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May 26

Converting between guitar tablature and pitched music notation

This question is mainly about music, but it involves mathematical procedures so that I post it here. Are the algorithms or mathematical methods to convert between guitar tablature and standard ("pitched") music notation? 113.178.24.91 (talk) 11:48, 26 May 2012 (UTC)

Probably better on the Wikipedia:Reference desk/Miscellaneous, but it is quite easy. The strings in standard tuning are EADGBE which are pitches of the open strings, each fret raises the note a 12th so the notes on the first fret are F, B♭, E♭, A♭, C F, see Guitar#Tuning for the notes on the other frets. So to translate a guitar tab simple find which fret each string is held on. Find the corresponding notes in the table and mark on the stave. Your typically playing chords so you'll have multiple notes sounding together.--Salix (talk): 12:35, 26 May 2012 (UTC)

The correspondence between the two systems isn't unique - going from staff notation to guitar tablature, you might be able to play the same note on several different strings (even ignoring different tunings) - maybe somebody has developed an algorithm to try and find a convenient fingering. In the other direction, tablature usually doesn't convey rhythm, and you have to make choices about the clef, keys, accidentals, etc., which are often largely a matter of taste. I'm pretty certain there will be software that can carry out these conversions (there is certainly software that can convert from MIDI data to staff notation, but it often comes out looking a little odd - being far too specific about timings, for example). 130.88.73.65 (talk) 13:44, 29 May 2012 (UTC)

May 27

Coupled pendulurms

Two identical pendulums of length $l$ are coupled by means of a spring of natural length $a$ and spring constant $k$. When the displacement angles $\theta _{1}(t)$ and $\theta _{2}(t)$ are small, the equations of motion are:
${\frac {d}{dt}}\theta _{1}+\Omega ^{2}\theta _{1}=K(\theta _{2}-\theta _{1})$
${\frac {d}{dt}}\theta _{2}+\Omega ^{2}\theta _{2}=-K(\theta _{2}-\theta _{1})$
where $\Omega ^{2}={\frac {g}{l}},K={\frac {k}{m}}$
Here $m$ is the mass of the bendulum bob and $g$ is the acceleration due to gravity.
Find the frequencies of the normal modes of oscillation. Give a simple justification of the above equations (a full derivation is not requried).

I'm not sure about the normal modes of oscillation; I get $\Omega$ and ${\sqrt {\Omega ^{2}+2K}}$
What's the logic behind these equations? Widener (talk) 18:22, 27 May 2012 (UTC)

They are just derived from considering the forces on the pendulum weights. Multiply through by m and by l and you have the force due to the displacement (assuming it's small) and spring on each mass and mass × acceleration.--JohnBlackburne^words_deeds 00:12, 28 May 2012 (UTC)

May 28

Convergent Power Series

Consider a smooth function $f:I_{x}\to \mathbb {R}$ where $I\subseteq \mathbb {R}$ is an open interval. Define a new functions $\rho [f]$, such that $\rho [f](y)$ is the radius of convergence of power series

$\sum _{k\geq 0}{\frac {f^{(k)}(y)}{k!}}\,(x-y)^{k}\,.$

What is known, and what has been written about, these functions? Please supply as many links and references as possible. — Fly by Night (talk) 17:28, 28 May 2012 (UTC)

The radius of convergence is the distance between y and the nearest singularity of the analytic continuation of f in the complex plane. Bo Jacoby (talk) 05:57, 29 May 2012 (UTC).

In particular, it is a 1-Lipschitz function, since it is the distance function to the boundary of some open set of $\mathbb {C} .$ --pma 20:29, 29 May 2012 (UTC)

This is only true if you assume that f is real analytic. For smooth f, it's more complicated. For instance with f the standard $e^{-x^{-2}}$ example, $\rho (0)=\infty$ even though f has no analytic continuation in any neighborhood of 0. Sławomir Biały (talk) 20:38, 29 May 2012 (UTC)

Though if the term radius of convergence is restricted to meaning the radius in which the series converges at all, with no regard to what it converges, treating the series as defining a complex-valued differentiable function is adequate. And given that if the function isn't analytic it can't be locally described by a power series, it doesn't seem very meaningful to use any other definition of the term.--Leon (talk) 21:41, 1 June 2012 (UTC)

Please supply as many links and references as possible. — Fly by Night (talk) 00:42, 3 June 2012 (UTC)

May 30

Are there any other non-Euclidean geometries?

