<yosemite37> Hey, PlanckWalk, for this one ” A straight trail leads from the Alpine Hotel, elevation 8000 feet, to a scenic overlook, elevation 11,100 feet. The length of the trail is 14,100 feet.” Do you have two buil<Nereid> those are some fucking tall buildings <yosemite37> But if you have two buildings you cant really find any angles <Nereid> only one building is mentioned <yosemite37> so what exactly does “scenic overlook” mean <yosemite37> oh nvm i got it <yosemite37> Nereid: Would the drawing look like this? http://i.imgur.com/YooyT.png <Nereid> what <Nereid> there are 3 lengths in the problem and you interpreted them all wrong. <yosemite37> Doesn’t elevation 8000 mean the building is 8000 ft tall? <Polytope> no <Nereid> have you ever seen a building that was anywhere near 8000 feet tall? <Nereid> I sure haven’t. <Nereid> and apparently elevation means something totally different for the scenic outlook. <Polytope> elevation is height above sea level <kororaa> lol <Polytope> and an overlook is “An elevated place that affords an extensive view” (of scenery, for example) <Nereid> overlook. <Polytope> outlook is a mail program <Nereid> nereid is an idiot who can’t type. <Lewis29> “Show that a group that has only a finite number of subgroups must be a finite group.” I was trying to consider the union of the subgroups, but cannot think of anything useful.. <yosemite37> Polytope: So 8000 is just for the building at like ground lvl? <Polytope> yeah, you’re not going to build a skyscraper in the mountains <beck> does anybody knows how to prove that a function from an interval to a surface , is a geodesic in the surface? <Nereid> beck: you need to know more about the function, probably. <yosemite37> oh wow the “trail” for this problem is a slanted line <Nereid> Lewis29: I can’t think of a straightforward way to do it, but <myrkraverk> Is there anything that can help me selecting k points from an m * n map? Just selecting ( r mod m, r mod n ) where r is a taken from /dev/random selects the same points way too often. <somiaj> Lewis29: Contradiction would be my approach, assume it is infinte and then start looking at cyclic subgroups. <phoenixstew49> http://mathbin.net/88957 <somiaj> Lewis29: seems to me you could prove every infinte group has an infinte number of cylic subgroups — your statement is the contrapositive <myrkraverk> I was hoping something clever in math could make it “easier.” <beck> Nereid, if i have g:[a,b] -> S with S a smooth manifold <Lewis29> somiaj: I don’t know if the group is cyclic to begin with <beck> isnt there any criteria for g in S like the derivatives to be something special? <Nereid> what the heck is a geodesic in a smooth manifold? <Lewis29> or does that necessarily matter <Nereid> you need a metric <Nereid> beck: ever heard of the geodesic equation <Nereid> Lewis29: so if your group is infinite, then it might contain an element of infinite order, or it might not. <Nereid> in either case see if you can get the desired result. <Lewis29> alright <Nereid> this is the first thing that comes to my mind. <beck> Nereid, well I have the dot products in the tangent space <beck> the fundamental forms,.. <somiaj> Lewis29: cyclic subgroups, look at the group for every a in G <Nereid> beck: so it’s more than a smooth manifold, it’s a riemannian manifold. <Nereid> beck: again, geodesic equation <beck> Nereid, yes im sure you are right <PlanckWalk> myrkraverk: You’d want something more like (r mod n, r div n) <beck> but the problem is , that I did not write something in my notes <PlanckWalk> Where r is in [0, nm) <beck> telling me something about the second derivative of the g:[a,b]-> Rm <Nereid> beck: for the third time, geodesic equation <beck> and I have the definition of “g is a geodesic is a critical point of the energy functional” <beck> Nereid, ok <Lewis29> somiaj: I feel like I’m handwaving, but the union of all those cyclic subgroups would equal G if I ran through every element of G <myrkraverk> PlanckWalk, Hmm, let me try that. <phoenixstew49> In http://mathbin.net/88957 do I have: a. x/-4 or b. -x/4; and the final result being: a. x = -12 or b. -x = 12 and swap to x = -12; i did a. on paper but i cannot figure out why and that is what I am t<beck> and some christoffel