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September 21

Ok, so we have the trigonometric sine, and then the hyperbolic sine...

The trigonometric sine is related to the unit circle, while the hyperbolic sine is related to the unit hyperbola. So, can there be sines for other conic sections as well? The parabolic sine and elliptic sine, perhaps?

Thanks for satisfying my mathematical urges 0_0 --Ķĩřβȳ ♥Ťįɱé Ø 06:48, 21 September 2006 (UTC)[reply]

The unit circle is described by the parametric equation x = cost, y = sint. The unit hyperbola is described by x = cosht, y = sinht. More generally, the equation x = acost, y = bsint describes an ellipse (so there is no need for separate "elliptic" sine and cosine) and the equation x = acosht, y = bsinht describes a hyperbola. Since the equation x = at, y = bt² describes a parabola, the natural candidates for "parabloic" sine and cosine are cospt = t, sinpt = t², which aren't at all interesting. So no. -- Meni Rosenfeld (talk) 09:36, 21 September 2006 (UTC)[reply]

Not quite — there's nothing "natural" about choosing x = t, y = t²; x = t³, y = t⁶ would do equally well.

If you use the hyperbolic functions to parametrise the hyperbole, by setting, say,

\gamma (t)=(\sinh(t),\cosh(t))

, the normalisation property is that

|\gamma '(t)|=|\gamma (t)|

(and the equivalent holds for cos and sin).

I'm going to leave solving this for the parabolic curve as an exercise for the reader, but would like to caution about the choice of "focus" in a parabola. Using the origin might not be the right thing.

RandomP 14:47, 21 September 2006 (UTC)[reply]

Not directly relevant to your original question, but there are some elliptic functions: sn, cn, and dn covered in the article Jacobi's elliptic functions that you might find interesting. Madmath789 10:04, 21 September 2006 (UTC)[reply]

There are also an infinite family of "sines" and "cosines" based on q-exponentials. --HappyCamper 15:05, 21 September 2006 (UTC)[reply]

Parabolic trigometry arises in the Kleinian geometry described in the book by I. M. Iaglom, A simple non-Euclidean geometry and its physical basis: an elementary account of Galilean geometry and the Galilean principle of relativity, which bears the same relationship to "Newtonian spacetimes" as Minkowski geometry bears to Lorentzian manifolds, for which see the book by Misner, Thorne, and Wheeler, Gravitation. More technically:

E^1,1 is a plane geometry (defined by an indefinite but nondegenerate quadratic form) which can serve as the model for tangent spaces to a two dimensional Lorentzian manifold,
E^1,0 is a plane geometry (defined by a degenerate quadratic form) which can serve as the model for tangent spaces to a two dimensional Newtonian spacetime,
E² is a plane geometry (defined by a positive definite quadratic form) which can serve as the model for tangent spaces to a two dimensional Riemannian spacetime.

There are algebraic formulations in which these three geometries arise from three kinds of Cayley-Klein algebras (a generalization of the complex number field considered as a two-dimensional real algebra). I wrote a Wikipedia article about this stuff once but (sigh) some ignorant mathcrank munged it out of existence. Unfortunately the book by Iaglom (a translation from the Russian) is rather hard to find; the only remaining copy which I know of is in the collection of the research library of the Los Alamos National Laboratory. You might be able to get it on interlibrary loan, even through your public library system, if you happen to live in the Western U.S. It's a suprisingly elementary book (originally written I think for Russian high school students), and parabolic trigonometry is significantly easier than circular or hyperbolic trig. (If you've seen books by Jaglom or Yaglom, this is the same author; at different times his name has been transliterated in many ways, which leads to all kinds of confusion).---CH 21:44, 21 September 2006 (UTC)[reply]

There are two Yagloms: A. M. Yaglom and I. M. Yaglom. I don't know their relationship, but they have published together.[1][2] --Lambiam ^Talk 22:38, 22 September 2006 (UTC)[reply]

What about sine and cossine which when: x = sqcos t, y = sqsin t, they define a square? ☢ Ҡi∊ff⌇↯ 04:17, 22 September 2006 (UTC)[reply]

The rule of 17

What is the rule of 17? It appears to be used in design considerations for equal temperament. --152.62.109.163 10:29, 21 September 2006 (UTC)[reply]

When placing frets on an equal tempered guitar (e.g.) the distance to the next fret (a minor second interval) can be calculated by multiplying the octave length (in inches, e.g.) with 1 minus the 12th root of 2 (1 - 2^(1/12), or approx. 0.059463. This is the same as dividing by 16.817, which is sort of close to 17. Maybe that's it.---Sluzzelin 13:40, 21 September 2006 (UTC)[reply]

Close, but no cigar. And two explanations I found on the web state that 17.817 is the 1/12th root of 2, which is nonsense.