If you take Euclidean geometry and delete the parallel postulate, are the hyperbolic geometry and the elliptic geometry the only two possible geometries? Is it not possible to have, say, a finite number but more than one straight lines intersecting a given point and straight line? Or, is it possible to have a point through which there exists infinite straight lines intersecting another given straight line, and another point in the same space where there are none? Widener (talk) 16:59, 30 May 2012 (UTC)

In the latter case you lose a defining property of the big three, that one neighborhood is like another (I don't know the name of this property). There are useful geometries without that property, such as the geometry of the physical universe, with varying spacetime curvature. —Tamfang (talk) 18:49, 30 May 2012 (UTC)

Oh, Tamfang, the word you're probably looking for is homogeneity. --Trovatore (talk) 19:36, 30 May 2012 (UTC)

Sure, let's go with that. —Tamfang (talk) 20:41, 30 May 2012 (UTC)

I think the issue you're running into here is the contrast between classic axiomatic geometry (such as Euclidean, Lobachevskyan, or the original version of Riemannian geometry), where all geometric notions are axiomatized in and of themselves, and the more modern sort of geometry, where geometric notions are defined in terms of notions from a larger theory such as real analysis or set theory. These larger theories have the flexibility to accommodate varying curvature and so on and still prove interesting stuff, whereas if you took one of the classic axiomatic geometries and simply dropped an axiom so as to allow for varying curvature, you probably couldn't prove very much at all.

Anyway, Widener, for the most general framework that's still a reasonably straightforward generalization of the stuff you're talking about, see semi-Riemannian geometry (where the "semi" part allows for stuff like the time–space distinction in relativistic spacetime). There are certainly even more general geometric frameworks but they get less obviously connected to the classic theories. --Trovatore (talk) 19:04, 30 May 2012 (UTC)

I'd argue that a more natural "most general framework" is the notion of a manifold with projective connection. Roughly, this is any space in which the notion of "straight line" makes sense, and looks like the familiar projective geometry on small scales. Sławomir Biały (talk) 13:01, 31 May 2012 (UTC)

The question is discussed rather thoroughly at Non-Euclidean geometry#Axiomatic basis of non-Euclidean geometry. Looie496 (talk) 19:22, 30 May 2012 (UTC)

Honestly, no, that doesn't strike me as a thorough discussion at all. Unless I missed it, it doesn't discuss the case Widener asked about, where the number of parallels may vary from place to place. Still, a useful link. --Trovatore (talk) 19:35, 30 May 2012 (UTC)

There are plenty of other geometries in mathematics, which differ more and more from Euclidean geometry. There are projective geometries, which includes a set of points at "infinity" - so in the projective plane, for example, any pair of lines intersects in just one point. There are finite geometries, with a finite number of points and lines. There is the taxicab geometry, in which a circle has four corners etc. etc. Gandalf61 (talk) 09:54, 31 May 2012 (UTC)

I think Sławomir has the best answer in the spirit of Widener's last case. I was going to drop in to mention arbitrary manifolds, but stopped because I didn't know the right words /definitions necessary for projective connection :) SemanticMantis (talk) 18:16, 31 May 2012 (UTC)

Q. What is the technical term for the innermost edge of a curved plane?

Q. What is the technical term for the innermost edge of a curved plane?

For a real-life example, imagine a race car taking the shortest path possible through each curve in a gran prix course.

I have searched via Wikipedia and Google and failed to find an answer to what I thought would be a rather simple question.

Thank you in advance for satisfying my curiosity!

19:37, 30 May 2012 (UTC) — Preceding unsigned comment added by Pstreeter (talk • contribs)

How about "inside edge", or "inside lane", in the car case ? StuRat (talk) 19:48, 30 May 2012 (UTC)

For motorsports the term is Racing line. A mathematician would think of this a curve which minimises distance.--Salix (talk): 20:20, 30 May 2012 (UTC)

The racing line minimizes only the time needed to complete the course however and not necessarily the distance traveled. Whether or not these two things coincide will depend on the shape of the race course. For an oval course, the distance traveled will be further than the most minimum distance which is the path of the innermost lane or the track's concave edge. --Modocc (talk) 04:08, 31 May 2012 (UTC)

Euclidean geometric planes are always flat with zero curvature everywhere. Hyperbolic planes are not flat, but these are, of course, not typical planes! However, the edges and surfaces of figures can be either convex or concave. With an oval race track, its innermost edge is concave and the outer edge is convex. --Modocc (talk) 04:08, 31 May 2012 (UTC)

