• Mathematician Chat v. 3.999...
    1,232 replies, posted
Discrete math is just loads of fun
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[IMG]https://latex.codecogs.com/gif.latex?%5Cinline%20%5Ctextup%7BFix%7D%5C%3Ba%5Cin%20%5Cmathbb%7BC%7D%5C%3B%5Ctextup%7Bwith%7D%5C%3B%7Ca%7C%3C1%5C%3B%5Ctextup%7Band%20consider%7D[/IMG] [IMG]https://latex.codecogs.com/gif.latex?f_a%28z%29%20%3A%3D%20%5Cfrac%7Bz-a%7D%7B1-%5Cbar%7Ba%7Dz%7D[/IMG] [IMG]https://latex.codecogs.com/gif.latex?%5Ctextup%7BProve%20that%7D%5C%3Bf_a%28z%29%5C%3B%5Ctextup%7Bmaps%20the%20unit%20disk%7D%5C%3BD%5B0%2C1%5D%5C%3B%5Ctextup%7Bto%20itself%20in%20a%20bijective%20fashion%7D[/IMG] I've tried some approaches but I can't get myself across the finish line, any hints?
Which part are you having trouble with: showing it maps the disk to itself, showing it's one-to-one, or showing it's onto? I've got the first two, working on the last. [editline]9th September 2015[/editline] Think I got em all. My reasoning for onto is... handwavy, but obviously true. Just not quite sure how to arrive at it algebraically. First show that it maps the disk to the disk, i.e. prove that |z-a|/|1-conjugate(a)z| is less than or equal to 1. Use |a-b| = |a|^2 + |b|^2 - 2 Re(a*conjugate(b)). For one-to-one, it follows directly from the definition. For onto, use the definition. Find an expression to determine z such that |z-a|/|1-conjugate(a)*z| = b for b in the unit disk. Once you find one, all you have to do is show that the denominator does not blow up in the unit disk (this is where I got slightly handwavy). [editline]9th September 2015[/editline] Actually I just realized how to make it not so handwavy.
How does one not handwavy?
[QUOTE=JohanGS;48652921]How does one not handwavy?[/QUOTE] Well I was having trouble figuring out where the modulus of the denominator of the inverse function zero was for the onto proof, but then I realized that it's equivalent and much easier to just find where the denominator itself was zero (since there's only one point with modulus zero). But of course you might not even have that issue. If you post your proof so far it'll be easier to help
I have nothing worth showing. The first point did cross my mind. The second I could assume was true, but it didn't feel rigorous to assume so (how or from what would this become clear?). The third I haven't tried at all. [editline]10th September 2015[/editline] I guess I could show that with |a| = 1, the boundary maps on itself and then show that a point on the disk maps onto the disk and use the fact that it's conformal. [editline]10th September 2015[/editline] Or could I? (will try tomorrow, today is thermodynamics day)
I'll abbreviate complex conjugation with cnj() and less than/greater than or equal to with leq/geq. Alright, I'll do the first part in detail (showing that it maps the disk to the disk), you want to show that, given |z| leq 1, |(z-a)/(1-cnj(a)*z)| leq 1. Now the absolute value splits up across a division, so you want to show that |z-a| leq |1- cnj(a)*z|. Then you can use that hint I gave to turn it into showing that 1 + |conjugate(a)|^2 - 2*Re(a*cnj(z)) geq |z|^2 + |a|^2 - 2*Re(z*cnj(a)). If you puzzle through a couple geometrical facts to put bounds on the conjugates and real parts showing up in this inequality, you can show that it's equivalent to |z|^2 leq 1. But this is obvious, since we're mapping from the disk. I'll try not to spoil the other parts. Start from the definitions of one-to-one and onto (definitely don't assume it's one-to-one, it's important to be able to prove it incontrovertibly). [editline]10th September 2015[/editline] I don't really remember anything about conformality so I'm doing everything the more elementary way :v:
Should be easy enough to show bijectivity, suppose w = f(z) and solve for an inverse. Without any paper I think you get the same formula but with -a You've already shown that this maps the unit disc to itself, and the inverse is unique by construction.