The true explanation depends on a little physics, a little perceptual psychology, a little mathematics, and artistic taste.

A vibrating guitar string is not nearly as stiff as a metal bar, nor even a piano string; so half a string produces a frequency twice as high as a full string. Perceptually, a double frequency sounds similar to the original, and very harmonious. In terms of a musical scale, the higher note is said to be an "octave" above the lower one, from the eight steps in a Western major scale. (The notes of a C-major scale are C, D, E, F, G, A, B, c.) The scales we use in Western music are originally based on simple frequency relationships; for example, a "fifth" (C to G) was a ratio of 3:2, a "fourth" (C to F) was a ratio of 4:3, and a major "third" (C to E) was a ratio of 5:4. However, around the time of Johann Sebastian Bach composers and performers on instruments like the harpsichord began to find this tuning decidedly inconvenient. The problem was that what sounded good for C-major sounded awful for G-major or F-major. Thus a transition was made to a chromatic scale where the ratio between any two adjacent notes was exactly the same, with twelve steps in an octave. If an "A" had a frequency of 440 Hz, then the ratio of "C" to "B", say, had to be the 12th root of 2, approximately 1.059463:1. The fifth became 1.498:1 instead of 1.5:1, the fourth became 1.335:1 instead of 1.333:1, and the major third became 1.2599:1 instead of 1.25:1. Every interval except the octave has been perturbed a little, but now the intervals work equally well in every key.

The guitar is a fretted string instrument, and (to a first approximation) the ratio of full string length to fretted string length gives the frequency ratio of shorter to longer. Thus a fret halfway along the string doubles the frequency, producing a pitch an octave above the unfretted pitch. Where do we place the other frets? We expect the fifth to have a fret at about 1/3, producing a ratio of 3:2, but we really need something systematic.

If the "E" string, say, has a length of L = 650 mm, what proportion p of its length should we remove (by fretting) to produce an "F" note, the next chromatic note above it? Well, if we remove L/p we are left with L−L/p, and the ratio of L to this should be the twelfth root of 2, 2^1/12. The answer is easily found, and happens to be

2^1/12⁄(2^1/12−1) ,

which is approximately 17.8172 — hence the "rule of 17". (Musicians are not necessarily the best mathematicians!) And because the ratios are all equal, the next fret should cut off the same proportion of the remaining length, and so on. --KSmrq^T 15:22, 21 September 2006 (UTC)[reply]

KSmrq's answer is, of course, nearly equivalent to Sluzzelin's: 16/17 is approximately 2^-1/12, and 18/17 is approximately 2^1/12 (though Sluzzelin was closer, I wouldn't bet on many musicians realising that).

RandomP 15:43, 21 September 2006 (UTC)[reply]

Hey! :-/ —Bromskloss 10:21, 22 September 2006 (UTC)[reply]

Vincenzo Galilei was quite fond of 18/17 (and so were luthiers of his time), but there are even better rational approximations... 53/50, 71/67, etc... (I put a list up at Talk:Semitone recently.) - Rainwarrior 19:35, 21 September 2006 (UTC)[reply]

Probability Q

I dont understand how the correlation coefficient relates to finding the mean and variance of a summation. For example, Men make 40,000 a year with SD of 12,000, women make 45,000 with SD of 18,000. Now assuming I am given an Rsquared (CC) of 0.7 between male and female earnings, how does this affect the mean and variance of total household income for dual earner families? I assume the mean would be 85,000, but then I am afraid of multiplying the variances and square rooting, because this doesnt involve the 0.7...help?

Edit: I attempted to use the equation: Var(X+Y) = VarX + VarY + 2 Covariance (XY) but to find the covariance using the Correlation Coefficient (CC) I need to use the eqn: Corr(XY) = Cov(XY)/((SDX*SDY)) but that gave me a covariance of 151.2 million (because it becomes: 0.7 = Cov/(12000*18000)

...Help?

