Wikipedia:Reference desk/Archives/Mathematics/2008 July 17

= July 17 =

Fractional derivative
Are there any 'real world' applications, meaningful quantities/measurables, etc of fractional derivatives? (that you know of)? thanks.87.102.86.73 (talk) 00:35, 17 July 2008 (UTC)


 * might be of interest. --Tango (talk) 00:56, 17 July 2008 (UTC)


 * Wow. Where to begin. Let's see. Fractional derivatives are found in the integrodifferential equations corresponding to Levy processes (see Feynman-Kac formula for the duality), which are often used to model a variety of different physical and financial processes (see the article). Existence in a fractional Sobolev space might mean that you can derive higher regularity (and thus better bounds for numerical convergence, etc.) for a function. Fractional powers of the Laplacian can manifest as Dirichlet-to-Neumann operators (see this paper) for other partial differential equations, which means that you can estimate, say, the heat flow through a wall without solving the Laplace's equation inside the wall. There's lots more fun to be had. G'luck! RayAYang (talk) 01:54, 17 July 2008 (UTC)


 * Do we have an article on the p-laplacian, Δp? We have biharmonic equation, but I didn't see any p other than 1 and 2. JackSchmidt (talk) 20:19, 17 July 2008 (UTC)


 * Doesn't look like it. Do you want to create one? RayAYang (talk) 23:27, 18 July 2008 (UTC)


 * I'll add it to my to-do list, unless someone beats me to it. siℓℓy rabbit  (  talk  ) 01:58, 19 July 2008 (UTC)


 * Thanks! It seems like every month some guest speaker is talking about p-laplacian this or that, so I often have some little bit of trivia that might be worth adding, but nowhere near enough perspective to even start a stub. JackSchmidt (talk) 04:05, 19 July 2008 (UTC)

Is timescales mathematics
Is timescales mathematics? Would it be acceptable as a mathematical document? 122.107.219.245 (talk) 14:10, 17 July 2008 (UTC)


 * That article is discussing an area of mathematics, yes. I don't know what you mean by "acceptable as a mathematical document". --Tango (talk) 16:24, 17 July 2008 (UTC)


 * If you're asking whether you should use it as a reference or citation in a paper, then the answer is probably no. However, it has links at the bottom that may be acceptable, such as the one to New Scientist. Black Carrot (talk) 23:13, 17 July 2008 (UTC)

Naive Bayes classifier for non-mutually-exclusive classes?
How does a naive Bayes classifier change if some of the classes are not mutually exclusive? Neon Merlin  17:27, 17 July 2008 (UTC)

choice of weights for linear regression
A question on choice of weights for weighted linear regression. I need to solve for x a problem of type $$ Ax = B $$ where $$ A $$ is size m by n, $$ x $$ is size n by 1, and $$ B $$ is size m by 1, m >> n > 1, given the weights vector $$ W $$ of length m. In other words, I would like to minimize $$\sum_{i=1}^{m} W_i \left( B_i - \sum_{j=1}^n A_{ij}x_j \right) ^2 $$. My question is: what is the appropriate choice of weights when different points i, i = 1...m, have different variances of $$ B $$ and different numbers of counts $$ C $$ that went into determination of the variances of $$ B $$ ? Let me explain my question in a bit more detail. I measure $$ B $$ for any given $$ A $$, many times. $$ A $$ is a discrete variable. $$ A $$ values may be assumed exact for all practical purposes. Measured value of $$ B $$ varies from measurement to measurement even for exact same $$ A $$. Each possible value $$ A_i $$ of $$ A $$ is sampled $$ C_i $$ times, $$ 2 < C_i < 10^5 $$. Note that $$ C_i $$ varies by 4 orders of magnitude. What is the best choice of $$ W $$ in that situation? I have tried three choices so far, none of which is really good. Choice 1. If I use $$ W_i = 1 / var(B_i) $$ I bias my regression towards points that may have low variance beacuse thay have been poorly sampled. Indeed, the lower is $$ C_i $$ the higher is the variance of variance of B, obviously; so for some points with low $$ C_i $$ the value of $$ var(B_i) $$ will come out very low by chance. That is bad. Choice 2. If I use $$ W_i = C_i $$, I bias my regression towards points with high $$ C_i $$, but these points not necessarily have lower $$ var(B_i) $$; also, any two points with same large $$ C_i $$ will have the same weight even though I can reliably claim that $$ var(B) $$ is larger in one than in the other. That is also bad. Choice 3. I can use Bayesian average instead of mean when evaluating $$ var(B_i) $$, but, first, I do not know if that is legitimate, and, second, the choice of constant for the Bayesian average is quite arbitrary. So, to repeat my question, what should I use for $$ W $$ ??? Please help! Thank you in advance, --OcheburashkaO (talk) 22:31, 17 July 2008 (UTC)


