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September 15

Specific parametrization of en ellipse

Define an ellipse in the usual way, with x(t) = a*cos(t), y(t) = b*sin(t). Does there exist a re-parametrization such that x = f(s), y = g(s) and (df / ds)^2 + (dg / ds)^2 = 1, and if so how can it be found and what is it? 24.255.17.182 (talk) 04:20, 15 September 2015 (UTC)[reply]

Or more generally finding such a parametrization for any curve. 24.255.17.182 (talk) 04:29, 15 September 2015 (UTC)[reply]

Of course there is. Not for any curve, but for many of them. ;)
The derivatives-related equation just says the (x, y) = (f(s), g(s)) point moves with a constant speed as it travels along the curve (variable s is time here). --CiaPan (talk) 07:51, 15 September 2015 (UTC)[reply]

This is called the arc length parameterization. If r(t) is your parameterization, define s(t) to be the arc length between r(0) and r(t). Then

r(s^{-1}(t))

is a re-parameterization of the sort desired. It's usually not possible to calculate, though.--Antendren (talk) 08:34, 15 September 2015 (UTC)[reply]

I see... so for an ellipse you would have to calculate the inverse of the elliptic integral? 24.255.17.182 (talk) 14:34, 15 September 2015 (UTC)[reply]

For ellipses, you end up with elliptic functions. —Kusma (t·c) 09:30, 15 September 2015 (UTC)[reply]

For WP coverage, see Differential_geometry_of_curves#Length_and_natural_parametrization, for external sources, you might find these helpful [1] [2] [3].

Compass equivalence theorem – in Polish?

Anybody knows what is the name of the Compass equivalence theorem in Polish? --CiaPan (talk) 08:25, 15 September 2015 (UTC)[reply]

Kompas Twierdzenie równoważności. --31.177.99.241 (talk) 22:22, 15 September 2015 (UTC)[reply]

Don't waste my time, please. Use your Polish language knowledge (if any) instead of automatic Google quasi-translator.
That something does not make much sense in Polish due to grammar inconsistency. Anyway even if it was grammatically correct, it's nonsense — this kind of 'compass' does not translate to 'kompas'... --CiaPan (talk) 06:11, 16 September 2015 (UTC)[reply]

I couldn't find any specific name for this theorem in Polish. As our article about it says, the theorem is equivalent to Proposition II of Book I of Euclid's Elements, so perhaps the only way to refer to it in Polish is "drugie twierdzenie księgi pierwszej Elementów", but it yields only one Google hit. — Kpalion^(talk) 16:09, 16 September 2015 (UTC)[reply]

Thank you. The same I found myself. I feel it would be nice to have an article in pl-wiki under a proper name of the theorem, however it seems I would have to invent the name myself—and that is forbidden, as WP:OR

sin, cos, tan when θ is 0° or 90°

What does it mean actually? In a right triangle there is one 90°, so θ cannot be 90° or 0°. --31.177.99.241 (talk) 22:18, 15 September 2015 (UTC)[reply]

There are at least three ways to answer this, all producing the same result.

1. You can choose to define the trig functions of these specific angles as limits. sin 0° means the limit of sin x as x approaches 0.

2. You can choose to define the trig functions as a numerical value computed by summing an infinite series. (Summing an infinite series also involves a limit. See here for specific series that will do the job.)

3. You can say that the right triangle can be degenerate, with three vertices and three sides but two vertices are the same point, two sides are the same line, and the third side has length 0.

