<antonfire> About the same way I would show it directly in person <Keshi> Yeah, but how is that? <antonfire> Using my psychic powers, of course. <antonfire> moses: think about how I showed that (0,1) was not compact. <Keshi> antonfire: ? <antonfire> I constructed an open cover of (0,1) which had trouble near the point 0. <wartech0> oh… <Jef91> “Prove the sum of the distances between all the pairs of vertices in every tree with an odd number of vertices is even.” <moses> i agree <moses> you could never get to zero <wartech0> I am going to kill webassign <Jef91> Any ideas as a jumping off point? <wartech0> murder it <moses> you would always be 1/n away from it <antonfire> Points close to 0 were under the cover, but to get arbitrarily close to 0, I needed infinitely many of the covering sets. <wartech0> I can’t believe I payed 50 bucks for this shit <moses> yes <antonfire> Any finite subset stopped short of 0, and so failed to contain some element of the set. <antonfire> You can do something very similar in general. <antonfire> if E is not closed, that there is a point x outside E whose every neighborhood intersects E. <antonfire> x plays the role of 0 in this case. <moses> how would that sam sub cover change if you let the new set in question be [0,1]? <moses> same* <antonfire> You want to construct a bunch of open sets that get arbitrarily close to x, but any finite collection of them doesn’t get close. <peppermint> sup guys <antonfire> moses: { (1/n, 1] | n in N } is simply not an open cover of [0,1]. <moses> that is true <antonfire> well <antonfire> neither is { (1/n, 2) | n in N } <moses> well yes <antonfire> Which is a better statement, since the things in it are actually open. <peppermint> some dumbass told me that any random sequence of real numbers can be the coefficients of a power series that has some nonzero radius of convergence <antonfire> ppilatus: what do you mean by “random sequence” <peppermint> i told him he was a faggot, but i’m not sure <moses> i was referring to the left side of the interval <antonfire> peppermint: to prove him wrong, you need a sequence that grows faster than any exponential function. <peppermint> yeah, thats what i thought <antonfire> There’s a pretty well-known one that does this, or you can contrive some random shit to do it. <peppermint> yeah lol <moses> if i have [0,1] is an open cover (0,1/n) for n in N? <antonfire> moses: no. <antonfire> 0 is not in any of those open sets. <peppermint> anyway, i once thought that the laplace transform of a sequence should have some bearing on how much it converges <antonfire> neither is 1, for that matter. <moses> oh fuck <peppermint> but it needs to be some kind of weird like discrete analog of the laplace transform <moses> so the points in the set have to be IN the open sets <antonfire> in at least one of the open sets <antonfire> otherwise it’s not an open cover <antonfire> just a bunch of open sets <Kydun> Anton: Remember when we got something like y = .3 ? I remember we then multiplied R2 and added to R1 but i forget by what or why <moses> for [0,1] an open cover is (-3,1)U(0,2) <frelleck> peppermint, if you’re allowed to skip powers of x you can do it <frelleck> i.e. if your sequence is a_n you can choose x^(b_n) to give a_n x^(b_n) a positive radius of convergence <moses> and you can definitly construct an open cover using some subcover from the open cover, although <antonfire> Kydun: when you get confused, think about the corresponding linear equations. <moses> i am now wondering how big the open cover is defined to be <frelleck> otherwise just take a_n = n! and the radius of convergence is zero <antonfire> moses: do you mean {(-3,1), (0,2)}? <moses> how come that unioned interval does not work? <antonfire> moses: oh it does. <antonfire> but it’s pretty clearly not what you meant <antonfire> since (-3,1) u (0,2) is just (-3, 2). <antonfire> So {(-3,2)} is also an open cover of [0,1]. <antonfire> Not a very interesting one, but an open cover neveretheless. <moses> well yes, but its hard to construct multiple subcovers <antonfire> Well, {(-3,1), (0,2)} is not that interesting either. <antonfire> Here’s an interesting one: { (x-1/100, x + 1/100) | x in R } <antonfire> Find a finite collection of those which covers [0,1]. <moses> ok question: when you use {} are you designating an interval inside of a set? <moses> or a set which consists of an interval? <antonfire> most of the times I’ve used it, I’ve used it do denote a set of intervals. <moses> ok <antonfire> I