 * Maybe I'll be back later, but weights proportional to reciprocals of variances is the usual thing. 75.72.179.139 (talk) 02:59, 18 July 2008 (UTC)

You wrote:
 * Each possible value $$ A_i $$ of $$ A $$ is sampled $$ C_i $$ times, $$ 2 < C_i < 10^5 $$.
 * Each possible value $$ A_i $$ of $$ A $$ is sampled $$ C_i $$ times, $$ 2 < C_i < 10^5 $$.

Did you mean each value of Bi? You said observations of A could be taken to be exact. If individual observations of B can be taken to be equally uncertain, then using weights proportional to the number of observations makes sense. But apparently this equality of uncertainty does not hold, according to what you're saying. More later........ Michael Hardy (talk) 03:20, 18 July 2008 (UTC)


 * I'm curious; (without giving too much away :-) what is the nature of the measurement that sampling counts vary so much?
 * And a little confused; in the cases I'm familiar with, A is the variable to be solved for. That is, finding an $$A$$ that minimizes $$\sum_{i=1}^{m} W_i \left( B_i - \sum_{j=1}^n A_{ij}x_j \right) ^2 $$.  It's not a constant until you're done.   Saintrain (talk) 16:48, 18 July 2008 (UTC)


 * I do not think choice 3 is applicable here: as I understand it, you do not have any "prior mean" at hand.  I also don't like the arbitrary constant, but I think that if you did have a prior mean, then using the Bayesian mean instead of the true mean when calculating the variance should rather much reduce the chance of getting an unrealistically low estimate for the variance (although not totally eliminate it, of course...), which would help make choice 1 a bit more viable. But until you get this prior mean somehow, the Bayesian mean is not applicable.


 * Regarding choice 1 vs. choice 2, I prefer choice 2. This is for two reasons.  First, the "bad thing" you mention for choice 1 strikes me as worse than the "bad thing" you mention for choice 2.  The choice 1 bad thing amounts to a small chance that the results will be way incorrect.  The choice 2 bad thing amounts to an admission that you are not fully using all the information available to you.  Assuming that this application has a decent tolerance towards not getting the absolutely best possible answer, I would far rather not use all the information I have than to risk getting egregiously bad results some of the time.  The second reason is that, even if you don't get unlucky with your estimates for the variances (i.e., even if your estimates of the variances all turn out to be correct), I still dislike the use of the weight $$ 1 / var(B_i) $$;  if there is a substantial difference (i.e., orders of magnitude) in the true variance from one i to another i, then you're effectively ignoring those is for which the true variance is rather larger, even if your value of $$ C_i $$ is large enough to get useful information from those samples.  But you can attribute this to personal dislike of this weighting;  I would like to hear from user 75 if (s)he knows some reasons why this weighting is used.


 * Regarding options outside of the three choices given, I'm going to suggest a quick-and-easy linear combination of 1 and 2:
 * $$ W_i = k \cdot \lambda \frac 1 {var(B_i)} + (1 - \lambda) C_i $$
 * where $$ \lambda $$ is an appropriately chosen function of $$ var(B_i) $$ and $$ C_i $$ that reflects how certain you are of your value for the variance (closer to 1 means more certain), and where k is just a constant (think of it as necessary to make the units come out right). Of course, I've just changed the problem to choosing a function $$ \lambda $$ (and an arbitrary but important constant k!), but I don't think the results are too overly sensitive to your choice of $$ \lambda $$... for example:
 * $$ \lambda = \begin{cases} 0 & \text{if } C_i < 30 \\ 1 & \text{otherwise} \end{cases} $$
 * is easy and not totally unreasonable.