[QUOTE=Joey90;48654515]Should be easy enough to show bijectivity, suppose w = f(z) and solve for an inverse. Without any paper I think you get the same formula but with -a You've already shown that this maps the unit disc to itself, and the inverse is unique by construction.[/QUOTE] I figured this was probably the easier was when I realized I basically had to find the inverse to prove onto. Is there anything you need to be careful about? Onto should be taken care of when you find it. If the original map is not one to one the inverse will not be well defined. But does showing that the original function maps the disc to itself guaranteed that the "inverse" you compute will be defined everywhere in the unit disk and not blow up?
[QUOTE=JohnnyMo1;48656124]I figured this was probably the easier was when I realized I basically had to find the inverse to prove onto. Is there anything you need to be careful about? Onto should be taken care of when you find it. If the original map is not one to one the inverse will not be well defined. But does showing that the original function maps the disc to itself guaranteed that the "inverse" you compute will be defined everywhere in the unit disk and not blow up?[/QUOTE] Existence of a unique inverse on the range of a function is enough to show that it is bijective on that range (I think this is trivial?) all we need to do is check the domains. As the inverse is effectively the same function we've already shown that it's well defined on the disc and it's image is contained in it. Thus every point in the disc must be mapped onto by it's inverse [i]which is also in the disc[/i], i.e. the range is the whole disc and the function is bijective there. You do still need to prove the first part though - that it's well defined and the image is contained in the disc.
You guys seem to know about quantum stuff, is it possible we'll ever use that knowledge for teleporting items? I understand living beings would be more involved but with quantum information shouldnt it be kinda straightforward to send say a block of iron a few miles away, faster than the speed of light?
So I just took a 2 question calc quiz and was feeling pretty confident about it when I handed it in. About 30 seconds later I realized I made a vital mistake on one of the questions and I was kinda freaking out. Tell me its going to be ok
[QUOTE=cody8295;48661097]You guys seem to know about quantum stuff, is it possible we'll ever use that knowledge for teleporting items? I understand living beings would be more involved but with quantum information shouldnt it be kinda straightforward to send say a block of iron a few miles away, faster than the speed of light?[/QUOTE] No. Quantum mechanics does not allow faster-than-light travel or communication.
[QUOTE=shadowdude14;48661672]So I just took a 2 question calc quiz and was feeling pretty confident about it when I handed it in. About 30 seconds later I realized I made a vital mistake on one of the questions and I was kinda freaking out. Tell me its going to be ok[/QUOTE] When you say vital mistake, do you mean something like "I accidentally used the wrong numbers in the question" or "I used the wrong method/approach". When I'm marking exams, the former gets a small penalty whereas the latter gets a big one.
[QUOTE=JohnnyMo1;48662087]No. Quantum mechanics does not allow faster-than-light travel or communication.[/QUOTE] Care to elaborate why?
[QUOTE=cody8295;48663070]Care to elaborate why?[/QUOTE] Well, quantum teleportation schemes that actually work rely on classical communication channels, which are limited by the speed of light by relativity. Then there's quantum entanglement, which looks to a casual observer like it should be able to send things faster than light, but it actually doesn't. You can't send information just by observing a state. Let's say Alice and Bob each have half of an entangled pair with state either spin up or spin down with 50/50 probability. First they might agree, "Okay, spin up is 1, spin down is 0," and try to send binary data like that, but then they realize that they don't control the outcome, so any message they send is random. So instead they agree, "Bob will observe his state at noon. If Alice has observed her state first, that's sending a 1 to Bob. If the state was unobserved and Bob collapses it, that's a 0." But upon trying that, they realize that Bob cannot tell if the state has been observed already. He measures either spin up or spin down... but he has no way of knowing whether Alice observed it first unless he goes and asks Alice. If he could clone the state at noon, he could measure it a bunch of times. If half are up and half are down, he knows it's unobserved, but if they're all either up or down, Alice has observed it already. However, there is a theorem in quantum mechanics which prevents cloning quantum states.
