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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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).[reply]

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

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)[reply]

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)[reply]

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(n²k_ncos(nt))| x {a²+(Sum(k_ncos(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)[reply]

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)[reply]

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).[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

'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)[reply]

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

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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

....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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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