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May 26

Wall Street and proof theory -- wtf?

This has to be a joke; at least I hope it is.[1] It just appeared on Slashdot.[2] I do know that Wall Street has been interested in typed functional programming for a while, and proof theory and type theory are closely related. But sheesh. Maybe this is another bogus trend like when they were trying to hire chess grandmasters as traders. 69.228.170.24 (talk) 02:06, 26 May 2010 (UTC)

That looks like somebody's blog and it has only been submitted to slashdot, not accepted as far as I can see. As a wikipedian I would reject it as no reliable sources. Dmcq (talk) 08:05, 26 May 2010 (UTC)

Well you never know what'll be useful. An expert in cryptogams from the Natural History Museum proved invaluable in the work at Bletchley Park in WWII :) Dmcq (talk) 08:55, 26 May 2010 (UTC)

I didn't know that story and it took me a minute to track it down: AskOxford: The Hidden Heroes. Until I read it, I thought maybe your cryptogam link was a typo. Neat story! -- Coneslayer (talk) 14:36, 26 May 2010 (UTC)

Well I wasn't proposing to put it in an article. But looking at it again it looks clear that it's a cute joke. The first thing I had thought of was that it had something to do with program verification, and then I didn't look any further. 69.228.170.24 (talk) 14:31, 26 May 2010 (UTC)

I'm attempting to verify this--it seems plausible. I'm coming to the conclusion that such a system--an endless variety of such systems in fact--could be designed. 146.96.40.12 (talk) —Preceding undated comment added 03:12, 27 May 2010 (UTC).

It's actually on the front of slashdot now. 69.228.170.24 (talk) 05:53, 27 May 2010 (UTC)

differential equation

x=fn(t) , d²x/dt²=kx^-2sgn(x)

Is there a known or obvious solution - (I couldn't find a list on wikipedia - perhaps there is such a page?). Thanks. 77.86.125.207 (talk) 02:13, 26 May 2010 (UTC)

or even x=fn(t) , d²x/dt²=k/(x²+a²)sgn(x)

It's supposed to represent oscillation offset a from the centre of an inverse square field.

If there's a description of how to go about finding a solution that would be just as good. (my maths education as far as it went didn't cover this sort of differential, and I'm dubious that I actually have the knowledge to do so) - so links to explanations are good. Thanks again.77.86.125.207 (talk) 02:41, 26 May 2010 (UTC)

If x'=0 when t=0 I'm fairly sure that the equation for x(t) is roughly of the form Sum(k_ncos(nt)) and so:

|Sum(n2kncos(nt))| x {a2+(Sum(kncos(nt))2} = B

where a,B are constants, sums are from n=1 to infinity. Don't know where to go from here.77.86.125.207 (talk) 05:10, 26 May 2010 (UTC)

dx/dt is an integrating factor, that is when you multiply both sides of the original equation by it then each side is the derivative of a simple expression. Integrating gives a first order differential equation which is solved by separation of variable. The details get a bit hairy and with the sgn(x) in there you need cases for the sign of x, so I don't feel like writing out a full solution, but knowing the integrating factor should get you started.--RDBury (talk) 09:21, 26 May 2010 (UTC)

It is the limiting case of a Kepler orbit with vanishing angular momentum. Close to the singular point x=0 the motion has unlimited speed and acceleration, unlike an oscillation. Choosing a suitable unit of time simplifies the equation to either $\scriptstyle d^{2}x/dt^{2}=x^{-2}$ or $\scriptstyle d^{2}x/dt^{2}=-x^{-2}.$ Bo Jacoby (talk) 11:17, 26 May 2010 (UTC).

Thanks for your support, I'll see if I can make further progress. Cheers. 77.86.125.207 (talk) 17:32, 26 May 2010 (UTC)

Extension of implications of Normal Topological Spaces

Hi RDesk,

I was hoping you'd be able to help me try and finish off this Topological Spaces result. First, I'm using the definition that a T. S. is 'normal' if for any pair of disjoint closed subsets A, B, there exist disjoint open subsets U, V containing A, B resp.

I want to prove that for any normal T. S. 'X', with A, B disjoint closed subsets in X, there exist open sets U, V containing A, B resp. such that the closures of U and V are disjoint.

I tried to go ahead as follows: suppose U and V are the minimal open sets containing A and B, and suppose w is in the intersection of U and V. Then for every open set W containing w, W contains a point u in U and v in V - so I want to show that W also contains a point a in A (similarly b in B), so that w must be in the closure of A and B, thus in A and B (closed sets) so A, B are not disjoint and we have a contradiction.

To do this I assumed W contains no point a in A, but contains u in U - then I tried to show that the closure of W, Cl(W) must also have no point in A either, so that Cl(W) is a closed set disjoint from A, and then the union of Cl(W) and U_c (the complement of U in X) is the union of 2 closed sets, so closed and contains no point in A; thus U\Cl(W) is an open set which contains A contradicting minimality. However, even assuming this is the correct way to do things, I can't see how to show Cl(W) contains no point in A - if indeed that is true.

Could anyone please provide any suggestion to where I might be going off track or what I might be missing? All help hugely appreciated - thanks! Estrenostre (talk) 10:47, 26 May 2010 (UTC)

Let A, B, U, V be as given. Then A and U^c are disjoint and closed so you get new open sets P, Q with A ⊆ P and U^c ⊆ Q. Similarly there are open sets R, S with V^c ⊆ R and B ⊆ S. When you draw out all the relationships you shold be able to show that P and S have disjoint closures.--RDBury (talk) 14:47, 26 May 2010 (UTC)

Btw, your approach has an issue in that there is in general no minimal open set containing A. You can talk about minimal closed sets and maximal open ones but the other way doesn't work.--RDBury (talk) 14:52, 26 May 2010 (UTC)

'Reflection invariant' Set

Hi,

Is there a term for/area of study of the type of bounded sets (specifically in $\mathbb {R} ^{n}$) such that $x\in U\,\Rightarrow$ $-x\in U$? (Where U also contains 0).

Specifically I'm looking at defining a norm on $\mathbb {R} ^{n}$ such that for an open bounded U as above, U is exactly the set of points of norm less than 1 - however, I'm also interested as a general matter, so don't worry about things being too irrelevant!

