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March 29

This function as an approximation to tetration?

Hi, what do you think of this function as an approximation to tetration?

${\ ^{b}a={a^{a^{b} \over {a}}}}$

(I hope the formula shows up correctly - let me know if it doesn't.) The two sides of this equation seem to equal each other with certain values. Please try it and let me know what you think. Have you seen this equation (or a similar one) anywhere else? If so, where? I really want to look at websites with further information on this and similar equations. Thank you. 4.242.147.228 (talk) 02:20, 29 March 2008 (UTC)

Well, $a^{a^{b} \over {a}}=a^{a^{b-1}}$. So (assuming b>=2 - the equation is trivially satisfied for b=1, and if b=0, it's only satisfied for a=1 from what I can see), if we take base a logs on both sides, we get $\ ^{b-1}a=a^{b-1}$, and again gives $\ ^{b-2}a=b-1$, so that would seem to be a necessary and sufficient condition for your equality to hold. I can't see any particular reason for that to be the case other than simple coincidence. --Tango (talk) 02:41, 29 March 2008 (UTC)

OK, thanks. I think I didn't read the tetration article thoroughly enough. I will go back and read it some more. You can consider this question answered for now. 4.242.147.174 (talk) 02:54, 29 March 2008 (UTC)

Function or Not?

I was having fun with the program Grapher, which comes with Mac OSX.5 (or Leopard), and I typed in y=x^-1, and got two curved lines that were reflected across the line y=-x. I was wondering 1) What is this called when a function comes up with two lines and 2) Is it a function at all? In theory, the two lines will never have the same y, but different x values because neither can ever tough the y-axis. This thought may be wrong too, but I figured that zero can't work because 0^-1 = 1/0 and you can't divide anything by zero. Could that alone make this not a function? I don't know. --Xtothe3rd (talk) 02:49, 29 March 2008 (UTC)

Great question! It is indeed a function. You correctly note that the domain of the function (all the numbers which could be put into the function) does not include x=0, however functions are allowed to have restricted domains. The only real criteria is that for any given input (within its domain), a function must not have more than one output (y value).

The function $y=x^{-1}$ does have asymptotes along both the horizontal and vertical axes, and is generally described as a hyperbola(although note that most for most hyperbolas, y is not a function of x. I don't know of a general term for a function made up of multiple disjoint parts, as in this case. But many interesting functions have that property. Hope this helps! --TeaDrinker (talk) 03:29, 29 March 2008 (UTC)

A topologist would call it "disconnected", if that helps. --Tango (talk) 15:19, 29 March 2008 (UTC)

This is a Piecewise continuous function. It is composed of two pieces, each of which is continuous. -- Meni Rosenfeld (talk) 21:02, 29 March 2008 (UTC)

Yes, it is, but that term also applies to a continuous function that is in once piece, so it doesn't really narrow it down. --Tango (talk) 21:55, 29 March 2008 (UTC)

Actually, 1/x is continuous everywhere, since the only point it isn't continuous at is a point it isn't defined at. I would use "piecewise continuous" to refer to something like a step function. --Tango (talk) 21:59, 29 March 2008 (UTC)

For the sake of discussion, a specific function being referred to as "piecewise continuous" can be taken to be discontinuous, since otherwise we would have called it continuous.

1/x is continuous in its domain, but not in $\mathbb {R}$. It is piecewise continuous in $\mathbb {R}$ (if you allow being undefined at transition points). -- Meni Rosenfeld (talk) 00:15, 30 March 2008 (UTC)

Of course, it's worth noting that 1/z is continuous and meromorphic on $\mathbb {C}$, with a simple pole at 0. 1/z is most certainly a function, although its continuity is partly because in complex analysis, all infinities are identical. 1/x just seems odd at first if you don't know about functions which have poles. Remember though, the polynomials you are used to dealing with also have poles, just they occur at x = infinity. -mattbuck (Talk) 02:56, 30 March 2008 (UTC)

Indeed - which leads to the natural and intuitive view of 1/z as a half-turn rotation of the Riemann sphere about an axis through +1 and -1. Gandalf61 (talk) 09:12, 30 March 2008 (UTC)

Glm link function choice

Howdy, does anyone know of a general test to choose a link function (among some restricted class--even if it is just two) for generalized linear models? For instance, in a binomial problem, is there a general way to decide if a logit link or a probit link is more appropriate, without dramatically influencing the inference (that is, testing both and taking the one which fits better)? I only know one special case (Box-Cox power transforms for Gaussian models can be realized as choices between link functions) where there is a general way to select a link (and even that could have some influence on inference). Thanks, --TeaDrinker (talk) 03:35, 29 March 2008 (UTC)

In a case like yours, changing the link will most likely have very little effect on any inference, unless you are playing the accept/reject a p-value game, and one links is just over the cutoff, and the other is just under. But things like confidence intervals, etc for parameters are not likely to change a lot. The most salient difference between the two links you mention is that the probit is in some sense a stronger one, in that the asymptotes to ±∞ occur faster and "harder". This property makes it easier to deem data as being outlying due to unusual contributions to the deviance, although using the logit will usually flag the same data, just not as obviously.

For binomial data, there is a theoretically attractive reason to chose the logit, See the generalized linear models article for details, or a good textbook (e.g., McCullagh & Nelder). But that does not imply that link works better for a particular application, it only means some math works out nicer.

It is in general dangerous to use one's data to direct the choosing of a model, then using such model for inference. Essentially, the process of making the modeling choice is not accounted for in the calculations of Types I and II error; the nominal rates can be way off. Baccyak4H (Yak!) 03:42, 30 March 2008 (UTC)

Thank! --TeaDrinker (talk) 20:54, 1 April 2008 (UTC)

Nicomachus, Introduction to Arithmetic

I cannot seem to find a fee, online copy of Nicomachus' "Introduction to Arithmetic". I find it hard to believe that there is no copy.

Do you know where I can find one?

Thanks so much,

R.B. 74.127.254.155 (talk) 05:45, 29 March 2008 (UTC)

It appears that archive.org has a copy, and is in fact linked to in the external links of Nicomachus, Though I don’t know what language it is in. GromXXVII (talk) 11:09, 29 March 2008 (UTC)

It's Greek to me...  --Lambiam 13:50, 29 March 2008 (UTC)

name of generalized cross product construct

In the Cross product#Higher dimensions section they talk about a form of generalizing the cross product to n dimensions, like

$\bigwedge (\mathbf {v} _{1},\cdots ,\mathbf {v} _{n})={\begin{vmatrix}v_{1}{}^{1}&\cdots &v_{1}{}^{n+1}\\\vdots &\ddots &\vdots \\v_{n}{}^{1}&\cdots &v_{n}{}^{n+1}\\\mathbf {e} _{1}&\cdots &\mathbf {e} _{n+1}\end{vmatrix}}$

Is there a specific name for this construct and are there any references about it? The other parts of that section talk about the wedge product, but I went to the article on wedge product doesn't really seem to talk about the above construct. Thanks, --131.215.166.126 (talk) 09:50, 29 March 2008 (UTC)

I wouldn't swear by it, but I think that is the wedge product, just expressed in a different way. It certainly satisfies the properties mentioned in the introduction to wedge product to do with linear independence. --Tango (talk) 15:22, 29 March 2008 (UTC)

It's not exactly the same thing as the wedge product. The wedge product of n vectors in a vector space E of dimension n+1 can be identified with a linear form on E once you choose an orientation on E. To recover a vector of E, you need in turn an Euclidean structure on E. If you make the "right choices", the vector you get after all these indentifications is indeed the same, but as you can see, it's not very canonical. I'd simply call this construct the cross product of n vectors. Bikasuishin (talk) 23:14, 29 March 2008 (UTC)

March 30

x + 2 = x + 2

What is the solution to this equation? Does it depend on what x represents? Certainly the solution is broader than "x is any number." It could also be any abstract quantity, any vector space - the space has to be closed under addition of a scalar 2. Maybe I'm just rambling--but what is the most precise and pedantic solution to such an equation? What is the solution to the equation x + yi = x + yi, where i is the imaginary unit? Is it sufficient to just say that x and y are elements of the real field? —Preceding unsigned comment added by 124.191.113.222 (talk) 07:13, 30 March 2008 (UTC)

While you could certainly find something more general if you wanted to, a relatively general setting for an equation like this is a unital ring. Essentially, you have a set with operations that behave like addition and multiplication, and there is a multiplicative identity (which we happen to call 1). You can then interpret 2 as 1 + 1. Since rings contain additive inverses (i.e. there is an element called -1 such that 1 + (-1) = 0, the additive identity), any element of the ring would satisfy x + 2 = x + 2. A more general setting would be any abelian group, but in this case there would be no element that would necessarily be called 2. The short story is, it depends on how you define "+" and "2". If you don't have a (right) additive inverse for 2, the equation might be non-trivial. 134.173.93.127 (talk) 07:28, 30 March 2008 (UTC)

Whenever x+2 is defined, x is a solution to x+2=x+2. Besides that, the equation contains no information about x. The equation x + yi = x + yi is satisfied whenever x + yi is defined. x and y can be complex numbers or quaternions or elements in any complex vector space. Bo Jacoby (talk) 08:26, 30 March 2008 (UTC).

I think I messed up in my question. The fact is we can say x is a real number for the first one; someone can object and say "it can also be a complex number;" someone can object and say, "more generally, it can be any quarternion." I really like the first answer. I thought the "unital ring" pretty much covers all bases - neat. But as for the second equation, I messed it up: if x + yi = a + bi and a = x and y = b, then I'm wondering what the solution for x and y is (probably just real numbers). —Preceding unsigned comment added by 124.191.113.222 (talk) 09:07, 30 March 2008 (UTC)

In common mathematical usage, we can take an equation like $x+2=x+2$ and assume it is clear from context that x should be a real number (or whatever), and then the set of solutions will be the set of all real numbers. But to deal with questions such as yours, we have to be more precise with what we do. We can't just take an equality in vacuum and assume that it itself says something about the realm we are discussing. A specification of where x comes from is required. Thus, you cannot treat a formula like $x=x+2$ as a valid equation. You need to say something like "$x=x+2$, where $x\in \mathbb {R}$", or "$x=x+2$, where $x\in \mathbb {Z} _{2}$" or "$x=x+2$, where x is a cardinal number". For the first case, there are no solution; For the other, there are. Implicit in my last example is that unital rings don't cover all settings where equations like those might make sense. Indeed, wherever you have addition and you have 2 you can discuss them.

The same goes for equations like $x+yi=2+3i$. You need to specify over which structure you are working; It needs to have addition, multiplication, 2, 3 and i. You can also require the unknowns to belong to some subset. Thus you can talk about "Solve $x+yi=2+3i$, where the operations are taken over the field of complex numbers, and x and y are real". -- Meni Rosenfeld (talk) 09:52, 30 March 2008 (UTC)

Indeed. Another way to say much the same thing is to ask what you mean by "+" and "2". We use the symbol "+" for lots of different things (addition of natural numbers, addition of integers, addition of rational number, addition of real numbers, addition of complex numbers, addition of quaternions, addition of integers modulo 3, addition of vectors, addition of matrices, etc, etc, etc), we also use "2" to mean lots of things (the natural number, the integer, the rational number, the real number, the complex number, the quaternions, the integer modulo 3, the integer modulo 4, etc, etc, etc). Once you narrow down what you mean by those symbols, it should be obvious what values x can meaningfully take. (I think there may a small element of confusion over whether it's meaningful to add an integer and a real number - it depends on how you define and construct them. The integers may be a subset of the reals, or they may just be isomorphic to a subset of the reals. In the former case, the addition makes sense, in the latter, I'm not sure it does. I may be being overly pedantic here to the point of actually being wrong...) --Tango (talk) 11:22, 30 March 2008 (UTC)

Assuming that at least the symbol "=" has its usual meaning, it is safe to say that any equation whose two sides are identical expressions – whether x + 2 = x + 2 or ξ∬η = ξ∬η – is either meaningless, or vacuously true for all unknowns in the domain of the operations. Which of the two, and what the domain of the operations is, depends on the context. The normal rule is that whoever produces the equation should also supply the context in which it is to be interpreted – with a lot of latitude if that context can reasonably be assumed because of standard notational conventions.  --Lambiam 20:15, 30 March 2008 (UTC)

You can do that, or you can take the domain to be absolutely everything and define meaningless statements to be false. That's the approach I was taking. --Tango (talk) 21:38, 30 March 2008 (UTC)

The domain of 'absolutely everything' is a moving target, leading to Russell's paradox. Bo Jacoby (talk) 13:23, 31 March 2008 (UTC).

