242.120 (talk) 22:13, 15 March 2011 (UTC)

powerball jackpot

What it the statistical odds that the next powerball jackpot will exceed $500 mil before being won? —Preceding unsigned comment added by 75.170.239.188 (talk) 17:08, 15 March 2011 (UTC)

It's going to be very difficult to calculate, as the size of the jackpot depends on how many tickets were sold, as well as how many times it rolls over (related to the number of tickets sold). The number of tickets sold varies, and depends somewhat on the size of the jackpot (larger jackpots mean more tickets sold). You really can't make the calculation without specifying some model of ticket sales. All that said, the largest Powerball jackpot so far was $365 million, so the chance that it'll hit $500 million on the next jackpot win is for all practical purposes indistinguishable from zero. -- 140.142.20.229 (talk) 22:51, 15 March 2011 (UTC)

It shouldn't be too difficult. Assuming a random distribution, the number of tickets sold is proportional to the chances of it being a roll-over. — Fly by Night (talk) 23:29, 15 March 2011 (UTC)

Continuous compound interest

Every calculus book I read has examples on "continuously compounded" interest. Does any financial instrument in the real world behave this way? My bank account sure doesn't. Our sections on continuous compounding don't seem to give any real examples. Staecker (talk) 17:38, 15 March 2011 (UTC)

I doubt there would be any financial institutions that do. The more often interest is calculated, the better it is for the customer. But I'm sure that there are examples of continuous exponential growth in nature. — Fly by Night (talk) 18:13, 15 March 2011 (UTC)

I know of at least one (funny) example. A bank was offering continuous compounding on it's savings account, and had a TV ad featuring an exhausted accountant who was recalculating the interest on your account "continuously", never sleeping. This may have just been a gimmick by one bank, though.

In general, I don't see the advantage. If the banks wants to give a higher interest rate, just offer a higher periodic interest rate instead. Using complex math that the majority of their customers can't understand is just going to cause confusion and frustration among the customer base and the tellers who are asked to explain it all.

Also, if the bank ends up paying more when offering 10% interest compounded continuously, than 10% interest compounded monthly, but the customer is unaware of the difference, it's not good for the bank. They may also include the "effective annual yield", but skeptical customers may ignore that and use the lower nominal rate to choose their bank.

Now, in the cases of businesses that charge you interest, I would think that continuous interest would be a way to squeeze a bit more out of the customers without appearing to do so. However, the argument of not wanting to alienate customers incapable of understanding the calculations still applies. StuRat (talk) 20:55, 15 March 2011 (UTC)

I'd hardly call it "complex math". The formula is A(t) = Pe^rt where A(t) is the amount in the bank after time t, P is the principal investment and r is the annual interest rate. — Fly by Night (talk) 21:10, 15 March 2011 (UTC)

Perhaps not complex for the average person here, but it is for the average bank customer. They'd just think the bank was making that formula up and want to know where the heck "e" came from. StuRat (talk) 07:42, 17 March 2011 (UTC)

I assume that no ordinary bank accounts offer this- I was wondering about more complicated financial instruments. As far as I understand, banks like to manage accounts according to occasional "transactions", each of which changes the balance. My bank deposits my interest every time they compound it in a transaction. Continuous compounding would mess up this whole system. But I wonder if there are some esoteric financial gadgets not typically owned by ordinary people that work by continuous compounding. There are certainly plenty of crazier and much more complicated things out there. Staecker (talk) 22:33, 15 March 2011 (UTC)

Apparently credit card interest is typically compounded daily. That's pretty close to continuous. Presumably there are practical constraints on having instruments compounded more often than daily. Sławomir Biały (talk) 23:07, 15 March 2011 (UTC)

There's a patent on sub-day compounding, but really continuous compounding may still be free ... –Henning Makholm (talk) 23:45, 15 March 2011 (UTC)

They're not financial instruments per se, but models for valuing stock options generally treat the risk-free interest rate as continuously compounded. The assumption makes sense in the context of the model, since the point is to arrive at an option's value rather than calculate the interest at a particular point in time. —Preceding unsigned comment added by 12.186.80.1 (talk) 20:28, 17 March 2011 (UTC)

Thanks- that's a good answer. Staecker (talk) 22:45, 17 March 2011 (UTC)

Bayesian probability

Are there any introductory probability texts that give solely Bayesian probability without providing or requiring a background in classical probability? (And obviously if so, what are some) Thanks. 72.128.95.0 (talk) 23:31, 15 March 2011 (UTC)

Jaynes' book is probably what you want. I haven't read it myself though. -- Meni Rosenfeld (talk) 09:06, 16 March 2011 (UTC)