The technical term for the shortest path between two points, given a set of constraints, is a geodesic. Looie496 (talk) 23:18, 31 May 2012 (UTC)

Not sure if this is pedantic, but a geodesic is only locally a shortest path. To illustrate, if we take the space of $\mathbb {SO} (3)$, and take the start and finish points to be the identity transform, a motion of constant angular velocity that consists of a rotation by $2\pi$ (about some axis through the origin) from start to finish still maps out a geodesic path, though it is not the shortest path.--Leon (talk) 21:35, 1 June 2012 (UTC)

June 1

Ergodic theory

The article mentions nothing about its history like when was it found and who found it? How far we have reached in proving this theory? Basically what is our progress about it?65.128.168.2 (talk) 03:22, 1 June 2012 (UTC)

Not an answer to your questions, but you may want to check out theory (mathematics). Mathematicians do not "prove" a theory. The word "theory" in math refers to a general body of ideas and theorems, not a "something which may or may not be true and needs to be proved or refuted". Staecker (talk) 12:43, 1 June 2012 (UTC)

The article does have a little history. It says "The first result in this direction is the Poincaré recurrence theorem". -- Poincaré_recurrence_theorem explains that it was published by Henri_Poincaré in 1890. SemanticMantis (talk) 14:15, 1 June 2012 (UTC)

Probability

If there is a set that contains 10 members, called set A, which contains {a, b, c, d, e, f, g, x, y, z} and there is a subset of set A that has three members, called set B, containing {x, y, z} and there is another subset of set A called set C, which contains 3 random members of set A, then the chance that one or more members of B are also members of B is 3/10 + 3/9 + 3/8, which equals 108/360 + 120/360 + 125/360. Right? And when you add them up, you get 353/360, which is approxiomately 98 percent. So the chance that a member of set B is a member of set C is almost 98 percent. Is that all right? Legolover26 (talk) 16:08, 1 June 2012 (UTC)

Does that 98% feel right ? Ball park estimate - the probaility that a random member of A is not in C is about 2/3. So the probability that we choose 3 members of A (maybe with repetitions) and none of them is in C is approx. (2/3)³, which is 8/27 or more than 1/4. That's a long way from 2%. Gandalf61 (talk) 16:31, 1 June 2012 (UTC)

There are (3!10) = 120 C-sets. (Here (3!10) is the binomial coefficient.) The number of C-sets having no element in common with set B is (3!7) = 35. The number of C-sets having one or more element in common with set B is 120−35 = 85. The chance that one or more members of C are also members of B is 85/120 = 71.833%. Bo Jacoby (talk) 16:44, 1 June 2012 (UTC).

I believe you. I didn't think that really made sense. It's kind of like that problem where I insist that if you flip 3 coins, the chance that all 3 of them will be on the same side is 1/2, because when you do that, there are always 2 coins that end up on the same side, and the chance that the third coin will be the same is 1/2 exactly. Legolover26 (talk) 17:08, 1 June 2012 (UTC)

No, if you flip 3 coins there are 8 possible outcomes, all are equally likely, and only two have all 3 coins on the same side, so the probability is 1/4. The side with at least two coins and the side of the remaining coin aren't independent, so you can't multiply the probabilities together like that. Hut 8.5 18:59, 1 June 2012 (UTC)

In your original proposition, where you say "the chance that one or more members of B are also members of C is 3/10 + 3/9 + 3/8" you are counting some possible outcomes more than once. Your sum should be

P(c₁ in B) = 3/10 = 108/360

P(c₂ in B & c₁ not in B) = (3/9)(252/360) = 84/360

P(c₃ in B & c₁,c₂ not in B) = (3/8)(168/360) = 63/360

So the correct probability is 108/360 + 84/360 + 63/360 = 255/360 or 85/120 as quoted by Bo jacoby. - Sussexonian (talk) 09:05, 3 June 2012 (UTC)

June 2

it's thirty years since I took Diff.Eq.

and I've never used it since, and I didn't get this far anyway.

I want to find a function u(t) such that $u(0)=0$ and $2(u')^{2}=3+\cos(2u)$. How do I go about it?

(For an artwork based on the lemniscate of Bernoulli, I want a parametrization such that the path "speed" is constant.)