symbols <Lewis29> (I think) <Nereid> Lewis29: it would, but so what? <Muhis> phoenixstew49, (-x)/4 = x/(-4) = -(x/4) … why? Because -(x/4) = -1*(x/4) = -1*x/4 = (-x)/4, and on the otherhand -1 = 1/(-1) so -1*(x/4) = 1/(-1)* (x/4) = x/(-4) <Muhis> Said simply: it doesn’t matter whether the – sign is in numerator or denominator * Lewis29 shrug <Nereid> Lewis29: do consider the two cases though. <Nereid> what you said is useful. <somiaj> Lewis29: no, you just count the number of those sub groups you have. <somiaj> Lewis29: you could have more subgroups, but the main thing is you have enough cylic subgroups to give you an infinte number of subgroups <phoenixstew49> Muhis: okay, so does -x = 12 == x = -12 or is that just a miscalcuation on my part in the process to find x <Muhis> phoenixstew49, and as you see: you get x = -12 and -x = 12 but these are exactly same solutions. If you multiply both sides of -x = 12 by -1, you get -1*(-x) = -1*12 x = -12 * Lewis29 thinks for a minute <phoenixstew49> Muhis: ahh now i see it <myrkraverk> PlanckWalk, It seems better than it was, but still selecting the same points quite often. <myrkraverk> I’m hoping I can lessen the need for special casing the same point. <PlanckWalk> myrkraverk: Random selections will often hit the same points <phoenixstew49> Muhis: thankyou, i could not think of the number to use that would balance the equation to be x = -12 if i had -x = 12 (being -1, which makes perfect sense now that you’ve pointed it out to me) <PlanckWalk> If you have N points, you’ll typically get the first repeat on the order of sqrt(N) points. <myrkraverk> Right now I’m working with m = n = 9 and k = 10. I’ve gotten one random point = same point selected 10 times. <Muhis> phoenixstew49, np, these are very common problems beginners tend to have with solving equations at first <PlanckWalk> You mean all 10 points were the same? <PlanckWalk> Or 10 times out of what? <myrkraverk> Typically I get 4 and 5 points. <myrkraverk> PlanckWalk, all points the same. <PlanckWalk> Sounds like broken code <myrkraverk> PlanckWalk, or broken /dev/random 
<PlanckWalk> /dev/random is supposed to be used very sparingly <myrkraverk> Hmm, maybe I should change to /dev/urandom then. <PlanckWalk> E.g. to generate just enough entropy for secure key generation. <myrkraverk> But that’s not much better. <myrkraverk> PlanckWalk, I’m generating minesweeper maps. <PlanckWalk> Oh, for that you may as well just seed a userspace PRNG with the clock or something. <myrkraverk> I may, but I’m not sure I want to. <PlanckWalk> Either way, you should be getting a lot more distinct points. <Polytope> just pick each square with a fixed probability <PlanckWalk> Yeah, it’s easily small enough for that. <myrkraverk> http://paste.lisp.org/+2QJ8 <myrkraverk> Erm, how would I do that? <PlanckWalk> You can even adjust that for constant number of selected squares <myrkraverk> erm? <myrkraverk> I don’t really know anything about probability – which means vague hints go over my head. <Polytope> loop through all of the cells, and generate a random number r in [0,1] for each cell. place a mine in the cell if r < m/c, where m is the number of mines remaining to be placed, and c is the number of cells <myrkraverk> Ok, let’s try that. <Polytope> I’m not sure if this is really better than just picking random cells, and rechoosing when you get a collision. <rking> So.. simple problem that I’m missing.. how would you map something like .3-.8 to be closer to .0-1? <rking> It’s a greyscale gradient that I want to increase contrast on.. I know there must be something simple with pow() or whatnot. Part of the problem is that its not strictly .3-.8, that is only a guess. <myrkraverk> Polytope, distincly worse it seems <Polytope> depends on the percentage of mines <myrkraverk> probably, or I coded it worse. <myrkraverk> *wrong <rking> I am looking for a way to f(.2) => closer to 0, f(.5) = .5, f(.8) => closer to 1 <freecandy> You’re still on this, rking. <rking> freecandy: Til I get it. =) <Polytope> how much closer <rking> Polytope: Ideally I could tweak it to increase/decrease that relationship. <rking> So what I’m trying to do is a CG candle flame. I have a correct gradient that makes the top and bottom of it translucent