The point where "classical probability theory" starts to differ from "Bayesian probability thoery" is when you get into applications, and at that point "classical probability theory" ceases to consist only of probability theory, whereas the point of Bayesianism is that it's all to be reduced to probability theory. So as long as it's only probability theory, it's consistent with both Bayesianism and frequentism. Michael Hardy (talk) 16:04, 16 March 2011 (UTC)

Is there a straightforward example application where the Bayesian and frequentist approaches give differing answers? 75.57.242.120 (talk) 10:35, 18 March 2011 (UTC)

There's no shortage of papers that debunk frequentism and give examples. Here's an example that came up in a search: [3] (I haven't read it). -- Meni Rosenfeld (talk) 11:44, 18 March 2011 (UTC)

March 16

2nd order difference equations vs 2nd order differential Equations

I need to clarify the distinction between difference and differential equations.

Am I correct in saying that given a characteristic equation for a second order difference or differential equation

$\lambda ^{2}+a\lambda +b=0$

The differential equation is stable if both roots are negative, but the difference equation is stable if the absolute value of both roots is less than one?

A differential equation will have a saddle point equilibrium if there is one positive and one negative root, the difference equation will have a saddle point if there is one root less than one in absolute value and the other root greater than one in absolute value?

Both the differential and difference equations will have complex roots (spirals) if the discriminant is negative?

Both the differential and difference equations will have pure imaginary roots (centres) if a=0 and b is negative. 118.208.133.15 (talk) 01:19, 16 March 2011 (UTC)

In the difference equation y_n+2+ay_n+1+by_n = 0 substitute y_n = λⁿ to get λ²+aλ+b = 0.

In the differential equation y' '(t)+ay'(t)+by(t) = 0 substitute y(t) = e^λt to get λ²+aλ+b = 0.

The λs have different interpretations in the two cases. Bo Jacoby (talk) 12:43, 18 March 2011 (UTC).

Probability

Prize: 5

Participant: 14 people (include my friend and i)

What is the probability of either of us win the prize? (and both of us win as well?)

Thanks!

218.188.16.142 (talk) 01:43, 16 March 2011 (UTC)

Assuming that each of the participants is equally likely to win a prize (i.e. this is something like a random draw, not a game of skill) and no participant can win two prizes, then the probability that exactly one out of you and your friend wins a prize is

${\frac {2\times 5\times 9}{14\times 13}}={\frac {90}{182}}\approx 49.5\%$

and the probability that you both win a prize is

${\frac {5\times 4}{14\times 13}}={\frac {20}{182}}\approx 11\%$

so the probability that one or both of you win a prize is just over 60%. Gandalf61 (talk) 09:12, 16 March 2011 (UTC)

[ec] There are ${\binom {14}{5}}=2002$ ways to distribute 5 prizes among 14 people. Of these, $2\cdot {\binom {12}{4}}=990$ give exactly one of you two a prize, and ${\binom {12}{3}}=220$ give both of you a prize. Thus the probability that at least one gets a prize is ${\frac {990+220}{2002}}={\frac {55}{91}}\approx 0.604$, and the probability that both get one is ${\frac {220}{2002}}={\frac {10}{91}}\approx 0.110$. See also Combination. -- Meni Rosenfeld (talk) 09:16, 16 March 2011 (UTC)

Wow Thanks a lot guys! I can never understand probability :S 218.188.16.142 (talk) 01:54, 17 March 2011 (UTC)

Don't tell yourself (nor anybody else) that there is something you can never understand. Tell yourself (and anybody else) that of course you can understand it when having it properly explained. Keep asking questions. Bo Jacoby (talk) 11:54, 17 March 2011 (UTC).

March 17

Derivatives

Ok this is a really elementry question that has bugged me years ago and has come back now that I'm concentrating on calculus again. Let f:U->R, where U is open in R, the real numbers. When we say the derivative at x we take the limit of the usual quotient as the usual increment h tends to 0. But the term f(x+h) is not defined for all h. So by the derivative at x do we take the limit of the function k:A->R, where A is the set of all h such that x+h is U, and A doesn't contain 0, and k is defined by k(h) = (f(x+h)-f(x))/h. What I'm trying to get at is that the usual limiting quotient has an ill defined domain. Thanks. Money is tight (talk) 13:54, 17 March 2011 (UTC)

Since U is open, for all x in U, there is an open neighborhood V of U containing x. Now f(x+h) is defined for h in an open subset containing 0. Invrnc (talk) 14:16, 17 March 2011 (UTC)