I'll be happy with the first few nonzero terms of a Fourier series. —Tamfang (talk) 02:09, 2 June 2012 (UTC)

You could just throw it on Wolfram Alpha. It gives solutions in terms of Jacobi's amplitude function, though... I couldn't get it to give anything more.

I've just played around with the parametric equations given in the article for the lemniscate, and they're a pretty decent approximation of constant speed. Can't you use that? — Kieff | Talk 02:27, 2 June 2012 (UTC)

It varies by a factor of √2. It's usable, yes, but my design will look that much better if most of that variation is squashed. I've been fiddling with $u(t)=t+a\sin(2t)+b\sin(4t)$, which is pretty good, but I thought why not try for whole hog? —Tamfang (talk) 02:58, 2 June 2012 (UTC)

You could also go the easy way and numerically approximate the total length of the curve by exhaustion to the desired precision. Then divide that arclength into smaller steps of the desirable resolution, and use these step sizes to "walk" the parameter t until you reach each step size, then store that value of t. Do this for the whole curve and you'll have a bunch of t values that will be approximately the same distance from each other on the curve. This is pretty easy to do and it'll only require a lookup on the table. For even greater precision, you can throw in some interpolation for intermediate values. Point being, closed forms for normalized parametrizations of curves are not really necessary and there are simpler methods to do it if you just want to use it somewhere. I've done it several times. — Kieff | Talk 19:00, 2 June 2012 (UTC)

I'm considering something similar, yes. —Tamfang (talk) 19:35, 2 June 2012 (UTC)

It works surprisingly well. —Tamfang (talk) 17:10, 3 June 2012 (UTC)

We also have an article Numerical methods for ordinary differential equations... It might be of some use too... --Martynas Patasius (talk) 00:49, 3 June 2012 (UTC)

Try $u(t)=1.1981407859366775\ t+0.08626709480287766\sin(2.3962743565815936\ t)+0.0018673707746179953\sin(2\cdot 2.3962743565815936\ t)$ (it's very good even without the last term). -- Meni Rosenfeld (talk) 19:21, 2 June 2012 (UTC)

Good heavens. How did you come up with that? —Tamfang (talk) 19:50, 2 June 2012 (UTC)

Taking the nonelementary closed form expression of the function as a starting point, I first found the period by using Mathematica to numerically solve $u(2T)-2u(T)+u(0)=0$ (choosing the correct root by inspection). Then the linear coefficient is ${\frac {u(T)-u(0)}{T}}$. After correcting for the period and the linear term, I ended up with a simple periodic function for which I computed the Fourier series.

PS the way to solve differential equations which involve only u' and u is to first write it as ${\frac {du}{dt}}={\sqrt {\frac {3+\cos(2u)}{2}}}$ which means ${\frac {dt}{du}}={\sqrt {\frac {2}{3+\cos(2u)}}}$. You can then find t as a function of u by integrating, and then u as a function of t by inverting. In this case both these operations result in nonelementary functions. -- Meni Rosenfeld (talk) 20:02, 2 June 2012 (UTC)

June 3

Phase trajectories

How do you prove that a phase trajectory is closed?

For a concrete example, suppose the energy $E={\frac {1}{2}}{\dot {x}}^{2}+{\frac {1}{2}}x^{2}+{\frac {\epsilon }{4}}x^{4}$ for some system, where ε is some constant. I want to prove that all phase trajectories are closed, but I don't know how. I can show that for any (positive) value of E, there's a value of x such that dx/dt is zero. But does this *prove* that the phase trajectories are closed? If it does, I don't see how. 65.92.7.168 (talk) 02:32, 3 June 2012 (UTC)

A phase trajectory isn't necessarily closed. The pendulum energy is $E=m{\frac {{\dot {x}}^{2}}{2}}-mg\cos x$. Solutions for $E>mg$ are unlimited and thus not closed. Solutions for $-mg<E<+mg$ are closed because the set $\{(x,{\dot {x}}):E=m{\frac {{\dot {x}}^{2}}{2}}-mg\cos x\}$ is a limited, one dimensional manifold. The solutions for $E=-mg$ are stable equilibria. The solutions for $E=+mg$ include unstable equilibria, and unclosed limited trajectories. Bo Jacoby (talk) 11:34, 3 June 2012 (UTC).