while the body of it emits colored light — but the effect is too shallow. You can barely<rking> Actually my mapping is more like: .2, => 0, .5 => 1 <rking> Sorry, I screwed the statement of it up. <Polytope> I guess it’s worse. placing N/2 mines in N cells should take N*ln(2) tries on average, and this is less than N. <rking> I am being stupid tonight. <somiaj> rking: linear mappings are the easiest, so you have two points (0.2, 0) and (0.5, 1) — find the equation of a line between those two points <rking> somiaj: Hrm. Well, I don’t have full control over the incoming 0.2 and 0.5 — they may be really more like 0.123 and 0.657 — What I need is a way to march the values that are below a certain midpoint closer t<rking> “below a certain midpoint closer to 0.0 and above that midpoint closer to 1.0″ <rking> I guess I can guess and hack it. pow(n,2)*2 and keep tweaking the constants. <freecandy> rking, are you aware of the expression “moving the goalposts?” <rking> freecandy: Not in this context, but yes. <freecandy> How the fuck was anyone supposed to guess “oh, .2 and .8 aren’t really numbers, just variables disguised as numbers” when you say ” I am looking for a way to f(.2) => closer to 0, f(.5) = .5, f(.8) =<rking> Yes, I see. <rking> I know that part of my problem in all of this is the problem statement. <rking> All I can really say about it is that it is an increase contrast function whose domain and range are 0-1 <freecandy> What are you given. What is the general form of the thing you want. What is the test to determine if thing you are given back is sufficiently close to what you want. Pretend you’re paying a programmer to sh<rking> freecandy: Let me think about the clearest way to express this. 10 min. <Polytope> freecandy: I guessed that <freecandy> congrats <Polytope> rking: a function that maps the interval [a,b] to [c,d] is given by f(x) = c + (d-c)*(x-a)/(b-a) * Lewis29 thinks he may need a number theory book to supplement abstract algebra <pwca> I have a collection of subsets A_n of a metric space X. there is a sequence such that a_n is in A_n and Di(A_n) = sup{d(x,y) : x, y in A_n} goes to zero as n goes to infinity. <pwca> do I need X to be complete for this to imply that a_n converges? <rking> freecandy, Polytope: Here are my input and outputs: http://pasteall.org/pic/26341 <pwca> oh, and A_(n+1) is a subset of A_n. <rking> Achieved literally in Gimp by applying heavy Brightness+Contrast. So that’s what I’m looking for, I suppose: the precise algorithm for Brightness&Contrast. <Lewis29> Polytope: cool, I remember needing to derive that function to show [a,b] and [c,d] have the same cardinality <pwca> seems to me that since sup{d(a_n,y) : a_n, y in A_n} <frelleck> to say that, you need X to be complete <meingtsla> gic: Sounds right <frelleck> I think it’s ||y||^2 actually <pwca> frelleck: so the fact that this goes to zero doesn’t imply that there is some limit point in A_n? <frelleck> y/||y|| <meingtsla> gic: Actually no it is in the denominator as frelleck says <pwca> I guess I see where completeness comes in, then. <gic> meingtsla: thanks. but i do not understand why there is a ||y||^2 in the householder reflection I – 2yy’/||y||^2 <pwca> since that is to ensure that there exists a limit? <frelleck> no pwca, if you write it out what you have is equivalent to it being a cauchy sequence <frelleck> so yes to ensure there’s an actual limit you need completeness <meingtsla> gic: Consider projecting (2,2) onto (2,0) if you will, to check your formula <frelleck> everything is within distance epsilon of y_n for N > n <gic> frelleck: thanks. meingtsla thanks <frelleck> so for m,n > N, d(y_m,y_n) <frelleck> there’s no y you can specify such that d(y_m,y) < epsilon <frelleck> without completeness <freecandy> rking: I asked for three things. You gave me two, in the form of a png with zero logic to guide the selection of a formula. <pwca> thanks, frelleck. <rking> freecandy: I’m missing, “the test to determine if thing you are given back is sufficiently close to what you want”? <freecandy> I gave up. <rking> freecandy: Thanks anyway (I mean it). <PlanckWalk> rking: As far as I know, GIMP uses a simple linear formula for brightness and contrast adjustment <rking> PlanckWalk: But it