(edit conflict) I don't see what you gain by changing from f to k—the function f is defined only on the set U, and the function k is defined only on the set A (which is simply U − x), so it seems to me that if you're uncomfortable with f you should be equally uncomfortable with k. But you shouldn't be uncomfortable with either of them. Since the set U is open, for every point x in U there is an open interval containing x that is completely contained within U. Since you are considering a limit as h approaches zero, you can simply focus on that open interval and ignore everything outside of it. In other words, it is true that f(x + h) is not defined for all h, but it is defined for all h that are close enough to 0, so the limit as h approaches 0 is well defined. Just ignore the values of h that are too far away from 0—you wouldn't care about them in the limit anyway. —Bkell (talk) 14:23, 17 March 2011 (UTC)

That's exactly my problem, the domain of the limiting quotient is ill defined if we just say its domain is some open ball containing x and in U. I know the resulting limit is the same but I like definite definition. Money is tight (talk) 14:37, 17 March 2011 (UTC)

Okay, I'll try to be more definite. Since $x\in U$, there exists some $\eta >0$ such that $(x-\eta ,x+\eta )\subseteq U$. We are examining the limit $\lim _{h\to 0}{f(x+h)-f(x) \over h}$, which is defined to be equal to a real number $L$ if and only if, for every $\epsilon >0$, there exists some $\delta >0$ such that $\left|{f(x+h)-f(x) \over h}-L\right|<\epsilon$ whenever $|h|<\delta$. For a particular value of $\epsilon$, if one value for $\delta$ satisfies this condition, then any smaller positive value for $\delta$ will also satisfy this condition. So, if we are aiming to prove that the limit is $L$ using a delta-epsilon proof, we can always choose $\delta \leq \eta$; then we will never run into a problem where the difference quotient is undefined. —Bkell (talk) 16:22, 17 March 2011 (UTC)

Monty Hall problem implications

I can't get two issues. First of all, it would be naturally to assume that it's not profitable for the host to offer some door (since the TV show would loose the car), as such the player will most likely choose any door, not suggested by host. Secondly, why the host would open door 3 if I chose door 1? Or they are not labelled? Thanks. —Preceding unsigned comment added by 89.76.224.253 (talk) 15:24, 17 March 2011 (UTC)

It's the rules of the game. The host has to offer you a choice, he has to open one of the remaining doors (and of those one that does not have the car), and he then has to offer you the choice of keeping "your" door or changing to the remaining one. How the doors are labeled is irrelevant - what is important is that the host knows where the car is, and never opens the car door. --Stephan Schulz (talk) 15:35, 17 March 2011 (UTC)

Keep in mind that these are the rules of the simplest game that results in the usual answer to the problem. They are not the rules by which Monty actually played. --Trovatore (talk) 01:48, 18 March 2011 (UTC)

Also, gameshows are happy when contestants win. Winners make for better ratings, and the increase in ad revenue more than makes up for the cost of the prizes.--203.97.79.114 (talk) 11:27, 18 March 2011 (UTC)

Invariance of Differential Forms

I've got a quadratic differential form ω = A du² + 2B du dv + C dv², where A, B and C are functions of u and v. It's a fact that ω is invariant under orientation preserving diffeomorphisms of uv−space. I'm interested in the function g(u,v) = AC − B², which is the determinant of the matrix representation on ω. Am I right in saying that g would not be invariant under orientation preserving diffeomorphisms of uv−space? I'm assuming that the functions A, B and C will transform one way, and that du and dv will transform another way; so that they cancel each other's transformations. Or, is it possible that g is invariant under orientation preserving diffeomorphisms of uv−space? (I really, really hope it is). — Fly by Night (talk) 16:50, 17 March 2011 (UTC)

Assuming I understand the question, the invariant gadget would be the volume form ${\sqrt {g}}du\wedge dv$. Again see the article metric tensor. Sławomir Biały (talk) 17:17, 17 March 2011 (UTC)

Stated another way, g is an invariant density of weight 2, meaning that it transforms by the square of the Jacobian determinant of the diffeomorphism. In particular, if the diffeomorphism is symplectic, then g is invariant as a scalar. Sławomir Biały (talk) 17:34, 17 March 2011 (UTC)

Concretely: In plain old R^2, consider the diffeomorphism that multiplies each coordinate by 2. This also doubles du and dv, so in order to preserve ω each of the coefficients has to shrink to a quarter of its former value. Then $g$ shrinks by ${\tfrac {1}{16}}$, so not invariant. –Henning Makholm (talk) 17:50, 17 March 2011 (UTC)

No, ω is not the volume form. I just used g to denote its matrix's determinant. It's more of a general question. I have an explicit form but, as I said, it's not the volume form. It's kernel directions identify the asymptotic directions. Actually, if the surface is given as the graph of a function then