What I meant was, I would like to know how to prove that a phase trajectory is closed if it really is closed. For instance, in the example I gave above I can see that all the phase trajectories are closed just by graphing them. But I'd like to know how to prove it. 65.92.7.168 (talk) 15:46, 3 June 2012 (UTC)

Fair scoring

This question was suggested by tennis scoring. Suppose, in a certain game, if player A plays first (e.g. serves in tennis) then player A wins with probability p and player B wins with probability 1 − p. Conversely, if player B plays first then player B wins with probability p and player A wins with probability 1 − p (i.e. the players are evenly matched). A match consists of a series of games. In the first a games, player A plays first. Then player B plays first in the next b games, then player A plays first in the next b games, and so on, with players alternating playing first for b games in a row. The match finishes when one player has at least c wins and is d wins ahead of the other player. For example, in a tennis set without a tie-break, a = 1, b = 1, c = 6, d = 2. In a tennis tie-break, a = 1, b = 2, c = 7, d = 2. The scoring system is fair if both players have a 50-50 chance of winning the match irrespective of the value of p. I assume the system must be fair for both the above tennis examples (though even this is not completely blindingly obvious to me), but what other fair values are possible for a, b, c and d? What are the general conditions? 86.181.203.150 (talk) 02:58, 3 June 2012 (UTC)

Intuitively almost no such system can be fair by your definition. You are requiring that the probability of either player winning must made a constant function of p for some choice of a small number of parameters, and further that these be natural numbers. Fairness would be possible if you were allowed to choose a specific p to achieve this (e.g. p=0.5), or to have an additional source of randomness (e.g. flipping a fair coin to choose who starts). For almost all values of p, one of the players will have a non-zero advantage for any given set of values for the parameters. In tennis, the system is chosen to make it "approximately fair", primarily through the increased values of c and d: if d≤a or d≤b−a, it is clear that one player will be certain to win unfairly when p=0 or p=1, while c helps to even things a bit by ensuring that more than a few random variables are summed. — Quondum^☏ 06:47, 3 June 2012 (UTC)

I am not convinced that your intuition is correct. I think you will find that the tennis systems are exactly fair, for any value of p, according to my definition.* I would certainly value a second (or third) opinion on that. 86.148.154.177 (talk) 11:56, 3 June 2012 (UTC) *We can ignore p = 0 and p = 1, where the match would go on forever.

After a closer look, I must concede your point. A match is fair when terminating is restricted to when both players have started an equal number of games. Various combinations of a, b and d ensure this (e.g. a=1, b=1 or 2, d even; c is immaterial). — Quondum^☏ 13:47, 3 June 2012 (UTC)

"a=1, b=1 or 2, d even; c is immaterial" looks good! Do you think there are any other fair possibilities? 86.148.154.177 (talk) 22:00, 3 June 2012 (UTC)

Just to be clear, d being even does not guarantee that the number of games played is even (for example, a tennis set can end 6-1 or 6-3) and obviously if the number of games played is odd then one player must have started more than the other. However, if d is even and the number of games played is odd, then the winning margin must be odd and hence greater than d and so playing one extra game couldn't affect the result. So I agree than Quondom's examples are fair. 60.234.242.206 (talk) 00:33, 4 June 2012 (UTC)

(edit conflict with previous message) I do not understand "A match is fair when terminating is restricted to when both players have started an equal number of games." Take the case a = 1, b = 1, c = 6, d = 2. If player A wins the first 6 games then player A wins. If player B wins the first game, and player A wins the next 6 games then player A wins. However your statement is interpreted, whether referring to number of starts prior to or including the terminating game, I don't see how it can be true in both those cases. Possibly I am misunderstanding something. 86.148.154.177 (talk) 00:40, 4 June 2012 (UTC)

I seem to be guilty of extrapolating from simple cases. If termination is restricted to when an equal number of games (including the terminating game) are started by each, the game is fair (though I do not imply that this is a requirement). My assertion about which values ensure this is not true (including about c), as pointed out. I'd have to make a more detailed analysis to determine fairness in the case of c>2, the other values being kept as for tennis. I'll give what I find later. — Quondum^☏ 08:16, 4 June 2012 (UTC)

