Wikipedia:Reference desk/Archives/Mathematics/2006 September 23

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Identities

What are identities involving nested hyperbolic functions and inverse trigonometric functions called? --HappyCamper 04:43, 23 September 2006 (UTC)

Say, something of this sort: $\tanh ^{-1}(\sin 2x))\equiv 2\tanh ^{-1}(\tan x).$

This is also equal to ${\mbox{gd}}^{-1}(2x),$ which uses the inverse Gudermannian function. I don't think the species has been named; perhaps call them "inverse-Gudermannian identities"? --Lambiam^Talk 15:57, 23 September 2006 (UTC)

Ah, I did not spot that quite so quickly. Actually, this came from a foreign language textbook that I found, there were other interesting identities invovling inverse sines and hyperbolic cosines, which was what prompted this question. --HappyCamper 20:32, 23 September 2006 (UTC)

Questions about the shape of the earth and distances

I was thinking, what is the relationship to a torus and a sphere? That is, let's assume that we have a torus and a sphere with equal surface area and equal volume. So let's say that the earth was a torus, with two equators, and the north pole and south pole coalesce into each other, in the middle. And let's say that the continents and oceans of the earth were proportionally mapped onto the torus. So my question would be, let's say someone wanted to get from Alaska to South America. In a spherical earth, a person would have to fly a plane south from Alaska to reach to South America. However, in a toroidal earth, a person could theoretically travel north, through the pole, and arrive at South America, approaching it from the south. Let's ignore the affects of weather. So would it be a shorter distance? And is there a formula to decide whether or not a distance from random points A and B on a torus would be longer or shorter than the proportional points A' and B' on a sphere? Thanks. --Ķĩřβȳ♥ŤįɱéØ 06:05, 23 September 2006 (UTC)

Let's begin by saying that there is not torus with equal surface area and volume as a sphere. Even if we just stick to "same surface area" (and lesser volume), I do not understand what your transformation from the sphere to the torus is, but I am inclined to believe that distances will typically be longer. -- Meni Rosenfeld (talk) 06:13, 23 September 2006 (UTC)

Ok, they have the same surface area then. I mean it the earth to look like this: [1]. --Ķĩřβȳ♥ŤįɱéØ 06:54, 23 September 2006 (UTC)

Here are some thoughts:

• The image is misleading, because the "second equator", the line where the poles are stitched together, has a discontinuity, where in the image it looks continuous.
• Suppose the earth has a radius of R, and the torus has an inner radius of a and an outer radius of b. Then:
  • The distance around the equator is 2πR on earth, (b+a)π on the torus.

What was I thinking? It's 2bπ. -- Meni Rosenfeld (talk) 08:01, 25 September 2006 (UTC)

• • The distance from the south pole, through the equator, to the north pole is πR on earth, π(b-a) on the torus (and of course, just the distance from the equator to a pole is half of that).
  • The surface area of earth is 4πR², of the torus is (b²-a²)π.
  • The distance around latitude θ is 2πRcos(θ) on earth, ((b-a)cos(2θ)+a+b)π on the torus. (a nicer formula: 2π(asin²θ + bcos²θ)
• We would want the first three of these to be equal for earth and the torus, but they are easily seen to be incompatible. If the area is to be equal, then at least one of the first two distances must be greater on the torus.
• The fourth distance, for θ = 90°, is 0 for earth but (b-a) for the torus. Not nice. This also means that whatever our choice for a and b, there will be near the poles points which are very close on earth but distant on the torus (and the ratio can be as large as we please).
• All that aside, regarding your original question: Yes, since we can "magically" leap from the south pole to the north pole, there will necessary be points which on earth are distant (one near the north pole, one near the south pole) but will be close on the torus.
• About a general formula, it could be complicated, but I have given above examples of some specific cases. Of course, since they are all circles, the corresponding arcs will have proportional lengths.