Unless you use New Foundations, which avoids the paradox of having a universal set.  --Lambiam 00:01, 1 April 2008 (UTC)

Probability of selection - both with and without replacement

I took probability and statistics in college, but I can't find my book. I take one-half a tablet of a prescription medicine every day. If I shake out a whole tablet, I take 1/2 and put the other half back. If I shake out a 1/2 tablet, I take it. If the container starts out with n tablets, how many selections must be made before the probability of shaking out a 1/2 tablet reach 0.5? Or, in other words, Starting with a container of "n" green balls, select a ball from the container. If the ball is green, remove it and replace it with a red ball. If the ball is red, remove it without replacement. How many selections must be made until the probability of selecting a red ball reaches 0.5? —Preceding unsigned comment added by Johnerbes (talk • contribs) 20:04, 30 March 2008 (UTC)

This is a lightly disguised version of the birthday problem. Algebraist 20:50, 30 March 2008 (UTC)

If you say so... --Tango (talk) 21:39, 30 March 2008 (UTC)

Apparently I need to be clearer (alcohol always dulls my clarity-detectors). The n tablets in the container correspond to n days in a (modified) year. Taking a tablet out and breaking it in half corresponds to putting a new person in the room with a random birthday. Getting a tablet you've already broken corresponds to adding a new person with a birthday you've seen before. Thus all the calculations in the article I linked apply, replacing 365 with n. Algebraist 22:00, 30 March 2008 (UTC)

But doesn't removing the half-tablet after you take it out again mess that up? --Tango (talk) 22:22, 30 March 2008 (UTC)

Eg. For n=3, I've worked out (by just drawing a probability tree) this problem gives probabilities of 1/3, 4/9 and 11/18 after 1, 2 and 3 picks respectively. The birthday problem with a 3 day year gives probabilities of 0, 1/3, 7/9 with 1, 2 and 3 people in the room respectively. Where's the correspondence? (It is possible I just can't add up, of course.) --Tango (talk) 22:42, 30 March 2008 (UTC)

At least one of us is very confused about what the problem is. For n=3 and 1 pick, we're taking one tablet out of our container of three whole tablets, and want to know the probability that the tablet we've taken is a half-tablet. Surely this is zero? In general, removing the half-tablets doesn't matter since everything's decided once we've picked a half-tablet anyway. Algebraist 23:04, 30 March 2008 (UTC)

I think that's just an off-by-one error - I mean after one pick, the probability for the next pick is 1/3. A rather odd way to phrase it, now I think about it - just shift all my numbers by 1! However, more importantly, I think we've interpreted the problem differently. I think you interpreted it as how many times do you have to pick a tablet before the chance of having had at least one half tablet is greater than 50%. I interpreted it as how many times do you have to pick a tablet before the probability of the next pick being a half-tablet is greater than 50%. From the phrasing of the question, either interpretation seems possible, although yours does seem slightly more sensible (mine requires considering lots of different branches, each with a different probability after k picks, and taking a weighted average, which is a pretty strange thing to do). Yours is also an easier calculation (it is, indeed, just the birthday problem). --Tango (talk) 23:19, 30 March 2008 (UTC)

Ah, I see. Another downside of your version is that it corresponds to a variant that is not treated in the birthday problem article, and I am going to bed. Unless you or someone else feels like solving this version, I suggest awaiting clarification from the OP. Algebraist 23:30, 30 March 2008 (UTC)

And I read it as looking for the expected number of picks before the number of remaining halves exactly equals the number of remaining wholes. (If this is actually what Johnerbes meant, he may not have realized that this condition is in some cases not reached until the container is empty.) From a starting position of w wholes and h halves, my model is f(w,h)=0 if w=h, otherwise f(w,h)=1 + [w/(w+h)]*f(w-1,h+1) + [h/(w+h)]*f(w,h-1) and my answer is f(n,0).

For n=1 through 6 I get 2, 1, 11/3, 35/12, 40487/7500, 83612/16875, and then the numbers start getting ugly. We obviously need some clarification on the original question, but I notice that from a pretty simple definition I've generated a sequence that doesn't appear in OEIS. Is it interesting enough? --tcsetattr (talk / contribs) 00:49, 31 March 2008 (UTC)

While your interpretation is certainly an interesting question and it makes a little more sense than mine, the fact that the word "expected" (or any synonym, eg. average) does not appear in the statement of the problem suggests to me that it's not what the OP had in mind. I can only guess, though. We've managed to come up with 3 interpretations, all getting completely different answers - I think I'll go with Algebraist's plan and go to bed and hope the OP clarifies things. --Tango (talk) 01:15, 31 March 2008 (UTC)

OP here. Although the problem procedure is well defined, I'm not sure that the question is well defined or if I am even asking the right question. The problem is best defined as starting with a container of "n" green balls and selecting from the container until it is empty. If a green ball is selected, it is replaced with a red ball. If a red ball is selected, it is removed. As an example, if you start with 100 green balls, the container will be empty after exactly 200 selections, as 100 green will have been removed (and replaced by 100 red), and the 100 red will also have been removed. Starting with 100 green, the probability of selecting red on the second selection (following the first red replacement) is 1/100. Likewise, on the 200th selection, the probability of selecting red is 1/1. At one extreme, if the first 50 random selections were all green, the container would then contain 50 green and 50 red, so, the probability of selecting red must be less than 1/2 for the first 50 (or n/2) selections. I think that tcsetattr came closest to interpreting my question when he stated "And I read it as looking for the expected number of picks before the number of remaining halves exactly equals the number of remaining wholes." This may be a question that can't be answered. Let me redefine the question to ask: Starting with "n" green balls, what is the probability of selecting a red ball on selection "p". Example: Starting with "n" green balls, the probability of selecting red is 0 on the first pick, 1/n on the second pick and 1/1 on pick 2n. —Preceding unsigned comment added by Johnerbes (talk • contribs) 15:25, 31 March 2008 (UTC)

All right, I'm taking another shot at it. Given an initial state of w wholes and h halves, we want to know the probability that the pth pick will be a half. If we knew the state immediately before the p'th pick, it would be easy. But there are several possible states, depending on how many of the preceding picks were halves. We need to know the probability of each of those states before we can answer the main question. The general form of that question is: from a certain initial state, what is the probability that exactly r of the next p picks will be halves? Let that be the definition of a new function: f(w,h,p,r). A few base cases, and then the recursive rule:

f(w,h,p,r) = 0 if r<0 (you can't pick a negative number of half-pills)
f(w,h,p,r) = 0 if r>p (2 picks will never yield 3 halves)
f(w,h,0,0) = 1 (0 picks always yields 0 halves)
f(w,h,p,r) = [w/(w+h)]*f(w-1,h+1,p-1,r) + [h/(w+h)]*f(w,h-1,p-1,r-1)

Now we're ready for the main question, what is the probability that the p'th pick will be a half? We need to take the probability of each preceding state (given by f) and multiply it by the probability of picking a half-pill from that state, then add those together. The general description goes like this: We had w wholes and h halves. We made p-1 picks, and removed r halves. We must have also removed p-1-r wholes, replacing them with halves. We now have w-(p-1-r) wholes and h+(p-1-r)-r halves. The probability of reaching the current position was f(w,h,p-1,r). The probability of the next pick being a half is [h+(p-1-r)-r]/[w-(p-1-r)+h+(p-1-r)-r] = (h+p-1-2r)/(w+h-r). Put that all together and we get your requested probability function:

$g(w,h,p)=\sum _{r=0}^{p-1}{\frac {h+p-1-2r}{w+h-r}}f(w,h,p-1,r)$

Starting with 100 wholes and 0 halves, the probability that the p'th pick will be a half-pill is g(100,0,p). Actually computing these numbers takes a long time because the recursion in f is so heavy. After spending a few CPU-minutes on it, I see that the probability crosses 50% between p=90 (0.49917...) and p=91 (0.50257...). The 91st pick is the first one to have a greater than 50% chance to be a half-pill. Something more interesting happens at the end of the list though. g(100,0,199)=0.85782... meaning there's about a 14.22% probability that a single whole pill survives to the end, being broken only on the next to last day, counter-intuitive to me but confirmed by simulation.

I do not see a way to simplify the formula, and I don't see the resemblance to the birthday paradox either. Maybe I'm just not clever enough. --tcsetattr (talk / contribs) 08:10, 1 April 2008 (UTC)

The nature of mathematics

I just wanted to share my (probably quite ignorant) view on mathematics to establish what other people think about the matter.

It is my current belief that pure mathematics is a completely human construct that has no "reality" outside of our minds (on the assumption we are alone in the universe). My reasoning for this is that the basic concept of number, on which all mathematics is based, is surely a human construct. We regard, for instance, a tree on its own in the desert "one tree" even though this is a designation based on our unique perspective - its perhaps more accurately a collection of cells, or configuration of atoms and on all scales we can consider it to have a range of quantities aside from unity. The universe doesn't "count" trees.

The system of mathematics that arises from the numbers and operations we use to analyse our surroundings is, I consider, a "flawed" construct that has particular nuances and structure grown out of our requirements for the science. If we had no computing technology, for instance, would many areas of discrete mathematics be as rich as they currently are? Because we create rules that are suitable to our needs, it helps to explain the patterns we see so regularily in mathematics - perhaps there are no ODD perfect numbers because this result stems from the way we have constructed our traditional algebras?

It would be quite an interesting thing to consider that something as deep and beautiful as complex numbers are "encoded" in the set of rules we have created in the "game" of mathematics, waiting to be revealed at some point in time- suggesting that whilst mathematics may be a completely artifical construct, we can still investigate and "discover" parts of it that follow in direct conclusion. Any thoughts?

Damien Karras (talk) 20:36, 30 March 2008 (UTC)

While some would share your view that mathematics doesn't exist outside of our minds, I strongly disagree. I find it absurd that we would be able to invent anything that didn't exist already.

It is not correct that all of mathematics is based on the concept of number. Numbers are just one example of an abstract entity which mathematics can deal with. Regardless, I also don't agree that numbers don't exist in nature - take for example photons, to which there is no substructure as far as I know. Do you not agree that there is in nature the concept of "one photon, two photons, three photons..."? Even taking trees as an example, while there are definitely many ways and many scales in which you can investigate them - surely one of these scales is treating the tree as a whole unit and counting "one tree, two trees...". You don't need humans for that.

I am platonistic in the sense that I believe there is an abstract "world of ideas", exhibiting incomprehensible richness and containing any abstract idea there could possibly be (the word "be" here is tricky, since it can be associated with physical existence in our natural universe, but the world I describe is at a much higher level than it). When we set up axioms and definitions, we basically construct a window into this world. What we see through that window is the theorems that follow from them. These theorems exist without us - all we have done is decided to look at them.

I do agree that we have a choice of where to look and how to interpret what we see. The axioms we work with often are just carefully constructed windows through which we can see those ideas that we consider useful due to the specifics of our physical universe. If we chose to look elsewhere, we would see different things. But all those ideas exists in a way which transcend humanity, time, and our physical universe. -- Meni Rosenfeld (talk) 21:20, 30 March 2008 (UTC)

Well put. --Tango (talk) 21:42, 30 March 2008 (UTC)

I generally agree with Meni here, but I do think you touch on an interesting (and rather philosophical) point. Mathematics is a completely abstract construct that does not necessary have any actual impact on reality, but our observations have shown a great many times how mathematics has accurately predicted things in our universe. And that adds lots of credibility to the claim that mathematics does have a direct and significant relationship to our reality. The only real assumption that I find in mathematics (as a science) is the assumption that the human brain computes logic correctly, this is not necessarily true, but the amount of evidence supporting that assumption is enormous. A math-wiki (talk) 22:44, 30 March 2008 (UTC)

Maths on its own doesn't predict anything in our universe - maths combined with a model does. The fact that it works so often is a sign that our models are good, it's not a sign that maths is good. Modelling trees as natural numbers when you're planting them and chopping them down turns out to work rather well. It's not natural numbers that are good - they just are what they are - it's the model that's good. Had we modelled trees are integers modulo 3, we would have found out that it doesn't work well at all. That wouldn't mean that integers modulo 3 are bad, it's just a bad model. --Tango (talk) 22:49, 30 March 2008 (UTC)

On a tangent (I am far away from being a mathematician): There are some experiments with the cognitive capacities of animals which seem to imply that abstracting numbers is not solely a human achievement. I recently read about an avian (Alex, An African grey parrot researched by Irene Pepperberg, Ph.D. / Brandeis) who "invented" a sort of Zero concept. Alex, by the way was an acronym for Avian Leaning EXperiment. --Cookatoo.ergo.ZooM (talk) 23:40, 30 March 2008 (UTC)

I think I would tread a middle (or more nuanced) ground on this. I do think it is potentially fair to say that numbers don't exist outside of our minds. Let's take your example of photons. There may be photons, but to give them a number, to count them and say "three photons", that takes us (or some other like minded observer). I don't think nature in itself goes around counting photons. Once could argue that some event happens after a certain number of photons hit a target (or some such, physics isn't my field) and that nature was thus counting the photons, or energy levels, or somesuch. I don't see that as a natural conclusion though. Nature does what nature does, and the counting is our doing. I would suggest that the counting is something that occurs in our model of nature because that is how we understand nature; nature doesn't need to do any counting, but we do to keep track of things within our model. Let's be honest, counting is setting up a bijection, and I suspect nature doesn't bother with that, it just runs along as it is; the bijection is something we set up to hopefully elucidate those workings.

On the "world of ideas" and "what could possibly be", there is the difficult question (as you note) of deciding what can "be". Does logic proscribe what is possible? If so, what logic exactly? Do we accept things as possible even if they contradict the law of excluded middle? Constructivist mathematicians would, and a case can certainly be made that classical logic is rather too strict. So what else are we allowed to drop and still admit as possible? Can we accept something as able to "possibly be" if it contradicts the law of identity? Such a thing seems rather hard to conceive, but then so are many other things (like completed infinities); maybe we just need some examples and some practice. Given such freedom the "world of ideas" begins to seem rather too unconstrained and not just a little paradoxical (though what that means when we're willing to let go of logic is a question in itself). I tend to fall into the non-platonist camp here; it seems better to forgo the "existence" of this universal world of ideas, and simply accept those "local" worlds of ideas that we care to delineate/define ourselves -- thus local existence only within our minds, and no platonist absolute universal eternal existence.