Thanks a lot :-) Simba31415 (talk) 13:36, 26 May 2010 (UTC)

Minkowski's theorem may be of interest.--RDBury (talk) 14:54, 26 May 2010 (UTC)

I believe you are looking for a theorem of Kolmogorov. A subset U of R^n is the collection of x with ||x|| < 1 for some norm ||.|| if and only if U is open, convex, and closed under reflection. You need to add "convex" to your requirements in order to get a norm. To define the norm of x you take the largest number c so that x/c is not in U but x/d is in U for all d>c. To get the triangle inequality, you need U to be convex. Norm (mathematics) has something on this. Let me know if you want a book reference. JackSchmidt (talk) 15:00, 26 May 2010 (UTC)

You forgot "bounded". And nonempty, come to think of it. Algebraist 17:44, 26 May 2010 (UTC)

I think the best term you'll find for your condition is "symmetric about the origin" or similar. balanced is related though. Algebraist 17:05, 26 May 2010 (UTC)

You're quite right, convex is on the list, I missed it out by accident sorry! Ahh, all you exceedingly smart people do scare me sometimes ;-) Simba31415 (talk) 17:37, 26 May 2010 (UTC)

Standard Deviation in Normal Approximation of Binomial Distribution

Hello. If some binomial distribution is composed of many random independent trials and can be closely approximated to a normal distribution, why is σ² = npq? Thanks in advance. --Mayfare (talk) 20:22, 26 May 2010 (UTC)

Look at this: Binomial_distribution#Algebraic_derivations_of_mean_and_variance. Michael Hardy (talk) 22:20, 26 May 2010 (UTC)

....also, remember that a binomially distributed random variable is the sum of n independent Bernoulli-distributed random variables, the value of each of which is either 0 or 1. So the question is: why is the variance of a Bernoulli-distributed random variable equal to pq? Then recall that

${\text{var}}(X)=E(X^{2})-(E(X))^{2}.\,$

In this case, is X is either 0 or 1, X² is the same as X, so E(X²) is the same as E(X). So you get p − p² = p(1 − p) = pq. Michael Hardy (talk) 22:24, 26 May 2010 (UTC)

Graph of a function

How can I find a function from looking at its graph? This is assuming the function is of the lowest degree possible to create that graph (i.e, the graph doesn't do anything googly where I can't see it, either off the portion I can see or very small down) and the important parts (zeros and critical points) are all on the portion of the graph I can see. Of course for this I mean a function that is not simple like a conic or a linear function. Thanks. —Preceding unsigned comment added by 76.230.8.121 (talk) 20:51, 26 May 2010 (UTC)

By some accounts, the function is the graph, so you're done. Maybe you mean a closed-form formula or other expression for the value of the function, but you didn't say so. Then you say "lowest degree", which suggests you have in mind polynomials, but you didn't say that either (nor is it always appropriate to assume it's a polynomial). Certainly you can fit a polynomial through finitely many points. Michael Hardy (talk) 22:28, 26 May 2010 (UTC)

Yes I do mean a polynomail formula i.e. y=ax^n+bx^(n-1)+...+ψ. (where ψ is some arbitrary constant which I use to indicate that the function ends, but not at any particular degree as z might imply. 76.230.8.121 (talk) 23:20, 26 May 2010 (UTC)

If you have n+1 points you can find an n degree polynomial that fits the points just by taking the general form y = a_nxⁿ + a_n-1x^n-1 +...+ a₁x¹ + a₀ and plugging the values of x and y for each point. That gives you n+1 linear equations in n+1 unknowns (a₀,..., a_n) which you can solve using linear algebra. Is this what you're trying to do, or are you looking for tricks to make a quick guess based on a pictured graph? Rckrone (talk) 00:32, 27 May 2010 (UTC)

Did you see our articles on interpolation and especially on polynomial interpolation, the Lagrange polynomials and the Newton polynomials? However, be aware that polynomials are not particularly nice when used for interpolation as they may oscillate wildly between the sampling points. —Tobias Bergemann (talk) 07:27, 27 May 2010 (UTC)

If you are given a graph on paper, you can't determine the exact points without measurement errors. Then you have to use some form of regression analysis, typically using a least squares method. 130.188.8.12 (talk) 08:56, 28 May 2010 (UTC)

Ellipse

I my mechanics textbook, the polar equation for an ellipse is given as r = r₀/(1-εcosθ), but in most websites the formula is given as r = a(1-ε²)/(1+εcosθ). How are the two equivalent? 173.179.59.66 (talk) 23:30, 26 May 2010 (UTC)

Having 1-εcosθ in the denominator versus 1+εcosθ changes which way the ellipse is oriented. Assuming 0≤ε<1, then using 1-εcosθ puts the closest point to the origin at θ = π and the farthest at θ = 0 and using 1+εcosθ reverses that. In the numerator, they're just labeling the constant that determines the size of the ellipse differently. r₀ = a(1-ε²). The constant a measures the semi-major axis while r₀ is just the radius at θ = π/2 and θ = 3π/2. Using r₀ gives you a more concise equation, but a is a more meaningful value. Rckrone (talk) 00:44, 27 May 2010 (UTC)

Thanks. 173.179.59.66 (talk) 01:12, 27 May 2010 (UTC)

May 27

Calculating df for unpooled t-test

I have two samples, same size but unequal variance. I'm trying to find a formula to exactly calculate the df, not just the smaller of the n-1 as df, something as the exact result my TI-83 will give me. I can't find the formula for this anywhere though. I just need to show the formula for a presentation. Is there any sense in caring about the exact formula and if so, any assistance finding it? Thanks 66.133.196.152 (talk) 05:35, 27 May 2010 (UTC)

You might start by looking at Behrens–Fisher problem. Michael Hardy (talk) 15:08, 27 May 2010 (UTC)

Gregory-Madhava Series

Can somebody please move this thread here from Science Desk? I don't know how to do it properly. http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Science#Gregory-Madhava_Series --117.204.87.183 (talk) 18:02, 27 May 2010 (UTC)

(copy & paste? --pma 18:17, 27 May 2010 (UTC))

Somebody claims in an article that at Princeton University and some other such institutions Gregory series is now being taught under the name of Gregory-Madhava Series. Leibniz formula for pi#History has a glaring pov sentence. "To give the rightful place to this great mathematician, the series should be named 'mAdhava srENi' or Madhava Series." But that makes things a bit cloudy. What is the actual state of affairs? I don't know mathematics, by the way.--117.204.87.183 (talk) 17:20, 27 May 2010 (UTC)

I know this formula - and had forgotten it had a name at all - so much for mathematical glory. I doubt many people really care - http://www.google.co.uk/search?hl=en&q=%22Leibniz-Gregory-Madhava%22+formula&aq=f&aqi=&aql=&oq=&gs_rfai= suggests some people are remembering all three. The maths desk might be a better place to ask.77.86.125.207 (talk) 17:27, 27 May 2010 (UTC)

Considering the OP came from Kerala I guess it is another attempt to rename a series for a local hero. I don't know why people feel this urge to fix attributions and correct all the books. The article Gregory's series should definitely have a better history section and I fullagree with mentioning Madhava in the leader but the bit saying it is also called the Gregory-Madhava or Madhava-Gregory or whatever series would need attribution. It isn't generally called that as far as I'm aware and that's that basically. Perhaps they call it that in India? What's there is wishful thinking as far as I'm concerned. Dmcq (talk) 13:01, 28 May 2010 (UTC)

By the way Nilakantha Somayaji of the Kerala school is also credited with discovering the series. Dmcq (talk) 13:15, 28 May 2010 (UTC)

There are primacy claims around these mathematicians. The matter was treated at some length in a popular novel Francis_Ittykkora that came out in 2009 and many people seem to have taken the shabby claims in the novel for truth. (The page asks for a move as the first sentence itself shows in that article.) I would like to know if anybody teaching in Princeton Maths Department would attest to such a renaming:))--117.204.81.28 (talk) 17:33, 28 May 2010 (UTC)