I previously knocked up a small program that ran simulations, but it took ages and still didn't deliver accurate enough answers to be conclusive. I have now (or so I thought) figured out a recursive way to quickly calculate the probabilities to good accuracy (13 or 14 decimal places). The "a=1, b=1 or 2, d even; c anything" cases come out to 0.5 as expected, but I am getting some odd-looking results with a = 2, b = 4, c = any, d = 9, where, for all values of p that I've tried, the probability of player A winning comes out very close to 0.5 but not quite (e.g. 0.500003372970541 for p = 0.7). I assumed at first that this was loss of precision or numerical artefact, yet all the other checks I have done seem to suggest that the answer should be correct to 13 or 14 d.p. It is a bit of a puzzle at the moment as to whether this case should be exactly 0.5 or not. Possibly I have messed up in some way that I can't spot at the moment. Anyway, any insight into what the theoretical answer for that case should be would be interesting, as well as the general question. Working it out analytically is beyond my capability, I think. 86.181.201.159 (talk) 14:12, 4 June 2012 (UTC)

(ec) I was utterly gobsmacked by my results. For simplicity, define a variable x ranging from −1 to +1, with p=(1+x)/2 (an even function of x for a score's probability is unbiased as the reversed score will have an identical probability).

For a=1, b=1, c=3, d=2, probabilities are 3-0:(1+x−x²−x³)/8, 3-1:(3−2x+2x³−3x⁴)/16, A wins from 2-2:(3+2x²+3x⁴)/16, hence total A:1/2 (fair)

Note that probalities for B are as above with x changed in sign (and scores reversed)), so 3-0 and 0-3 do not have the same probability (assuming p>1/2, A will win 3-0 more often than B will win 0-3), but this is exactly balanced by B winning more often 1-3 against A's 3-1.

For a=1, b=1, c=4, d=2, we get similar results for p>1/2, with more 4-1 wins by A than 1-4 wins by B, again exactly balanced by B's 2-4 wins against A's 4-2 wins.

I have to admit that I find this behaviour highly counterintuitive (so much for my intuition). I wouldn't know how to generalize this short of doing the math for each set of parameter values, which I would write a program to do as it is pretty tedious.

An analytic result may be possible if it turns out that "adjacent" scores exhibit this balancing property. — Quondum^☏ 14:27, 4 June 2012 (UTC)

I've looked at simple cases with odd d (d=1 and d=3, and a=b=1), and it looks pretty sure that these are unfair, looked at analytically – and I expect your sims will corroborate this. I suspect that your use of large d merely brings the "unfairness" of unfair cases down. Do you have an application that makes a fair amount of effort worthwhile, or is this for general interest? — Quondum^☏ 16:54, 4 June 2012 (UTC)

Can one zero be larger than another?

As there are larger and smaller infinities, can there be said to be larger and smaller zeroes? - is a set of no cats smaller than a set of no elephants? If I subtract the set of no women from the set of all humans, do I get half-zero as the result...? Adambrowne666 (talk) 06:17, 3 June 2012 (UTC)

The concept of "zero" is derived from the concept of "addition". If the zero is supposed to be a two-sided identity for a certain binary operation your call "addition", then the zero is unique. It does not matter what "addition" is, it may be a set union in your example. It is the property of identity (a.k.a. neutral element) which is crucial. Incnis Mrsi (talk) 08:54, 3 June 2012 (UTC)

This triggers an interesting thought. The description of a binary operation with a two-sided identity element fits a central idempotent a of a ring R in relation to the corner ring aRa. Thus, one can almost construct a counterexample with zeros of different sizes: a (partially) ordered set of "zeros" given the "addition" operation with a series of sets in a subset-superset relationship chain (I know this "counterexample" fails in that it is not all the same set). — Quondum^☏ 10:36, 3 June 2012 (UTC)

Infinitessimal deals with a number of mathematical theories that support having quantities that are smaller than any finite quantity. And they're not without real applications, Surreal numbers for instance have been used in analysing games. Dmcq (talk) 09:03, 3 June 2012 (UTC)

Still the short and correct and elementary answer is: No, one zero cannot be larger than another. 0=0. (Even in nonstandard analysis the nonzero infinitesimals are nonzero.) Bo Jacoby (talk) 11:40, 3 June 2012 (UTC).

Something else that might interest you is how one limit can approach zero more quickly than another. Let's say you have two limits, 1/x and 2/x. As x goes to infinity, both approach zero. However, the second number is always twice as much as the first, for any given x. StuRat (talk) 23:33, 3 June 2012 (UTC)

Error due to "flat earth" approximation?

Maybe I should have posted this under Science, but since it's really just applied math, I'll post it here.