All of that said, I don't think that means we "have a choice of how we see things" in the free sense that all choices are equal. We are, ultimately, products of our biology, and that has a great deal to say about how we see and categorise (which is ultimately the start of mathematics in many ways) things. We, and most anything else like us (e.g. aliens across the galaxy that have even roughly the same scale-wise view of things), are necessarily predisposed toward certain abstractions due to their relative efficacy. We see a tree not a collection of atoms because that's productive for us. We count because it is a fairly canonical abstraction from such discrete categorisation, and efficacious in our internal model of the world. This gives some sense of universality, but doesn't provide for some external eternal platonic truth; instead we merely have efficacy for our situation. -- Leland McInnes (talk) 13:58, 31 March 2008 (UTC)

Heisenberg said: If a question cannot be asked experimentally, you do not need to answer it theoretically. Can the question, whether pure mathematics is a completely human construct that has no "reality" outside of our minds, be asked experimentally? Bo Jacoby (talk) 00:45, 31 March 2008 (UTC).

That's science. We're talking about either maths or philosophy (I'm not sure which!), so falsifiability doesn't really apply. --Tango (talk) 01:08, 31 March 2008 (UTC)

Bo Jacoby: You may be interested in indispensability arguments in the philosophy of mathematics.--droptone (talk) 11:43, 31 March 2008 (UTC)

The question is definitely not one of mathematics. Only if it is a scientific question of the real world do we have a hope of finding out. Which observation or experiment will tell us? If no observation will tell us, then the question is pure philosophy. Mathematics, like physics and economy and biology, is historically inspired by the reality outside of our minds. Bo Jacoby (talk) 15:25, 31 March 2008 (UTC).

You can take ZF as an axiomatization of set theory and consider the Axiom of determinacy (AD). As is well known, AD is independent of ZF: it does not follow from the axioms of ZF. It is a quite plausible statement, though, so we may want to add it as a new axiom to ZF, giving a new system ZFD = ZF+AD. Likewise, the Axiom of choice is a plausible statement that is independent of ZF, so we may want to add it, giving a new system ZFC = ZF+AC. Why not add both and obtain an even more powerful system ZFCD = ZF+AC+AD?

Well, ZFCD is a bit too powerful. In ZFCD you can also prove that 2 + 2 = 5. So unless ZF itself is terribly wrong, AC and AD cannot both be true. In any mathematical universe at least one of the two is false. Does it make sense to ask which one is false? Can we hope to conceive of some argument that finally settles the issue, like "AD definitely holds, and therefore AC is false" (or the other way around)? Any such argument has to be based on a new axiom or deduction rule not in ZF, but then how can we argue that that axiom or rule is sound? We know we can only prove it if we posit yet another axiom, and so on, ad infinitum.

There are consistent geometries in which the parallel postulate holds, and others in which it does not hold, and by now it has been generally accepted that it is meaningless to ask which of these possibilities is "true" and which is "false". Both Euclidean and non-Euclidean geometries can serve as useful models of parts of observable reality, but that does not make them in some sense more true. We should likewise accept that there are different consistent set theories, none of which is more true in any meaningful sense than any of the others. Unlike for the divergent geometries, there are no useful applications of set theory as a model of part of observable reality where the difference is testable by an experiment.

If a conclusion is necessarily true in mathematics, it is because we have gamed the rules such that the conclusion is inevitable. Let me count: 1+1+1+1+1+1+1+1+1+1+1 = 11. Now I know how many trees I have in my garden. But do the rules of arithmetic also apply if you want to count the number of waves in the Atlantic ocean on April 1, 2008? These rules hold for discrete systems whose individuals have a permanent identity. To the best of our understanding, nothing in physical reality is fundamentally discrete and permanent.

Although we can communicate mathematical thought – an ability that has to be innate – to other human beings, and apply this successfully to models of reality, the mathematical content itself is a mental construction, and the assignment of discrete persistent identities to, say, "trees", is a mental act reflecting an ultimately subjective (even if intersubjective) way of imposing a never fully fitting order on reality.  --Lambiam 23:21, 31 March 2008 (UTC)

March 31

micro economic development

1.A chemical manufacturer processes two chemicals C1 and C2 in varying propositions to produce three products A,B and C.He wishes to produce at least 150 units of A,200 units of B and 60 units of C.Each tonne of C1 yields 3 of A,5 of B and 3 of C.If C1 costs $40 per tonne and C2 costs $50 per tonne

(a)Formulate the problem as a minimisation LP model (5mks)

(b)Formulate the inverse or dual of the primal formulation in (a)above (1mk)

(c)Express the dual in (b)above as standard LP model (5mks)

(d)set up the initial simplex tableau for the solution to this problem(do not solve) (4mks)

(e)using the graphical method,identify the feasibility region for the solution in the LP model.(5mks) —Preceding unsigned comment added by 41.220.120.202 (talk) 09:01, 31 March 2008 (UTC)

First off, the reference desk will not do your homework, which this very much appears to be. Second, this is economics, not mathematics... I suggest you take it to the humanities desk or someting, as it's fairly likely no one here will understand the terms. -mattbuck (Talk) 09:11, 31 March 2008 (UTC)

Well, this looks like a standard linear programming exercise to me, in which case this would be the right desk to post it, except for the fact that it is obviously a homework assignment. If the questioner can show evidence that they have spent some time working on the problem, or at least thinking about what the various terms mean, then we might be able to provide some help. Our articles on linear programming, dual problem, simplex algorithm and feasible region may be good places to start. Gandalf61 (talk) 09:35, 31 March 2008 (UTC)

Yeah, this is maths, but we need to know where the OP is stuck in order to help. We're certainly not going to provide a complete solution. --Tango (talk) 14:24, 31 March 2008 (UTC)

$\cos {\frac {\pi }{2^{n}}}$cos(π/2ⁿ)

I'm looking for the general (non-iterative) non-trigonometric expression for the exact trigonometric constants of the form: ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$, when n is natural (and is not given in advance). Do you know of any such general (non-iterative) non-trigonometric expression? (note that any exponential-expression-over-the-imaginaries is also excluded since it's trivially equivalent to a real-trigonometric expression).

• Let me explain: if we choose n=1 then the term ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ becomes "0", which is a simple (non-trigonometric) constant. If we choose n=2 then the term ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ becomes ${\begin{aligned}{\frac {1}{\sqrt {2}}}\end{aligned}}$, which is again a non-trigonometric expression. etc. etc. Generally, for every natural n, the term ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ becomes a non-trigonometric expression. However, when n is not given in advance, then the very expression ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$ per se - is a trigonometric expression. I'm looking for the general (non-iterative) non-trigonometric expression equivalent to ${\begin{aligned}\cos {\frac {\pi }{2^{n}}}\end{aligned}}$, when n is not given in advance. If not for the cosine - then for the sine or the tangent or the cotangent.

Eliko (talk) 10:38, 31 March 2008 (UTC)

Not sure that this fits your definition of "non-iterative", but best I can offer is

$\cos \left({\frac {\pi }{2^{n}}}\right)={\frac {\sqrt {2+{\sqrt {2+\cdots {\sqrt {2}}}}}}{2}},$

where the expression on the right hand side has n nested square roots. You can derive this by iteratively applying

$2\cos \left({\frac {x}{2}}\right)={\sqrt {2+2\cos x}}\,.$ Gandalf61 (talk) 11:11, 31 March 2008 (UTC)

This is an iterative expression. Eliko (talk) 11:20, 31 March 2008 (UTC)

It's derived iteratively, but I wouldn't say it was iterative itself. Perhaps it would help if you explained why you need it to be non-iterative, then we could better understand what you mean by the phrase. --Tango (talk) 14:27, 31 March 2008 (UTC)

Ok. I'll explain. Any solution for my original question could have been helpful in answering other questions, like: finding a non-trigonometric proof for the following algebraic, non-trigonometric claim: Every real interval includes a point x having a natural number n such that ${\begin{aligned}(x+i)^{n}\end{aligned}}$ is a real number. Note that this is an algebraic, non-trigonometric claim, so one may naturally expect that it may be proven by a non-trigonometric proof (note also that any proof based on an exponential-expression-over-the-imaginaries should also be excluded since such an expression is trivially equivalent to a real-trigonometric expression). Eliko (talk) 15:17, 1 April 2008 (UTC)

If you want the number in a familiar, that is algebraic, form, this is the best you're gonna get. Each constant is expressed as a simple, finite nesting of radicals. Pretty much by the nature of square roots, no smaller formula will do. Black Carrot (talk) 15:56, 31 March 2008 (UTC)

I'm looking for the closed formula if any exists. However, if you say now that "this is the best I'm gonna get" - then I have to believe you. Anyways, thank you for your reply. By the way, Any solution for my original question could have been helpful in answering other questions, like: finding a non-trigonometric proof for the following algebraic, non-trigonometric claim: Every real interval includes a point x having a natural number n such that ${\begin{aligned}(x+i)^{n}\end{aligned}}$ is a real number. Note that this is an algebraic, non-trigonometric claim, so one may naturally expect that it may be proven by a non-trigonometric proof (note also that any proof based on an exponential-expression-over-the-imaginaries should also be excluded since such an expression is trivially equivalent to a real-trigonometric expression). Eliko (talk) 15:17, 1 April 2008 (UTC)

The value of arg(x + i) as a function of real x is strictly monotonically decreasing. Let a and b be real numbers with a < b, and take integer n such that

n (arg(a+i) − arg(b+i) ≥ 2π.

This implies that there is a value x in the interval [a,b] such that

n arg(x+i) ≡ 0 (mod 2π).

Then Im (x+i)ⁿ = sin n arg(x+i) = 0.  --Lambiam 00:05, 2 April 2008 (UTC)

A sine snuck in at the end there - does that count as non-trigonometric? --Tango (talk) 00:18, 2 April 2008 (UTC)

Tango is correct: this is a trigonometric proof since it uses trigonometric expressions, such as Sin (and Arg). Eliko (talk) 01:12, 2 April 2008 (UTC)

I don't agree that arg is a trigonometric function. The modulus and argument of a complex number z ≠ 0 are simply real numbers M and A such z = Me^iA, where M ≥ 0 and A is only determined modulo 2π. If you want to avoid sin, you just need a few more steps: n arg(x+i) = 2kπ for some integral k. Putting M = mod(x+i) and A = arg(x+i), we have:

(x+i)ⁿ = (Me^iA)ⁿ = Mⁿe^inA = Mⁿe^i2kπ = Mⁿ,

which is real.  --Lambiam 10:03, 6 April 2008 (UTC)

Tango was correct. Look:

Arg is defined - either:

• by a trigonometric function, e.g. by stating that every real x satisfies: Cos(Arg(x)) + iSin(Arg(x)) = x; or:
• by an exponential-function-over-the-imaginaries, e.g. by stating that every real x satisfies: exp(iArg(x))=x.

However, since the beginning I've made it clear that I'm looking for a proof which does not assume the very existence of any trigonometric function, and since the beginning - I've also made it clear: "note that any exponential-expression-over-the-imaginaries - is also excluded, since it's trivially equivalent to a real-trigonometric expression".

When I wrote that "Arg" is a trigonometric function - I just meant that Arg assumes the existence of trigonometic functions, e.g. the existence of an exponential-function-over-the-imaginaries (which is trivially equivalent to a trigonometric function).

Let me explain why I insist on a non-trigonometric proof (hence - on a proof not invovling the exponential-function-over-the-imaginaries):

There are many theorems which can be proven by both trigonometric means and non-trigonometric means, e.g: the theorem stating that: ${\begin{aligned}(-1)^{n}\end{aligned}}$ is either 1 or -1 for any natural n. Now look at the following theorem: "Every real interval includes a point x having a natural number n such that ${\begin{aligned}(x+i)^{n}\end{aligned}}$ is a real number"; Note that the expression ${\begin{aligned}(x+i)^{n}\end{aligned}}$ looks "naive", simply algebraic, non-trigonometric, and involving no imaginary exponent, so one may naturally expect that also this theorem may be proven by a non-trigonometric proof, just as the first theorem, having the naive simply algebraic expression: ${\begin{aligned}(-1)^{n}\end{aligned}}$. The abstract question is now: whether any such non-trigonometric proof does exist.

Eliko (talk) 13:14, 6 April 2008 (UTC)

FWIW, here's a JSTOR paper On Defining the Sine and Cosine that defines sine and cosine using this dyadic subdivision.

Nbarth (email) (talk) 00:32, 2 April 2008 (UTC)

No definition of functions is a proof for the very existence of functions satisfying the definition.

Unfortunately, I couldn't enter the JSTOR paper. Does it suggest an iterative definition or a closed definition? Eliko (talk) 01:12, 2 April 2008 (UTC)

More precisely, it's a construction, of sine and cosine; it does prove that the construction works and thus that the functions exist.

It is an iterative definition, so it doesn't satisfy your desiderata.

As mentioned by other previously, it's unlikely that a closed form solution exists; I would consider your question about "a real interval containing a real solution" to be fundamentally analytic, (it's rather like the Intermediate Value Theorem), not algebraic, so I don't find it surprising that one must use analytic methods.