Keep in mind that who actually discovered/invented it is not the issue for WP. Whatever terminology is being used most often in textbooks is the standard here, right or wrong. It's not out job to figure out what should be named after whom and fix it.--RDBury (talk) 18:34, 28 May 2010 (UTC)

People, don't you read the posts before you reply them? I was asking about a claim I saw somewhere that a long established formula has recently been renamed to accommodate some not so old discoveries. The claims are suspicious because there is little in the mainstream publications about this school. Primacy or invention is none of my concern and I didn't pose anything here to imply that it is. In my last post I specifically wanted to know if a new appellation has been in vogue at Princeton. I suppose WP couldn't be without some people from Princeton. However, two posters above hint at some nationalistic or primacy claim motive on my part which in fact is not there.--117.204.80.153 (talk) 20:31, 28 May 2010 (UTC)

Sorry about that. Half the theorems in maths are named after the wrong person and there's always people trying to rename them which as said above isn't Wikipedia's job. If you give a reference to the article about Princeton saying that it could possibly help. I did a quick google and saw a number of references to Leibniz-Gregory-Madhava series which is getting rather overlong! Dmcq (talk) 21:21, 28 May 2010 (UTC)

Unlike Astronomy where there is a body that has some say over nomenclature, there is no such body in mathematics. I.e. a theorem/series/whatever is called whatever people choose to call it. The usual way in which names change is that someone decides to use a new name in a paper/conference/etc (and makes a point why this name is justified) and others either follow suit or don't. If there are a number of recent papers using eibniz-Gregory-Madhava, than that *is* an alternative name. 213.160.108.26 (talk) 22:48, 28 May 2010 (UTC)

I edited the Leibniz formula for pi article. Bo Jacoby (talk) 13:03, 30 May 2010 (UTC).

Newton's F*dt=m*dv integral!

Resolved

Will appreciate your control for the following differentials:

Newton's F*dt=m*dv is written as

F*r*dt=m*r*dv (energy conservation equation).That is

1/2*m*Vr^2+m*gr*r+1/2*I*w^2=m*r*dVr (total energy,where g is variable,Vr is the radial velocity and Vp, the perpandicular velocity to Vr, does not affect the total energy)).Then from

Vr^2+2*gr*r+I*w^2/m=2*r*dVr a differential equation is written with (2*gr*r+I*w^2/m=K*dt^2)

dr^2+K*dt^2=2*r*d(dr) and this differential equation is solved as

r=-a*t*(t-tmax)+K where K=-a^2*tmax^2/(4*a+1)

On Cartesian the graph of (r) is a parabola. On Polar, the graph is a cardioidal looking spiral (when on a plane).

The orbit of the planets is this spiral with billions of spires.The amplitude of the spiral's rings are expanding and then after compressing. No sign of ellipse, no focus, no aphelion, no perihelion nor equality of swept out areas in equal interval of time. No Kepler laws according Newton's laws.(??!!).TASDELEN (talk) 19:31, 27 May 2010 (UTC)

This is not the first time you've asked this or a similar question in the past few weeks. Have you read Kepler's laws of planetary motion#Derivation from Newton's laws as suggested before? Please read through them first and come back here if you still have specific questions that you do not understand about that derivation. Marking as resolved for now. (p.s. This poster's sheer persistence in asking long-winded, borderline nonsensical questions on both Science and Maths RefDesks leads one to assume that either they don't have a grasp of English or that they want to be deliberately difficult. Either way we can;t really help them.) Zunaid 21:00, 27 May 2010 (UTC)

I've told you before, you need to count the d's: 1/2*m*Vr^2+m*gr*r+1/2*I*w^2=m*r*dVr. That can't be a true statement since one of the four terms has a d and the other three don't. --Tango (talk) 21:16, 27 May 2010 (UTC)

• No Tango.The statement (1/2*m*Vr^2+m*gr*r+1/2*I*w^2=m*r*dVr) is correct as it is the energy conservation expression. With the following correction (K/dt^2=m*gr*r+1/2*I*w^2) we write:

1/2*m*(dr/dt)^2+K/dt^2=m*r*d(dr/dt)/dt which is,after simplification:

r'^2+2*K/m/dt^2=2*r*r" a differential equation (with all the terms of the same size) having the solution:

r=-a*t^2+a*t*tmax+K where K=-a*tmax^2/4=Constant.In fact:

dr/dt=r'=-2*a*t+a*tmax

d(dr/dt)/dt=r"=-2*a

r'^2=4*a^2t^2-4*a^2*t*tmax+a^2*tmax^2

2*r*r"=-4*a*(-a*t^2+a*t*tmax+K)=4*a^2*t^2-4*a^2*t*tmax-4*a*K which is ending by:

4*a^2t^2-4*a^2*t*tmax+a^2*tmax^2+000000=4*a^2*t^2-4*a^2*t*tmax-4*a*K and after simplification

a^2*tmax^2=-4*a*K and K=-a*tmax^2/4 (as I have posed=Constant).

All this has one meaning: [r=-a*t*(t*tmax)+K] and the graph of this equation is a spiral on Polar plane.Not an ellipse.I think Kepler's laws of planetary motion#Derivation from Newton's laws should be revised for (Derivation) and should not be copy-pasted as icon. I insist: either Newton's law indicates my solution or the solution of Zunaid. I need mathematical proofs, confirmations, and not discouraging sentences like they don't have a grasp of English. What about your Turkish? Meanwhile, a difficult question about Kepler's area law: do you believe to the derivations on Wiki? Is Vp perpandicular velocity to Vr (radial velocity) variable or constant? Wiki says: variable. I say constant. Thanks.TASDELEN (talk) 13:10, 28 May 2010 (UTC)

The article should have a link to where it obtained the derivation from in the References section. Or check out the under-graduate text-book Calculus by James Stewart which actually has a problem exercise on this exact topic. I've done this derivation myself many years ago (I did mathematics as a core course of my degree and took astronomy as an elective) and can confirm that yes, under Newton's laws planetary orbits do come out as ellipses and not spirals. I'll be damned to do that derivation again, it requires a good grasp of vector calculus and a few lemmas regarding the dot product and cross product.

The reason I question your grasp of English is because it makes it difficult to understand your questions. I don't know if you're questioning the derivation of Kepler's Laws (ridiculous) or the accuracy of our article on that topic (entirely plausible). What do you mean "The orbit of the planets is this spiral with billions of spires.The amplitude of the spiral's rings are expanding and then after compressing".

