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March 26

Distribution of points on a sphere

I've been using this algorithm for coming up with a uniform distribution of points on an n-sphere (Python notation):

X = [random.gauss(0, 1) for i in range(n)]
t = math.sqrt(sum([i**2 for i in X]))
Y = [X[i] / t for i in range(n)]
return Y

Does the resulting Y distribution have a name, or can a closed form be derived for it? 70.190.182.236 (talk) 20:39, 26 March 2014 (UTC)

This is the unique rotationally-invariant probability measure on the sphere. I'm not sure it has a name as such, but it can be obtained as a symmetry reduction of the Haar measure on the group of rotations SO(n). The fact that your probability distribution has this property follows from the fact that if $X_{1},\dots ,X_{n}$ are iid (standard) Gaussian random variables, then the n-dimensional random variable $X=(X_{1},\dots ,X_{n})$ is normally distributed with pdf ${\frac {1}{(2\pi \sigma ^{2})^{n/2}}}\exp(-x^{T}x/2\sigma ^{2})$, which is rotationally invariant. Sławomir Biały (talk) 21:02, 26 March 2014 (UTC)

Note that this topic was discussed here before (although not specifically what to name the method you describe): See Wikipedia:Reference_desk/Archives/Mathematics/2007_August_29#Distributing_points_on_a_sphere and Wikipedia:Reference_desk/Archives/Mathematics/2010_April_30#Even_distribution_on_a_sphere. StuRat (talk) 21:16, 26 March 2014 (UTC)

I think those are about optimally distributing multiple points on a sphere, whereas the OP is asking about the probability distribution of a single point chosen "uniformly at random" on a sphere. Sławomir Biały (talk) 21:20, 26 March 2014 (UTC)

That's a very nice algorithm, thanks very much 70.190.182.236 and Sławomir. Dmcq (talk) 15:28, 27 March 2014 (UTC)

There might be a slight bias to the output due to limits of machine floating point operations. I'm not really an expert on such things, but I think it would be prudent to drop the cases where the norm t is either very small or very large. Sławomir Biały (talk) 17:08, 27 March 2014 (UTC)

I formatted the code for easier reading; I hope you don't mind. By the way I'd use

Y = [x/t for x in X];

no need for range there. —Tamfang (talk) 07:27, 28 March 2014 (UTC)

See also b:Mathematica/Uniform_Spherical_Distribution, which seems as sensible a name as any; it also gives explicit formulas in terms of θ and φ. --Tardis (talk) 08:46, 30 March 2014 (UTC)

phi := 2 ArcSin[Sqrt[Random[]]] (for colatitude) is weird, with an unnecessary sqrt; I don't understand the voodoo that got there (being unacquainted with Jacobians), and haven't worked out whether it matches the formula that I've always used: arcsin(1-2*random()), if the output of random() is between 0 and 1 – an application of Archimedes' theorem that the area between two planes both perpendicular to the axis of a cylinder is the same on the cylinder as on an inscribed sphere (if each plane cuts the sphere in at least one point). —Tamfang (talk) 04:50, 31 March 2014 (UTC)

March 27

Lottery probability

Hi

let's say in a lottery game one has to choose k numbers out of N. In such situation, it's obvious the probability to choose exactly the right k numbers is 1/C(N,k). But How do I find the chanaces to choose correctly just j (j<=k) of the k numbers (I am looking for the derivation of the answer)? Thanks, 212.179.61.122 (talk) 09:15, 27 March 2014 (UTC)

To be sure I understand: you choose k numbers, the lottery draws k numbers, and you win if you have at least j in common? In that case, let's start by finding the probability of having exactly j in common, which we'll do by counting the number of ways to have j in common. First, I select the common j from the k which were drawn: C(k, j). Then I fill in the remaining k - j from the undrawn N-k: C(N-k, k-j). So ${\frac {C(k,j)\cdot C(N-k,k-j)}{C(N,k)}}$ is the probability of getting exactly j in common. If you want at least j in common, then sum from j to k: $\sum _{i=j}^{k}{\frac {C(k,i)\cdot C(N-k,k-i)}{C(N,k)}}$.--80.109.80.78 (talk) 09:59, 27 March 2014 (UTC)

Thank you very much. Is it actually the hypergeometric distribution? 212.179.46.21 (talk) 11:22, 27 March 2014 (UTC)

Well, it's a special case. Using the notation of that page, you have K = n.--149.148.255.195 (talk) 13:54, 27 March 2014 (UTC)

March 28

1 x 10¹¹

I found the number "1 x 10¹¹" ; my question is, how is it different from "10¹¹" and "1.0 x 10¹¹" ?