doesn’t clip or anything. Values near 0 approach 0 (but don’t overlap), .5 stay near .5, and near 1.0 approach 1.0 <PlanckWalk> According to the source code, it does clip. <PlanckWalk> At least in my version. <PlanckWalk> (Which is probably fairly old by now) <PlanckWalk> There are other adjustment tools that don’t <rking> PlanckWalk: Maybe a tad, but my point is that in the UI if you drag the slider it seems nonlinear. <PlanckWalk> Brightness is linear, constrast isn’t. <rking> That sounds right. <PlanckWalk> More precisely, the adjustment made by the constrast control is linear in the original value, but nonlinear in the slider position. <PlanckWalk> For given settings, the transformation is linear. <PlanckWalk> (Apart from the clipping) <rking> PlanckWalk: But using your previous equation, I don’t get how to determine a,b and c,d. <PlanckWalk> You mean Polytope’s equation? <Polytope> if you are trying to stretch [.2, .5] to [.1, .5] then a=.2, b=.5, c=.1, and d=.5 <poorbot> Hello. <rking> Polytope: Using your equation that puts .1 at -0.0333 and .9 at 1.0333 … so, clip those off? <Polytope> .1 isn’t in the interval <Polytope> but you might decide that [0, .2] should be transformed to [0, .1] <moses> im having a really hard time trying to understand how that for E as a subset of R that every infinite subset of E has a limit point in E. <somiaj> What is E? <moses> E is a subset of R <moses> i said above <somiaj> any subset? <Polytope> it’s not true for every subset of R <somiaj> Because if E = Z, then you have a problem <Polytope> it’s true precisely when E is compact <moses> well This is from the Heine-Borel Theorem <moses> ^^ <somiaj> so E is compact? <Polytope> you can’t delete random words from theorems <moses> oh <moses> yeah <moses> E is most definitly compacrt <moses> sorry <moses> It really is compact <moses> that is part of the theorem <rking> Polytope: Thanks for your help, btw. <moses> anyone? <poorbot> http://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem <Home> can anyone see this <Muhis> hmm? <poorbot> Home, nope, something is wrong with your client. <Home> oh no <Home> thats not good <poorbot> Nope, check the hyperbolic fluxor. <Home> is (4-2x)/3 4/3 – (2/3)x too jumbled up to read <poorbot> Somewhat. <poorbot> Actually yes. <poorbot> Use more parenthesis and brackets. <Home> is (4-2x)/3 always going to equal 4/3 – (2/3)x <Home> i tested values -3 through 3 and they all were true <somiaj> Home: what is the rule for adding/subtracting fractions? <Home> i am now aware that theres a rule for that <Home> thank you for telling me <somiaj> Home: well how do you hadd 3/5 + 4/7? <somiaj> Home: actually simpler how do you add 3/5 + 1/5 <Home> wait didnt you mean for dividing a side with a fraction <Home> adding subtracting fractions is cakework <somiaj> well then you should know what you wrote is true <somiaj> you need a common denomintor to add fractions, but you can just eas easly split them up by the same principal <Home> im just trying to simplify equations with x and y but divided one side by 3 <Muhis> Home, then you surely know that a/c + b/c = (a+b/c) now read this from right to left! <somiaj> Muhis: put sthe closing ) in the wrong spot <Muhis> (a+b)/c* <Home> oh that proof equation is what i have on my notepad <Home> but with numbers and x and y <poorbot> Home, maybe I can tutor you in private, if you have more. <Home> well it was just one specific thing that has bothered me for a while <Home> im surprised i havent memorized that simple proof <Home> i think i saw a teacher divide one side by a number and then he skipped a step and i got confused <Home> i guess that stuff happens in pre calculus classes <Home> is it sane to do 100 practice problems a day <Home> thank you for the help <poorbot> Home, yeah 100 is nothing. Do more. <Home> oh i am so slow because i took a big break and started studying again <kororaa> + <t-mart> in doing a singular value decomposition, A = USV^t, where U’s columns are the eigenvectors of AA^t and V’s columns are the eigenvectos of A^tA (and S is the eigenvals along the diagonal). im having trouble wit<yosemite37> A ship leaves the port of Miami with a bearing of S80E and a speed of 15 knots. After 1 hour, the ship turns 90 degrees toward the