The fact is that Newton's laws do predict Kepler's laws as has been mathematically and experimentally confirmed for the last 300 years. Your claim "No sign of ellipse, no focus, no aphelion, no perihelion nor equality of swept out areas in equal interval of time. No Kepler laws according Newton's laws" means you've made a mistake somewhere. Your derivation/result doesn't suddenly overturns 300 years of actual observations. Zunaid 15:04, 28 May 2010 (UTC)

Proposals for changing articles should be on the talk page of the article, not on the reference desk. Please note that any change of this magnitude, i.e. saying Newton's derivation is wrong or that we go in a spiral round the sun, would need to be backed up by citations. Wikipedia does not accept WP:Original research in its articles. The reference desk is for asking questions, it is not for proposing your answers in contradiction to Wikipedia articles. If you have an original solution of something that is notable enough to be in Wikipedia it needs to be published in a paper first before the Wikipedia article can be changed to reflect your research. Dmcq (talk) 15:03, 28 May 2010 (UTC)

• NO for.."predict Kepler's laws as has been mathematically and experimentally confirmed for the last 300 years". According my solution, Newton's law of (F*dt=m*dv) resulted in to (r=-a*t*(t-tmax)+K). If this is rigth, Wiki's derivation of Newton's law is wrong. If this is wrong, I have to understand where is the mistake (mathematical lines are there), before saying Wiki's derivation is rigth. I hope you are there to answer similar questions. When commenting r=-a*t*(t-tmax), I think this is a spiral. Canonically, I must comment "this is an ellipse" to have your favour (?!). I am a mechanic, diesel engine repairer, but I am on the field to discuss Wiki's derivation .I do not approve for example Wiki's "Kepler's area law" derivation. And I sincerely asked you if Vperpandicular is constant or variable. Wiki says variable. I say constant. Canonically,I must not ask such (ridiculous) question, otherwise I am wrong, but to be canonic doesn't mean to have rigth. I agree your clause WP:Original research. That was a challenge of discussion before acting for "The reference desk is for asking questions, it is not for proposing your answers in contradiction(!) to Wikipedia articles. If you have an original solution of something that is notable enough to be in Wikipedia it needs to be published in a paper first before the Wikipedia article can be changed to reflect your research", which I agree totally, but peer-reviewers may be old minded canonics,and I prefered to discuss this matter with Wiki's peoples,hoping that they are more democratic.Are you? TASDELEN (talk) 18:49, 28 May 2010 (UTC)

Seemingly I am an old minded canonical. Dmcq (talk) 20:55, 28 May 2010 (UTC)

We can't point out precisely what your mistake is - your entire argument is complete nonsense. It is one big mistake. You need to go away and learn the relevant maths before trying to use it. I am going to delete any further questions from you on this topic, you are wasting everybody's time. --Tango (talk) 21:05, 28 May 2010 (UTC)

As an ordinary (and I hope democratic) Wiki person, I note that you didn't reply to my attempt to be kind to you on the [Science desk]. I agree with Tango, and we could probably find several million other people worldwide who agree with Newton & Kepler (plus a billion who don't know what it is all about). Can you not find a more profitable outlet for your abilities? Dbfirs 15:22, 29 May 2010 (UTC)

May 28

What were her speeds please

I want to know the speeds that Kitty O'Neill did when she did a top speed of 412 miles an hour on her quarter mile run. I dont know how to work out these things, so I dont know how long her run took and I dont know what her average speed was. Can someone tell me how to work it out please —Preceding unsigned comment added by 125.239.204.103 (talk) 22:36, 28 May 2010 (UTC) The time it takes to do the quarter of a mile is call a ET but I dont kow what Kitty O'Neils ET was.

Try Kitty O'Neil.--RDBury (talk) 02:36, 29 May 2010 (UTC)

That is a lovely storey but it doesn't tell me how to work it out because it doesn't have a ET. My Pop says never mind because he knows a school teacher and will find out how its is worked out. I forgot to put my name before 125.239.204.103 (talk) —Preceding undated comment added 04:18, 29 May 2010 (UTC).

WikiAnswers says that the ET for the 412 mph run was 3.235 seconds, which gives an average speed of 278.2 mph. Gandalf61 (talk) 11:21, 29 May 2010 (UTC)

May 29

Is Axiom of Extensionality Converse true (special formulation)?

This is the typical form of the axiom of extensionality in ZF: $\forall x\forall y[\forall z(z\in x\Leftrightarrow z\in y)\Rightarrow x=y]$. The converse of this form is clearly true, and in fact this is explained on the ZF page.

However, a second formulation is also given on that page: $\forall x\forall y[\forall z(z\in x\Leftrightarrow z\in y)\Rightarrow \forall z(x\in z\Leftrightarrow y\in z)]$.

The converse of that formulation is $\forall x\forall y[\forall z(x\in z\Leftrightarrow y\in z)\Rightarrow \forall z(z\in x\Leftrightarrow z\in y)]$. How would one go about proving this? Is it usually taken as an additional converse axiom in this formulation? I am a beginner at set theory so I am curious about this alternative formulation of this axiom. —Preceding unsigned comment added by JamesMazur22 (talk • contribs) 00:40, 29 May 2010 (UTC)

Suppose x and y are such that your hypothesis holds, $\forall z(x\in z\Leftrightarrow y\in z)$. Can you show that x and y must in fact be equal? Hint: try substituting {x} for z. --Trovatore (talk) 00:44, 29 May 2010 (UTC)

(With that said, the "second formulation of extensionality" strikes me as a fairly silly way of phrasing the axiom. Unless it appears in that form in some standard reference, I'd go ahead and remove it from the page.) --Trovatore (talk) 00:54, 29 May 2010 (UTC)

Why should it be removed? The proof that the converse is true is quite simple. Assuming the hypothesis, as suggested by Trovatore, there is a set A of which x is a member by the axiom of pairing. Using the axiom schema of specification, there is a set B, such that $p\in B\Leftrightarrow p\in A\land \forall z[z\in p\Leftrightarrow z\in x]$. x is clearly a member of B (the double implication operator is reflexive). Under our hypothesis, y must be a member of B as well. Therefore, $y\in A\land \forall z[z\in y\Leftrightarrow z\in x]$. Using conjunction elimination we come to our final conclusion, $\forall z[z\in y\Leftrightarrow z\in x]$. Q.E.D. JamesMazur22 (talk) 13:49, 29 May 2010 (UTC)

I certainly had no mathematical objection to it. The proofs are trivial and there's no particular need to rehash them. My objection is that it looks like an overly clever attempt to make the axiom into a cute formal expression, with no real connection to its usage or meaning. Kind of thing just gets on my nerves a little bit. But I haven't looked up the rationale Algebraist mentions. Maybe I'll get around to it. --Trovatore (talk) 18:41, 29 May 2010 (UTC)

The article Zermelo–Fraenkel set theory gives some references for this version, as well as a reason for using such a perverse thing. Axiom of extensionality gives a bit more detail, without references. Algebraist 14:14, 29 May 2010 (UTC)

Maple 13

What is the output of "int(sin(cos(x)),x);" on your Maple 13 for Windows? The output on my Maple 13 for Windows is $\int sin(cos(x))\,dx$ without calculating. I also tried Maple 13 for Mac, and the output was a calculated result. Anything wrong on my computer?--百楽兎 (talk) 13:59, 29 May 2010 (UTC)

What that means is that Maple cannot find a simpler form for the indefinite integral than the integral itself. My own conclusion is the same, that there is no simpler expression in terms of standard functions. Dmcq (talk) 15:11, 29 May 2010 (UTC)

Perhaps your original problem was sin^-1(cos(x)) or sin(cos^-1(x))? Dmcq (talk) 15:20, 29 May 2010 (UTC)