don’t know what you mean by “an interval inside of a set” or “a set which consists of an interval”. <moses> well my prof/graduate student he just uses () to denote intervals <antonfire> so do I. <moses> and we are talking about sets of intervals <antonfire> (x – 1/100, x + 1/100) is an interval. <moses> so {} seemed proper <moses> i just wanted to confirm <antonfire> It’s an open interval of length 1/50 centered at x. <antonfire> So basically, I want you to find a collection of open intervals of length 1/50 which cover [0,1]. <peppermint> frelleck: who the hell does that <frelleck> well I wasn’t sure what you were asking <frelleck> if you just want a series with a radius of convergence zero it’s easy <peppermint> thats retarded its like hurrr a_n increases too fast lets slow it down by padding it with ton’s of zero’s <frelleck> people ask retarded questions here <frelleck> but yes that’s what’s going on <moses> hmm do i pick specific x in R ? <wartech0> net split is whats going on <antonfire> moses: sure. <moses> hmmm but its the same x <moses> its the same x right? <antonfire> sounds like you don’t even know what { (x – 1/100, x + 1/100) | x in R } means. <rads> antonfire: so I did n=1, n=2, n-3 for the problem I was talking about. I came up with the formula (2n-1)(2n-3)(2n-5)… <rads> I’m not sure how this links back to permutations and combinations though <antonfire> rads: does it have to? <moses> i think i means for any x in R i plug in the R and then that becomes my interval <moses> it* <moses> err i plug in the x <rads> antonfire: not necessarily — I just assumed it would since it’s in the homework for that chapter. I was expecting there to be a more general solution <antonfire> moses: how big is the set { (x – 1/100, x + 1/100) | x in R}? <Guest89715> on GL(F_2), when I do matrix multiplication…what happens when theres a 2? <antonfire> Can you name an element of that set or two? <moses> its infinite <antonfire> Guest89715: 2 = 0. <Guest89715> antonfire: thats what I thought, thanks <antonfire> Guest89715: for that matter, 1982309810 = 8230918837426 <moses> (-1/100,1/100) (199/100,201/100) <fpqc> moses, the elements of that set are in bijection with elements x in R <antonfire> moses: can you name a bunch of elements of that set which, together, cover [0,1]? <Kydun> Anton: I’m blazing through these matrices now, you the man <moses> ok <fpqc> duh <moses> (-1/100,1/100),(99/100,101/100) <moses> holy shit <pascals> fuck this proof man <moses> i am not smRT 
<antonfire> moses: that only covers a small portion of [0,1]. <rads> antonfire: basically I feel like I don’t really understand the problem. I did the n=1, n=2, n=3 and saw the pattern, but I don’t know where it came from <antonfire> rads: do you know why there are n! permutations of a set? <antonfire> of a set of n elements, I mean. <rads> I think of it as taking it step by step. the first one you have n, the second one n-1 choices, etc… <antonfire> there’s a similar argument for this situation. <rads> it gets confusing when you’re picking two at once though <antonfire> Let’s say you’re the 2n’th person. You walk into the party, and pick someone to pair up with <antonfire> How many ways are there to do that? <moses> how about the union from a=0 to a=100 of “the set”_a <redbowl> in discrete math, what’s the “-” simple called? as in set A-B <antonfire> moses: you’ll have to be more clear than that. <antonfire> redbowl: set difference <antonfire> redbowl: or relative complement. <rads> if there are 2n people, there would be 2n-1 ways <redbowl> thanks <rads> this is starting to make sense now <antonfire> rads: your buddy is standing next to you, it’s his turn to pair up. How many ways can he choose someone to pair up with? <moses> ok lets call the set you made A the open cover would be {(-1/100,1/100)U…U(99/100,101/100)} <pascals> does anyone else find counting proofs to be incredibly difficult? <rads> so the next person walks in, the previous two are already paired up, so there are actually 2n-3 choices now? (-2 for the pair and -1 for the person who walked in) <antonfire> rads: yup <antonfire> “and so on” <rads> that’s a very simple way of thinking about it. thank you <redbowl> is the upside down triangular dots if, or then? <antonfire> redbowl: it’s “I don’t want to be taken seriously” <redbowl> ? <antonfire> moses: (a), you mean commas where you put Us. <antonfire> (b) it’s unclear what the … is supposed to mean. <moses> ok lets call the set you made A the open cover would be {(-1/100,1/100),(0/100,2/100),(1/100,3/100),…,(99/100,101/100)} <antonfire> moses: there you go. <hmmmm> quick question <antonfire> So, {(x – 1/100, x + 1/100) | x in R} is an open cover of [0,1]. You just found a finite subcover. <antonfire> Consisting of 101 elements. <moses> that is sick as fuck <hmmmm> so i have z, w in C. |z| <moses> ty bro <antonfire> hmmmm: it’s also true that |w| – |z| < |z – w| <hmmmm> hmmm? <hmmmm> how would i concisely make that argument though? <antonfire> make what argument? <antonfire> that |w| – |z| < |z – w|? <hmmmm> yeah <antonfire> Just say “blah blah, so |z| – |w| <hmmmm> i wonder if that’ll fly well <antonfire> Because it’s pretty much the same argument, with the role of z and w reversed. <hmmmm> this prof is a rigor nutjob <antonfire> If you’re worried about it, write the whole thing out again. <antonfire> with z and w reversed. <hmmmm> i can hand in a 100% perfect proof and then find at least 5 red marks in various places throughout <hmmmm> so something as simple as this has the potential to be hazardous <antonfire> That suggests it’s not actually 100% perfect. <Dvorac> hmmmm: They won’t mark that. <hmmmm> it’s stupid stuff like “not enough work” <hmmmm> “what’s this? show this” <antonfire> That suggests it’s not 100% perfect. <hmmmm> it’s perfect to me when i wrote it! <GutenTag> If I have vector x and vector y, each representing heights of n people, one in x in meters and y in feet, I am to prove that the correlation is = 1. This is obvious, of course, since they measure the same <antonfire> That suggests it’s crap. <Dvorac> antonfire is correct <GutenTag> cov(x,y) == std(x) * std(y) <antonfire> GutenTag: x is a scalar multiple of y. <antonfire> that’s pretty much all you need to know. <hmmmm> nothing sinks my spirits lower than when i handed in homework thinking “yep, this is 100%. there’s no way anything can be wrong, my work is impeccable” and then get it back with all these red marks everywhere<hmmmm> for stupid stuff too <hmmmm> they’re just playing hardball <redbowl> whats the theorem that says if two sets are subsets of eachother ,they are the same set? <antonfire> If it’s for stupid stuff, you shouldn’t give a shit. <hmmmm> redbowl that’s not a theorem <antonfire> I suspect it’s for stuff that’s not actually stupid. <redbowl> hmmmm: what is that then? <antonfire> redbowl: that’s pretty much the definition of “the same set” <hmmmm> the definition of set equality <GutenTag> antonfire: I realize that, but in the problem definition, it says to refer to formula 2.10, which is the corr(x,y) == cov(x,y) / [std(x) * std(y)], and the problem is asking for a proof. <antonfire> GutenTag: yes. <antonfire> You should prove that if x is a multiple of y, then cov(x,y) / (std(x) * std(y)) = 1. <antonfire> This is not very difficult if you’re familiar with what cov and std actually denote. <Dvorac> redbowl: Set equality by using a double containment argument is what you are thinking of. It follows from the axiom of extensionality <GutenTag> I see. Well, yes, I have cov(x,y) = Sxy = i/(n-1)* E(k=1 to n) (x(k) – x-bar)(y(k) – y-bar) <GutenTag> oops, that should be 1/(n-1) <antonfire> GutenTag: at some point, you’re going to need to use the fact that, you know, x is a multiple of y. <GutenTag> antonfire: so, I should write either X as some scalar s*y ? <antonfire> yes. <hmmmm> also how do i show that is a line in C? <antonfire> in this case, s = .3048, but it really doesn’t matter. <hmmmm> do i get it in the form of a * x + bi ? <redbowl> If I’m asked to “prove that R n (S-R) is an empty set”, would a paragraph explaining it be a proof? <hmmmm> redbowl, nope <antonfire> redbowl: depends on how well you explained it. probably not. <antonfire> It should convince a skeptical person who is familiar with the definitions. <redbowl> I said “By definition (S-R) doesn’t contain R, so R n (S-R) doesn’t contain R <redbowl> seems pretty logical to me <Kydun> b <hmmmm> … <hmmmm> well <antonfire> A = [0,2] doesn’t contain B = [1,3] so A n B doesn’t contain B. <antonfire> so clearly A n B is empty amirite <hmmmm> that is true, but they’re probably looking for an argument where you assume the opposite, then let an arbitrary variable be in that, and draw up a contradiction <hmmmm> what’s wrong with that? <antonfire> They’re looking for an argument that’s not stupid, at least. <hmmmm> so uh anybody know the answer to my question? <Dvorac> redbowl: Let x be in R n (S-R). Then x is an element of R and x is an element of S-R… or you could