# Wikipedia:Reference desk/Archives/Mathematics/2010 March 2

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# March 2

## Hyperbolic Line Reflection, and Isometry between H/D

Hi all,

How would I find a diffeomorphism which preserves area on the square |u|<1, |v|<1 with the 2 Riemannian metrics ${\displaystyle {\frac {du^{2}}{(1-v^{2})^{2}}}+{\frac {dv^{2}}{(1-v^{2})^{2}}}}$ and ${\displaystyle {\frac {dv^{2}}{(1-v^{2})^{2}}}+{\frac {du^{2}}{(1-v^{2})^{2}}}}$ (i.e. the diffeomorphism maps from the square with one Riemannian metric to the square with the second Riemannian metric). I really have no idea how to approach this one, since I have basically no experience with diffeomorphisms, and the problem just says 'show one exists', without any hint as to what it might be!

In a similar geometrical vein, how would I show that ${\displaystyle g(z)={\frac {z-a}{z-{\bar {a}}}}}$ defines an isometry from the upper half plane H to the disc D, where ${\displaystyle a\in H}$? I know that the distance in H is given by ${\displaystyle d(x,y)=2tanh^{-1}(|{\frac {x-y}{1-{\bar {x}}y}}|)}$, and I tried substituting ${\displaystyle x={\frac {z_{1}-a}{z_{1}-{\bar {a}}}},y={\frac {z_{2}-a}{z_{2}-{\bar {a}}}}}$ into the distance formula in H, and I ended up with ${\displaystyle 2tanh^{-1}(|{\frac {z_{1}-z_{2}}{(z_{1}-Re(a))+(z_{2}-Re(a))}}}$, and setting ${\displaystyle z_{1}=Re(a)+re^{i{\theta }_{1}},z_{2}=Re(a)+re^{i{\theta }_{2}}}$, we get ${\displaystyle d=2tanh^{-1}(|tanh({\frac {{\theta }_{1}-{\theta }_{2}}{2}})|)=|{\theta }_{1}-{\theta }_{2}|}$, i.e. the argument between the 2 points z1 and z2 when taken around the circle of centre Re(a); but where do I go from here? I have a feeling I'm very close - but is that really the correct distance between z1 and z2 in the Disc model?

Many many thanks! Delaypoems101 (talk) 02:35, 2 March 2010 (UTC)

## HOW TO DRAW

Please Explain in detail HOW TO DRAW ANGLES MEASURING 70 degrees and 40 degrees WITHOUT USING PROTRACTOR. Also explain me whether all the angles can be drawn without using protractor. If not what type of angles can be drawn without using a protractor.Kasiraoj (talk) 04:53, 2 March 2010 (UTC)

If you don't have a protractor, what do you have? A straightedge and compasses? If so, see Straightedge and compass#Constructible angles. As for your specific examples, I don't think either can be constructed. I know 20 degrees can't (that article mentions that is it impossible to trisect ${\displaystyle {\frac {\pi }{3}}=60^{\circ }}$) and you can easily get from either 70 degrees or 40 degrees to 20 degrees (30 is easy to construct and 70-30=40, you can then bisect 40 to get 20). --Tango (talk) 05:08, 2 March 2010 (UTC)
Neither is possible by classical (precise) methods, but if your ruler has accurate markings you can use Angle trisection for an approximate method. (e.g. for 70 degrees, construct two 60 degree angles with a common arm, bisect one to get 30 degrees, then trisect this to give ten degrees next to the sixty degrees). This method will not be accepted by Pure Mathematicians, but may well be taught in Engineering as a good approximation for most practical purposes. There may be even more accurate methods, but mathematicians can prove that none of them can be exact in a super-precise mathematical sense. Surprisingly, it is possible to construct 69 degrees exactly (I'm working on the method). Dbfirs 08:21, 2 March 2010 (UTC)
Why do you say pure mathematicians wouldn't accept angle trisection by compass and marked ruler? Euclid might not have done, but pure maths has moved on quite a bit since him. Algebraist 13:23, 2 March 2010 (UTC)
Because, from a "perfect precision" point of view, all trisection methods involve measuring a length, and this inevitable relies on the accuracy of the ruler and the trisecter's estimating ability. There is a sense in which Euclidean constructions depend only on the perfection of ideal compasses and straight edge, and not on the user's ability to estimate. Dbfirs 14:36, 2 March 2010 (UTC)
You can trisect an angle without even the ruler and compass :) See [1] for instance. Dmcq (talk) 13:27, 2 March 2010 (UTC)
Yes, I like that method, but it still involves estimating a length doesn't it? Dbfirs 14:36, 2 March 2010 (UTC)
No estimation, but it is a Neusis construction like the various classical methods. Dmcq (talk) 14:47, 2 March 2010 (UTC)
Thanks for the link, but may I quote from it "Neusis became a kind of last resort that was invoked only when all other, more respectable, methods had failed." I still consider that it involves a form of visual estimation. Dbfirs 16:14, 2 March 2010 (UTC)
Depends on your standards. It is more difficult than extending a compass from one point to another though that also involves estimating. Dmcq (talk) 18:51, 2 March 2010 (UTC)
Yes, good point! Dbfirs 19:37, 2 March 2010 (UTC)
Am I correct in thinking that the only (non-Neusis) constructible integer angles are those that are multiples of three degrees? Dbfirs 19:53, 2 March 2010 (UTC)
Yes. Algebraist 23:42, 2 March 2010 (UTC)

## Riddle, where is the mistake in ALL x = 0 ?

Hello you all over the world. Please excuse my poor English, I'm a froggy. I'm puzzled by the folowing enigma that shows that ALL x=0! I can't find the mistake though I think it must be just a small trick. For your answers, please consider that I should be able to anderstand an advenced level in maths (4 years in university but it was 30 years ago).