Mathematica can't find a simpler expression either. Robinh (talk) 19:15, 29 May 2010 (UTC)

But Maple 13 for MAC can. That's why I confused. --百楽兎 01:18, 30 May 2010 (UTC)

What exactly did Maple 13 for Mac output? -- Meni Rosenfeld (talk) 08:07, 30 May 2010 (UTC)

Also, what does Mac13 do with Int( -sin(y) / sqrt(1-y^2), y ) ? - DVdm (talk) 10:06, 30 May 2010 (UTC)

What does this have to do with anything? You should wait for the OP to confirm that Maple receives magical integration powers from the Mac (rather than just having a mistake in the input) before hijacking the thread to solve your integral. -- Meni Rosenfeld (talk) 10:37, 30 May 2010 (UTC)

_{But I have to turn in my homework tomorrow morning! - DVdm (talk) 12:00, 30 May 2010 (UTC)}

He was just substituting y=cos(x) in the integral. Not that I think it'll do much good. Dmcq (talk) 12:11, 30 May 2010 (UTC)

Oh, I see. Sorry for the unfounded allegations. I obviously didn't suspect it was "homework", rather something you needed for some personal investigation. They don't give non-elementary integrals as homework.

I still don't understand why this query was important, though. -- Meni Rosenfeld (talk) 12:38, 30 May 2010 (UTC)

Forgive me for asking, but you are joking, right? DVdm (talk) 12:59, 30 May 2010 (UTC)

No. Which part? -- Meni Rosenfeld (talk) 13:20, 30 May 2010 (UTC)

I asked the OP what the output was for debugging purposes. You then asked about a seemingly unrelated integral (it didn't occur to me spontaneously that it's the same integral, substituted). I currently don't know why you asked that, but at the time I wrongly assumed that you were hoping that Maple has some unknown tricks and wanted to leverage it for your own purposes. The "magical powers" bit was humor of sorts, meant to show that this hope was unjustified. Then I saw Dmcq's explanation and your own reply, which I thought meant to use sarcasm to show why my assumption was absurd. -- Meni Rosenfeld (talk) 13:33, 30 May 2010 (UTC)

Ok, you are getting close. I'll leave it to you to figure out my real, deeper intensions. Cheers :-) - DVdm (talk) 13:42, 30 May 2010 (UTC)

I cleaned up the indentation of this nice dialogue for clarity. Bo Jacoby (talk) 06:56, 1 June 2010 (UTC).

Huh? You messed it up. Replies should be one level deeper than what they're replying to. See Wikipedia:Indentation. I've restored it to the original correct form. -- Meni Rosenfeld (talk) 07:05, 1 June 2010 (UTC)

I like the little bit of outdenting, it reminds me of the Mouse's Tale. Dmcq (talk) 09:15, 1 June 2010 (UTC)

What does F(x|theta) mean?

What does this mean; $F(x\mid \theta )\,\!$ ;in the article Admissible decision rule?--Wikinv (talk) 14:46, 29 May 2010 (UTC)

See Conditional probability density function and Conditional probability. DVdm (talk) 14:51, 29 May 2010 (UTC)

why cant we integrate (sinx)^2 and (cosx)^2 without using cos2x

why cant we integrate (sinx)^2 and (cosx)^2 without using cos2x identity and does this extend to (tanx)^2 with another identity? —Preceding unsigned comment added by Jenny1004 (talk • contribs) 15:42, 29 May 2010 (UTC)

Certainly one can integrate sin²x dx without using any double-angle formula:

$\int \sin ^{2}x\,dx=\int (\sin x)(\sin x\,dx),$

so let

$u=\sin x\,$

$dv=\sin x\,dx$

and then we get

$du=\cos x\,dx$

$v=-\cos x\,$

Now integrate by parts:

${\begin{aligned}\int u\,dv=uv-\int v\,du&=-\sin x\cos x+\int \cos ^{2}x\,dx\\&=-\sin x\cos x+\int 1-\sin ^{2}x\,dx\\&=-\sin x\cos x+x-\int \sin ^{2}x\,dx\end{aligned}}$

Next, add the integral to both sides:

$2\int \sin ^{2}x\,dx=-\sin x\cos x+x+{\text{constant}},$

and finally

$\int \sin ^{2}x\,dx={\frac {-\sin x\cos x+x}{2}}+{\text{constant}}.$

Michael Hardy (talk) 16:46, 29 May 2010 (UTC)

Differentiating a power of tan gives a next higher order polynomial in tan which is quite interesting. You can use that together with the integral of tan to get the integral of any power. No tan(2x) is needed. Dmcq (talk) 17:17, 29 May 2010 (UTC)

Hey,no way.....int\sin^2x is x/2-sin2x/4 So the above method wont hold good!!Rohit.bastian (talk) 11:22, 1 June 2010 (UTC)

But $\sin 2x/4=\sin x\cos x/2$, so they're the same. Staecker (talk) 11:55, 1 June 2010 (UTC)

OK, let me put the bottom line this way:

${\begin{aligned}\int \sin ^{2}x\,dx&={\frac {-\sin x\cos x+x}{2}}+{\text{constant}}\\[12pt]&={\frac {x}{2}}-{\frac {\sin(2x)}{4}}+{\text{constant}}.\end{aligned}}$

They're both the same thing. See trigonometric identity. Michael Hardy (talk) 13:42, 1 June 2010 (UTC)

Making 2+2=5

It was cited as one of the dumbest Yahoo! Answers questions of all time. Now I ask it here. Keeping as close to definitions in the usual fields as possible, is there an algebra or can we redefine Z such that 2+2=5? SamuelRiv (talk) 17:30, 29 May 2010 (UTC)

2 + 2 = 5, heh. --MZMcBride (talk) 17:48, 29 May 2010 (UTC)

That's it! Screw you guys - I'm leaving the project! ;) SamuelRiv (talk) 18:05, 29 May 2010 (UTC)

See Modular arithmetic. Compute modulo one. $\scriptstyle e^{2\pi i(2+2)}=e^{2\pi i\cdot 5}.$ Bo Jacoby (talk) 18:35, 29 May 2010 (UTC).

Depends on what you mean by "5". If you are willing to define 3 + 1 = 5, then, yes, you can have 2 + 2 = 5. (But note that then 5 + 1 ≠ 3 + 3, but rather 5 + 1 = 2 + 3, and you'll have to come up with a new symbol for "2+3".) However, if you want 1+1=2, 2+1=3, 3+1≠5 and 2+2=5, there's no way of doing it without breaking associativity of addition: 2+2 => 2+(1+1) => (2+1)+1 => 3+1, so if 2+2=5, that implies 3+1=5. -- 174.24.200.38 (talk) 21:18, 29 May 2010 (UTC)

If you round all the terms of 2.4+2.4=4.8 to the nearest integer, you get 2+2=5. That's the least convoluted way of doing it, I think. --Tango (talk) 21:25, 29 May 2010 (UTC)

Bo's response sets all integers = 1, effectively (mod 1), but it's a neat illustration. IP's is just symbolic, and I'm looking for a modification of Z (or Q or R or whatever) that doesn't necessarily have to preserve any of the field or group axioms. Tango - that's a pretty neat idea - it looks like you could make an algebra like ZxZ and have a function taking it to Z that looks to human eyes like a neat trick. But of course, it's still a trick where the actual math is done in the second dimension - the decimal place.