What's wrong with a covariance of 151.2 million? -- Meni Rosenfeld (talk) 19:17, 21 September 2006 (UTC)[reply]

Well...that would mean that the new variance of the total household income, using Var(X+Y) = VarX + VarY + 2 Covariance (XY), would be IMMENSE...and that would also mean that the Variances of the individual X and Y would barely factor in, which seems counterintuitive. Is the 151.2 million an accurate intermediate step? ChowderInopa 22:02, 21 September 2006 (UTC)[reply]

Var(X)+Var(Y) = ~~2169~~ 468 million, so how can you say that would "barely factor in"? I'm not sure what you mean by "Rsquared (CC)". Correlation coefficients can be negative, so what is being squared here? --Lambiam ^Talk 00:42, 22 September 2006 (UTC)[reply]

ChowderInopa, you have forgotten that the variance is the square of the SD. Btw, Lambiam, I got VarX + VarY = 468 million. -- Meni Rosenfeld (talk) 07:03, 22 September 2006 (UTC)[reply]

You are correct; I copied the wrong number into my mental calculator. --Lambiam ^Talk 11:17, 22 September 2006 (UTC)[reply]

You are correct, I was using the standard deviations, which was why I was confused as to the small contribution to the final answer. Oh, and for the R squared CC, i thought i remembered from a stats class that the R^2 and Correlation coefficient are the same? or is that just R and the CC? Finally, my confusion stems from the fact that I can not visualize how such a huge variance for the total family income makes sense... I have always wondered the practical point (aside from calculations) of the variance...SD i understand, as showing the centeredness of the data, but variance is such a huge, (in my mind, completely abstract) number, compared to the rest of the data. Seriously, what use is it to know that the variance of family income is hundreds of millions? ChowderInopa 18:05, 22 September 2006 (UTC)[reply]

Variance is the square of the standard deviation. This also means that the units in which it is measured are the square of the units of the SD (and the data). So if the SD is 10,000 USD, then the variance is 100 million "USD squared". You cannot compare the 100 million with the 10,000 because they are not measured with the same units (the same way you cannot compare a mass of 100 kilograms with a length of 2 kilometers). Variance, just like SD, measures the centeredness (or lack thereof) of the data, but using a completely different scale from the data itself. -- Meni Rosenfeld (talk) 18:23, 22 September 2006 (UTC)[reply]

Histogram

Does any one know why a histogram is named such?--Willworkforicecream 17:51, 21 September 2006 (UTC)[reply]

The basic meaning of Greek histos is: "something that has been made to stand upright", with specific meanings of "mast", "beam" and "loom". The -gram part means: "something that has been written". Together: "something written with beams". --Lambiam ^Talk 19:17, 21 September 2006 (UTC)[reply]

Thanks.--Willworkforicecream 13:38, 22 September 2006 (UTC)[reply]

Displacement, Velocity, Acceleration, Jerk, ...

The first derivative (with respect to time) of displacement is velocity, the second derivative is acceleration, and the third derivative is jerk. Is there a word for the fourth derivative of displacement? —Mets501 (talk) 20:02, 21 September 2006 (UTC)[reply]

This is discussed in the Jerk article. Chuck 20:30, 21 September 2006 (UTC)[reply]

Wow I'm blind! I thought I read the whole article... Thanks anyway Chuck :-) —Mets501 (talk) 20:49, 21 September 2006 (UTC)[reply]

Ah! I didn't know there was a word for it, but I have many times thought about it when going by metro – realising that the engineer never took it into account, but should have. —Bromskloss 10:16, 22 September 2006 (UTC)[reply]

…and I love the "snap, pop, crackle"! :-) —Bromskloss 10:22, 22 September 2006 (UTC)[reply]

Bijection from Z^2 to Z

Sorry for the notation in the subsection header. I'm looking for a simple bijection $\phi :\mathbb {Z} \times \mathbb {Z} \leftrightarrow \mathbb {Z}$ . By the fact these two sets have the same cardinality, we know some bijection must exist, but I'm wondering if there's a fairly simple function that exhibits this property.