For all x belonging to Z we have the identity :

eix = (eix)1 = eix(2π/2π) = (ei2π)x/2π = 1x/2π = 1

but knowing that eix=cos x + i sin x it brings

cos x = 1 and sin x = 0 , that is x=0 FOR ALL x ! Of course, in fact the correct conclusion should be x=2kπ, k in Z, but anyway the trouble is the same.

Thank you for your help. In a hurry to read your answers. Joël DESHAIES-Rheims-France---82.216.68.31 (talk) 13:01, 2 March 2010 (UTC)

Bonjour Joël. It's because ${\displaystyle a^{bc}=(a^{b})^{c}}$ is not always true when a,b,c are complex numbers. Have a look at the complex numbers secton of Exponentiation Tinfoilcat (talk) 13:13, 2 March 2010 (UTC)
Well, it's quite simple: raising to power is not commutative in complex numbers. Additionaly, and even more important, raising to power with non-natural exponent is NOT a function—it may be a multi-function (think eg. of (complex)√1), so it is important to carefully analyse which of possibly many values you take.
Compare (complex) (1^2)^(1/2) and (1^(1/2))^2. --CiaPan (talk) 13:20, 2 March 2010 (UTC)

Hello, I'm the asker. Thank you for clear explanations, I got it.Joël DESHAIES.Rheims-France---82.216.68.31 (talk) 15:50, 2 March 2010 (UTC)

That cos x = 1 and sin x = 0 does not imply that x = 0. Rather, it implies that x is an integer multiple of 2π, i.e. x ∈ {0, ±2π, ±4π, ...}. Michael Hardy (talk) 15:54, 2 March 2010 (UTC)
Hello Michael, please see my 9th line at the top, I wrote this remark. My title is x=0 because it's more "dramatic", more thrilling! Joël.--82.216.68.31 (talk) 16:17, 2 March 2010 (UTC)
Hi Joël. Note that the point here is not specific of complex numbers. You may do the same fallacy, and maybe see it better, with -1=(-1)2/2=[(-1)2]1/2=1. The point is that a left inverse to a map need not to be a right inverse to it; see left and right inverses. --pma 17:10, 2 March 2010 (UTC)

## Logistic Function

I was looking at the logistic function. This looks like cumulative distribution function of some sort. I decided to discover which distribution corresponded to the logistic function. If p(t) is the distribution that we want, then the logistic function and p(t) are related by

${\displaystyle \int _{-\infty }^{x}p(t)\ {\mbox{d}}t={\frac {1}{1+e^{-x}}}.}$

Differentiating both sides of the equation by x gives

${\displaystyle p(x)={\frac {e^{-x}}{(1+e^{-x})^{2}}},\ \ {\mbox{i.e.}}\ \ p(t)={\frac {e^{-t}}{(1+e^{-t})^{2}}}}$

We can show that this is right by integrating this last expression for p(t) with respect to t with t ≤ x. Doing this, we arrive back at the first equality. What kind of a probability distribution is this p(t)? It looks like a bell-shaped curve but the formula doesn't look like the expression for the Gaussian distribution. Any ideas? •• Fly by Night (talk) 16:30, 2 March 2010 (UTC)

See Logistic distribution. -- Meni Rosenfeld (talk) 16:44, 2 March 2010 (UTC)
Ahhh... D'OH! •• Fly by Night (talk) 16:52, 2 March 2010 (UTC)

## Strange proof

Let S=1+2+4+8+... Then, 2S=2+4+8+16+... Subtracting produces

 2S=  2+4+8+...
- S=1+2+4+8+...
---------------
S=-1


This is clearly absurd, but where's the error in this proof? --J4\/4 <talk> 16:37, 2 March 2010 (UTC)

The error is in the subtraction. Since S is infinite, you can't subtract ${\displaystyle 2S-S}$.
Alternatively, if your definition of infinite sums doesn't allow convergence to infinity, the first error is writing S=1+2+4+8+... to begin with.
Note, however, that this is only absurd in the (extended) reals. It is perfectly valid in, say, the 2-adics. -- Meni Rosenfeld (talk) 16:42, 2 March 2010 (UTC)
But since that's the same logic that's used to derive the formula S=a/(1-r) for an infinite series, doesn't that mean that the formula is invalid as well, since it's also an infinite series? --J4\/4 <talk> 16:46, 2 March 2010 (UTC)
What I meant is not that this is an infinite series but that the sum is infinite. If ${\displaystyle |r|<1}$ then ${\displaystyle \sum _{n=0}^{\infty }ar^{n}}$ converges to a real number, and the above manipulations are valid. -- Meni Rosenfeld (talk) 16:52, 2 March 2010 (UTC)
See 1 + 2 + 4 + 8 + · · ·. Gandalf61 (talk) 16:56, 2 March 2010 (UTC)
You can only use the standard arithmetic rules to manipulate infinite series if they converge. Yours doesn't, so the rules don't apply. (If you use a different topology, in which it does converge, then your answer is correct - the 2-adic numbers have one such topology.) --Tango (talk) 19:58, 2 March 2010 (UTC)
See also Divergent series#Theorems on methods for summing divergent series. Zunaid 09:22, 3 March 2010 (UTC)

There is no error in your argument, and your result should not be considered absurd. You think that because a finite sum of positive numbers is positive, then so should a infinite series of positive numbers also be positive, but it is not necessarily so. Computing the binary expansion of a nonnegative integer S, the least significant bit first, write S=a+2((Sa)/2) where a=0 if S is even and a=1 if S is odd. So 11 = 1+2+0+8+0+0+0+0+... and −1 = 1+2+4+8+16+... It is perfectly sensible. Bo Jacoby (talk) 23:20, 3 March 2010 (UTC).