• Mathematician Chat v. 3.999...
    1,232 replies, posted
I have never finished high school due to illness; currently doing a study but math is not in the classes, cannot do a side-class to this to get my degree in maths so.. I was wondering can someone give me a good alternative to start from start to end in English? I mean; I do understand math & everything in Dutch, but don't get me started on English mathematics any help / website / advice would be great
[QUOTE=Siemz;46620031]I have never finished high school due to illness; currently doing a study but math is not in the classes, cannot do a side-class to this to get my degree in maths so.. I was wondering can someone give me a good alternative to start from start to end in English? I mean; I do understand math & everything in Dutch, but don't get me started on English mathematics any help / website / advice would be great[/QUOTE] What are you finding difficult about it? Most of our maths is symbol-based, do you guys use different mathematical symbols to us?
[QUOTE=agentalexandre;46618892]Can someone explain the topologist's sine curve to me? I'm having trouble understanding how a connected set is not necessarily path-connected.[/QUOTE] This is very handwavy, but for intuition's sake you can think of it as being because there are no "breaks" in the space, so it's connected, but the rapid oscillation near the origin means that trying to connect the origin to another point would require a path of infinite length.
What kind of rule or way of thinking should I use when realizing which x that makes x^(2n+1)/(n+1) from 1 to infinity converge?
[QUOTE=JPlus;46620669]What are you finding difficult about it? Most of our maths is symbol-based, do you guys use different mathematical symbols to us?[/QUOTE] Some things are written completely different and some things even as simple as + , - , times 'x' , divided by etc are done completely different Not to mention naming of things It's mostly that I've been teached differently from when I was little, so switching is pretty hard. But not being done learning I think it is better to start over then to half finish something and leave it with that
How is + done different then? Just curious. Also you can just look up the English word on Wikipedia and then switch to the Dutch page, so you have without doubt the right translation. Oh and "early" difficulties in mathematics can very much be due to new (symbolic) notation, you'll get used to it.
I was doing all the stars on khan academy so they deleted the star images and I have to take the long way around.
[QUOTE=Matthew0505;46613688]In an antiderivative of a polynomial you can replace each x^n with (1/(n+1))x^(n+1), is there any rule like this for an antidifferential?[/QUOTE] In a way. For continuous functions, the antiderivative is calculated by doing an integral. For discrete functions, you do a summation instead - so you should look at results for those. Unfortunately summation is never as neat as integration (typical of discrete functions vs continuous) To calculate the antidifferential of x^n we need to calculate the sum of x^n, which is given by [url]http://en.wikipedia.org/wiki/Faulhaber%27s_formula[/url] - not very appealing. However we can do better if we look at a similar, but less neat function: x(x-1)...(x-n+1) - Note that there are n terms, so this is like x^n, but with each term shifted by one (in some sense this is dues to the discreteness) In this case we can easily sum, which gives us something like (1/(n+1)) x(x-1)...(x-n) - Note how similar this is to the integral of x^n, except with the terms shifted in the same way as above. If you want to get really clever, making this continuous is a bit like changing the 'shifting by one' to 'shifting by h' and then shrinking h. Thus the shifting collapses down to x^n integrating as you expect. (Proof: Exercise) There are nice results for anything that sums well, you could try a^x for example (sums to a geometric progression).
[QUOTE=JohnnyMo1;46621079]This is very handwavy, but for intuition's sake you can think of it as being because there are no "breaks" in the space, so it's connected, but the rapid oscillation near the origin means that trying to connect the origin to another point would require a path of infinite length.[/QUOTE] I'm not sure I understand this. Why is a path of infinite length a problem?
[QUOTE=Krinkels;46624857]I'm not sure I understand this. Why is a path of infinite length a problem?[/QUOTE] Because we generally require a path to be some continuous mapping of [0,1] into the space (or, equivalently, [a,b]) so we can do fun things like the fundamental group. I know it's handwavy (I don't even think length is as well defined in a topological space as it is in R^n) but that's how I've always pictured it. Now that I'm looking harder into it, I'm having a hard time even telling how far off base it might be. Similarly I'm not sure I understand what this has to do with connectedness: [quote]To see why it's connected: draw an open ball around any point on the graph or at the origin. The vast majority of points in the ball will not be on the curve.[/quote] Take, for instance, all points in R^2 with rational coordinates. That's certainly not connected but it satisfies what you said.