Maybe I'm more interested in a different question: can we *construct* a general closed algebra over something looking like addition that makes it such that 2+2=5 without 5=4. Upon thinking about it: I want a+a=b but b!=2a. Ugh, that's bad. The answer, if I want this structure to be at least homomorphic to one of the usual fields, is to introduce another operator * such that a+a = b != a*a, or in other words, 2+2 != 2*2. I guess in such a case * is arbitrary, though it has to be closed and probably distributive-associative, and it can only work over this "scalar" magma (so let's say we can't select elements arbitrarily from R - actually, that should be a rule in general, because R lets us do anything). SamuelRiv (talk) 18:39, 30 May 2010 (UTC)

Consider a random variable a, say, the outcome of flipping a coin. (Number of heads = 0 or 1). Then a+a may be the outcome (0 or 1 or 1 or 2) of flipping two coins, while 2a is twice the outcome (0 or 2) of flipping one coin. Then a+a is not equal to 2a. So

(0.5±0.5)+(0.5±0.5)=1±0.7

while

2·(0.5±0.5)=1±1 .

Bo Jacoby (talk) 07:20, 1 June 2010 (UTC).

Question about 3SAT and NP-Complete

When determining the complexity class of an algorithm solving 3SAT, would you consider the input size to be the number of literals or the number of variables? I ask since given n variables the max number of literals is a cubic polynomial in n. 67.163.183.146 (talk) 19:32, 29 May 2010 (UTC)

A power of 3 difference in the size makes no difference as far as being NP complete is concerned. Dmcq (talk) 19:58, 29 May 2010 (UTC)

I'm aware of that, though, I suppose putting NP Complete in the title was a tad unclear; I'm just curious which you would use for input size; sorry about the confusion:) Thanks:) 67.163.183.146 (talk) 20:23, 29 May 2010 (UTC)

I'd use the number of clauses rather than the number of variables as that seems to be what matters most in transforming it into other problems. Dmcq (talk) 22:43, 29 May 2010 (UTC)

May 30

Complex analysis questions

Plot of the function exp(1/z), centered on the essential singularity at z=0. The hue represents the complex argument

Are all bijective holomorphic functions taken over the entire complex plane linear? What about over the entire Riemann sphere? Also, does anyone know how this image was made? 74.14.109.234 (talk) 00:47, 30 May 2010 (UTC).

Yes and yes. Algebraist 00:50, 30 May 2010 (UTC)

Thank you. 74.14.109.234 (talk) 02:21, 30 May 2010 (UTC)

Bijective meromorphic functions on the Riemann sphere are linear fractional transformations. That's the more interesting fact. Then requiring them to be holomorphic merely says the denominator cannot vanish at any finite point, so the denominator is constant; hence the function is affine. Michael Hardy (talk) 02:55, 30 May 2010 (UTC)

What do you mean by how the image was made? It is a plot of $\exp(1/z)$, with a color function described in the image's page. It was created by a software that can create plots of functions (possibly Matlab). If you want more specific details you'll have to ask the person who created it, Functor Salad. -- Meni Rosenfeld (talk) 06:41, 30 May 2010 (UTC)

It has already been asked on his talk page and there is no answer. I don't need to know which program was used for this specific image, just how this type of plot could be done. 76.67.75.115 (talk) 16:34, 30 May 2010 (UTC)

In the highest generality, to make such a plot you go over each pixel, calculate the corresponding value of the function, calculate from it the required values of hue, saturation and luminance, and color the pixel accordingly. Some software will have functionality to facilitate this process, the syntax of which will depend on the software. Are you asking for an example for such a software, and the syntax for this software? -- Meni Rosenfeld (talk) 17:56, 30 May 2010 (UTC)

I was asking for a piece of software that could do this, but I found Sage. 76.67.79.184 (talk) 00:52, 31 May 2010 (UTC)

Algebraic expressions for the standard trigonometric functions

I noticed in the article for the hyperbolic functions that each function has an equivalent algebraic expression (such as the one for the hyperbolic sine, $\sinh(x)={\frac {e^{x}-e^{-x}}{2}}$). Do expressions like these exist for the regular trig functions (sine, cosine, tangent, etc.)? I realize that they can be represented by their respective Taylor series, but I'm looking to see if there exists some sort of a finite formula, like the aforementioned. — Trevor K. — 01:12, 30 May 2010 (UTC) —Preceding unsigned comment added by Yakeyglee (talk • contribs)

$\cos x={\frac {e^{ix}+e^{-ix}}{2}}$

$\sin x={\frac {e^{ix}-e^{-ix}}{2i}}$

Michael Hardy (talk) 02:53, 30 May 2010 (UTC)

If you looked at the definition of e^ix you would find that it is

$e^{ix}=cis(x)=\cos(x)+i\sin(x)$

then if you replace x with -x

$e^{-ix}=cis(-x)=\cos(-x)+i\sin(-x)$

$e^{-ix}=cis(-x)=\cos(x)-i\sin(x)$

Postscript: cis(x) is the notation I used while I was doing electrical engineering. It allows me to record on paper the lecture notes very quickly. 139.130.1.226 (talk) 20:52, 1 June 2010 (UTC)

Exam question thats stumping me...

Hi everyone, I was in an exam, and I came across this question that I wrote down because I had no idea how to solve it:

6sin(x) - 8cos(x) = 5

Having received my marks back from the exam, I know the answer is x = 180n + (-1)^n 30 degrees + 53 degrees 8 minutes. I have no idea why! Can anyone help? 110.175.208.144 (talk) 05:41, 30 May 2010 (UTC)

Hint: Rewrite the equation as $10(0.6\sin x-0.8\cos x)=5$. -- Meni Rosenfeld (talk) 06:33, 30 May 2010 (UTC)

Still not following, I'm afraid :( I don't really see how that helps me... 110.175.208.144 (talk) 06:52, 30 May 2010 (UTC)

Can you verify that the stated numbers are solutions to the equation? Can you draw the graph of f(x) = 6sin(x) - 8cos(x) - 5 ? Can you by brute force crudely estimate a solution to f(x) = 0 ? Perhaps Meni's hint shows a shortcut. (Sorry, we are not going to give the detailed answer away). Bo Jacoby (talk) 07:06, 30 May 2010 (UTC).