$\omega ={\frac {f_{uu}\,\operatorname {d} u^{2}+2f_{uv}\,\operatorname {d} u\,\operatorname {d} v+f_{vv}\,\operatorname {d} v^{2}}{|f_{uu}f_{uv}-f_{uv}^{2}|^{1/4}}}\,.$

M. Spivak (1999), Comprehensive introduction to differential geometry, vol. 3 (3 ed.), p. 85, ISBN 0-914098-72-1— Fly by Night (talk) 18:00, 17 March 2011 (UTC)

I was referring to g, not ω, above. The form ${\sqrt {g}}\,du\wedge dv$ is a volume form in the uv-plane, and it is invariant assuming that the quadratic differential form ω is. In any event, if you need to regard g itself as invariant, then you need to regard it as a density (that is, a section of a certain line bundle), not a scalar function. Sławomir Biały (talk) 18:15, 17 March 2011 (UTC)

(edit conflict) You've lost me, what is g above? What is ω above? I don't need g to be invariant; it either is or it isn't. I would like it to be, and it would mean I've done what I want to do. What does it mean to regard g as a density? A section of a line bundle is a vector field... Like I said, you've lost me. — Fly by Night (talk) 18:18, 17 March 2011 (UTC)

$g=AC-B^{2}$ is the determinant of the matrix representation of the quadratic form ω in the uv coordinate system, per the notation you established. This transforms by the square of the Jacobian determinant of the coordinate transformation. So it is not an invariant scalar, but is an invariant section of the line bundle $\wedge ^{2}T^{*}\mathbb {R} ^{2}\otimes \wedge ^{2}T^{*}\mathbb {R} ^{2}$. See density bundle. Sławomir Biały (talk) 18:23, 17 March 2011 (UTC)

Should there really be four copies of the cotangent bundle, or should it be $\scriptstyle \wedge ^{2}T\mathbb {R} ^{2}\,\otimes \,\wedge ^{2}T^{*}\mathbb {R} ^{2}$? — Fly by Night (talk) 19:04, 17 March 2011 (UTC)

It's four copies if the cotangent bundle, by dimensional analysis of the original coefficients of omega. Sławomir Biały (talk) 19:09, 17 March 2011 (UTC)

Binary operation

the binary operation ₴ is defined on a set A=(4,6,8,) by a ₴ b= ½ (a-b) if a= 4,b=6 and c=8, evaluate the following; a ₴ (b+c) —Preceding unsigned comment added by 41.218.231.127 (talk) 17:14, 17 March 2011 (UTC)

What does ₴ do to a and b? It subtracts the second from the first, and then halves the result. If you want to know what a ₴ (b + c) is then subtract the second argument from the first, and then half the result. That will give you the general answer. Then just substitute in the values given for a, b and c. — Fly by Night (talk) 18:02, 17 March 2011 (UTC)

Having said that, I think you might have given the wrong set for A. A binary operation must be closed, i.e. it takes two elements of A and gives you an element of A as the answer. If A = {4, 6, 8, …}, then we have problems, for example 4 ₴ 6 = ½(4 – 6) = –1. But –1 is not in A. If we extend A to the set of all even numbers we still have problems because 2m ₴ 2n = ½(2m – 2n) = m – n; which need not be even. Even if we extend A to the set of all whole numbers we still have problems because m ₴ n = ½(m – n); which is a whole number if and only if m and n are both even numbers. The best you can get away with is A = Q, i.e. A is the set of rational numbers. — Fly by Night (talk) 03:18, 18 March 2011 (UTC)

It is sufficient if the members of the set A are binary fractions. A = {a2⁻ⁿ : a is an integer, n is a positive integer}. Bo Jacoby (talk) 11:24, 18 March 2011 (UTC).

Rigorous theory of "div, grad, curl, and all that"

Standard textbooks on vector calculus sweep under the carpet the fact that the expressions for things like div, grad, curl, Laplacian, etc. are invariant under rotations, and other things of interest to mathematicians but in which engineers are happy to just trust the mathematicians. Which are the most elementary among books giving a fully rigorous account of such matters? Michael Hardy (talk) 18:09, 17 March 2011 (UTC)

I've had good experiences with Marsden and Tromba's textbook "Vector calculus", although I probably wouldn't think of it as "fully rigorous". It seems to have about the right amount of rigor for a solid undergraduate treatment. Unfortunately, I don't have a copy on my shelf, so I can't say if it discusses precisely these issues. Sławomir Biały (talk) 18:20, 17 March 2011 (UTC)

Thank you. Michael Hardy (talk) 03:49, 18 March 2011 (UTC)