south. After 2 hours, maintaining the same speed, what is the bearing to t<yosemite37> I’m not sure but would you do 80 – arctan(2)? <poorbot> Sorry what’s S80E? <nayfie> draw a picture <yosemite37> 80 degrees east of south <poorbot> I don’t know what S or E meant. I see. <yosemite37> The part that confuses me is where it says he turns 90 degrees, how exactly do you know if after the turn you will be in 4th or 3th quadrant? <yosemite37> any idea? <freecandy> Have you drawn a picture? <yosemite37> I have it drawn before the ship makes the 90 degree turn <freecandy> Can you see which way is south? <freecandy> You just need a direction now, not a distance. Draw a ray. <yosemite37> Yeah but will the ray pass into the 3rd quadrant or stay in the fourth? <freecandy> It will pass into the third. <freecandy> Look at it this way: if she were headed due east and turned 90 degrees towards the south, the ray would be parallel to the y-axis, no? <freecandy> So zoom out really really far until they’re really really close and rotate it just a tiny bit. Boom, you’re in the 4th quadrant. <freecandy> Erm, 3rd. <yosemite37> freecandy: so theta would be arctan(2)? <freecandy> How did you get that? <freecandy> I’m not saying it’s wrong, btw. <freecandy> Also, you need a theta bearing dictionary here <yosemite37> no i did it wrong sorry, i used hours, i need to use the knots <yosemite37> since it says maintaining the same speed, both rays have a distance of 15 knots? <freecandy> Rays have infinite distance. I was using finding the ray as an intermediate step towards finding the segment. <yosemite37> oh right, what I was trying to say was would both of the legs of the right triangle be that much <freecandy> 15 knots? <yosemite37> ye <freecandy> 15 knots long? <freecandy> “speed of 15 knots” knots are a measure of speed, not distance. <yosemite37> oh sorry <peterhil> How can I find the smallest subset that covers all integers in some range? The input is the factors of the integers in that range. <yosemite37> i thought it was distance since it also mentioned hours in there <freecandy> peterhil: ask the original question. <freecandy> distance=rate*time <peterhil> freecandy: The measure comes from old way to measure distances in the sea, by having a rope with knots in certain distances from each other… that distance being around 514 meters. <freecandy> Thanks for the history lesson, guy. Now let’s talk about your factors. <peterhil> I’m making an additive sound synthesis program, and I want to find the smallest amount of samples needed to cache all complex samples. Their frequencies are in fractional parts of the sampling rate. <freecandy> Think of us as North Korea and South Korea, you can pick who you are. The only way we’re going to engage is if you bring your front line to or beyond the DMZ. Your first attempt tried to go beyond, but your<peterhil> For example, if I have the complex samples for 1 Hz at 44100 KHZ sampling rate (1/44100 ratio), i have 44100 samples that also happen to cover several other frequency ratios, like 2/44100 = 1/22050, 3/44100 <peterhil> ok <freecandy> Ah, so you’re looking for an lcm. <freecandy> So if we want 5 Hz, 6 Hz and 7 Hz sounds, we’d want 210 Hz sampling. But if it’s 6, 7 and 8, 168 would suffice, right? <peterhil> Not exactly, but that helps. I want to cache those samples, but caching them all isn’t feasible. So, I wondered if it would be feasible to only store the samples for 1.0/some_prime_number and the fewest numb<yosemite37> freecandy so 15 knotts per hour, the first segment is for 1 hour and the second is for 2 so 15 and 30? <freecandy> Yes, yosemite. <freecandy> I don’t know what you mean by samples. Do you mean it in the popular sense of just some random crap I grabbed from some song, or something else? <poorbot> A 6-sided number cube, with faces numbered 1 through 6 is to be rolled twice. What is the probability that the number that comes up on the first roll will be less than the number that comes up on the second r<poorbot> Well. Let’s see. 1/6 * 1/6 if you roll it twice, what should I do to keep going? I am lost. <peterhil> It’s kind of related to how many rationals there are below 1/1 (or alternatively 1/2) that have a denominator limited below 44100.
Feb 162012