If using a "flat earth" approximation when surveying land, about how big are the errors caused by this approximation? What I mean is this: I have a survey map of some land. It uses the metes and bounds system. If I interpret the "metes" as vectors, I notice that the vectors don't quite add up to zero. There is an error of a few inches. Maybe it's a data entry error on my part, but anyway, I'm just wondering what kind of tolerance there should be for this sort of thing. 75.37.236.219 (talk) 11:28, 3 June 2012 (UTC)

Errors due to curvature of the earth are likely to be swamped by the non-planar nature of even the "flattest" piece of ground, unless you are dealing with a salt flat, for example. Even then, the piece would have to be several miles in size for the earth's underlying curvature (for a radius of c. 4000 miles) to be significant. The closure error on your survey will be due to measurement errors. A search on "surveying closure error" will give some approaches to reconciling things.109.148.243.127 (talk) 15:22, 3 June 2012 (UTC)

What is "significant"? One inch? Six inches? Ten feet? Please clarify. 75.37.236.219 (talk) 16:37, 3 June 2012 (UTC)

Assuming a perfect sphere for the earth, two points 1 mile apart along the curved surface are only 0.999999997 miles apart in a straight line, the difference in the two lengths being about 2 ten-thousandths of an inch. A bit more detail: if two points on the surface of a sphere of radius r subtend an angle theta at the centre, then the curved distance is r.theta and the straightline distance is 2.r.sin(theta/2). The piece of land would have to be many miles in size for the difference in distance to be as much as 1 inch.109.148.243.127 (talk) 18:05, 3 June 2012 (UTC)

At 1km from the tangent point where a plane intersects the Earth's surface, the height difference is about 6371(1-cos(1/6371))km or about 3 inches. -Modocc (talk) 20:39, 3 June 2012 (UTC)

One comment to add is that, while the deviation per mile or kilometer is small, depending on the type of surveying you do, this could be cumulative. So, if you're only off a tenth of an inch for one property, it's no problem, but if you use the ending boundary of each as the starting point for the next, and measure, say, 100 properties in a row, each with a 1/10th of an inch deviation in the same direction, now you're off 10 inches, and this begins to be a problem. To avoid this, you should try to go off a fixed reference each time, rather than measuring from iffy boundary lines. StuRat (talk) 23:29, 3 June 2012 (UTC)

How do I calculate date of birth knowing how old someone was on certain dates?

I am doing a bit of genology research. There is census data that shows the age of particular person when the data was colected. To give an example: Census 1. takes place in 1795, the person is 14 years old; 2. takes place in august 1811, he is 29; 3. on 10 february 1816 he is 33; 4. 1 august 1826 - he's 43; 5. 1 march 1834, the guy is 51. Simply substracting age from year will give varied results (from 1781 to 1783 in this case). I am not that good at maths, I would probably use some online age calculator, but it occured to me that using all data points the estimation would be more accurate. And also there are some other considerations - some of censuses are dated with a particular date, some are not, some people have their ages listed in fractions (say person is 9 3/4 i.e. 9 years and 9 months old) and all the dates are old style. ~~Xil (talk) 22:46, 3 June 2012 (UTC)

Certainly the more precise figures are preferable. Just listing an age, in years, during a given year, provides almost 2 years of wiggle room. For example, if somebody is listed as 1 in 2012, they might have been almost 2 on Jan 1, 2012, meaning they were born, say, on Jan 2, 2010. Or, they might turn 1 right on December 31, 2012, meaning they were born December 31, 2011. Also, some people will round up, so list their 7 month old as age 1, making things even worse. And this doesn't even consider those who are intentionally lying about ages. StuRat (talk) 23:21, 3 June 2012 (UTC)

The data provide boundaries for your estimate.

• On some date in 1795, NN's age was 14; this could mean he turned 14 on the day of the census, or 15 the next day, or anything between. The census could be on any day of the year. So his birthdate could be 1780-Jan-02, or 1781-Dec-31, or any day between.
• Age 29 in 1811 Aug similarly gives a range 1781-Aug-02 to 1782-Aug-31.
• Age 33 on 1816-Feb-10 gives 1782-Feb-11 to 1783-Feb-10.
• Age 43 on 1826-Aug-01 gives 1782-Aug-02 to 1783-Aug-01.
• Age 51 on 1834-Mar-01 gives 1782-Mar-02 to 1783-Mar-01.