[QUOTE=JohnnyMo1;48663240]Well, quantum teleportation schemes that actually work rely on classical communication channels, which are limited by the speed of light by relativity. Then there's quantum entanglement, which looks to a casual observer like it should be able to send things faster than light, but it actually doesn't. You can't send information just by observing a state. Let's say Alice and Bob each have half of an entangled pair with state either spin up or spin down with 50/50 probability. First they might agree, "Okay, spin up is 1, spin down is 0," and try to send binary data like that, but then they realize that they don't control the outcome, so any message they send is random. So instead they agree, "Bob will observe his state at noon. If Alice has observed her state first, that's sending a 1 to Bob. If the state was unobserved and Bob collapses it, that's a 0." But upon trying that, they realize that Bob cannot tell if the state has been observed already. He measures either spin up or spin down... but he has no way of knowing whether Alice observed it first unless he goes and asks Alice. If he could clone the state at noon, he could measure it a bunch of times. If half are up and half are down, he knows it's unobserved, but if they're all either up or down, Alice has observed it already. However, there is a theorem in quantum mechanics which prevents cloning quantum states.[/QUOTE] Thanks for explaining, it's always interested it
Actually to append that, it disallows cloning of [I]arbitrary[/I] quantum states. I have to admit I'm not entirely sure how cloning a carefully chosen quantum state is prevented, but this is generally considered a settled issue. I may look into that, but no one believes entanglement can send information, so I'm sure it is prevented somehow.
What's happening over in the maths thread? I am just about to start a new job, and am trying to plan my return to finish my B. Mathematics at some point. Might see how things turn out for the near future though.
I am seriously having trouble with Mahalanobis Distance and I can't find any good resources online that explain it
[QUOTE=OrDnAs;48670178]I am seriously having trouble with Mahalanobis Distance and I can't find any good resources online that explain it[/QUOTE] [url]https://en.wikipedia.org/wiki/Mahalanobis_distance#Intuitive_explanation[/url] The last sentence here (of course, in the context of the entire section) does a pretty good job of making me think I understand what it is. [editline]13th September 2015[/editline] If I've understood it right, the Mahalanobis Distance is somewhat like a normalised distance if you have a non-uniform distribution.
The formula was giving me trouble (because it's really complicated to calculate a covariace matrix with only a calculator). I just needed to normalize the distances and then all set.
now, how the FUCK am I supposed to calculate it in my next test.
[QUOTE=OrDnAs;48671291]now, how the FUCK am I supposed to calculate it in my next test.[/QUOTE] Pen and paper, buddy. The way a mathematician does!
[QUOTE=sltungle;48671354]Pen and paper, buddy. The way a mathematician does![/QUOTE] It's a big chore even using SPSS or Excel
My math college starts soon (12 days) and I can't wait :D. I am studying practical math.
so i'm an idiot and would be very grateful if someone could help me out with statistics I have to choose a population with a size of at least 50, and there needs to be 3 variables for me to study (1 qualitative and 2 quantitative). Originally, I chose to have my population be the six continents (with all the countries in them exceeding the minimum size of 50). My qualitative variable was "Gender", and my two quantitative variables were "Percentage of Smokers" and "Average Life Expectancy". My professor OK'd this part and let me proceed. So, I need to make a table with 3 columns (Variables 1, 2, and 3) and at least 50 lines of data. This is the example I was given (the population is the 100 members of the U.S. Senate): [IMG]http://puu.sh/kg6sM/af367590a7.png[/IMG] The problem is, it would end up looking like "Continent | Gender | Percentage of Smokers | Average Life Expectancy |" and continents don't have a gender. But my professor said I was okay to go ahead with this setup earlier so what the shit, this doesn't make sense. I don't know what to do. If I were to make the continents the qualitative variable, what would my population be? "all the people in the world" maybe? or the amount of countries that make up every continent? i hate this :( EDIT: i think i get it now, the countries of the world that make up the continents are my population
Has anyone here learned the Trachtenberg system of basic arithmetic? I just got the book and it looks very impressive, even though we all have calculators in our pockets now I still thought it would be a cool skill to learn. The only problem I'm having is that while everything is easy to do on paper, it's not easy to do any of it in your head which rather defeats the purpose. I'm gonna have have to learn memory techniques as well.
Complex analysis is pretty trippy. I like.
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