[QUOTE=JohnnyMo1;46627861]Because we generally require a path to be some continuous mapping of [0,1] into the space (or, equivalently, [a,b]) so we can do fun things like the fundamental group. I know it's handwavy (I don't even think length is as well defined in a topological space as it is in R^n) but that's how I've always pictured it. Now that I'm looking harder into it, I'm having a hard time even telling how far off base it might be.[/QUOTE] - In any metric space you need any path to be bounded, since otherwise by applying the distance (from a particular point) you end up with a continuous function [0,1] -> R which is unbounded. This contradicts the Boundedness theorem. - His argument was missing a few steps, but the crux is right. The curved part is clearly connected (as the continuous image of a connected space) and the origin is connected (as a single point), all we have to do is show no open sets containing them can be disjoint - cue his argument. (There are some details here, but they're fairly easy to fill in from the definitions)
[QUOTE=Joey90;46630630]- In any metric space you need any path to be bounded, since otherwise by applying the distance (from a particular point) you end up with a continuous function [0,1] -> R which is unbounded. This contradicts the Boundedness theorem.[/QUOTE] Do we have to talk about unbounded paths? All the interesting stuff on the topologist's sine curve is pretty clearly happening within a ball of finite radius at the origin. Are you using bounded in a different sense than I think? [QUOTE=Joey90;46630630]- His argument was missing a few steps, but the crux is right. The curved part is clearly connected (as the continuous image of a connected space) and the origin is connected (as a single point), all we have to do is show no open sets containing them can be disjoint - cue his argument. (There are some details here, but they're fairly easy to fill in from the definitions)[/QUOTE] I'm still not seeing what relation that has with points not being on the curve. We don't even need to consider that there are any points not on the curve. Maybe I'm misinterpreting his argument. [editline]3rd December 2014[/editline] moral of the story is I'm getting rusty and I want to start grad school already
I should really RTFQ [QUOTE=JohnnyMo1;46631552]Do we have to talk about unbounded paths? All the interesting stuff on the topologist's sine curve is pretty clearly happening within a ball of finite radius at the origin. Are you using bounded in a different sense than I think? [/QUOTE] You were talking about curves of infinite length. A path cannot be unbounded in the global sense - i.e. it cannot go infinitely far away from the end points, which is what I was saying. Sadly it's not what you were asking :v: It [i]can[/i] however, have infinite length if it has infinite local behaviour. For a concrete example, imagine a line that zig-zags between (1/n, 0) and (1/n, 1/n) for n = 1 to infinity. Each segment has length >2/n, so the total length is > sum of 2/n, which is infinite. It doesn't even have to have a pathological endpoint, but it must be infinitely 'wiggly' - so for example any fractal curve has infinite length. For something even closer to our example try x.sin(1/x) [QUOTE=JohnnyMo1;46631552] I'm still not seeing what relation that has with points not being on the curve. We don't even need to consider that there are any points not on the curve. Maybe I'm misinterpreting his argument. [/QUOTE] I totally misread (again...) I skimmed some words and assumed he was trying to say the ball has infinitely many points on the curve. (Which is pretty much the opposite of what he did say!) More sleep required probably.
[QUOTE=Joey90;46632419]I should really RTFQ You were talking about curves of infinite length. A path cannot be unbounded in the global sense - i.e. it cannot go infinitely far away from the end points, which is what I was saying. Sadly it's not what you were asking :v: It [i]can[/i] however, have infinite length if it has infinite local behaviour. For a concrete example, imagine a line that zig-zags between (1/n, 0) and (1/n, 1/n) for n = 1 to infinity. Each segment has length >2/n, so the total length is > sum of 2/n, which is infinite. It doesn't even have to have a pathological endpoint, but it must be infinitely 'wiggly' - so for example any fractal curve has infinite length. For something even closer to our example try x.sin(1/x) I totally misread (again...) I skimmed some words and assumed he was trying to say the ball has infinitely many points on the curve. (Which is pretty much the opposite of what he did say!) More sleep required probably.[/QUOTE] Right, okay. What I'm getting from this is basically I think all the stuff I was saying was correct :v: [editline]3rd December 2014[/editline] Good, I was a little worried I was losing it.
Yeah I'm very confused about this. So from [url]http://en.wikipedia.org/wiki/Connected_space[/url], a space is connected if it cannot be represented as the union of two or more disjoint nonempty open subsets. I was trying to say that the curve had no nonempty open subsets, hence it could not be such a union. I've made a mistake somewhere. By the same reasoning I can show lots of disconnected things are connected. What exactly have I done wrong?
