# Wikipedia:Reference desk/Archives/Mathematics/2010 January 26

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# January 26

## Inverse Trigonometric Identity Prove

Hello. How would you prove $\arctan m + \arctan n = \arccos \frac{1 - mn}{\sqrt{(1 + m^2)(1 + n^2)}}$? Thanks in advance. --Mayfare (talk) 02:11, 26 January 2010 (UTC)

Take the tangent of both sides. The lhs can then be evaluated using the tangent angle addition formula; you should get $\frac{m+n}{1-mn}$ after canceling tans with arctans. Let the messy expression inside arccos be X, and define θ=arccos(X), or X=cos(θ). Then use $\tan^{2}(\theta)+1=\sec^{2}(\theta) \,$. You know sec(θ)=1/X. A bunch of messy algebra confirms that the rhs equals the lhs. --COVIZAPIBETEFOKY (talk) 02:30, 26 January 2010 (UTC)

I like your proof. Do we have to apply tangent on both sides since that would be crossing the left and right sides? --Mayfare (talk) 02:54, 26 January 2010 (UTC)

You always have to do the same thing to both sides, otherwise the two sides would no longer be equal. --Tango (talk) 04:32, 26 January 2010 (UTC)
Alternative geometric approach: draw a right angled triangle PQR with base 1 and height m. Using same base but going in opposite direction (i.e. "down" instead of "up", draw a right angled triangle PQS with base 1 and height n. You now have a triangle PRS with sides PR = √(1+m2), PS = √(1+n2) and RS = m+n. The angle opposite side RS is arctan(m)+arctan(n). Now apply the cosine rule. Gandalf61 (talk) 14:28, 26 January 2010 (UTC)

## Cross ratio equal to 1/2tan^2 of spherical distnce

Hi all,

I wasn't sure whether to ask this here or not, I was hoping to sort it out myself but after 4 days thinking about it and so far had no such luck, it's clear that I'm not going to make any further progress without a little help - I don't think I can quite get my head around the geometry that's in play.

If u,v$\in \mathbb{C}$ correspond to points P, Q on $S^2$, and d denotes the angular distance from P to Q on $S^2$, show that $-\tan^2(\frac{d}{2})$ is the cross ratio of the points $u, v, \frac{-1}{u^*}, \frac{-1}{v^*}$, taken in an appropriate order (which you should specify). (The star denotes complex conjugation - I'm not sure how to do the 'bar' in latex!)

Now I'm useless at geometry, but if I recall correctly, $\frac{-1}{u^*}$ would correspond to the stereographic projection of the point (-P), right? And likewise with v - other than that however, I really can't see a smart way to do this. I certainly don't want to try all 6 permutations of the 4 points and see what pops up on the cross ratio, but at the same time I can't see intuitively where the $-\tan^2(\frac{d}{2})$ could have come from in order to try and work out how to take the cross ratio to get the desired result. I'm aware that the angle between the 2 points P and Q is equal to d, but I can't seem to work anything out in the complex plane rather than the sphere. Please help!

Many thanks in advance! Mathmos6 (talk) 04:22, 26 January 2010 (UTC)

Cross-ratio is a projective invariant i.e. it is unchanged by projective transformations. So you can rotate $S^2$ to place u and v at convenient points without changing the cross-ratio. For example, you can place u at the origin and v at a point x on the real axis. Then -1/u* is at infinity, -1/v* is at -1/x, and the cross-ratio is -x2 (I think). Gandalf61 (talk) 11:38, 26 January 2010 (UTC)

## I was wondering...

Is there a set numbers considered as the infinitesimal set? --Neptunerover (talk) 07:09, 26 January 2010 (UTC)

No. You can talk about the subset of infinitesimals within some number system, but that's all. An infinitesimal is a number that, however many times you add it to itself, will never give a result greater than 1 (or any other finite number). What infinitesimals exist will depend on what number system you are working in. In all the commonly used systems (ie. subsets of the complex numbers) the only infinitesimal is 0 (which is usually explicitly excluded when talking about infinitesimals). --Tango (talk) 07:43, 26 January 2010 (UTC)
I was thinking of within the set of rational numbers. --Neptunerover (talk) 07:55, 26 January 2010 (UTC)
If you wish to see concretely an infinitesimal among rationals, call it x and just add it to them as a tascendental extension. You'll find the ordered field of rational functions Q(x), where the order is such that 0<x<q for any positive rational q, and in general for two elements, f<g means that f<g as functions in a right neighborhood of 0.--pma 09:28, 26 January 2010 (UTC)
Thank you. That darn f of x stuff is the semester I had an algebra teacher who taught straight out of the textbook, which lost me from the beginning. I don't believe those books are intended to make sense on there own, otherwise there would be no need for the class. --Neptunerover (talk) 09:41, 26 January 2010 (UTC)
One should emphasise: that is among rationals, not within rationals. Q(x) is not a subset of Q. Zero is the only infinitesimal within the rationals. --Tango (talk) 10:00, 26 January 2010 (UTC)
More emphasis than that would be aragoto style. --pma 11:20, 26 January 2010 (UTC)