Tom Apostol's old two-volume calculus book covered this stuff pretty rigorously. Again though, it's been a while (high school) since I looked at it, so I don't remember what was in it. 75.57.242.120 (talk) 10:38, 18 March 2011 (UTC)

I checked both Apostol and my own recommendation, and neither of them seems to discuss the behavior of div, grad, and curl under rotations. The best I could find aimed at an elementary level is (don't laugh) Schaum's outlines of vector analysis. They prove in detail that the gradient of a radial function is a rotationally-invariant vector field, and leave it as an exercise to show that the curl and divergence of a rotationally invariant vector field are rotationally invariant, and that the expression for the gradient is rotationally invariant. Sławomir Biały (talk) 13:20, 18 March 2011 (UTC)

March 18

Calculus theorems

What are some degenerate cases of vector, multivariable, or just higher calculus theorems that can be applied to elementary (2-dimensional/on a plane) calculus and can solve problems easily that would be difficult to solve with elementary calculus alone? I know that many higher theorems are only generalizations of lower theorems so I would like to exclude these. Thank you for your responses. 72.128.95.0 (talk) 01:33, 18 March 2011 (UTC)

There is a nice application of Green's theorem to calculate area. Is this the sort of thing you have in mind? Sławomir Biały (talk) 02:40, 18 March 2011 (UTC)

Unrecognised mathematical notation

Hi. Please see:

http://img145.imageshack.us/img145/5386/formula.gif

This is part of a formula generated by the WolframAlpha website. I don't understand these elements:

• The diagonal line of three dots at the end of line 1.
• What looks like a superscripted numeral 1 at the start of line 2.
• What look like fraction bars but with no numerators or denominators in lines 4 and 5.

Any ideas? 86.177.106.14 (talk) 03:57, 18 March 2011 (UTC)

I think they're all just bad formatting resulting from trying to split a long expression into several lines. Can you give us the query that generated this result? -- Meni Rosenfeld (talk) 07:24, 18 March 2011 (UTC)

See this. 86.183.3.118 (talk) 12:52, 18 March 2011 (UTC)

Tensors

This is a general problem: How does one go about investigating the effects of rotating coordinate axes regarding tensors? For example, what does a 2-fold symmetry about all 3 orthogonal axes translate to in tensors and suffix notation? Thanks —Preceding unsigned comment added by 131.111.222.12 (talk) 10:47, 18 March 2011 (UTC)

The whole point of tensors is to abstract away from a particular coordinate system. So to represent things like that you have to use the representation of a tensor in a particular coordinate system in which the symmetry you are interested in holds. Dmcq (talk) 11:54, 18 March 2011 (UTC)

Thanks Dmcq, but how does one represent a tensor in a particular coordinate system? (sorry, I am new to this...) —Preceding unsigned comment added by 131.111.222.12 (talk) 15:57, 18 March 2011 (UTC)

An order-k tensor in a space with n dimensions can be represented as an array of n^k numbers. For example, a 2nd-order tensor in a 3-dimensional space is represented as a 3x3 matrix of 9 numbers. Gandalf61 (talk) 16:26, 18 March 2011 (UTC)

Take a look at the Covariance and contravariance of vectors article, and in particular the Use in tensor analysis subsection. — Fly by Night (talk) 19:01, 18 March 2011 (UTC)

Converting a Difference in GPS co-ordinates to a distance?

If you have two GPS co-ordinates, x & y is there an easy way to convert (x - y) into a distance in kilometers? Or does the curvature of the Earth make this a difficult problem? --CGPGrey (talk) 20:18, 18 March 2011 (UTC)

See Great-circle_distance. I don't exactly think this qualifies as "easy", but at least it explains how to compute the distance. Gscshoyru (talk) 20:23, 18 March 2011 (UTC)

It depends a lot on how far apart they are. If you're talking about locations within, I don't know, a few hundred miles of each other, and you don't need extreme accuracy, you can probably do it in your head. Just remember that one nautical mile is one minute of arc. So to get the north-south distance, look at the difference in latitudes in degrees, multiply by 60, and that's the answer in nautical miles. Add another ten percent to get regular miles, or maybe 70 percent to get kilometers.

For east-west distance it's a little more complicated, but only a little — you have to multiply by the cosine of your latitude.

Now you have two edges of a right triangle, and you just use the Pythagorean theorem. --Trovatore (talk) 20:29, 18 March 2011 (UTC)

Ah, I was a little low on the conversion factor. Add fifteen percent to nautical miles to get miles, or 85 percent to get kilometers. --Trovatore (talk) 20:31, 18 March 2011 (UTC)