Unless one of us has blundered, we have a problem: he can't be born after 1781-Dec-31 (else he couldn't turn 14 before the end of 1795) or before 1782-Aug-02 (else he'd be 44 on 1816-Aug-01). But I hope you see the principle. —Tamfang (talk) 23:36, 3 June 2012 (UTC)

Not all 5 data points can be true, because Point 1 is consistent only with Point 2. But there is a small sliver of time where Points 2 to 5 are all true. These 4 data points are all consistent with the birth occurring between 2-31 August 1782. This would mean that in 1795, depending on exactly when, he was 12 or 13, not 14. -- ♬ Jack of Oz ♬ ^{[your turn]} 23:52, 3 June 2012 (UTC)

Can't this discrepancy be expained by the fact the dates are old style? The censuses are consistent with each other in giving age on previous census. However the 1795 census data has date mising, it is possibe the actual census itself took place on a bit different date. Anyways - I am already aware that by simply substracting the estimiate can be off by two years. What I am going for is that each data point limits the posibilities to the point where we can say that, if we exclude the first data point, he was born in August 1782. And I need to do this for several persons, so figuring out a way for computer to do it is what I need, for which I need a formula and I am not really that good at maths ~~Xil (talk) 02:05, 4 June 2012 (UTC)

I double checked the ages are right, the 1795 census lists age at previous census (which apperently took place in 1782) as well - then he was one. At any rate I need to reduce the estimiate to within a year, because I have a ton of records with similarily named people on a spreadsheet that I want to sort by birth year to identify different individuals. One person having different birthdays creates a mess, I could of course take a guess, but some records acctualy have the real birth date, so I might not notice I missed ~~Xil (talk) 03:05, 4 June 2012 (UTC)

Where was old style (Julian) dating still in use after 1752? —Tamfang (talk) 03:45, 4 June 2012 (UTC)

Sweden (it finally managed to convert in February 1753 after crazy goings on for 52 years), Russia, Greece, Japan, Korea, China and probably parts of the Ottoman Empire. -- ♬ Jack of Oz ♬ ^{[your turn]} 04:10, 4 June 2012 (UTC)

In this case it is from the "revision lists" of the Russian Empire. ~~Xil (talk) 05:00, 4 June 2012 (UTC)

Here's your formula. Given an integer age and a range of possible recording dates, if we assume that the age was rounded down to N years on the recording date, EARLYDATE is N+1 years before the day after the first possible recording date, and LATEDATE is N years before the last possible recording date. Each of your age records will give you its own EARLYDATE and LATEDATE. The actual birthdate (if all assumptions are valid) lies between the latest EARLYDATE and the earliest LATEDATE, inclusive.

With fractional ages, you'll need to decide during what part of the year a child's age is expressed as 9½ ... —Tamfang (talk) 05:40, 4 June 2012 (UTC)

I wouldn't mind having more exact date where possible, if you go only by years you could end up with wrong result if it lands somwhere near the new year. Just saying that it doesn't need to be extremely accurate and 1795 data point, which has exact date missing, might still be taken into account. And anyways I don't see how this is something to feed to my computer, instead of doing logic myself. So it would take a formula to calculate date x years (and maybe also x months prior) and if I get the diffrence between two calendars right, it should throw in an extra day on turn of the century. And then something to automaticaly estimiate latest early date and earliest late date ~~Xil (talk) 07:54, 4 June 2012 (UTC)

June 4

definition of set

set is a well defined collection of objects. but is {1,2,a,b} can be called a set? do the elements in the set need not have any relation among them? — Preceding unsigned comment added by Sateesh shivaraju (talk • contribs) 15:53, 4 June 2012 (UTC)

Well, they do have a relationship - they're members of the same set. There could also be other relationships. For example, the set of items listed by Sateesh shivaraju is {1,2,a,b}. But my guess is that you're approaching this from a computer science perspective, where numbers and characters have different "types" and containers are typically limited to only holding a single "type" of object. That, however, is a limitation of computers, not of mathematics. The divisions in mathematics aren't as "predefined" as those in computer science. To some extent you could argue that all members of a set are of the same kind, but that's a little facile, because in mathematics you can shrink or expand the definition of "the same kind" to include or exclude whichever objects you feel like at the time. For example, the set I gave above only contains items of the type "things that can be listed by Sateesh shivaraju". Such mathematical "types" don't have to be rigidly predefined like the integer/float/string types in programming. -- 71.35.105.132 (talk) 17:07, 4 June 2012 (UTC)