## Normed Spaces

I want to find out the largest c such that $\lVert\alpha_1e_1 +\alpha_2e_2\rVert \ge c(|\alpha_1| + |\alpha_2|)$ where ${e_1,e_2}$ is the standard basis of $R^2(R)$ endowed with the standard norm. How should I proceed? I am interested in the general way of finding out such a c in case of arbitrary bases or normed spaces. Thanks.-Shahab (talk) 08:41, 26 January 2010 (UTC)

I assume you want c independent from α1 and α2. So you want to compare the L1 and L2 norms in R2. Use the Cauchy-Schwarz inequality with the two vectors (1,1) and (α1, α2):
1| + |α2|=1|α1| + 1|α2| ≤ (12 + 12)1/2(|α1|2 + |α2|2)1/2=21/2 (|α1|2 + |α2|2)1/2. Recall that the CS inequality is an equality whenever the two vectors are linearly dependent (here it means |α1|=|α2|); this tells you that c:=2-1/2 is the largest. To compare two Lp norms use analogously the Minkowski inequality. In general, the problem of computing the best constant of a normed space embedding may be a very hard problem in the calculus of variations. --pma 09:05, 26 January 2010 (UTC)
Can you elaborate on why c:=2-1/2 is the largest part, not on equality in CS iff vectors are LD but how you use this to conclude that c:=2-1/2 is the largest. I intutively understand it but I am having trouble deriving it rigorously. Thanks for the rest.-Shahab (talk) 10:25, 26 January 2010 (UTC)
You want the largest c with the property that: for all $\alpha_1$ and $\alpha_2,$ there holds $\lVert\alpha_1e_1 +\alpha_2e_2\rVert \ge c(|\alpha_1| + |\alpha_2|).$ The number $2^{-1/2}$ has this property; any number larger than $2^{-1/2}$ has not, because the property fails when tested with (1,1).
We can also say it in another way: since the inequality is trivially satisfied for (α1, α2)=(0,0), you may forget about (0,0) and ask it just for all (α1, α2)≠(0,0). You may then divide the RHS by the positive quantity$|\alpha_1| + |\alpha_2|,$ and if you think a little you see that what you are looking for is exactly the greater lower bound of the numbers $\scriptstyle\frac{\lVert\alpha_1e_1 +\alpha_2e_2\rVert }{(|\alpha_1| + |\alpha_2|)}$ among all (α1, α2)≠(0,0). Here, we even have a minimum, which is reached by (1,1). --pma 11:03, 26 January 2010 (UTC)
Thanks. It's all clear now. Have a nice day-Shahab (talk) 11:10, 26 January 2010 (UTC)

## A loxodromoid?

Maybe you do have more specialists here? In the Reference desk of the Finnish Wikipedia we got a question like this; I translate it here:
"Does a loxodrome have a 2D counterpart (or is it even possible for it to have such one)? A spiral is very similar, indeed, but, on a circle, its other end crosses the circumference while a loxodrome is wholly located on the surface of a sphere turning towards the poles of it."
I suppose, this is a matter of the qualities of the polarities of a sphere and a circle? --Watsimous (talk) 17:42, 26 January 2010 (UTC)

There is a generalization here (if you speak French as well as English and Finish), but it's a loxodrome defined on surfaces other than a sphere, not a loxodromic surface. It may be possible to invent something like that but question is whether anyone has seriously proposed it as a useful concept and studied it. Mathcurve.com is, imo, pretty complete with this type of thing so if it isn't there the answer is probably no.--RDBury (talk) 20:02, 26 January 2010 (UTC)
The only thing I can think of would be to omit, say, y from the parametric equations for a normal loxodrome; then it will trace a sinusoidal path within a circle and terminate at two opposite points. That's assuming that your correspondent means by a "2D counterpart" a loxodrome-like curve within the circle instead of on the sphere, rather than a 2-manifold based on the loxodrome. --Tardis (talk) 21:33, 26 January 2010 (UTC)
Thank you, guys! As far as I understand, these answers are pretty much complete. --Watsimous (talk) 18:42, 28 January 2010 (UTC)