• I think 10¹¹ is the exact number 100,000,000,000.
• I'm sure that 1.0 x 10¹¹ indicates more significant figures than 1 x 10¹¹. 1.0 x 10¹¹ has an uncertainty of 0.05 x 10¹¹, it is between 95 billion and 105 billion.
• 1 x 10¹¹ is between 95 billion (closer to 100,000,000,000 than to 9 x 10¹⁰ = 90,000,000,000) and 150 billion (closer to 100,000,000,000 than 2 x 10¹¹ = 200,000,000,000).

Is that usage/interpretation, esp. the latter, correct? - ¡Ouch! (^{hurt me} / _{more pain}) 07:21, 28 March 2014 (UTC)

This all seems consistent with Significant_digits#Scientific_notation. But really, this stuff can be context dependent, and there may be different customs of notation used by different authors. If we cant to be clear, we can always just use explicit plus or minus symbols, e.g. $10^{11}\pm 500$, rather than writing out $1.000000000\times 10^{11}$ to make use of a convention. SemanticMantis (talk) 15:27, 28 March 2014 (UTC)

The bands on either side due to limited precision are usually symmetrical, so on that basis 1 x 10¹¹ would be between 50 billion and 150 billion.→86.173.216.149 (talk) 19:11, 28 March 2014 (UTC)

March 30

Continuously differentiable function

Hi,

This is a homework question, but I'm hoping just for a hint or two, or confirmation I'm doing it correctly. We were given a piecewise function, defined by f(x)= x^3 if x>=0, -x^3 if x<0. The first part was to prove that f was differentiable, and then it asks what order is f continuously differentiable. I've been finding the derivatives of f by using the definition at x=0, and I've been checking continuity of the derivative function using the definition of continuity, i.e. left side limit=function value=right side limit. Then I repeat for the next derivative (f', f⁽²⁾x, f⁽³⁾x). This seems very tedious- is there another way of doing it? Once I get to f⁽³⁾x=0, I can conclude that f is continuously differentiable of infinite order, right? Or am I missing something...

Thanks in advance! 27.32.104.185 (talk) 12:11, 30 March 2014 (UTC)

If you get f⁽³⁾(x)=0 in this case then you're doing something wrong. Also, for real functions, differentiable, twice differentiable, thrice differentiable, ..., infinitely differentiable and analytic are all different, of which your problem is partial example. For complex functions things are (ironically) a bit simpler.--RDBury (talk) 16:05, 30 March 2014 (UTC)

You might be taking an excessively tedious approach. The derivatives are (or should be) easy to find whenever x ≠ 0 without resorting to the tedious approach, and you can make more intuitive inferences at x = 0 from this approach. —Quondum 18:25, 30 March 2014 (UTC)

Sorry, I meant f⁽⁴⁾(x)=0. I know apart from x=0 the power rule suffices, but what other way could you find f'(0), etc other than using the definition of the derivative on both sides? Can we assume f'(0)=lim f'(x) as x approaches 0? Could you explain what you mean by intuitive inferences? 129.94.8.150 (talk) 00:53, 31 March 2014 (UTC)

I mean that simply by plotting the derivative of successive orders you'll quickly get a feel for what's what. And you would not be correct to say that f⁽⁴⁾(0)=0. To see that, you need to find f⁽³⁾(x). —Quondum 02:24, 31 March 2014 (UTC)

• The approach is perhaps excessively tedious in the sense that an advanced mathematician looking at this would immediately know the answer without having to do anything -- but the tedious approach is what you are expected to do as a student. But if you are coming to the conclusion that the function is continuously differentiable of all orders, you've made a mistake -- f⁽³⁾ is not continuous. Looie496 (talk) 02:37, 31 March 2014 (UTC)