--65.95.178.150 (talk) 00:19, 16 September 2015 (UTC)[reply]

As you can see e.g. by looking at the article sine, there are multiple possible definitions of the trigonometric functions. The right triangle definition only makes sense for positive acute angles, but this definition can be replaced by the unit circle definition to apply to any real angle. (And then this could be further replaced by a definition that makes sense for any complex number, though in this case you lose the natural connection to geometry.) Of course the two definitions are designed to agree when they both make sense. --JBL (talk) 00:19, 16 September 2015 (UTC)[reply]

Notice that tan 90 does not mean anything, since it is sin 90/cos 90, which is 1/0.--Scicurious (talk) 02:16, 16 September 2015 (UTC)[reply]

"Is not a real number" and "does not mean anything" are not synonyms. --JBL (talk) 03:47, 16 September 2015 (UTC)[reply]

If you're up for a bit of advanced math, you might be interested in the concept of analytic continuation. Basically, it's a process by which you take some function that's defined over only a limited domain, and then find a "more general" function which is defined over a bigger domain, but is identical to the original function in that sub-domain where the original function is valid. A classic example of this is the gamma function, which can be considered to be the analytic continuation of the factorial function. By their original definition, factorials are only defined for positive integers. For positive integers, the gamma function produces the same results as the factorial function, but because of how the gamma function is defined, it also produces values for non-integer real numbers as well. Frequently, mathematicians will actually redefine the original function to be it's analytical continuation, so it doesn't suffer from the domain limitations the original definition has. For example, take exponentiation. In the "original" definition, exponentiation is repeated multiplication: xⁿ = x*x*x ... *x*x. But how would you compute terms with arbitrary real exponents? What does 0.5134 of a multiplication by x even mean? Well, it turns out that there's this other function (the inverse of the anti-derivative of 1/x) which has all the same properties as exponents. So we can then just *define* this function as e^x, for any arbitrary real number "x", and then use that definition and the the properties of exponents to calculate arbitrary exponents. (A property you really want for an analytical continuation which is to be redefined as the actual function is that multiple lines of reasoning all point to the same definition. It's like that for exponentiation - there's various ways to get to the definition of exponentiation for arbitrary real exponents, and they all give the same answers. I just illustrated one path.) A similar thing happens with the trig functions. The "original" definitions all dealt with right triangles, so under those definitions, they're really undefined outside of the domain of 0° to 90°, excluding endpoints. But what you can do is find some other function which has the same values in the 0° to 90° range, and has all the properties which you expect the trigonometric functions to have (e.g. varying smoothly, having the same relation between sin(x) and cos(x), ...), and then *redefine* the trigonometric functions to be that new definition. It matches the original definition inside the original domain, it has all the properties which the original function has, but it also is defined for values where the original definition is silent. -- 160.129.138.186 (talk) 17:52, 16 September 2015 (UTC)[reply]

I don't think your usage of the term "Analytic continuation" in this context is correct. This term is used when you have a function which is already analytic in some domain, and you extend it to be analytic in a bigger domain. And this is done more or less only in the context of complex analysis. A function on the integers can't be analytic (except maybe with some degenerate sense of the word).

Specifically, the gamma function isn't the only analytic function which extends the factorial - you can easily find others, this one is just the simplest. And the extension of exponentiation would have lost most of its punch if you couldn't use power laws to define exponentiation with rational exponents. -- Meni Rosenfeld (talk) 20:35, 16 September 2015 (UTC)[reply]

The PDF of Children in Lake Wobegon

In Lake_Wobegon, "all the children are above average", which is of course impossible for the probability distributions in common use; however, math is sometimes surprising, with different magnitudes of infinity and infinite sets with zero measure. So, it possible to have a probability distribution (say, one with infinite variance) where every element is greater than the expected value? OldTimeNESter (talk) 23:30, 15 September 2015 (UTC)[reply]

No. Undergrad analysis exercise: a positive integrable function integrated over a nontrivial interval gives a positive number. --JBL (talk) 00:21, 16 September 2015 (UTC)[reply]

Maybe there are no children. -- BenRG (talk) 06:07, 16 September 2015 (UTC)[reply]

Maybe the average is taken over all the people in the camp, and they make a point of hiring the worst camp staff they can find? Or go one step further, take it over all organisms in the camp grounds, bacteria are terrible at just about everything worth measuring for a child, so they bring the average WAY down. MChesterMC (talk) 08:42, 16 September 2015 (UTC)[reply]

It's a stretch, but the closest thing I can think of - if you take a distribution with undefined mean such as the Cauchy distribution, then the average could be said to be simultaneously any real number (much the same way as 0/0 could be said to be any real number). For every child, his score is higher than at least one instantiation of the average.