 * However, I'm sure there are better definitions of W lurking out there waiting to be found. Eric.  83.173.235.68 (talk) 20:30, 20 July 2008 (UTC)

Rational Angle
Is the angle between two faces of a regular tetrahedron, arccos(1/3), a rational multiple of pi? If not, how would you prove it? Black Carrot (talk) 23:21, 17 July 2008 (UTC)
 * Typing "arccos(1/3)" into my calculator, dividing by pi and pressing the fraction button fails to give a fraction, which would suggest it's not rational (it could be rational with a denominator too large for my calculator to handle, I suppose). I wouldn't know where to start with proving it analytically. --Tango (talk) 00:01, 18 July 2008 (UTC)
 * One thing to be careful about with many calculators is that they have much more precision when dealing with fractions. That is for instance, the CAS in the TI line will do fractions with hundreds of digits, but if you start out with a decimal they only store around 9-15 digits, depending on the model – so getting a fraction from a decimal can be tricky.
 * I don’t remember what model, but I used a calculator once that I actually got to round “0.3333333333333” or so to 1/3, so all of them might not be entirely accurate either. GromXXVII (talk) 11:14, 18 July 2008 (UTC)

I don't think that arccos(1/3) is a rational multiple of pi. And one way to prove it would be by contradiction. Assume that this arccos(1/3)/pi is rational and get a contradiction like pi being a rational number also.--A Real Kaiser...NOT! (talk) 04:26, 18 July 2008 (UTC)


 * It isn't. Let &theta;=arccos(1/3).  Then the complex number
 * $$\alpha = e^{i\theta} = \frac{1}{3}+i\frac{2\sqrt{2}}{3}$$
 * has degree 2 over the rationals, and so cannot be a root of any cyclotomic polynomial since the minimal polynomial of &alpha; is not cyclotomic. siℓℓy rabbit  (  talk  ) 06:00, 18 July 2008 (UTC)
 * How does one tell that the polynomial is not cyclotomic? 24.227.163.238 (talk) 17:03, 18 July 2008 (UTC)
 * It's pretty obvious. There are only three cyclotomic polynomials of degree two, since there are only three integers whose totient function is two: 3, 4, and 6.  The corresponding cyclotomic polynomials are
 * $$\Phi_3(z) = z^2+z+1,\quad\Phi_4(z) = z^2+1,\quad\Phi_6(z)=z^2-z+1.$$
 * whereas minimal polynomial of &alpha; is 3z2&minus;2z+3.  siℓℓy rabbit  (  talk  ) 18:53, 18 July 2008 (UTC)
 * Is arccos(1/3) an irrational algebraic [i.e. not transcendental] multiple of pi (or pi^2 sqrt(pi) 1/pi etc)? I.e. could you describe it using a finite number of operations on integers and pi? --Random832 (contribs) 18:40, 18 July 2008 (UTC)

All I can say is that your number "arccos($$\tfrac{1}{3}$$)" is not a rational multiple of $$\pi$$, nor is any other rational number. $$\pi$$ is an irrational number, so any multiple of $$\pi$$ will be irrational as well. Earthan Philosopher (Talk) 19:30, 19 July 2008 (UTC)
 * arccos(1/3) is not rational (if it were, e would be algebraic). Algebraist 19:34, 19 July 2008 (UTC)

It still wouldn't be a rational multiple of $$\pi$$ then because it wouldn't be rational at all. Earthan Philosopher (Talk) 19:42, 19 July 2008 (UTC)
 * The sentence 'a is a rational multiple of b' means 'a/b is rational'. Algebraist 19:59, 19 July 2008 (UTC)


 * Right, I intended "a rational multiple of pi" to mean "pi times a rational number." Thanks for the help. Black Carrot (talk) 03:31, 20 July 2008 (UTC)