Nbarth (email) (talk) 07:22, 2 April 2008 (UTC)

More precisely, the statement:

Every real interval includes a point x having a natural number n such that $(x+i)^{n}$ is a real number.

is not purely algebraic: the use of interval means you're using the order structure on the reals, hence their topology. Thus the statement is fundamentally a topological / analytic statement, and one should expect an analytic proof. Indeed, as it's a statement about iterating a function of complex variables, it's properly a question in (discrete) complex dynamics. I'm no expert in this field, but if you wish to pursue this specific question further, that's where I would recommend turning.

Nbarth (email) (talk) 09:51, 2 April 2008 (UTC)

Yeah, it probably has to be based on an analytic proof, due to the interval. However, must its analytic proof be based on trigonometric constructions (or any equivalent analytic construction being sufficient for the very existence of any trigonometric function)? Note that the naive (simply algebraic) expression $(x+i)^{n}$ doesn't make us "recall" trigonometric properties, does it? this is not the case in the alternative definitions for the trigonometric functions you've found in JSTOR, is it? Anyways, thank you for your proposal of referring to: complex dynamics.

Eliko (talk) 13:44, 2 April 2008 (UTC)

$(x+i)^{n}$ is a complex exponential, so it is closely related to the behavior of the exponential function; De Moivre's formula is the most natural way of seeing why trig functions naturally come in. More generally, considering complex numbers in polar coordinates is very natural, as seen in the multiplication of complex numbers and Euler's formula.

Thus it's not surprising that when looking at multiplying complex numbers, you get trig functions.

Nbarth (email) (talk) 23:27, 16 April 2008 (UTC)

Also, for the question about $(x+i)^{n}$, you could express it in terms of binomial coefficients (the imaginary part is a sum of the terms with odd powers of $i$), so it's possible that there's a combinatorial proof.

Nbarth (email) (talk) 23:27, 16 April 2008 (UTC)

Further, the geometric POV gives a stronger result: for any given interval $[a,b]$, for all n sufficiently large ($n\geq N$: not just a single n), $S=\left\{(x+i)^{n}\mid x\in [a,b]\right\}$ contains a real number — the general bound is $N={\frac {1}{\arg(a+i)-\arg(b+i)}}$: since exponentiation multiplies argument, after this point S wraps all the way around the circle (contains points of all arguments).

Nbarth (email) (talk) 23:27, 16 April 2008 (UTC)

• You used the "arg" function, as User talk:Lambiam did (on this page), but (as I explained to him) the "arg" function mustn't be used here. Why? See above (my explanation to Lambiam), and see also my next comment.
• As I've indicated above, I'm not looking for the trigonometric proof, but rather for an algebraic (non-analytic non-trigonometric) proof. e.g. for a combinatorial proof. I still don't know whether any such combinatorial proof exists (after having expressed $(x+i)^{n}$ in terms of binomial coefficients).
• I could certainly find a non-trigonometric proof if I could find a series $(a_{n})$ satisfying the following three properties:

1. The series $(a_{n})$ converges to zero (e.g. if it were the series $2^{-n}$, but not necessarily).
2. The series ${\displaystyle }$ is expressible algebraically (e.g. if it were the series $2^{-n}$, but not necessarily).
3. The series $\cos(a_{n})$ is expressible algebraically (e.g. if $(a_{n})$ were the series $\arccos(1/n)$, but not necessarily).

If you can find such a series, which satisfies all of the preceding three properties, I'll be very grateful.

Eliko (talk) 08:23, 17 April 2008 (UTC)

A metrisation theorem

I want to prove the following version of the Urysohn metrization theorem: A normal second-countable Hausdorff space is metrisable. I'd be grateful for a cross-check of this proof.

Let X be such a space. Let $\{U_{n}\}$ be a base for the topology. Let $(U_{n_{i}},U_{m_{i}})$ be an enumeration of all pairs of elements in this base such that ${\overline {U}}_{n_{i}}\subset U_{m_{i}}.$ For each i let $f_{i}$ be a continuous function satisfying $0\leq f_{i}\leq 1$ and such that $f_{i}$ is 0 on ${\overline {U}}_{n_{i}}$ and 1 on the complement of $U_{m_{i}}.$ Let

$d(x,y)=\sum _{i=1}^{\infty }{\frac {1}{2^{i}}}|f_{i}(x)-f_{i}(y)|.$

We observe that given $x\in X$ and some open set $U_{m}$ in the base containing x, there exists another set $U_{n}$ in the base such that $x\in U_{n}\subset {\overline {U}}_{n}\subset U_{m}.$

Then the following assertions hold:

1. d is a metric on X. We let $X_{d}$ denote X viewed with the metric topology.

2. All open subsets of X are open in $X_{d}.$ Specifically, given an open set U in the base and a point x in U, there exists $\epsilon >0$ such that the ball $B_{\epsilon }(x)\subset U.$

3. All open subsets of $X_{d}$ are open in X. Specifically, given B open in $X_{d}$ and x in B, there exists an open subset U of X such that $x\in U\subset B.$

For (1), it is clear that d is nonnegative symmetric, and the triangle inequality is also immediate. Now if $x\neq y$ we can find an open set $U_{m}$ containing x and not containing y. Applying the observation above we can then find $U_{n}$ such that $x\in U_{n}\subset {\overline {U}}_{n}\subset U_{m}.$ Thus in the sum for d there will be some i for which $|f_{i}(x)-f_{i}(y)|=1$, so that $d(x,y)>0.$ Thus d is a metric on X.

For (2), find $U_{n}$ with $x\in U_{n}\subset {\overline {U}}_{n}\subset U$ and consider the corresponding i in the sum for d. If $y\not \in U$ then $|f_{i}(x)-f_{i}(y)|=1,$ and hence $d(x,y)\geq 1/2^{i}.$

For (3), let $\epsilon >0.$ For each i there exists an open set $V_{i}$ containing x such that $f_{i}(V_{i})\subset B_{\epsilon }(f_{i}(x)).$ Hence for $y\in V_{i}$ we have $|f_{i}(x)-f_{i}(y)|<\epsilon .$ The intersection $V_{1}\cap \cdots \cap V_{n}$ is an open set containing x. For y in this intersection, we have

$d(x,y)\leq \sum _{i=1}^{n}{\frac {\epsilon }{2^{i}}}+\sum _{i=n+1}^{\infty }{\frac {1}{2^{i}}}|f_{i}(x)-f_{i}(y)|.$

It follows that we can make $d(x,y)$ as small as we wish by taking n sufficiently large. That is, given $\delta >0$ we can find an open subset V of X containing x with $V\subset B_{\delta }(x).$  — merge 13:31, 31 March 2008 (UTC)

I can't see any problems. Good work! Algebraist 14:26, 31 March 2008 (UTC)

Thank you very much!  — merge 14:31, 31 March 2008 (UTC)

I find (2) a little unclear; I find your notation is a little confusing. You've ordered the base as $U_{n}$ and then ordered pairs of bases. So to each base element $U_{n}$ there exist many (probably countably infinite) pairs $(U_{n_{i}},U_{m_{i}})$. Hence the phrase "consider the corresponding $i$" is ambiguous; there are many such $i$'s. Furthermore, if you're proving that "there exists $\epsilon$ so that $B(x,\epsilon )\subset U$," there should be some explicit connection between $\epsilon$ and your ${\frac {1}{2^{i}}}$ (and note that the ${\frac {1}{2^{i}}}$ isn't well-defined until you tell us how you choose $i$). The ideas of the proof all seem ok; it's just that I find the notation a little confusing. SmaleDuffin (talk) 17:34, 31 March 2008 (UTC)

The 'corresponding i' is the i corresponding to the pair $U_{n}\subset U$. Algebraist 17:48, 31 March 2008 (UTC)

Ah, ok. Thank you. So: Let $U$ be a base set; then there exists numbers $n_{i}$ and $m_{i}$ so that $U=U_{m_{i}}$ and $x\in U_{n_{i}}\subset {\overline {U_{n_{i}}}}\subset U_{m_{i}}$. Are there other indices that work as well? Do there exist numbers $n_{j}$ and $m_{j}$ so that $U=U_{m_{j}}$ and $x\in U_{n_{j}}\subset {\overline {U_{n_{j}}}}\subset U_{m_{j}}$ with $i\neq j$? If this is possible, consider the following construction. Let, for each $k$, $U_{m_{k}}=U$ and $x\in U_{n_{k}}\subset {\overline {U_{n_{k}}}}\subset U_{m_{k}}$, and let $y_{k}$ be a point not in $U$. Then $d(x,y_{k})>{\frac {1}{2^{k}}}$, and so $\lim _{k\to 0}d(x,y_{k})=0$. This certainly doesn't prove that $x$ is a limit point of points outside of $U$, but the argument provided doesn't seem to rule this out. There should be some argument (probably involving $\epsilon$) that shows that the infimum of $d(x,y_{k})$ is strictly greater than zero. Of course, it's been a long while since I've thought about these sorts of things, so I could very well be missing something obvious. SmaleDuffin (talk) 19:08, 31 March 2008 (UTC)

Huh? $\lim _{k\to 0}d(x,y_{k})=0$ doesn't appear to make sense (k is ranging over the natural numbers, yes?) and if you meant $\lim _{k\to \infty }d(x,y_{k})=0$ then it doesn't follow from $d(x,y_{k})>{\frac {1}{2^{k}}}$ and isn't true, either. All you need to do (given x and U) is pick an i such that x is in $U_{n_{i}}$ and $U_{m_{i}}=U$, and then by construction, if z is in ${\overline {U}}_{n_{i}}$ and y is outside U, then f_i(z)=0 and f_i(y)=1, so d(z,y) is at least 1/2ⁱ. I really don't see what the problem is here. Algebraist 19:33, 31 March 2008 (UTC)

Never Mind; I figured out where I was confused. Thanks. SmaleDuffin (talk) 20:35, 31 March 2008 (UTC)

microeconomic development

A chemical manufacturer processes two chemicals C1 and C2 in varying propositions to produce three products A,B and C.He wishes to produce at least 150 units of A,200 units of B and 60 units of C.Each tonne of C1 yields 3 of A,5 of B and 3 of C.If C1 costs $40 per tonne and C2 costs $50 per tonne

(a)Formulate the problem as a minimisation LP model

(b)Formulate the inverse or dual of the primal formulation in (a)above

(c)Express the dual in (b)above as standard LP model

(d)set up the initial simplex tableau for the solution to this problem(do not solve)

(e)using the graphical method,identify the feasibility region for the solution in the LP model. —Preceding unsigned comment added by 41.220.120.202 (talk) 09:01, 31 March 2008 (UTm

C1 C2

 A  B   C

A-150
B-200
C-60
SUBJECT TO
C1=3A+5B+3C
C2=5A+5B+C

C1+C2>OR =150A+200B+60C

C1=3A+5B+3C
C2=5A+5B+C

I think that you accidentally omitted one clause from the original assignment, namely:

Each tonne of C2 yields 5 units of A, 5 of B, and 1 of C.

Your model doesn't look right. I assume that by "A" you mean the number of units of A, and by "C1" the number of tonnes of C1. It is a bit confusing to let A, B etcetera stand both for commodities and for quantities, but let's stick with that for now. You express C1 and C2 in terms of A, B and C, but it should be the other way around:

A = 3C1 + 5C2

etcetera. Or are these equations already meant to be the dual problem?

I hope this will help you to go further.  --Lambiam 23:47, 31 March 2008 (UTC)

Following Lambian's remark on a missing premise, the primal could be stated as

${\underset {c_{1},c_{2}}{\min }}40c_{1}+50c_{2}$

...Subject to

$(A=)3c_{1}+5c_{2}\geq 150$

etc.

Within this formulation, there are 2 variables and 3 constraints.