[QUOTE=JohnnyMo1;46633374]Right, okay. What I'm getting from this is basically I think all the stuff I was saying was correct :v: [editline]3rd December 2014[/editline] Good, I was a little worried I was losing it.[/QUOTE] Not quite - you were using the path having infinite length as a contradiction (which it's not), the problem is that the path doesn't converge, so cannot possibly be continuous. [QUOTE=Krinkels;46634259]Yeah I'm very confused about this. So from [url]http://en.wikipedia.org/wiki/Connected_space[/url], a space is connected if it cannot be represented as the union of two or more disjoint nonempty open subsets. I was trying to say that the curve had no nonempty open subsets, hence it could not be such a union. I've made a mistake somewhere. By the same reasoning I can show lots of disconnected things are connected. What exactly have I done wrong?[/QUOTE] What are the open subsets on the curve? As a subspace of R^2, by [url=http://en.wikipedia.org/wiki/Subspace_topology]definition[/url] the open sets are intersections of open sets in R^2 with the curve - of which there are plenty! The problem is the disjoint and covering part.
[QUOTE=Joey90;46635012]Not quite - you were using the path having infinite length as a contradiction (which it's not), the problem is that the path doesn't converge, so cannot possibly be continuous.[/QUOTE] So in your previous post, you're saying an "infinitely wiggly" path [I]will[/I] have infinite length, but that infinite length won't be sufficient to ensure you can't have a path going through the wiggly section? I'm a little confused by what you mean by it doesn't need a pathological endpoint. Do you just mean the curve can continue just fine past the origin, and we still lose path-connectedness? (seems pretty obvious)
I have a problem that I need help with and I can't find the answer googling. I have three fractions with different denominators that I need to turn into one fraction. I understand the idea of finding a common denominator but I can't see it because the fractions have awkward denominators in relation to each other. Is there some kind of trick to solving this that I haven't learned? here are the fractions: (1/1000) + (1/2200) + (1/4700). (I can't solve the fractions before adding them because I'm working on resistors in parallel and need to flip the new fraction before solving it.
I dunno that there's anything tricky about this one, unless maybe you haven't done it with three numbers. Done in LaTeX because formatting fractions on the forum sucks: [IMG]http://i57.tinypic.com/20u4j88.png[/IMG]
[QUOTE=JohnnyMo1;46635787]So in your previous post, you're saying an "infinitely wiggly" path [I]will[/I] have infinite length, but that infinite length won't be sufficient to ensure you can't have a path going through the wiggly section? I'm a little confused by what you mean by it doesn't need a pathological endpoint. Do you just mean the curve can continue just fine past the origin, and we still lose path-connectedness? (seems pretty obvious)[/QUOTE] In relation to the path-connectedness, we just need to show a path (= curve) doesn't exist from the origin to the rest of the graph. The only option is the whole space, however this isn't a path as it's not continuous. What I was getting at is that the infinite length is [i]not[/i] the problem, as paths [i]can[/i] have infinite length. The pathological endpoint thing just meant the 'infinite length' doesn't have to be around a particular point (which is true for counterexamples in other situations) - a fractal curve has infinite length at [i]all[/i] points.
[QUOTE=JohnnyMo1;46637230]I dunno that there's anything tricky about this one, unless maybe you haven't done it with three numbers. Done in LaTeX because formatting fractions on the forum sucks: [IMG]http://i57.tinypic.com/20u4j88.png[/IMG][/QUOTE] almost as sneaky as the factor theorem i am watching you mr spy
So I just took the Putnam yesterday. Suuuuper hard, but I was able to solve two, which surprised me, since I've never been able to solve one on my own before. Having all that time really helps, I suppose. Fingers crossed that I got them right, because if I did, that might put me in the top 500, which means I get my name published. Also, here's one of the questions I wasn't able to get, for you masochistic bastards. I figure you might enjoy it. [QUOTE] Suppose that [; f ;] is a function on the interval [1, 3] such that [; -1 \le f(x) \le 1;] for all [; x ;], and [; \int^3_1 f(x) \,dx = 0 ;]. How large can [; \int^3_1 \frac{f(x)}{x} \,dx ;] be? [/QUOTE]
[QUOTE=Xeloras;46659894]So I just took the Putnam yesterday. Suuuuper hard, but I was able to solve two, which surprised me, since I've never been able to solve one on my own before. Having all that time really helps, I suppose. Fingers crossed that I got them right, because if I did, that might put me in the top 500, which means I get my name published. Also, here's one of the questions I wasn't able to get, for you masochistic bastards. I figure you might enjoy it.[/QUOTE] Oh yeah I tried that one too. An optimal f must be decreasing. I suspected that f = 1 on [1,2] and -1 on (2,3] would give the solution but I am by no means sure. If I hadn't spent so much time on this one I might have gotten the Lipschitz continuity one.