do R n (S-R) = (R n S) – (R – R) = … <Muhis> Well… if x belongs to R n (S-R), it follows that x belongs to R and it belongs to S-R, but it belonging to S-R implies it belongs to S but does not belong to R… so x belongs to R and it does not belong to R<redbowl> what is this proof called? Proof by contradiction? <antonfire> redbowl: yes. <Dvorac> hmmmm: Think of C as R^2. Help? <hmmmm> nope, not even in the slightest <kerio> Dvorac: wait, no <hmmmm> maybe it’ll help people help me if i give the entire question <Dvorac> kerio? <hmmmm> Let a, b in C with b != 0. Prove that the set L = {z in C : Im((z-a)/b) = 0 } is a line in C <kerio> Dvorac: a complex line is just t * (a+bi) <RustyShackleford> Dvorac: i’m assuming you use dvorak? <kerio> with t in C <RustyShackleford> i’ve thought about trying it <kerio> RustyShackleford: the benefits are neglegible, apparently <Dvorac> RustyShackleford: Actually… <ski> (.. hm, it really is just an ordinary negation proof, not a full contradiction proof) <RustyShackleford> i’ve heard so <RustyShackleford> i’d have to relearn how to type <kerio> hmmmm: oh, they mean an actual line? <antonfire> kerio: that would be all of C. <hmmmm> yeah <kerio> antonfire: unless you’re in C^n, yeah <kerio> what’s a line in R? * hmmmm shrugs <hmmmm> i was given no definition of “line” <antonfire> kerio: the set L is clearly not a complex line. <antonfire> hmmmm: do you know how to check if a subset of R^2 is a line? <kerio> antonfire: then it’s probably C as in R^2 <hmmmm> nope <antonfire> hmmmm: a line in R^2 is a set of the form { (x,y) in R^2 | ax + by = c }, where a and b are not both zero. <hmmmm> okay <antonfire> A line in C, then, is a set of the form { x + iy in C | ax + by = c }, where a and b are not both zero. <hmmmm> right <hmmmm> unfortunately it just claims that b != 0 <hmmmm> says nothing about a <antonfire> hmmmm: … <hmmmm> it’s a pretty good guess that the variables are a and b for that reason <antonfire> … <antonfire> That’s a stupid guess. <antonfire> they’re a and b because I didn’t want to think of other letters <hmmmm> those are the standard letters <antonfire> I wasn’t even paying attention to whether a and b happened to be used in the thing you wrote. <GutenTag> Ok, I’ve thought about this proof (that corr == 1, when I have vector X measured in feet, and vector Y measured in meters, both of n people). I understand that all x(k) = s*y(k), where s is a scalar multip<hmmmm> it’s not a concidence though <antonfire> hmmmm: I just fucking told you it’s a fucking coincidence <hmmmm> okay, whatever <antonfire> Jesus christ <Dvorac> … <hmmmm> you’re right <Dvorac> Proof by …coincidence? <GutenTag> haha <antonfire> Stop reading into the mysticism of the symbols and start doing math. <hmmmm> no, i was just saying that this “coincidence” is leading me to what i’m supposed to end up with <hmmmm> more like a hint <Dvorac> It’s not. <kerio> hmmmm: whenever you see complex numbers in your case, consider them as vectors of R^2 <hmmmm> well i’ll get back to you when i finish this <antonfire> GutenTag: that’s what you’re trying to show, yes. <RustyShackleford> what exactly does dx(du, dy) represent when you are solving integrals? <antonfire> GutenTag: except you switched x and y at some point. <pascals> so this definition, a leaf vertex is just a vertex with 1 edge? <RustyShackleford> i understand that it tells you to what variable you are integrating in respect to <antonfire> pascals: that’s the usual definition of a leaf of a tree, yes. <Dvorac> pascals: a leaf vertex is just a vertex with degree 1 <GutenTag> oh, so I did. Hmm. I’m not sure what to do next. But, since you’ve assured me I’m on the right track, I will think about it some more. <RustyShackleford> why does the process of evaluating an integral cause this to disapear? <Drew|> anybody familiar with spherical coordinate system? <antonfire> GutenTag: it’s trivial if you know some properties of cov and std, slightly less trivial but still easy if you only know their definitions, and obviously impossible if you don’t even know what cov and std m<Dvorac> Drew|: Ask the question and then we’ll look. <theseb> Drew|: just ask your question <theseb> Drew|: don’t tap dance around it <Drew|> Sketch the solid described by the given inequalities: p <Drew|> p being rho <GutenTag> antonfire: I’m trying to recall if I can “factor out” my s. <antonfire> GutenTag: if you can’t remember, think about the definitions. <theseb> Drew|: plot cross sections in various planes
Feb 142012