-- Braveorca 23:55, 21 September 2006 (UTC)[reply]

How about a square spiral with two arms? Melchoir 00:05, 22 September 2006 (UTC)[reply]

Well, I thought of that one (the conventional "put them on a string like so and then straighten the string"), I was wondering if there was something like

\phi (x,y)=xy-x+y

- a simple, easy-to-define function. - Braveorca 02:26, 22 September 2006 (UTC)[reply]

There will not be a continuous map, in the sense that the map is continuous on the underlying RxR, like the polynomial you mention. I recall that this follows from the different dimensionalities of the two sets (points in ZxZ have too many neighbors to map continuously down to the line). There are discontinuous mappings that are simple. E.g.

(\sum _{n=0}^{\infty }2^{n}x_{n},\sum _{n=0}^{\infty }2^{n}y_{n})\leftrightarrow \sum _{n=0}^{\infty }2^{2n}x_{n}+\sum _{n=0}^{\infty }2^{2n+1}y_{n}

which just interleaves the bits of x and y in the forward direction and demultiplexes them in the reverse direction. -- Fuzzyeric 04:51, 22 September 2006 (UTC)[reply]

Fuzzyeric, your example does not work. The assignment you've described is not a bijection. The reverse (right-to-left) function is not one-to-one, so the forward (left-to-right) function is not a surjection. For example, 1/6 is a repeating fraction 0.001...₂ in binary. When demultiplexed, it gives a pair of (0.01...₂, 0) = (0.1₂, 0) = (1/2, 0). This in turn interleaved gives 1/2, which is obviously not the starting number, 1/6.

You need either some extra restrictions for this method to work, or just redefining it. The latter can be done in a simple way—just interleave strings of ones terminated by zero, rather than single digits. This protects against 'gluing' multiple ones into an infinite string, and as a result guarantees a one-to-one mapping in both directions.

Examples:

(1/3, 14/15) = (0.01...₂, 0.1110...₂) = (0.010...₂, 0.1110...₂) maps to 0.0111010...₂ = 0.011101...₂ = 29/63

(1/3, 1/3) = (0.01...₂, 0.01...₂) = (0.010...₂, 0.010...₂) maps to 0.0010...₂ = 0.001...₂ = 1/6

(1/3, 55/64) = (0.01...₂, 0.110111₂) = (0.010...₂, 0.11011100...₂) maps to 0.0110101110100...₂ = 1507/3584

CiaPan 16:14, 22 September 2006 (UTC)[reply]

Ooops, sorry, I confused integer numbers with reals. Of course you method works fine for integers, as their binary representation as always finite. CiaPan 16:35, 22 September 2006 (UTC)[reply]

Uh... the question before us concerns integers, right? Melchoir 16:28, 22 September 2006 (UTC)[reply]

Melchoir is right that this is a question about integers, not rationals. Although, CiaPan, is right for the wrong reason. The recipe I gave only works for half of ZxZ because it doesn't handle the signs well. A better method would poke the sign of x into the least significant bit ("-" → "1", "+/0" → "0"), then take the sign of y to be the sign of the result. And this should patch up the omission. -- Fuzzyeric 16:40, 22 September 2006 (UTC)[reply]

Map integer numbers onto naturals with

m:\mathbb {Z} \to \mathbb {N}

given by

m(z)={\begin{cases}2z&{\textrm {for}}\ z\geq 0\\-2z-1&{\textrm {for}}\ z<0\end{cases}}

then use bijection

b:\mathbb {N} ^{2}\leftrightarrow \mathbb {N}

defined as

b(n,m)=\left((n+m)^{2}+n+3m\right)/2

CiaPan 06:25, 22 September 2006 (UTC)[reply]

It's interesting to note that CiaPan's map runs sequentially through the diagonals with slope -1, in the order (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ... . The other step does fold, spindle, and mutilate Z, though... -- Fuzzyeric 01:35, 23 September 2006 (UTC)[reply]

Yes, that's exactly what I wanted when inventing this function (of course I'm sure there are and were hundreds people, who invented it before me!). And the m(z) mapping is essentialy the same which you described above: it shifts an absolute value by one bit and inserts the sign into the ones' position. --CiaPan 06:22, 27 September 2006 (UTC)[reply]

My favourite bijection from the square of a countable set to itself uses the set

X=\mathbb {Z} ^{+}=\{1,2,\ldots \}

. Then define a bijection

f:X\times X\to X

by

f(m,n)=2^{m-1}(2n-1)

.

Doctorbozzball 11:15, 23 September 2006 (EDT)