[QUOTE=Krinkels;46660598]Oh yeah I tried that one too. An optimal f must be decreasing. I suspected that f = 1 on [1,2] and -1 on (2,3] would give the solution but I am by no means sure. If I hadn't spent so much time on this one I might have gotten the Lipschitz continuity one.[/QUOTE] Awesome, I'm so happy to hear someone else took it! I suspected the same when I first touched it, but then I figured that when you divided it by x, you'd probably end up with zero again, so it probably wouldn't be the largest you could get. Then I thought possibly sin(x) or cos(x), but the integral for those are disastrous, and I'd probably end up having to represent them with a Taylor series. I know I got the one with the Taylor series of (1 - x + x^2) * e^x. I had a pretty rock solid proof for that one. And I think I got the one where you were representing things with a base ten over expansion. That one was surprisingly easy, because all the number really had to do was not have a zero in it to have a unique over expansion. But those are the only two I'm pretty sure I had a good base on.
[QUOTE=Xeloras;46663261]Awesome, I'm so happy to hear someone else took it! I suspected the same when I first touched it, but then I figured that when you divided it by x, you'd probably end up with zero again, so it probably wouldn't be the largest you could get. Then I thought possibly sin(x) or cos(x), but the integral for those are disastrous, and I'd probably end up having to represent them with a Taylor series. I know I got the one with the Taylor series of (1 - x + x^2) * e^x. I had a pretty rock solid proof for that one. And I think I got the one where you were representing things with a base ten over expansion. That one was surprisingly easy, because all the number really had to do was not have a zero in it to have a unique over expansion. But those are the only two I'm pretty sure I had a good base on.[/QUOTE] I think I got the Taylor series one too. I solved it pretty quick but then forgot how to rigorously justify that the series for xe^x is x times the series for e^x. I spent an hour just using induction to derive the series. The 'find the determinant' one was also surprisingly easy. The trick is to see that it's row equivalent to a diagonal matrix. I'm kicking myself about the over-expansion though. Could've been an easy (1*10^1)+(0*10^0). Anyway, for the integral, when you divide by x you don't get zero again. If f is how I described it, then the integral of f/x is ln(4/3) because you divide by x before integrating. It's basically a continuous version of a weighted average: points to the left of the interval contribute more to the integral than those to the right. Perhaps you need the calculus of variations to solve it?
"Spectral sequences are a powerful book-keeping tool for proving things involving complicated commutative diagrams. They were introduced by Leray in the 1940’s at the same time as he introduced sheaves. They have a reputation for being abstruse and difficult. It has been suggested that the name ‘spectral’ was given because, like spectres, spectral sequences are terrifying, evil, and dangerous. I have heard no one disagree with this interpretation, which is perhaps not surprising since I just made it up."
-snip-
Could anyone help me figure out the radius of convergence? [IMG]http://latex.codecogs.com/gif.latex?%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bx%5E%7Bn%5E2%7D%7D%7Bn%21%7D[/IMG]
[QUOTE=JohanGS;46669520]Could anyone help me figure out the radius of convergence? [IMG]http://latex.codecogs.com/gif.latex?%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bx%5E%7Bn%5E2%7D%7D%7Bn%21%7D[/IMG][/QUOTE] With something like this, it's either 0,1 or infinity [sp]very rarely scaled by a constant[/sp] In this case you can easily use the ratio test to show convergence <1 and divergence >1 For completeness, it's also clear that it converges for |x| = 1.
Okay so I'm not the biggest fan of general mathematics since I don't usually see their applications very well, physics is more of my thing. I do have to say however that seeing trigonometric equations apply to Simple Harmonic Oscillators is really neat and makes me realize that as much as I bitched about it, what I (re)learned in Calculus this quarter is actually [I]supremely[/I] relevant. and then when its linked to the unit circle and differentiation and offsets good lord it got really neat really fast
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