The dual should be something like

${\underset {A,B,C}{\max }}150A+200B+60C$

...Subject to (something like)

$f(A,B,C)\leq 40$

$f(A,B,C)\leq 50$

(Three variables and two constraints).Pallida  Mors 18:02, 1 April 2008 (UTC)

April 1

Socks

You have a basket containing a mix of short socks and long socks. How many socks do you need to pull out to be certain of having a matching pair? --67.185.172.158 (talk) 05:24, 1 April 2008 (UTC)

Pulling out three socks would be enough!A Real Kaiser (talk) 07:10, 1 April 2008 (UTC)

If you like long words, see pigeonhole principle. --Tango (talk) 09:00, 1 April 2008 (UTC)

Well, assuming there are only short and long socks, and that all short and long socks are identical, and that there is no difference between left and right sock... -mattbuck (Talk) 01:17, 2 April 2008 (UTC)

This reminds me of a similair question but with gloves instead of socks. Then is a matched two gloves of the same hand? Or a pair of one of each left and right? Mensa had this in a test and the answer was the probability for pullign out two identical, not one of each left and right....--Dacium (talk) 02:11, 3 April 2008 (UTC)

April 2

Infinities

I know infinity is pretty big, and you cant get bigger. But what if you raised infinity to the power infinity? What happens then/ —Preceding unsigned comment added by 79.76.250.241 (talk) 00:52, 2 April 2008 (UTC)

Well... there are different infinities. It's also rather deceptive. For instance, say we start with the counting numbers (1,2,3,etc). Now, it's quite clear that these are a strict subset of the integers (that is all whole numbers, positive or negative). But we can label the integers by the counting numbers. Let teh 1st integer be 0, the 2nd be 1, the 3rd be -1, the 4th be 2, the 5th be -2, etc. Then it is clear that we can just keep on counting to infinity, and that thus they have the same cardinality - essentially, there are the same number. We can do the same with the rationals. However, we cannot do this for the real numbers - see Cantor's diagonalization argument. This infinity is much bigger. Welcome to maths. -mattbuck (Talk) 01:14, 2 April 2008 (UTC)

Your question has already been treated by Georg Cantor in the 19th century. Generally, the concept of "infinity" has two different meanings: infinite ordinal number, and infinite cardinal number. If you mean "infinite ordinal number", then what you call "infinity to the power infinity" is traditionally symbolized: $\omega ^{\omega }$. See here for a deeper discussion on $\omega ^{\omega }$, as well as on its full meaning. A parallel concept of the desired power exists also for the infinite cardinal number. Eliko (talk) 01:28, 2 April 2008 (UTC)

I think it's a bit misleading to say that infinity has two meanings, since that suggests it has only two meanings. For instance, the phrase "infinity to the infinity" makes sense as a limit, in which context it's unambiguously infinity. On the other hand, in the nonstandard reals infinity to the infinity is different from both infinity and any interpretation in terms of cardinality or ordinality of sets. Black Carrot (talk) 08:45, 2 April 2008 (UTC)

When I wrote "two meanings" I meant "at least two meanings". Anyway, thank you for your comment. Eliko (talk) 13:03, 2 April 2008 (UTC)

Sorry, Kakarot, but that wouldn't be "infinity to the infinity, but rather "infinity to the infinity'th" or "[...] infinity'th power." Or you could be right as you stand. I don't really know. Anyway, here's a picture of what's being dealed with here:

${\infty }^{\infty }$ flaminglawyerc 21:32, 2 April 2008 (UTC)

"infinity to the infinity" and "infinity to the infinity'th (power)" sound like different ways of saying exactly the same thing to me. What distinction are you trying to draw? --Tango (talk) 00:42, 3 April 2008 (UTC)

It is linguistic nitpicking: just like we don't vocalize the written expression 3³ as "three to the three", the fiery jurist objects to voicing ∞^∞ as "infinity to the infinity". We do say "three to the power three". The intention of Black Carrot was clear enough. To the nitpicker: the past participle of the verb to deal is dealt. More importantly, recasting the question as ∞^∞ completely ignores the earlier contributions pointing out that there are several notions of "infinity" in mathematics, for most of which the symbol ∞ is not appropriate.

In how many ways can we make a sequence of length 6 if each element of the sequence is picked from a set with 10 members (for example the set of the numbers 0 through 9)? The answer is 10⁶. In general, there are N^L different sequences of length L if each element belongs to a given set of size N. So an interpretation of "infinity to the power infinity" is: the number of different sequences of infinite length, each of whose elements belongs to a given set of infinite size.

In this interpretation we are talking about cardinalities, which measure set sizes. The simplest kind of infinite sequence is one whose elements can be indexed by natural numbers, and the simplest infinite set is again that of the natural numbers. So we are talking about the size of the set of infinite integer sequences with only nonnegative elements. The "size" (cardinality) of the natural numbers is denoted by ℵ₀ (pronounced like "Aleph 0"), the smallest infinite cardinal. So then the question can be formulated as: how big is ℵ₀^ℵ₀?

For finite sequences, a somewhat surprising result (surprising until you get used to it) is that the length does not make a difference:

ℵ₀² = ℵ₀³ = ℵ₀⁴ = ... = ℵ₀.

Once you are used to this, the next surprising thing is that in the limit it does make a difference:

ℵ₀^ℵ₀   = ℵ₁   > ℵ₀.

ℵ₁   ℵ₀^ℵ₀ is also the size of the set of real numbers, and is therefore known as the cardinality of the continuum. It is also denoted by ${\mathfrak {c}}$ or ℶ₁. --Lambiam 08:32, 3 April 2008 (UTC)

Did it become standard to assume the continuum hypothesis when I wasn't paying attention? Matthew Auger (talk) 21:09, 3 April 2008 (UTC)

While we don't say "three to the three", we do say "e to the x", not "e to the x-th". Both ways are clearly used in some contexts. I think in the context of infinity, "infinity to the infinity" is the clearest way to phrase it. Exactly what it means depends very much on what you mean by "infinity". (And Matthew is right - you probably shouldn't assume the continuum hypothesis without saying you're doing so.) --Tango (talk) 21:17, 3 April 2008 (UTC)

Goody! I like linguistic quibbles. My rationalization would be that adding a -th suffix assumes we're dealing with the ordinal properties of infinity, which I was explicitly trying to move away from. I was treating infinity as an ideal element and exponentiation as a notational convenience, referring to an abstract binary operation. So, it was quite acceptable to phrase it that way. Black Carrot (talk) 09:39, 4 April 2008 (UTC)

Prime ideals in a ring

I'm struggling to prove something about prime ideals for an assignment: Given R, a nontrivial commutative ring with unity, use Zorn's lemma to show that the set of prime ideals has minimal elements with respect to inclusion, and that any prime ideal contains at least one such minimal prime ideal.

I've managed to show the existence of minimal prime ideals, but I'm struggling to figure out how I can show that every prime ideal contains a minimal prime ideal. I tried doing it by contradiction (assume that every prime ideal contains a smaller, nontrivial prime ideal) but I can't see any problems with that assumption that would give me a contradiction. The "turtles all the way down" notion seems a little counterintuitive, but not actually contradictory. Can anybody give me some pointers? Thanks, Maelin (Talk | Contribs) 01:34, 2 April 2008 (UTC)

Well, if R is also an integral domain, (0) is a minimal prime ideal. So that case is simple, and then assume you have zero divisors which gives you something else to work with, that may or may not be useful with what else you have. GromXXVII (talk) 11:03, 2 April 2008 (UTC)

What about applying Zorn’s lemma kind of backwards. That is, viewing the partial ordering as containment instead of inclusion. Then if there is an “upper bound” which in this case is a prime ideal contained in all prime ideals, Zorn’s lemma gives you precisely that every prime ideal contains a “maximal” prime ideal, which in this case is a minimal prime ideal. I’m not sure how to show that there is such an “upper bound” though, or if it even exists, because in general the intersection of prime ideals is not a prime ideal. GromXXVII (talk) 11:33, 2 April 2008 (UTC)

The intersection of a decreasing chain of prime ideals is prime, though, so it's indeed an application of Zorn's lemma. Bikasuishin (talk) 12:22, 2 April 2008 (UTC)

As a matter of interest, how did you manage to prove the existence of minimal prime ideals without proving the second part? To me they seem to both require essentially the same application of Zorn. Algebraist 14:54, 2 April 2008 (UTC)

Tennis superiority

This should be slightly more complicated than it seems.

A tennis tournament has every participant playing every other participant. No ties are allowed. A player A is superior if for every other player B, either A beats B or there is a third player C such that A beats C and C beats B. Prove that if there is only one superior player then he or she beats every other player. Reywas92^Talk 01:37, 2 April 2008 (UTC)

Here are a few steps towards the goal: Assume that A is the only superior player, but A' beats A. Then for any player B in the tournament, either A' beats A beats B, or there is another player C(B) such that A' beats A beats C(B) beats B (the (B) to explicitly state that the choice of C depends on the choice of B). From there, if you can prove that A' must be superior, you reach a contradition. Confusing Manifestation(Say hi!) 03:12, 2 April 2008 (UTC)

Algebra

Okay, I help with this question here, but I really don't get it.

Given any three natural numbers, show that there are two of them, a and b, that a^3b-b^3a is divisible by thirty. I only understood to the even/odd part. Thanks, 76.248.244.196 (talk) 01:42, 2 April 2008 (UTC)

a^3 b - b^3 a = a b (a - b) (a + b), and to be divisible by 30 you need it to have factors of 2, 3 and 5. You say you understand the odd/even bit, so let's move on to divisible by 3. When Keenan talked about modulo arithmetic, what he means is talking about numbers in terms of their remainder when divided by various numbers. In the case of "modulo 2", there are two types of number - odd and even. In the case of "modulo 3", there are three - multiples of three, numbers one more than a multiple of three, and numbers one less than a multiple of three. We can write these as 3n, 3n + 1 and 3n - 1 respectively, where n is an arbitrary whole number.

This may be the long way of doing it, but consider all three cases for each of a and b, and what they mean for the factors a, b, a - b and a + b. For example:

If a = 3n + 1 and b = 3m - 1, then neither is divisible by 3, but a + b = 3n + 1 + 3m - 1 = 3n + 3m = 3(n + m), which is. If you go through all nine combinations, you will eventually see that in each case, at least one of a, b, a - b and a + b will be divisible by 3.

The "divisible by 5" part is what requires the pigeonhole principle. Again, for the sake of enlightenment, go the long way and consider all of the possibilities modulo 5 (so let a = 5n, 5n + 1, 5n + 2, 5n + 3 and 5n + 4, or if you're a fan of symmetry, 5n, 5n + 1, 5n + 2, 5n - 1 and 5n - 2 is equivalent). You'll see that there only some combinations are guaranteed to make one of the factors divisible by 5, so what you have to prove is that, given three numbers, there are always 2 that will fall into one of those combinations (for example, if at least one of them is divisible by 5 you're home and hosed). Hope that helps, Confusing Manifestation(Say hi!) 02:57, 2 April 2008 (UTC)

Thanks a lot! I fully understand now, but it will be a little difficult to put all the possibilites in a proof. Thanks, 76.248.244.196 (talk) 20:40, 2 April 2008 (UTC)

Listing all possible cases is considered a valid, if slightly ugly, method of proof. For an extreme case, check out the Four color theorem. Also, like I said, my method goes the long way but there are several shortcuts along the way depending on how comfortable you get with the modular arithmetic - for a start, you can simplify things by simply stating at various points "modulo 5" and then drop out all the 5n terms. And the 5 x 5 table in the last step comes down to a single statement involving one trivial case and one where there are two possible categories for the pigeonhole principle. Confusing Manifestation(Say hi!) 22:59, 2 April 2008 (UTC)

For showing that 5 can be made a factor by choosing a and b: consider the remainders modulo 5 of the three numbers, giving three numbers from the set {0,1,2,3,4}. If any of the three is 0, we are done (because of the factor ab). If any two are the same, we are also done (because of the factor a−b). Otherwise, we have a subset of three different numbers of the set {1,2,3,4}; only one of those four is missing. We need to show that we can pick a & b such that the final factor a+b is 5. If the missing fourth number is 1 or 4, take those giving remainders 2 & 3, otherwise take those giving 1 & 4. To complete the proof, show that for any pair of numbers a & b the number is a multiple of 6.  --Lambiam 09:09, 3 April 2008 (UTC)

Geometry

In triangle ABC, draw angle bisectors AD and CE, where D is on BC and E is on AB. If angle B is 60 degrees, show that AC=CD+AE.

I've figure out that if the intersection of AD and CE is F, then <CFA and <EFD are 120 degrees and <DFC and <EFA are 60 degrees, but I'm not sure what's next. I'm reposting from here, but I'm not sure how to use the law of cosines on it, as I don't know any sides. Thanks, 76.248.244.196 (talk) 02:21, 2 April 2008 (UTC)

Also draw the third bisector. What point does it go through (other than B)? Figure out the sizes of some new angles that have appeared now. Do you spot any congruent triangles?  --Lambiam 09:21, 3 April 2008 (UTC)

Distribution of a maximum of multivariate normal

Does anyone know (or know a reference) on the distribution of max(X) if X is a multivariate normal random vector with various means (can be assumed to have constant, ie $\sigma ^{2}I$, variance). Simulation tells me it is gamma-like, but I haven't been able to prove it. Any ideas? Thanks, --TeaDrinker (talk) 02:28, 2 April 2008 (UTC)

Unless I have made a mistake or misunderstood your question, this is a straightforward calculation:

$F(x)=P(\max X\leq x)=P\left(\bigwedge _{i=1}^{n}X_{i}\leq x\right)=\prod _{i=1}^{n}P(X_{i}\leq x)=\prod _{i=1}^{n}\Phi \left({\frac {x-\mu _{i}}{\sigma }}\right)$

$f(x)=\sum _{i=1}^{n}{\frac {1}{\sigma }}\Phi '\left({\frac {x-\mu _{i}}{\sigma }}\right)\prod _{j\neq i}\Phi \left({\frac {x-\mu _{i}}{\sigma }}\right)$

It's possible that this is harder if the covariance matrix is not diagonal. -- Meni Rosenfeld (talk) 15:08, 2 April 2008 (UTC)

It is more complicated if the components are not independent. By the way, a standard notation for the pdf Φ' is φ.  --Lambiam 09:27, 3 April 2008 (UTC)

Thanks! I tried this approach, but didn't see any slick simplifications (which I was hoping for--the iid case, of course simplifies nicely, but perhaps this one is just irreducibly ugly). Thanks again, --TeaDrinker (talk) 16:50, 2 April 2008 (UTC)

Square digit chain

As in this problem. Why can't there be other loops? Can someone show me a proof that all possible chains will end in 1 or 89? 70.162.25.53 (talk) 03:43, 2 April 2008 (UTC)

It is not overly clever, but note that if you start with a four digit number, the largest number you can get is 324, and in fact, any number with four or more digits must decrease (and does so very rapidly until there are only three digits). Thus every number eventually becomes a number smaller than 325. It is therefore sufficient to simply check all the numbers 2 through 324 to verify that 1 and 89 are the only possible outcomes. It is not my area of expertise but I don't see a slick solution. --TeaDrinker (talk) 05:03, 2 April 2008 (UTC)

Actually, the bound can be brought down to 162. To see why, consider any x<324 that largest digit square sum you will find is 166, and then consider any number less than 166, that one that will produce the largest digit sum is 99, not 159 81+81=162, vs 1+25+81=26+81=107. so with that you can refine the bound to 162. Another hint, you can replace 89, by any of the other numbers after the first 89 in the sequence they provided, because those numbers are part of the loop. Additionally, any number who's sum converges to any of those number in that loop will go to 89, and every other number in that loop. A math-wiki (talk) 06:30, 2 April 2008 (UTC)

Going beyong squares, we can generalise this problem to iterating the sum of any power of a number's digits, and indeed generalise further to any base. Using similar arguments to the above, we can show that there is always some N such that the set of integers {1...N} is mapped to a subset of itself, and any integer greater than N is mapped to a strictly smaller integer. So all chains eventually enter the region {1...N} and once inside they never leave it. The long-term behaviour of any chain must therefore be to settle into a loop or to reach a fixed point (which is only a loop of length 1). Loops and fixed points can then be found by a finite number of trials.