Somewhat more direct hint: The fact that the exam is asking this question implies that the syllabus presumably included List of trigonometric identities#Angle sum and difference identities. Qwfp (talk) 07:09, 30 May 2010 (UTC)

I can verify that the stated numbers in the answer are correct, as it is in both the textbook and the teacher's marking notes. And I know that you're not going to give the detailed answer :) I just need to get past this first step, methinks... And if it helps, Qwfp, the general formula we are given in class was "In general, the solution for sin(x) = sin(y) is given by x = 180 x n + (-1)^n x y where n is an integer". What's messing with my head is where they got to the 30 degrees and 53 degrees 8 minutes from... 110.175.208.144 (talk) 07:21, 30 May 2010 (UTC)

You can write $0.6\sin x-0.8\cos x$ as $\sin(x-\theta )$ where $\sin \theta$ is .8; I assume that's where the 53° come in. The angle whose sine is 5/10 = .5 is easy to find. You then have the sines of two expressions being equal. The expressions themselves being equal is one solution but others are obtained from identities which I'm assuming were covered in your class.--RDBury (talk) 07:27, 30 May 2010 (UTC)

So after RDBury's step I'll end up with sin(x - theta) = sin 30? EDIT: Never mind, I solved it :) Thanks heaps, guys :) 110.175.208.144 (talk) 07:38, 30 May 2010 (UTC)

Prove

Hello. Is there an easy way to prove: (${\overrightarrow {u}}$ × ${\overrightarrow {v}}$) ⋅ ((${\overrightarrow {v}}$ × ${\overrightarrow {w}}$) × (${\overrightarrow {w}}$ × ${\overrightarrow {u}}$)) = (${\overrightarrow {u}}$ ⋅ (${\overrightarrow {v}}$ × ${\overrightarrow {w}}$))²? Using algebraic vectors would be a nightmare. Is there any property of dot or cross products that can simplify this identity? Thanks in advance. --Mayfare (talk) 12:42, 30 May 2010 (UTC)

According to Triple product, you have $a\cdot (b\times c)=b\cdot (c\times a)$ and $(a\times b)\times (a\times c)=(a\cdot (b\times c))a$. Using these, the derivation is fairly straightforward. -- Meni Rosenfeld (talk) 13:09, 30 May 2010 (UTC)

Matrices to the power n in an n-dimensional complex vector space

Hi everybody,

Is it necessarily the case that if A and B are complex n x n matrices and (AB)^n = 0, then (BA)^n must also be 0?

I feel like I should be able to come up with a counterexample but having looked at the case n=2 I can't spot anything - this is just one of 5 parts of a question intended to take half an hour or less on an exam, so it shouldn't take long ("state a counterexample or provide a proof") - I tried looking at the rank briefly to produce a proof but I'm not convinced it's true, so I'd appreciate any suggestions you could make.

Thanks a lot, 131.111.185.68 (talk) 16:39, 30 May 2010 (UTC)

This is true. Proof outline: The only eigenvalue of AB is 0; the only eigenvalue of BA is 0; Cayley–Hamilton theorem. -- Meni Rosenfeld (talk) 18:07, 30 May 2010 (UTC)

Thankyou Meni, once again and indeed as always you've been a great help :) 82.6.96.22 (talk) 18:17, 30 May 2010 (UTC)

You're welcome :) -- Meni Rosenfeld (talk) 18:40, 30 May 2010 (UTC)

Have also a look to Characteristic polynomial#Characteristic polynomial of a product of two matrices --pma 21:00, 30 May 2010 (UTC)

May 31

Notational Inconsistencies in Vectors

I'm confused about the use of $|$ in notations for the length of a vector. According to the Dot product article:

$|\mathbf {a} |={\sqrt {\mathbf {a} \cdot \mathbf {a} }}$

However, the Norm_(mathematics)#Euclidean_norm article writes:

$\|\mathbf {x} \|:={\sqrt {x_{1}^{2}+\cdots +x_{n}^{2}}}$

So my question is in a couple of parts:

1. Is there a difference between ||a|| and |a|? Is one using the wrong notation?
2. If not, what's the difference between the two?

• The dot product article says: a.a is the square of the length of a (which is written as ||a|| on the Norm article).
• This implies that:

$|\mathbf {a} |={\sqrt {\|\mathbf {a} \|^{2}}}$

which would imply that ||a|| and |a| are the same.

Thanks in advance --Philipwhiuk (talk) 12:22, 31 May 2010 (UTC)

There's no real difference between the two, it's just a different notation for the same thing. Some people reserve |x| for (real or complex) absolute value and denote more general norms with ||x||, some don't bother and use the simpler |x| for all norms.—Emil J. 12:36, 31 May 2010 (UTC)

Many thanks for your quick reply EmilJ. --Philipwhiuk (talk) 12:39, 31 May 2010 (UTC)

... and, of course, $|\mathbf {a} |={\sqrt {\mathbf {a} \cdot \mathbf {a} }}={\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}$ for a 3-D vector $\mathbf {a}$. Dbfirs 16:38, 31 May 2010 (UTC)

Percentage increase

if I have four things, and I add another thing, is that usually called a 25% increase in things (a percentage of the original amount) or a 20% increase (a percentage of the final amount)? 81.131.69.251 (talk) 18:47, 31 May 2010 (UTC)

The former, i.e. a 25% increase. —Qwfp (talk) 18:51, 31 May 2010 (UTC)

Ta. 81.131.69.251 (talk) 18:53, 31 May 2010 (UTC)

Please help with a couple questions I've been stuck on for weeks! Curvature, projective modules & metric spaces

Hey everyone, I'm revising for my exams (hooray) at the moment, and having arduously completed about 10 years of past papers, I've been left with about 4 or 5 questions which I just can not for the life of me finish. I know you're meant to post all your working etc. here but as I said, these are the only questions left I can't get my head around, so there's very little working I've managed to achieve or else I would probably have been able to finish the problem with enough perseverance.

I appreciate I'm asking a for lot of help, so if you just have even 1 suggestion for any of these problems which could be of use then please do shout, the more of these I can get sorted before tomorrow the better, and of course I don't expect you to have to go out of your way to pick up my slack when I can't figure something out! :)

1. (Final part of a long question) Show that a finitely generated projective module over a principal ideal domain is free. I've shown that every free module over an arbitrary ring is projective, where here I'm using the definition that an R-module p is projective if whenever we have module homomorphisms $f:M\to N$ and $g:P\to N$, with f surjective, there exists a homomorphism $h:P->M$ with $f\circ h=g$.

2. Let (X,d) be a metric space with at least 2 points. If $f:X\to \mathbb {R}$ is a function, we define Lip(f) $Lip(f)\,=\,sup_{x\neq y}{\frac {|f(x)-f(y)|}{d(x,y)}}+sup_{z}|f(z)|$, provided this supremum is defined. Now with $X=\mathbb {R}$, suppose $(f_{i})_{i=1}^{\infty }$ is a sequence of functions with $Lip(f_{i})\leq 1$ and with the property $f_{i}(q)$ converges as i tends to infinity, for every rational q. Show the $f_{i}$ converge pointwise to a function f satisfying $Lip(f)\leq 1$. Suppose also $(f_{i})_{i=1}^{\infty }$ are any functions with $Lip(f_{i})\leq 1$. Show there is a subsequence $f_{i_{1}},\,f_{i_{2}},...$ which converges pointwise to a function f with $Lip(f_{i})\leq 1$. Where to start? I showed that Lip(X) is a vector space over $\mathbb {R}$ and that Lip is a norm on it, and then from then on I have no clue where to go. I tried to work out how to treat the irrational numbers in the first part and I presume it utilizes the density of the rationals, but I haven't been able to get anywhere at all from there - it was a hard year for Analysis and this was definitely the worst question, completely off the course.