We don't know what your teacher asked for in terms of work shown. They might prefer you to show an epsilon-delta proof of the derivative for all points in the domain (but I doubt it). Anyway, the "least work" is to show a that a certain derivative of f is not continuous. Demonstrating lack of continuity at a point rigorously can be bit tricky, but in this case your teacher probably doesn't want that either. As for "can we assume.." -- no, not in general, unless you've already shown that f' is cts at x=0. SemanticMantis (talk) 15:32, 31 March 2014 (UTC)

Thanks, I got it. And I realized the simple mistake I've been doing as I differentiated it. Thanks everyone. 27.32.104.185 (talk) — Preceding undated comment added 05:29, 1 April 2014 (UTC)

Another Curios "Coincidence" Between The Superellipse x³ + y³ = 1 And The Gaussian function

As before, my question is whether there's more to this than meets the eye, especially given the general relationship between Gaussian integrals and superellipses:

$\int _{0}^{1}{\sqrt[{2}]{1-x^{2}}}~dx~=~{\frac {\pi }{4}}$

$\int _{0}^{1}{\sqrt[{3}]{1-x^{3}}}~dx\approx {\sqrt {\frac {\pi }{4}}}~=~\int _{0}^{\infty }e^{-x^{2}}dx$

The error in that last approximation is of about ⁠1/344⁠ . For yet another "coincidence" between the two, see here. My only explanation so far is that all three functions (Gaussians, cubic superellipses, and hyperbolic secants) are bell-shaped, and therefore can be brought close to each other in a manner similar to that of Laplace's method. — 79.113.195.131 (talk) 14:37, 30 March 2014 (UTC)

Name for operation on convex polytopes

The cone of a convex polytope P in a vector space V can be defined as a the subset of R×V given by {(t, {1-t)p): p∈P, t∈[0, 1]}. I found it useful to generalize this as follows: Let P₁ and P₂ be convex polytopes in vector spaces V₁ and V₂. Define a convex polytope Q in R×V₁×V₂ by Q={(t, {1-t)p₁, tp₂): p₁∈P₁, p₂∈P₂, t∈[0, 1]}. This reduces to the cone of P₁ when P₂ is a single point. More generally, If P_i has v_i vertices, f_i faces and dimension d_i then Q has v₁+v₂ vertices, f₁+f₂ faces and dimension d₁+d₂+1. Does the operation (P₁, P₂)→Q (up to affine variations in the definition) have a name or standard notation? --RDBury (talk) 16:20, 30 March 2014 (UTC)

This looks like a join (topology). --Mark viking (talk) 16:31, 30 March 2014 (UTC)

Thanks, it looks like topologically it is the join. So I'll go with "polytope join" unless someone has a more standard name. I also found prismatoid which is slightly similar but not the same. The combinatorial equivalence class of the join depends only on P₁ and P₂ but for the prismatoid it depends on how they are oriented. --RDBury (talk) 19:17, 30 March 2014 (UTC)

April 1

Other extension of cone to 'hypercone'

The article Hypercone extends the concept of a cone z^2=x^2+y^2 to z^2=w^2+x^2+y^2. However, I'm interested in the other extension, w^2+z^2=x^2+y^2. Does it have a name?

While I can imagine the hypercone sections fairly clearly by doing t^2=x^2+y^2+z^2 (t being time giving shrinking then expanding sphere), I'm having problems with imagining the sections of the other extension. Any suggestions?Naraht (talk) 14:12, 1 April 2014 (UTC)

You should be able to visualize one set of parallel sections as a hyperboloid of revolution, scaling as with the sphere. —Quondum 14:28, 1 April 2014 (UTC)

The "other" has parallel sections corresponding to hyperboloids of revolution of one sheet with t=0 corresponding to the infinite double cone and the abs(t) increasing "expanding the hole" in the middle of the one sheet.

OTOH, the intersections for the hypercone if the variable with is treated like t is on the other side (z^2=x^2+y^2+t^2), then you get the hyperboloids of revolution of two sheets "inside" the cones...