Fixed points of these "powers of digits" functions are called narcissistic numbers. Considering sums of cubes of digits, for example, there are four fixed points (as well as the trivial fixed point 1), two loops of length 2, and two loops of length 3 [1]. Gandalf61 (talk) 11:21, 2 April 2008 (UTC)

Another loop is 0 → 0 (or is 0 no longer a number)?  --Lambiam 09:39, 3 April 2008 (UTC)

But no other number has a sequence that reaches zero, so it's kind of useless. — Kieff | Talk 10:18, 3 April 2008 (UTC)

Must true facts in mathmatics are "kind of useless". The issue was that the statement "EVERY starting number will eventually arrive at 1 or 89" is false. Unless I made a mistake, the answer to the question "How many starting numbers below ten million will arrive at 89?", counting only nonnegative integers as starting numbers, is 8581146. This is completely useless knowledge, and only an idiot or a mathematician would pose that question.  --Lambiam 11:29, 3 April 2008 (UTC)

So the ratio of numbers that arrive at 89 to numbers that arrive at 1 is approximately 8:1, and the 89 loop also has length 8. That's interesting ... Gandalf61 (talk) 22:59, 3 April 2008 (UTC)

The ratio is much closer to 6:1.  --Lambiam 23:27, 3 April 2008 (UTC)

Ah, yes, 6:1. I wonder if that ratio converges as the upper limit increases ... Gandalf61 (talk) 08:31, 4 April 2008 (UTC)

Apparently, the behaviour of the ratio is rather erratic. See OEIS:A068571 which gives the number of happy numbers (numbers which arrive at 1) less than or equal to 10ⁿ for n=1 to 21. An optimist might see signs of convergence to some proportion around 12%, but I wouldn't put money on it ! Gandalf61 (talk) 10:36, 4 April 2008 (UTC)

Area and Volume Ratios

I'm having trouble understanding Area and Volume ratios as the teacher didn't fully explain it at the time... does anyone have a basic summary of it? —Preceding unsigned comment added by Devol4 (talk • contribs) 06:46, 2 April 2008 (UTC)

A given 3 dimensional shape will have a surface area and a volume. The surface area to volume ratio is the surface area divided by the volume. That ratio is smallest for a fixed volume or surface area if the shape is a sphere. Does any of that help? If not, what specifically is confusing you? --Tango (talk) 12:39, 2 April 2008 (UTC)

Could it be about the fact that scaling a figure by a factor k will scale its area by a factor k² and its volume by a factor of k³ ? -- Xedi (talk) 19:18, 2 April 2008 (UTC)

For a simple example, if you have a cube whose sides have a length of 1cm, then its surface area A = 6cm², and its volume V = 1cm³. Then the ratio A/V = 6cm²/1cm³ = 6cm⁻¹. If instead we take a block of dimensions , A = 7cm² while V = 1cm³ as before, so now A/V = 7cm⁻¹.

We also have an article Surface area to volume ratio, which however is not about the maths but about the significance in physical chemistry and biology.  --Lambiam 09:58, 3 April 2008 (UTC)

In 5-Card Draw Poker, what are the odds of a royal flush without a wild card when 2 are in play?

A friend and I were playing 5-card draw Poker, using 2 joker cards as wild cards. This was in addition to the standard 52-card deck. We each drew 3 cards on our turns, and I ended up with a royal flush (A,K,Q,J,10 all of the same suit), without a wild card. I don't recall what my friend ended up with, so I hope it doesn't matter much. I do remember that he didn't get a wild card either. I know the odds of a royal flush are about 650,000 to 1, but that's without a draw, in which case the odds are higher. So, i've been wondering ever since what odds I overcame to draw this hand without getting one of the 2 wild cards that were in play. Thanks so much to whoever can answer this for me. MoeJade (talk) 09:22, 2 April 2008 (UTC)

Because of the draw, there isn't a unique answer - it depends on your decision making process. I would think the decision making process that would give the highest odds of a royal flush (but would probably not be a good strategy in practise) would be to keep only cards 10 or higher, and only cards of the suit that would leave you with the most cards. If we assume that strategy, we can calculate the odds. It's not a quick calculation though, and I don't have time right now. If no-one else tackles it before I find time, I'll give it a go. My method would be to calculate it separately in the cases of drawing no cards, 1 card, up to 5 cards, and weight each of them by the chance of that case occurring. --Tango (talk) 12:32, 2 April 2008 (UTC)

I was afraid of that. Unfortunately, I have only a basic understanding of statistics, so I quickly became lost trying to determine the odds. Don't spend too much time on this problem; I was hoping it wouldn't be too complicated, but if it is, just give me a ballpark range if you can. MoeJade (talk) 14:41, 2 April 2008 (UTC)

How does "very unlikely" suit you? While it's going to be more likely than getting it straight away, it will still be very unlikely. We can find a lower bound by finding the chance of getting a royal flush when dealt 10 cards (since that's the most cards that can involved in one hand of 5-card draw, so you can only get a royal flush if there is one somewhere in those 10 cards). That's an easier calculation. There are 4 choices of Ace, and once that's decided, the next 4 cards are determined, you can then have anything as the remaining 5, and all of those can appear in any order. So that's (4*1*1*1*1*(10!/5!)*47*46*45*44*43)/(52!/(52-10)!)=0.00039, about 1 in 2578. So, we know the odds are somewhere between about 650,000 to 1 and 2500 to 1. Probably closer to the higher number, but it depends on your strategy. --Tango (talk) 17:43, 2 April 2008 (UTC)

If you have four 8's and a 9, will you throw all 5 cards away? If you don't, the probability of a royal flush is 0. If you do, you increase that probability. Someone who's trying for a royal flush at all costs will have a slightly better chacne of making one than someone who's using a sensible strategy. --tcsetattr (talk / contribs) 20:47, 2 April 2008 (UTC)

Exactly. The strategy I described above should maximise your chances of getting a royal flush, but it's a pretty stupid strategy in a real game. Trying to calculate the odds for a realistic strategy is pretty much impossible - you would have to define that strategy precisely, and that's going to be very complicated. --Tango (talk) 21:04, 2 April 2008 (UTC)

When you say "complicated," how complicated do you mean? I would like to, even if it takes me a couple weeks to memorize, know how to get a royal flush. I could pwn some n00bs in online poker... flaminglawyer^c 21:20, 2 April 2008 (UTC)

If you're going to play draw poker by whatever strategy maximizes your probability of getting a natural royal flush, let me just say that I'd love to have you in my game :-) --Trovatore (talk) 22:42, 2 April 2008 (UTC)

I said "realistic strategy", I didn't say "strategy that will always get a royal flush". Such a strategy is obviously impossible. --Tango (talk) 00:37, 3 April 2008 (UTC)

It was a joke. The strategy is not very realistic for someone who does not want to lose most of the time.  --Lambiam 11:36, 3 April 2008 (UTC)

statistics

Hi, I am a teacher trying to determine validity of test questions. I understand what validity is. My question is what percent of content must be measured in order to make a test valid. For example if there are 100 vocabulary words a test has only one question about them then the test could not be accurately measuring knowledge. If there are ten questions about them does that 10% represent enough of a statistical population to ensure the validity of the testing device. Is there a minimum percentage needed to do so? Thank you. 206.219.72.83 (talk) 15:20, 2 April 2008 (UTC) I am not sure what symbols to show?

Do you really mean validity? I believe you mean precision and/or accuracy. Accuracy is not the issue if you picked the question(s) randomly: even with one question, if I knew (say) 80 of the 100 answers, the average proportion of right answers would be 0.80, as I would get it right 80% of the time, and wrong 20% of the time. But this alludes to the real problem: precision. For one question, your assessment of me will be off by at least 20% all of the time. But if you quantify that (lack of) precision correctly, your results can still be valid, even if they are not useful.

You may wish to look at the binomial distribution article for a followup. Baccyak4H (Yak!) 18:00, 2 April 2008 (UTC)

I interpreted it as a question about statistical significance. See sample size - that article should get you started, at least. --Tango (talk) 18:46, 2 April 2008 (UTC)

Assuming all questions are equally difficult, and the student scores P correct answers out of a total of N randomly selected questions, the best estimate for the fraction of items known to the student in the total universe of possible questions is P/N. However, this is only an estimate. Let c be a measure of confidence we want to achieve, say c = 0.95. If the true fraction of items known to the student is x, the probability of observing P or more correct answers out of N is:

$\sum _{k=P}^{N}{\binom {N}{k}}x^{k}(1-x)^{N-k}\,.$

As x gets smaller, this probability gets smaller, but if it goes below , it becomes unlikely small, so that means x is too small. Define x_low as the value of x for which the probability equals .

Likewise, the probability of observing P or fewer correct answers out of N is:

$\sum _{k=0}^{P}{\binom {N}{k}}x^{k}(1-x)^{N-k}\,.$

As x gets larger, this gets smaller, and eventually 0 (unless P = N). Define x_high as the value of x for which this probability equals . With some confidence we can say that x is in the range from x_low to x_high. This depends on N, P, and c (but not on the size of the universe of questions). Given fixed N and c, the worst case (the widest range) is when P is about one half of N.

If N is fairly large and P = ¹/₂N, we may approximate the two probabilities above by

$1-\Phi \left({\frac {{\tfrac {1}{2}}-x}{\sigma }}\right)$

and

$\Phi \left({\frac {{\tfrac {1}{2}}-x}{\sigma }}\right)\,,$

where

$\sigma ={\sqrt {\frac {x(1-x)}{N}}}\,$

and Φ is the cumulative distribution function of the standard normal distribution.

How large N should be depends on how high c must be, and how narrow the range [x_low, x_high] is required to be.

Mathematically this is not easy to handle, but since the critical area is around x = ¹/₂, we can slightly simplify things by just putting

$\sigma ={\frac {1}{2}}{\sqrt {\frac {1}{N}}}\,.$

A 2-sided confidence interval for c = 0.95 means that x_high−x_low is about 4σ. Suppose that we require this width to be at most 1/10. Then the simplified formula for σ tells us that N must be at least 400.  --Lambiam 12:37, 3 April 2008 (UTC)

Afraid of sets

Since I realised that not every carelessly defined set is acceptable, I have been afraid that I will break something, and I don't know enough to understand the relevant axiom in ZFC (if that's even the axioms to use!). When I can figure out a bijection between my proposed set and something I know is a set, it's OK I guess, but how about "the set of all functions from $\mathbb {R}$ to $\mathbb {R}$"? What are your suggestions for figuring out what is an acceptable set and what is not? Thanks. —Bromskloss (talk) 16:39, 2 April 2008 (UTC)

ZFC is as good an axiomatization of set theory as any other, and I think the axioms guaranteeing the possibility to construct certain sets are easy enough to understand. Basically there are a few legal operations you can perform on sets, such as power sets, subsets and binary unions and Cartesian products. The collection of functions from $\mathbb {R}$ to $\mathbb {R}$ is a subset of $P(\mathbb {R} \times \mathbb {R} )$ and thus a set. -- Meni Rosenfeld (talk) 16:57, 2 April 2008 (UTC)

Ah, a subset of $P(\mathbb {R} \times \mathbb {R} )$. Of course! Thanks. Speaking of ZFC being as good as any other, what difference does the choice of axioms do? Is ZFC the one everyone is using today? With other axioms, would the resulting mathematics similar at all? Would it still work for physics? Bromskloss (talk) 17:07, 2 April 2008 (UTC)

The choice of axioms can make a lot of difference - they basically define what a set is. You define a set differently and it will behave differently. ZFC is pretty widespread these days. There are a few people working on alternative axioms (eg New Foundations). The axiom of choice (the C part) isn't always used (it's independent of ZF, that is, all the other axioms, so you can choose to include it or not - it usually is, though, in my experience). The things you might think are sets which actually aren't are called proper classes - they're usually very big (for an appropriate and vague definition of "big", at least) collections, like "all sets". A set of functions between two sets is quite small by comparison, and is a perfectly fine set. Most things you'll come across work perfectly well as sets - just stay away from anything like "the set of all sets", and you should be fine. --Tango (talk) 17:23, 2 April 2008 (UTC)

In practice I think a lot of working mathematicians use NBG. It's lighter than ZFC, and the set/class distinction is good enough for pretty much anything non set-theorists are ever going to do. If you're not doing set theory you just need to be able to steer clear of the most glaring pitfalls/paradoxes, and sets and classes, or a fixed universe of sets, or any similar light-weight type theory is good enough for that. -- Leland McInnes (talk) 18:05, 2 April 2008 (UTC)

In practice, working mathematicians don't use a formal axiomatic system at all. They work as Platonists, whether they claim to be or not. --Trovatore (talk) 18:09, 2 April 2008 (UTC)

A fair call - Ultimately I simply meant that the sets vs. classes distinction that NBG uses is the only point that I've seen crop up in non set theorist mathematics with regard to set theoreti paradoxes. Thus working mathematicians take NBG's sets and classes and pretty muh ignore the rest of axiomatic set theory.