I hope this answers your question. -- Meni Rosenfeld (talk) 08:24, 23 September 2006 (UTC)

Yes, it answers many of my questions, but I don't understand one part. How is it the second equator discontinuous? --Ķĩřβȳ♥ŤįɱéØ 08:38, 23 September 2006 (UTC)

It is not continuous in the same sense as a sphere because you're mapping a sphere (the map of Earth) on a torus. The north pole is connected to the south pole in the torus, so if you did morphed Earth as a torus, you'd have a discontinuity in the landmasses. Here's a visualization:

That horizontal middle line is how it would look like in the second equator, and that's what I think he meant with discontinuity. Hope that helps. ☢ Ҡi∊ff⌇↯ 09:33, 23 September 2006 (UTC)

Yes, suppose the north pole would be painted blue and the south pole painted red. Then in the stitch, you'll see a discontinuous jump from red to blue. -- Meni Rosenfeld (talk) 10:03, 23 September 2006 (UTC)

In the most general terms, it's not possible to go from a surface without a hole, to one with a hole, without tearing the surface. StuRat 10:53, 23 September 2006 (UTC)

Iterated sine function

Is there a name and a documented use for an iterated sine function? Example (my notation): $\sin _{5}x=\sin \sin \sin \sin \sin x$

I've been playing with it and it's been kinda interesting: when i tends to infinity, $\sin _{i}x\times \left({\frac {1}{\sin _{i}\pi /2}}\right)$ approaches a smooth square wave $f(x)=\operatorname {sgn} \sin x$ (using the sign function), which looks better than one made by additive synthesis of sine waves. I was also wondering if there would be a way to get the triangle and sawtooth waves with a similar method. $\tan \sin x$ is pretty close from a triangle wave, but I can't find a way to make it converge to a true triangle wave. $\sin \tan x$ resembles a sawtooth wave, but it gets chaotic near the jumps... Still, pretty interesting behaviors. ☢ Ҡi∊ff⌇↯ 08:30, 23 September 2006 (UTC)

Well, the infinite iteration could simply reprsented as $\sin x=x$

sawtooth

And that could be represented as an infinite series: $\sin x,\sin(\sin x),\sin(\sin(\sin x))),\sin(\sin(...))$

So the iterated sine would be the nth term of this series, where n is the number of iterations. I don't think this has a name though.

I'm not sure about your second question though. My guess would be a piecewise function involving absolute value. I hope that helps --Ķĩřβȳ♥ŤįɱéØ 09:03, 23 September 2006 (UTC)

I'm not sure if I get your point. The infinite iteration doesn't fall back to simply $\sin x=x$. Just plot sin(sin(sin(...x))) yourself and take a look. About the second question, I really don't want to involve piecewise functions here. ☢ Ҡi∊ff⌇↯ 09:26, 23 September 2006 (UTC)

The iterated sine function converges pointwise (i.e., for each fixed x) to zero. I would not like to express this exactly as the infinite iteration could simply reprsented as $\sin x=x$; but there is a connection with the fact that that equation has no other (real) solution than zero. That makes the idea of Kieff to multiply with a 'normalising factor' rather interesting. (Note, that there is nothing magical connected with $\pi /2$ in this context. If you pick any number c strictly between 0 and $\pi$, and use $(\sin _{i}c)^{-1}\,$ as normalising factor, you'll get the same limit function.) The 'normalisation' counteracts a very slow convergence close to the origin.

I've no idea whether or not this is done - this is not exactly my branch of mathematics - but if it isn't, I think it would be worth doing. JoergenB 15:43, 23 September 2006 (UTC)

Experimentally, the normalizing factor appears to be asymptotically equal to ${\sqrt {\frac {i}{3}}}$. --Lambiam^Talk 16:33, 23 September 2006 (UTC)

How come you say any number c strictly between 0 and π will get the same limit function? I used π/2 because it is a point when the function is at its highest value, so the normalization would make its max be exactly 1 at any iteration. Does that detail become completely irrelevant when dealing with infinite iterations? ☢ Ҡi∊ff⌇↯ 17:17, 23 September 2006 (UTC)

OK, I'll answer, but I fear I can't without getting slightly technical.

Starting with c strictly between 0 and $\pi /2$ yields a positive number not greater than 1 in the next step, and continues with a strictly decreasing sequence of positive numbers:

$\sin c>\sin _{2}c>\sin _{3}c>\sin _{4}c>\ldots >0$

(since, by an exercise in elementary calculus, indeed $0<\sin x<x\,$ for $0<x<\pi /2\,$). Thus, this sequence has an infimum $L\geq 0\,$. Now, if L were strictly greater than 0, then for some i we would have $\sin _{i}c<\arcsin L\,$, yielding the contradiction $\sin _{i+1}c<L\leq \sin _{i+1}c\,$. Thus, instead, $L=0$, for any start value c in the open interval $(0,\pi /2)\,$.