I think a standard name would be ultrahyperbolic cone (or ultrahyperbolic null cone). More generally, you can consider the null cone associated to any nondegerenerate quadratic form. Over the reals, these are characterized by their signature. The (space like) cross sections are always (affine) quadrics. The null cross sections are actually just affine spaces. Sławomir Biały (talk) 19:39, 1 April 2014 (UTC)

Is my statistics book wrong?

Hi

I've encounterd the following question: "three balls are chosen randomly from an urn containing 7 black balls and one yellow balls. What is the probability that a yellow ball was chosen (at least one time)? The book claims the answer is C(7,2)/C(8,3) = 0.375. But doesn't this answer assumes the black balls are distinguishable, while they are not? I've run a simualtion of this problem in Python, and after 10,000,000 "experiments" the probability stabilizes on 0.330, not 0.375. Thanks! 212.179.46.23 (talk) 08:43, 2 April 2014 (UTC)

The question is ambiguous, but the "(at least one time)" condition implies to me that the balls are selected with replacement: i.e. choose one, put it back , choose another, etc. In that case the probability choosing at least one yellow is 1 - (probability of choosing all black) = 1 - (7/8)^3 ~= 0.33, as in your simulation. If the balls are chosen without replacement then the answer given in the book is correct: there are C(8,3) possible choices, of which C(7,2) include the yellow ball. AndrewWTaylor (talk) 09:05, 2 April 2014 (UTC)

Yep I wonder if the original question included the "at least one time" bit? If it did then it certainly was posed wrong as it implies replacement. Dmcq (talk) 09:09, 2 April 2014 (UTC)

Sorry - It didn't include "at least on time", it's my fault. Yet, doesn't the answer in the book assumes the black balls are distinguishable, while they are not? 212.179.46.23 (talk) 09:32, 2 April 2014 (UTC)

Whether or not the blackballs are distinguishable has no effect on the answer. In general, assuming objects are distinguishable won't affect probability, it will only affect questions that begin "In how many ways can you..."--80.109.80.78 (talk) 09:52, 2 April 2014 (UTC)

Here's a way to calculate the answer without assuming distinguished balls: You have a 1/8 probability of drawing the yellow ball with your first draw and a 7/8 probability of drawing a black ball. If you draw a black ball, there are 7 remaining balls, so you have a 1/7 probability of drawing the yellow ball on your second draw, and a 6/7 probability of drawing a black. If you drew a black for your first and second ball, you have a 1/6 probability of drawing the yellow with your third draw. So (1/8) + (7/8)(1/7) + (7/8)(6/7)(1/6) = 3/8 = .375.--80.109.80.78 (talk) 09:57, 2 April 2014 (UTC)

That's a nice way of looking at it - basically a Tree diagram (probability theory). Generalising it shows that with $r$ choices (without replacement) from $n$ balls, one of which is yellow, the probability of choosing the yellow is $r/n$. I'm not sure whether that result is surprising or obvious... AndrewWTaylor (talk) 18:12, 2 April 2014 (UTC)

It's obvious if you think about it backwards: you draw your $r$ balls first, then the yellow one decides which ball it's going to be.--80.109.80.78 (talk) 19:41, 2 April 2014 (UTC)

Equation to fit a curve

I found a table of measured emissivities vs temperature for a metal in an old text book. I plotted them on a graph at http://i60.tinypic.com/2cy2wj5.jpg. Is there a simple formula for e as a function of T that, with suitably chosen constants, will closely conform to this graph? Below about 1900 K, e = 0.0000338T^1.18 is an accurate fit and above about 2350 K, e = 0.00606T^0.5 is a close fit, with a smooth transition between 1900 and 2350 K. Unfortunately, the textbook does not give an explanation of why emissivity follows this curve, not does it have any math. 120.145.97.4 (talk) 13:24, 2 April 2014 (UTC)

Emissivity is in general a function of the material, the surface condition of the material, the temperature of the material, the orientation of the material surface relative to the pyrometer, etc. People have related the empirical thermal dependence of emissivity to the empirical spectral dependence of emissivity, but there is no general theory that I know of for this. With regard to the plot, it is a mice smooth curve; such curves can be well-fit by for instance splines. --Mark viking (talk) 18:46, 2 April 2014 (UTC)