As a side note, I'm more of a structuralist and never could entirely understand the platonists. I would prefer and welcome a category or topos theoretic foundation. -- Leland McInnes (talk) 01:13, 3 April 2008 (UTC)

I was using "Platonist" as an (admittedly imprecise) shorthand for "realist", which certainly includes structuralists -- in fact structuralism is the main branch of realism in contemporary foundational philosophy, in the sense of considering isomorphic structures to be the same, except when you need to distinguish between the way they're embedded in some larger structure. Note that structuralism is already a sufficiently strong form of realism to guarantee that, for example, the continuum hypothesis has a well-defined truth value. --Trovatore (talk) 02:39, 3 April 2008 (UTC)

That depends on the school of structuralism in question. The realist structuralists in mathematical philosophy, while hurdling the representation issue, still put themselves in what seems to me to be an untenable position. Count me in the same school as Bell, Mac Lane and McLarty I guess. -- Leland McInnes (talk) 12:29, 3 April 2008 (UTC)

Not directly, sure, but working mathematicians build on what's gone before, and that will be built on particular axioms (perhaps not historically, but someone will have formalised it at some point). Generally, mathematicians use whatever axioms are needed to do what they're trying to do (eg. if you need AC to prove your theorem, then you assume AC, otherwise you ignore it completely). The various (common) choices of axioms don't usually contradict each other in the kind of work most mathematicians do, so people don't generally worry about it. --Tango (talk) 18:36, 2 April 2008 (UTC)

"Someone will have formalized it at some point". Yes. But that does not mean that it is "built on axioms". The axioms are extracted from the mathematics, not the other way around. Oh, it's true that there's an interplay between them; the picture of the von Neumann hierarchy did not really come clear until after Zermelo's formalization, though in retrospect we can see that it was inherent in Cantor's work all along. --Trovatore (talk) 18:40, 2 April 2008 (UTC)

But until something is formalised, you can't be sure it's right (even then, Godel gets in the way, but lets not complicate things any further). While a lot of maths was (and, I guess, still is) invented/discovered without a formal axiomatic approach, it's only since it's been formalised that we've been able to be sure we're not talking nonsense (for example, see Russell's Paradox). So, while the maths itself may not be built on axioms, the proof that the maths is correct is. Whether you choose to distinguish between the two is a matter of definition of the word "built", I suppose. --Tango (talk) 18:54, 2 April 2008 (UTC)

You can never be "sure" it's right. Only "pretty sure". Essentially mathematics is an empirical science, subject to the limitations of human knowledge like any other; the differences are quantitative rather than qualitative. --Trovatore (talk) 19:02, 2 April 2008 (UTC)

Are you referring to Godel? If so, I already dismissed him as an unnecessary complication to this discussion. If not, then I disagree. By my understanding, assuming you're working in a consistent and complete framework, a mathematical proof is absolute. There is no need for any kind of experimentation. --Tango (talk) 19:18, 2 April 2008 (UTC)

Well, I could point out that it's a bit cavalier to dismiss as "an unnecessary complication" a proof that the thing you're assuming (depending, I guess, on what precisely you mean by a "framework") doesn't exist. But in a sense you're right; it is an unnecessary complication, because the error of Euclidean foundationalism should have been clear even in Euclid's time. It's an infinite regress -- you claim that your results are not subject to error because they are derived from your alleged foundation, but how do you know that the foundation itself, or your method of derivation, is not subject to error? --Trovatore (talk) 19:51, 2 April 2008 (UTC)

It's extremely cavalier, but you can't get bogged down in the details in every discussion. Our axioms are as good as it's possible to get them, so most of the time, it's easiest just to assume they are right. --Tango (talk) 20:55, 2 April 2008 (UTC)

That's fine by itself. But it's not a very convincing as part of an argument that formalization is some sort of absolute defense against error. My position is that the objects of study (natural numbers, real numbers, functions, sets) are the primary notion, and the axioms that describe them are the subordinate concept. You don't have to agree with that, but making the axioms primary on the grounds of certainty is not going to work. --Trovatore (talk) 21:15, 2 April 2008 (UTC)

How certain it is isn't really relevant. Without a formal axiomatic approach, you have very little certainty at all. You can't really prove anything about natural numbers until you know what a natural number is - the naive approach of "they're the numbers you count with - a 2 year old child could do it!" is just as dangerous as naive set theory and could well lead to equally paradoxical results. While we can't be completely certain we've got it right, what certainty we can have comes from the axiomatic approach. You can do an awful lot of maths without worrying about rigour, but there's always the chance you'll trip up like Russell did. An axiomatic approach doesn't eliminate that danger, but it gets pretty close. --Tango (talk) 23:44, 2 April 2008 (UTC)

No, the axiomatic approach is pretty much irrelevant to the question of eliminating the danger. Note that Russell's paradox was discovered in a completely formalized version of set theory, due to Frege, who thought he was formalizing informal Cantorian set theory but may well have been wrong about that (Wang Hao thought so -- it may depend on which moment in Cantorian thought you choose to look at). At best, axiomatics provides a way of, as it were, "localizing" the risk, saying, "if there's an inconsistency, it should show up here". When an apparent inconsistency shows up in informal argument, that's a useful thing to have. But confidence in the correctness of the results is fundamentally not about axiomatics. --Trovatore (talk) 00:04, 3 April 2008 (UTC)

Oh, I should say: irrelevant, except insofar as the formalization helps to find the errors. That is arguably what happened in the Frege-Russell case -- when Frege formalized his notion of set (which he thought was the same as Cantor's but he was probably wrong about that), he made it easier for Russell to find the error in the concept. This is a useful aspect of formalization. But the fact that a deductive system is formal, by itself, does not provide any extra confidence whatsoever in its results. --Trovatore (talk) 02:19, 3 April 2008 (UTC)

Certainly, a formal system can be just as wrong as any other system. However, it is possible to prove the consistency of a formal system (you need to use a different system to do it, of course, so it's far from perfect), it isn't possible to do so for an informal system. It's all just moving the risk around, as you say, but that can be quite handy. --Tango (talk) 12:43, 3 April 2008 (UTC)

It's an interesting thing to do for lots of reasons, but increasing confidence in the results is not really one of them, because to prove the consistency of a formal system you need to use a stronger theory (one even more "likely" to be wrong"). Tango, I really think you have the wrong idea about formal systems; you've let yourself be taken in by the errors of the formalists. Formal systems are more an object of study than they are a tool for deriving conclusions. --Trovatore (talk) 15:56, 3 April 2008 (UTC)

Are the ZFC axioms sufficient to rule out the possibility of a self-referential set, i.e., a set that contains itself as an element? -GTBacchus^(talk) 17:50, 2 April 2008 (UTC)

Yes. It follows from the axiom of foundation that there is no such set. --Trovatore (talk) 17:58, 2 April 2008 (UTC)

From the discussion above, I seem to be able to extract that the most commonly discussed axiomatic set theories will eventually lead to the same mathematics, once you get past the most fundamental levels. I.e., most concepts in one theory would have its counterpart in another. They would all be useful for physics. Is that correctly understood? Is it possible that some completely different theory would lead to useful mathematics that is out of reach of what we are using now? —Bromskloss (talk) 19:43, 2 April 2008 (UTC)

Yeah, I think all the common axiomatic systems would agree on anything used in physics (as long as you can construct the real numbers in all of them, you're pretty much sorted, I would think). As for your last question - probably depends on what you mean by "useful". Different choices of axioms will certainly lead to new and interesting maths. Whether it will have any uses, it's hard to say, but people seem to be able to find real world uses for the strangest of mathematical concepts (which are generally created because a mathematician was curious as to what would happen, so thought "why not find out?", and just gave it a go), so I think there's a very good chance someone will make use of any maths we can come up with. --Tango (talk) 20:55, 2 April 2008 (UTC)

Open loop transfer function

In my syllabus we've got study of control systems, and it says that that Open Loop transfer function for a normal feedback system is G(s)*H(s). Even if it is not a unity feedback system. I don't understand how it makes sense - open loops means that H(s) is disconnected from the system, yet how is it being considered? All my textbooks use G(s)*H(s), and I asked my professor, he could not come up with a satisfactory answer.

The example control system which I'm talking about is this - http://classes.engr.arizona.edu/ame558/public_html/topic1/bd2.gif That, incidentally, seems to be "standard" control system - almost all formulae and theorems are derived keeping that control system in mind.

Thanks! --RohanDhruva (talk) 19:50, 2 April 2008 (UTC)

"Open-loop", in this case, does not refer to the situation where you control your system without a feedback. Instead, you should unplug $F(s)$ from the circle (which calculates the difference) and consider the signal path from $E(s)$ to $F(s)$. That gives $G(s)H(s)$ (or $KG(s)H(s)$ as it seems to be written in the picture). They actually define it just like that, as $F(s)/E(s)$, on one of the accompanying pages. —Bromskloss (talk) 20:37, 2 April 2008 (UTC)

OK, if you did send in a signal $R(s)$ after unplugging $F(s)$ from the difference circle, you would be controlling the system without a feedback, but that's not the point. —Bromskloss (talk) 20:47, 2 April 2008 (UTC)

Thanks Bromskloss. I still don't understand the "logic" behind it being defined as $F(s)/E(s)$. In open loop, you mean to say, the signal $F(s)$ is not connected to a summing point, instead, it acts as a point to take the output from? If so, what happens to the output signal $C(s)$. The wikipedia page for Closed loop pole defines "Open loop transfer function" as -- The open-loop transfer function is equal to the product of all transfer function blocks in the forward path in the block diagram. Would that clarify anything? --RohanDhruva (talk) 02:10, 3 April 2008 (UTC)

I agree with you that it does not seem obvious that $F(s)/E(s)$ would be of any interest, but I think it can be useful when studying the porperties of the whole feedback system. I don't remember exactly how, though. In the mean time, $C(s)$ just sits there. —Bromskloss (talk) 07:03, 3 April 2008 (UTC)

Dilemna!

I don't know how to write this using <math> tabs, but I'll do my best to explain it. If you can, please write it in those "magic words."

The problem is:

There's a magic thing in algebra. Pick any positive whole number. Square it. Add [twice the original number]. Add 1. Now find the [square root] of that number. Subtract the original number, and the answer is 1.

Now that's fine and dandy. Even the reasoning isn't too hard.

${\sqrt {x^{2}+2x+1}}-x=1$

Find quadratic roots...

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

Meaning...

$x+1-x=1$

Meaning...

$1=1$

That's fine and dandy. But if you don't find the quadratic roots - ah! A dilemma. See:

${\sqrt {x^{2}+2x+1}}-x=1$

We have to square every term in order to get rid of the [square root sign]. So:

$x^{2}+2x+1-x^{2}=1^{2}$

Take out the zero pairs.

$2x+1=1$

Subtract...

$2x=0$

But alas! The number you are told to pick has to be a positive whole number. And zero isn't positive (or negative). So it shouldn't work. But try as hard as you like; as long as x is a positive whole number, it works. Can someone please explain this? flaminglawyer^c 20:50, 2 April 2008 (UTC)

It's just an algebraic error: $(a+b)^{2}=a^{2}+2ab+b^{2}\neq a^{2}+b^{2}$. --Tango (talk) 21:01, 2 April 2008 (UTC)

(Edit conflict.) (Fixed <math> notation.) You've made an error going from ${\sqrt {x^{2}+2x+1}}-x=1$ to $x^{2}+2x+1-x^{2}=1^{2}$. Note that $(a+b)^{2}$ does not simplify to $a^{2}+b^{2}$. —Bromskloss (talk) 21:04, 2 April 2008 (UTC)

(edit conflict squared) "We have to square every term in order to get rid of the [square root sign]. " Ummmm, no, it doesn't work that way ... Go ahead and add $x$ to both sides in place of that step, then square both sides (and most decidely not "every term"). --LarryMac | Talk 21:08, 2 April 2008 (UTC)

Oh, so you would square the sides. I thought it was every term. I get it now. (it was kind of new territory for me, since I am only taking Algebra 1) flaminglawyer^c 21:13, 2 April 2008 (UTC)

can u plz solve this?