Of course, if we're rigorous, then since the average is undefined, no child can truly be above it.

A less exotic interpretation is that the average is taken over children in general, not just children in the lake. The phrasing doesn't exclude this possibility. -- Meni Rosenfeld (talk) 09:49, 16 September 2015 (UTC)[reply]

They haven't specified in what measure (height, weight, intelligence, strength, etc) all the children are above average. If a different measure is used for every child it may well be that every child is above average in some measure. This is similar to how 90% of drivers think their driving standard is better than average. Quite possibly so, since everyone has a different opinion of what "better" means. 86.161.117.99 (talk) 21:58, 16 September 2015 (UTC)[reply]

I would not be surprised if there was some way use something like a supertask – an infinite set of discrete elements tending to some limit, such that the limit is also the expectation value. I think the expectation value of a random selection from the infinite set

\{1,{\frac {1}{2}},{\frac {1}{4}},{\frac {1}{8}},...\}

should be 0, but I'm really not sure. If this is the case though, the infinite number of Woebegon children could have IQs of 116, 108, 104, 102, 101, 100.5 and so on. Smurrayinchester 12:27, 17 September 2015 (UTC)[reply]

- (Assuming that I'm allowed to select randomly from an infinite set, and that the process – which will almost certainly give me a number with more digits than Graham's number – doesn't crash my computer/my brain/the universe) Smurrayinchester 12:32, 17 September 2015 (UTC)[reply]

There is no uniform distribution on a countably infinite set. Any distribution on the set in question assigns positive probability to some element and thus has positive expected value. This has nothing whatsoever to do with supertasks. --JBL (talk) 13:09, 17 September 2015 (UTC)[reply]

Actually, I think this is possible without any exotic ideas. All you need is that the statistic measured can take negative values (of arbitrarily large magnitude).

For every positive integer n, let the stat take the value

-n

with probability

{\frac {6}{\pi ^{2}n^{2}}}

. This is a proper distribution, its mean is negative infinity, and 100% of children have a finite stat, hence larger than the average.

If you're restricted to nonnegative values, all you can have is 100% of children being lower than the average. -- Meni Rosenfeld (talk) 13:23, 17 September 2015 (UTC)[reply]

It's quite possible for all the kids in Lake Wobegon to be above the state or national average, although extremely unlikely, unless LW has a very small population. It there is only 1 kid, there's a 50% chance. StuRat (talk) 16:50, 17 September 2015 (UTC)[reply]

It is possible, but requires either no children or infinitely many. (With no children, it's true vacuously.) You don't need a probability distribution to compute the mean of a sequence $a_{1},a_{2},\dots$ (of "childen"). More generally, if $(X,\nu )$ is a measure space and $f:X\to \mathbb {R}$ is a measurable function, then the upper mean of f is defined by

{\overline {f}}=\lim _{M\to \infty }\sup _{\nu (A)\geq M}{\frac {1}{\nu (A)}}\int _{A}f\,d\nu ,

the supremum taken over measurable subsets A of X, whose measure is at least M. Likewise, the lower mean is defined by

{\underline {f}}=\lim _{M\to \infty }\inf _{\nu (A)\geq M}{\frac {1}{\nu (A)}}\int _{A}f\,d\nu .

These are both well-defined extended real numbers. If the upper and lower means coincide, then this is the mean of the function f. For example, suppose that X is the set positive integers and ν the counting measure. Define the "child" function by $f(n)=-n$ . Then the average is $-\infty$ , and every child is above average. Sławomir
Biały 21:46, 17 September 2015 (UTC)[reply]

Thanks for the great answers; you guys rock! OldTimeNESter (talk) 01:43, 18 September 2015 (UTC)[reply]