In particular, for any two given (fixed) c and d in that interval; there is a natural number j, such that $\sin _{1+j}c<\sin d\,$ and $\sin _{1+j}d<\sin c\,$. Moreover, the sine function is strictly increasing in the interval from 0 to 1 (check its derivative!), whence these properties carry over to all higher iterations as well:

$(1)\ \ \sin _{i+2j}c<\sin _{i+j}d<\sin _{i}c,\ i=1,2,3,4,\ldots$.

If you've followed the argument this far, then you're ready for the main point: Every time we do the iteration, we are replacing the value of some x with this x times a factor ${\frac {\sin x}{x}}$, and these factors converge to 1 as x approaches 0. In fact, in this interval, $x-{\frac {x^{3}}{6}}<\sin x<x$, whence the factor is a number between $1-{\frac {x^{2}}{6}}$ and 1. Thus, also the j'th power of this factor tends to one. Summing up, if we compare the quotients of the terms in formula (1) with $\sin _{i}c\,$, then we find that the first quotient converges to 1, while the third one indeed is (the constant) 1; whence the squeezed middle quotient also converges to 1.

I hope I glossed over an appropriate amount of details (giving you just enough to fill in yourself) :-) JoergenB 18:59, 23 September 2006 (UTC)

For a triangle wave, try $\arcsin ~\sin x$ or $|x-\lfloor x\rfloor -1/2|$. For sawtooth, try $\arctan ~\tan x$ or $x-\lfloor x\rfloor$. – b_jonas 17:46, 23 September 2006 (UTC)

Thanks for those but they're not what I'm looking for right here. I'm trying to find smooth functions that approximate these ideal waveforms as some value increases, like the iterated sine with the square wave. Using the floor function is kinda bad in this case too, since it's a bit of a "hack". ☢ Ҡi∊ff⌇↯ 06:13, 24 September 2006 (UTC)

Well, the triangle one might be possible by modifying the square wave one you have... if you are transforming a sin into a sqaure gradually via some process, it seems that the triangle should be available, since the square is, I believe, $\sum _{i=1}^{\infty }{\frac {\sin {((2i-1)x})}{(2i-1)}}$, and the triangle the milder $\sum _{i=1}^{\infty }{\frac {\sin {((2i-1)x})}{(2i-1)^{2}}}$, a less severe transformation of your initial sin might get you there (or something with an equivalent spectrum, maybe). Saw waves might be harder, since you will need to find some way to generate the even harmonics as well. Do you know about Frequency modulation synthesis? It is similar to what you are doing, though it is not usually taken to infinity like what you are doing here. - Rainwarrior 18:20, 24 September 2006 (UTC)

But by the way, if you are trying to find a smooth function that approximates one of these, why not just use the Fourier series (the sums I just mentioned above) of them and stop after a certain number of harmonics? - Rainwarrior 18:30, 24 September 2006 (UTC)

Oh, I don't have a direct use for these stuff, it's just something I've been toying with, really. But the problem with Fourier series is that they have artifacts. The iterated sine gives a much better result than a square wave Fourier series, and it has no artifacts at all, so I thought "hey, maybe there are similar functions for the other waves, and they could be pretty handy"... That's pretty much it.

I already tried getting some info from the Fourier series, btw, but that didn't really help. :( ☢ Ҡi∊ff⌇↯ 20:53, 24 September 2006 (UTC)

What do you mean "artifacts"? A summation of sine waves is continuous... But, if you're interested in finding other curves which approximate various waveforms, why not construct a polynomial curve with suitable guidelines between 0 and 2pi? (For starters, make it pass through (0,0) and (2pi,0), and make both tangents at those points equal as well, then keep going, add point or tangent constraints until you have the curve you like.) - Rainwarrior 03:57, 25 September 2006 (UTC)

The artifacts I'm talking about is the Gibbs phenomenon. And no, a polynomial would take away all the fun of it (too easy), plus it would not be cyclic. ☢ Ҡi∊ff⌇↯ 07:49, 25 September 2006 (UTC)

Sorry for my stupidity, but what does an X with the L and the mirrored L around it mean? ChowderInopa 00:09, 24 September 2006 (UTC)

It denotes application of the floor function to the argument. For example, ⌊3.14⌋ = 3, because 3 is the largest integer that does not exceed 3.14. If we had a skyscraper with a floors at the height of every integer, and real numbers had to live at their own heights, then 3.14 would be living on the 3rd floor. --Lambiam^Talk 00:32, 24 September 2006 (UTC)