(x+1)^10 —Preceding unsigned comment added by 64.229.54.4 (talk) 21:35, 2 April 2008 (UTC)

It's equal to (x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1) --Carnildo (talk) 22:39, 2 April 2008 (UTC)

Check out Binomial expansion. And that can't be solved, because it isn't an equation, but it can, like I had linked, be expanded. Confusing Manifestation(Say hi!) 22:49, 2 April 2008 (UTC)

Thanks!!... it was easy:P...thanks! —Preceding unsigned comment added by 64.229.54.4 (talk) 23:24, 2 April 2008 (UTC)

${x}^{10}+10\,{x}^{9}+45\,{x}^{8}+120\,{x}^{7}+210\,{x}^{6}+252\,{x}^{5}+210\,{x}^{4}+120\,{x}^{3}+45\,{x}^{2}+10\,x+1$ --wj32 ^t/c 06:14, 3 April 2008 (UTC)

April 3

Stupid question

Hi there, kindly Mathematics help desk editors. I'm in the middle of studying and I can't figure out what the other two solutions to this equation are; I *have* checked my textbook and looked it up online, but I'm really confused :/ It'd be great if someone could explain if they had the time! :D

$sin2x={\frac {1}{2}}$ for $0\leq x\leq 2\pi$

I did the inverse sine of 1/2 (in rads) and got pi/6 and from there used the quadrant diagram (and divided by 2) to get x = pi/12 and x=5pi/12. However my textbook says that 13pi/12 and 17pi/12 are also solutions and I can't for the life of me figure out why. Any pointers in the right direction would be appreciated. Cheers, -- Naerii 00:57, 3 April 2008 (UTC)

The key thing to remember is that when x ranges from 0 to 2*pi, 2x ranges from 0 to 4*pi. When you use what you call the "quadrant diagram", you need to find all the angles up to 4*pi which have sine 1/2, and then half them, which will give you angles up to 2*pi, as required. (And remember: If you ask the question, you're stupid for a minute, if you don't ask the question, you're stupid for life!) --Tango (talk) 01:07, 3 April 2008 (UTC)

See also Inverse trigonometric function#General solutions. PrimeHunter (talk) 01:18, 3 April 2008 (UTC)

If only I could understand that article :( But thanks for both of your help, I can't believe I was so dense as to miss the 4pi thing haha. Thanks, -- Naerii 01:32, 3 April 2008 (UTC)

Accidental variance discovery

While considering generalizations of the Moore-Penrose pseudoinverse, in particular matrices of the form $C=(AB)^{-1}A$ for non-square A and B, I happened across a peculiar result. Generating matrices of a fixed size by choosing their elements uniformly from $[-{\frac {1}{2}},{\frac {1}{2}})$, I fixed B and generated two long lists of A matrices (300k each). I then evaluated the C matrices and took the sample variance of each element of C (treated as a random variable with each instantiation of C being a trial; clearly they are not independent). The result was two matrices of variances, one for each set; the oddity is that the two matrices satisfy $V'=DV$ for a diagonal matrix D.

Put simply, what does this mean about the dependence of C on A? The sample size is irrelevant; I observe the same effect generating just 3 Cs and taking the (somewhat silly) variances of $\{C_{1},C_{2}\}$ and $\{C_{2},C_{3}\}$. My initial suspicion was that the various C matrices themselves differed only in the application of a diagonal matrix, but this is not the case: neither the matrices that satisfy $C'=MC$ nor those that then satisfy $M'=NM$ are diagonal or even symmetric. Any thoughts? --Tardis (talk) 03:58, 3 April 2008 (UTC)

question about rational equations

When solving a rational equation, why is it necessay to perform a check? —Preceding unsigned comment added by 72.94.241.26 (talk) 15:07, 3 April 2008 (UTC)

You're talking about solving an expression of the form

${\frac {P(x)}{Q(x)}}=0$, right?

For $x^{*}$ to be a solution to this equation, it must be the case that $P(x^{*})=0$. But it's also necessary that $Q(x^{*})\neq 0$, for if that fails, then the expression ${\frac {P(x)}{Q(x)}}$ is undefined (it is not equal to zero in this context).

Hence, solving this equation amounts to picking the values of x that solve $P(x)=0$ while at the same time make $Q(x)\neq 0$. For example,

${\frac {x^{2}-1}{x-2}}$ has $\pm 1$ as solutions, but

${\frac {x^{2}-1}{x-1}}$ is only solved by -1, since Q(1)=0. Is this what you were referring to? Pallida  Mors 15:41, 3 April 2008 (UTC)

well kinda but i just wanted to know why it is necessary to do the check after you do the equation —Preceding unsigned comment added by 71.185.111.206 (talk) 16:40, 3 April 2008 (UTC)

Could you give example of a question and what you've been told to check? --Tango (talk) 16:55, 3 April 2008 (UTC)

It's good to check the answer to any problem, because different solution methods can introduce spurious answers (for example, if you square an expression, you get a positive and a negative answer in the end, but only the positive one may be an actual solution). Of course, you could also lose solutions in the process, and that's harder to detect unless you have some knowledge of how many solutions the problem should have (e.g. if you graph it). Confusing Manifestation(Say hi!) 22:17, 3 April 2008 (UTC)

See also Extraneous solution.  --Lambiam 22:59, 3 April 2008 (UTC)

Have a problem setting up equation to these questions

1. The cost, in millions of dollars, to remove x % of pollution in a lake modeled by C=6000/200-2x

 a. What is the cost to remove 75% of the pollutant?
 b. What is the cost to remove 90% of the pollutant?
 c. What is the cost to remove 99% of the poolutant?
 d. For what value is this equation underfined?

I can't figure it out how to solve this. please help —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 15:20, 3 April 2008 (UTC)

You have the cost function in terms of the percentage of the pollution you want to remove. C is the cost. x is the percent of the pollution you're removing. If you have x, how would you get C? –King Bee ^{(τ • γ)} 15:45, 3 April 2008 (UTC)

but how do i set uo the equations? —Preceding unsigned comment added by 71.185.111.206 (talk) 16:38, 3 April 2008 (UTC)

You don't need to set up any equations - you're given the equation in the question. --Tango (talk) 16:54, 3 April 2008 (UTC)

You have the equation — though surely it should be $C={\frac {6000}{200-2x}}$ which is written as "C=6000/(200-2x)". You need merely identify which quantit(y/ies) in the problem are to be used for x. (You are always asked for the cost, so you can't be being given C.) For (d), you might graph the function $C(x)$, or else note that it is a rational function, to find points where it is undefined. As an aside, you have been given a silly model for the cost: $C(0)\neq 0$, so it costs money to do nothing at all! I suppose that the everpresent cost might be the cost of bringing the equipment and people to the lake in question, but it'd still be more realistic (and a lot cheaper for thorough jobs) to use $D(x)=30\left(1-\operatorname {lg} (1-x\%)\right)$ (see binary logarithm), chosen to match the given formula at $x=0,x=50\%$. --Tardis (talk) 17:08, 3 April 2008 (UTC)

You are given the following:

The cost to remove  x%  is equal to .

This is supposed to hold for all possible values of the variable x. So, for example:

The cost to remove 54% is equal to .

The cost to remove 68% is equal to .

The cost to remove 93% is equal to .

Do you see the pattern? You can do the problems a, b and c in the same way. For d, remember that you are not allowed to divide by 0.  --Lambiam 23:16, 3 April 2008 (UTC)

Problem.

Theres a quadrilateral with corners A, B, C, and D. AB and CD are the long sides, and they are parallel. The length of BC is 8. The measure of angle C is 30, and the measure of angle A is 45. Find the perimeter of the object. I got $44+4{\sqrt {3}}+4{\sqrt {6}}$, but just about everyone else I have asked to solve this problem has gotten different answers. If someone could please draw this figure using the <math> parameters, I think that ppl would be able to solve it easier-ly. So: what is the real perimeter? flaminglawyer^c 17:34, 3 April 2008 (UTC)

Unless I'm missing something, those details don't determine a single shape - there's nothing stopping you stretching AB and CD, as long as you add the same amount to each. Am I missing something, or have you missed something out? --Tango (talk) 17:49, 3 April 2008 (UTC)

Agreed with Tango. Also, I don't think the math markup can draw figures. If you tell us what you did to reach your conclusion, and hopefully supply the missing information, we can probably help further. --LarryMac | Talk 18:07, 3 April 2008 (UTC)

OK: I'll try to draw it, but I'll have to rewrite the info, too. Here it is:

   A__________________________B
   /                         /
  /                         /
 /                         /
/_________________________/D
C

Line AB is parallel to line CD. Lines BD and AC are not parallel. The length of line BD is 8. Angle D is 30 degrees. Angle A is 45 degrees. From some law of mathemeatics, in a right triangle with angles measuring 45-45-90, the measure of the hypotenuse is ${l}{\sqrt {2}}$, where l is the length of either leg (not the hypotenuse). In a right triangle with angles measuring 30-60-90, the length of the longer leg is ${s}{\sqrt {3}}$, where s is the length of the shorter leg. The length of the shorter leg is half of the length of the hypotenuse. You solve this equation by dividing the figure into a rectangle and two right triangles. flaminglawyer^c 19:29, 3 April 2008 (UTC)

It might help to draw it with A and D being acute angles, since you've given that they are, and I'll add the numbers in too:

   A___________________B
    \45 |             |\
     \  |             | \ 8
      \ |             |30\
      C\|_____________|___\D

Have I drawn in the triangles in the places you meant? Using the triangle with the 30 deg angle, you can get that the height is ${\frac {\sqrt {3}}{4}}$, then using that and the other triangle, you can get that AC is ${\frac {\sqrt {6}}{4}}$ (this is all assuming I can do right angled trig in my head - the numbers may be wrong, but it's the method that's important). However, that's as far as you can go. There is no way to determine the lengths of AB or CD without more information, since I can draw another shape which still satisfies the information you've given but has different AB and CD:

   A_______________________________________B
    \45 |                                 |\
     \  |                                 | \ 8
      \ |                                 |30\
      C\|_________________________________|___\D

See? So, unless there is another piece of information you haven't given us, the perimeter is undetermined. --Tango (talk) 19:50, 3 April 2008 (UTC)

Flaming lawyer, are you sure you got the diagram right? I mean, a quadrilateral ABDC doesn't make as much sense as ABCD

 <nowiki>
   A___________________L___________B
    \45 |                  _______/|\
     \  |         ____8___/        | \ 
      \ | _______/                 |30\
      D\|/_________________________|___\C

and using this, we could probably work something out using BD = 8. -mattbuck (Talk) 19:52, 3 April 2008 (UTC)

In fact, using this, let us label the height of the shape as H, and the length of the top between the two triangles as L. Then $L^{2}+H^{2}=8^{2}$, the length of AD is $H{\sqrt {2}}$, AB = L+H, BC = 2H, $CD=L+{\frac {\sqrt {3}}{2}}H$. You will still get quite a few possibilities, but you are limited as to your choices. If nothing else, you'll get it as an equation in 1 variable, and can find the range of possibles. -mattbuck (Talk) 20:07, 3 April 2008 (UTC)

You get that L can range between 0 and 8, and everything is determined from there. That's only marginally better than the previous version - it's just the addition of an upper bound. There's still a continuum of possibilities. --Tango (talk) 20:13, 3 April 2008 (UTC)

? (Puzzle)

I was wondering: you know that puzzle, with the canoe and the river and the wolf, chicken, and chicken food (or some variation of it)? Is there any math behind that puzzle at all? Just wondering... flaminglawyer^c 19:40, 3 April 2008 (UTC)

Well, it's more a problem of logic, but since logic is the foundation of maths, I suppose you could claim there was something in it. -mattbuck (Talk) 19:48, 3 April 2008 (UTC)

I've seen the problem re-written in terms of graph theory, so in that sense there is maths behind it. Confusing Manifestation(Say hi!) 22:13, 3 April 2008 (UTC)

Someone asked for the solution to that problem, and a generalization if possible a few weeks ago on this desk, I'm too lazy to track it down, but it's in the archives somewhere. A math-wiki (talk) 22:15, 3 April 2008 (UTC)

It's here: Wikipedia:Reference desk/Archives/Mathematics/2008 March 22#Chicken, fox, and grain problem.  --Lambiam 22:57, 3 April 2008 (UTC)

The solution is: Take the chicken across first, go back and get the chicken food and take it across and bring the chicken back to the beginning with you, and then take the wolf across (coming back for the chicken last). 86.149.97.133 (talk) 22:34, 3 April 2008 (UTC)

See fox, goose and bag of beans puzzle. Gandalf61 (talk) 22:48, 3 April 2008 (UTC)

I've seen it used as a very simple example of operations research. A similar riddle asks how you can toast three slices of bread in the shortest time, assuming your toaster holds two slices but only does one side of each. It made more sense with older toasters. Black Carrot (talk) 09:02, 4 April 2008 (UTC)

Trigonometric equations

Hi there, any help solving this would be appreciated.

$2sin^{2}(x)-2sin(x)=cos^{2}(x)$

The only two identities we've been taught are:

$tan(x)={\frac {sin(x)}{cos(x)}}$ and $sin^{2}(x)+cos^{2}(x)=1$

I've tried rearranging any number of different ways but trig identities get me every time x.x How do I get the x by itself? Thanks in advance! —Preceding unsigned comment added by 86.149.97.133 (talk) 22:33, 3 April 2008 (UTC)

Express the $\cos ^{2}x$ in terms of $\sin ^{2}x$ and then rearrange. Then let, say, $y=\sin x$, and then you might see something interesting, unrelated to trigonometric functions. x42bn6 Talk Mess 22:36, 3 April 2008 (UTC)

Thanks!!! —Preceding unsigned comment added by 86.149.97.133 (talk) 22:40, 3 April 2008 (UTC) oh quadratics :D lot easier than it looks, thank you so much —Preceding unsigned comment added by 86.149.97.133 (talk) 22:42, 3 April 2008 (UTC)