3. Deduce (from the fact that the Gaussian curvature K of a surface locally given by the graph of an infinitely differentiable function $f:U\to \mathbb {R}$, where U is an open subset of $\mathbb {R} ^{2}$, is equal to $K={\frac {f_{xx}f_{yy}-f_{xy}^{2}}{(1+f_{x}^{2}+f_{y}^{2})^{2}}}$ - this was the first part of the question, so this is assuming I calculated that right!) that if $\Sigma \subset \mathbb {R} ^{3}$ is a compact surface without boundary, its curvature is not everywhere negative. Give with justification a compact surface in $\mathbb {R} ^{3}$ without boundary whose Gaussian curvature must change sign. I can see that the curvature is negative if and only if the second fundamental form is negative, but I'm not sure if they're looking for a geometrical argument here or an algebraic one - personally I'd prefer the latter if it really is a 'deduce' and not a 'show'. In addition, perhaps I'm getting confused, but I thought a compact surface had to be closed and bounded, in which case surely it necessarily has a boundary? I don't think I can do the last part without understanding what they mean by 'without boundary' - could we just for example attach a hemisphere to part of a hyperboloid (in a sortof 'icecream cone' fashion) to get a surface with both positive and negative curvature, or is there more to it than that?

As above, I'm sorry this post is so long and wordy, but I've spent weeks and weeks doing past papers and I've whittled it down to just these questions to ask for help on, so if you can suggest anything in any case which might be of use in any of the 3 questions please don't hesitate, even if you only have time for a quick comment. I would say that questions 2 and 3 are probably the most pressing, since I was able to tackle the majority of Q1 without any problems. Thankyou so so much in advance! :-) Otherlobby17 (talk) 18:49, 31 May 2010 (UTC)

I think I have some suggestions for part 2 of question 2 Suppose also $(f_{i})_{i=1}^{\infty }$ are any functions with $Lip(f_{i})\leq 1$. Show there is a subsequence $f_{i_{1}},\,f_{i_{2}},...$ which converges pointwise to a function f with $Lip(f_{i})\leq 1$. We just need to find a subsequence that converges for all rationals then use the first part. All the fi must have values within [-1,1] since their Lip is less than 1. By the Tychonoff theorem, [-1,1]^(omega) is compact; it's also first countable so it's sequentially compact. Now enumerate the rationals in the real numbers as q1,q2,q3... And write out the q1,q2,q3... horizontally and the f1,f2,f3... vertically like a multiplication table, and then evaluate each fi at qj.

Then this is a sequence of sequences in [-1,1]^(omega) (the rows are the elements of [-1,1]^(omega)), and so has a convergent subsequence. This subsequence is the one you want. I might be wrong I’m only a student. I think user:pma has expertise in this area Money is tight (talk) 04:16, 1 June 2010 (UTC)

June 1

Find extremum

I remember from calculus (a looongg time ago ;) that to find if a point is an extremum of a function you take increasinly higher order derivatives until the result is a nonzero constant. I've come across this again, but I'm stuck. How do you do that with something like sin(2x)? That will never differentiate to a constant! Of course I know that from the basic properties of the sin function that the extrema are at 45, 135, etc., etc. in my example, but the problem I have to deal with may be much more complex than that.

The higher order derivatives don't have to be a nonzero constant. They just need to have a nonzero value at the point you are evaluating. For $\sin(2x)$ at $x_{0}=\pi /4$, differentiating once gives $2\cos(2x_{0})=0$; once more gives $-4\sin(2x_{0})=-4<0$, so this is a local maximum. -- Meni Rosenfeld (talk) 18:33, 1 June 2010 (UTC)

Our article is here: higher-order derivative test.—Emil J. 18:36, 1 June 2010 (UTC)

Crop circles of pi and Euler's identity

How does this crop circle work? Kittybrewster ☎ 20:00, 1 June 2010 (UTC)

What crop circle? -- Meni Rosenfeld (talk) 20:46, 1 June 2010 (UTC)

As shown in google search +crop circle +eulers identity Kittybrewster ☎ 21:26, 1 June 2010 (UTC)

explanation here [3]. why are aliens communicating in ASCII and not Unicode ? 87.102.77.88 (talk) 23:17, 1 June 2010 (UTC)

Thank you. Very comprehensive response. Kittybrewster ☎ 05:05, 2 June 2010 (UTC)

As shown in google search +crop circle +pi +2008 +Wroughton. Kittybrewster ☎ 05:05, 2 June 2010 (UTC)

Maximum/Minimum Irradiance for "Human habitable" planet temperatures?

(Third time's a charm)
I keep trying to create a simple model for anyone who wants to speculate on a planet having
habitable temperatures for people (not extremophiles) based on Irradiance.
Take for example HD 38801 b:
Star Radius = 2.53 sol
Star Te = 5222 K
Stefan–Boltzmann constant, σ = 5.67051E-8
Semi-major axis = d, in this case 1.7
Eccentricity = e, in this case 0
Emissivity = ε, (Earth=0.62009)
Albedo = A, (Earth=0.3)

=((((R^2)*σ*(Te^4)*(1-A))/(4*ε*(d±(d*e))^2))^0.25)-273.15
So I can get a global annual average temperature for a planet,
If I assume a global average albedo and a global average emissivity

	Albedo
εmissivity	0.20	0.25	0.30	0.35	0.40	0.45
0.80	5°C	1°C	-4°C	-9°C	-14°C	-20°C
0.75	10°C	5°C	0.7°C	-4°C	-10°C	-15°C
0.70	15°C	10°C	5°C	0.3°C	-5°C	-10°C
0.65	20°C	16°C	11°C	5°C	-0.1°C	-6°C
0.60	26°C	21°C	16°C	11°C	5°C	0.5°C
0.55	33°C	28°C	23°C	17°C	11°C	5.4°C

Ignoring the perspective that single value Albedo and Emissivity are very
simplistic, since "Global Annual Average Temperature" is also but it exists:
A) at what levels of irradiance (max/min) does this become moot?
B) what combinations of Albedo & Emissivity are unrealistic or possible??

Just to be clear I am talking about people. The GAAT (14°C) is a surface temperature.24.78.167.139 (talk) 20:45, 1 June 2010 (UTC)

I came upon this chart in a search and realized this illustrates well the idea that there are a lot of possibilities but some are unrealistic. - 24.78.167.139 (talk) 20:45, 1 June 2010 (UTC)

This isn't really the right forum for that question. We cover mathematical physics but this seems more like general science or astronomy.--RDBury (talk) 00:45, 2 June 2010 (UTC)

"Astrophysics (...) is the branch of astronomy that deals with the physics of the universe, ..." 24.78.167.139 (talk) 03:33, 2 June 2010 (UTC)

June 2

weakly normalised

In some articles like this there are some assumptions that a function need to be weakly normalised. I didn't found any explanation for this. What it is mean? (This is not about something of Lambda calculus) --77.124.146.172 (talk) 05:34, 2 June 2010 (UTC)