Mathematician Chat V.floor(π) - General

[url=http://math.stackexchange.com/questions/74347/construct-a-function-which-is-continuous-in-1-5-but-not-differentiable-at-2?lq=1]Oh okay thanks, I think I got it now.[/url]

[QUOTE=ArgvCompany;44878946]I'm preparing myself for my first year of engineering physics. The math part of the first year will consist of two analysis courses (both single and multi-variable) and a linear algebra course. What is the best things I can do in order to prepare for the university-level math? I've taken an "introductory course to real analysis" before, if that matters.[/QUOTE] The flag implies that you're german but then you mentioned Sweden so I'll assume that's where you're from. As long as you can handle the previous math (as in the math/physics required) you should be good to go. You could always start reading up on things that will eventually come to get ahead, but just make sure you have a good foundation. Do the "sommarmatte". Where are you going to study by the way?

I was doing some GRE prep today. One problem was to evaluate sqrt(3 + sqrt(3 + sqrt(3 + sqrt(...)))). I wasn't sure how. When I saw the solution I felt like a dumbass.

Do the minimal polynomials of sqrt(3), sqrt(3+sqrt(3)), ... 'approach' x^2 - x - 3 in some way? Also, someone mentioned to me today that there was some weird way to evaluate the factorial at negative integers. Does anyone know what he was talking about?

I don't really know anything detailed about it, but could it be the Gamma function?

It cannot - the gamma function has poles at the negative integers, and at zero since the argument's shifted over by one. I thought the same was true of any interpolation of the factorial, so maybe the solution comes from combinatorics rather than analysis.

[QUOTE=Krinkels;44944772]Do the minimal polynomials of sqrt(3), sqrt(3+sqrt(3)), ... 'approach' x^2 - x - 3 in some way? Also, someone mentioned to me today that there was some weird way to evaluate the factorial at negative integers. Does anyone know what he was talking about?[/QUOTE]Not in the classic sense - it's obvious what the other roots of the minimal polynomial must be, e.g.: [;\pm \sqrt{3};] [;\pm \sqrt{3 \pm \sqrt{3}};] etc. Most of these roots won't 'converge' to anything (they'll only converge when in some sense there's an infinite number of 'leading' pluses or minuses) so the resulting polynomials will probably be a mess, however they will have roots closer and closer to the real ones. You could try plotting a few and seeing what they look like... I would imagine the roots get clustered around the 2 roots of the quadratic, but have many between them as well. I don't think there's any way to evaluate factorial at negative integers... You can extend it to complex numbers by using the Gamma Function but it has simple poles at negative integers (so you could, for example, extend it to [;\mathbb{C} _ \infty;]) You could come up with some other analytic function that does not have simple poles there, but it would be pretty non-standard.

im failing geometry

[QUOTE=Dub!;44945097]im failing geometry[/QUOTE] Hey if it makes you feel better I failed half my Electronics Engineering program.

Persistence is more valuable than intelligence. Stick with it and you'll do fine.

Say you're on an infinite grid of squares you want to travel on. You can only move to squares that: -share an edge with the square you're on -have no lava on them -you haven't been on before. There is probability [i]p[/i] that a square has lava on it. What is the probability that you can travel [i]n[/i] squares? What if the infinite grid is replaced with a solid where every face shares an edge with four other faces?

[QUOTE=Krinkels;45021417]Say you're on an infinite grid of squares you want to travel on. You can only move to squares that: -share an edge with the square you're on -have no lava on them -you haven't been on before. There is probability [i]p[/i] that a square has lava on it. What is the probability that you can travel [i]n[/i] squares? What if the infinite grid is replaced with a solid where every face shares an edge with four other faces?[/QUOTE] That sounds fairly hard, unless there's a trick I'm missing (quite possible)... What I'd be interested in is whether n->infinity converges to a non-zero value. It's clearly monotonic decreasing, but is there a non-zero probability of 'escaping' to infinity? (Intuitively it feels like there is, but I'm not sure if that's easier or harder to prove!)

[QUOTE=JohnnyMo1;44942715]I was doing some GRE prep today. One problem was to evaluate sqrt(3 + sqrt(3 + sqrt(3 + sqrt(...)))). I wasn't sure how. When I saw the solution I felt like a dumbass.[/QUOTE] Is this the right approach? [sp]Let S = sqrt(3 + sqrt(3 + sqrt(3 + sqrt(...)))). Then S = sqrt(3 + S). S^2 = 3 + S. S^2 - S - 3 = 0. S is the positive root of the quadratic S^2 - S - 3 = 0.[/sp]

[QUOTE=Joey90;45025760]That sounds fairly hard, unless there's a trick I'm missing (quite possible)... What I'd be interested in is whether n->infinity converges to a non-zero value. It's clearly monotonic decreasing, but is there a non-zero probability of 'escaping' to infinity? (Intuitively it feels like there is, but I'm not sure if that's easier or harder to prove!)[/QUOTE] I think not: [IMG]http://i.imgur.com/OJ3KCsm.png[/IMG] Maybe I'm wrong, but I was hoping that the limit was a more interesting function.

[QUOTE=Krinkels;45096151]I think not: [IMG]http://i.imgur.com/OJ3KCsm.png[/IMG] Maybe I'm wrong, but I was hoping that the limit was a more interesting function.[/QUOTE] I think there are a few dodgy things going on there, unless I'm misunderstanding... [;f_k = P($there exists some k-path$);] [;g(k,n) = \;$number of k-paths with n adjoining squares at the end$;] then you're then saying [;P($there exists some (k+1)-path$) = P($there exists some k-path$) . P($a k-path is extendable to (k+1)$);] by using the g's. There are a few problems... The first is mixing up what is fixed - the f's are averaging over the whole sample space of punctured grids, but the g's are choosing a specific punctured grid. If you want to use the g's in an overall probability, you need to average them over the sample space as well, sadly this isn't as simple as averaging them individually: E(f(X)) =/= f(E(X)) Additionally, the formula isn't even right in individual cases - if there is more than one k-path, you need to have the probability that at least [i]one[/i] of the paths is extendable. The other big problem is assuming everything is independent. Unfortunately not the case: for example 2 k-paths that end the same are going to have the same extendability. The first and second issues might be solved with a construct like: [;h(k,n) = P($exactly n k-paths$);] But you can't form a meaningful recurrence because any statement about the extendability of k-paths to (k+1)-paths is tied to the exact set of paths you are talking about. (And as mentioned above, it's not good enough to just take the 'average' - though I don't even know how you would do that either!) I'll probably keep thinking about it, but I stick by my statement that it seems hard.

Has anyone here seen a proof that polynomials are the unique functions which can be differentiated to zero that isn't just "integrate 0 as many times as you want and you'll get a polynomial?"

I saw a special case once. Here's the link: [url]http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial[/url] At the moment I can't think of any counterexamples when [0,1] is replaced with R, or when f is finitely differentiable, but I'm sure if it were true then it would have been proved.

[QUOTE=Krinkels;45115896]I saw a special case once. Here's the link: [url]http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial[/url] At the moment I can't think of any counterexamples when [0,1] is replaced with R, or when f is finitely differentiable, but I'm sure if it were true then it would have been proved.[/QUOTE] I don't [I]expect[/I] there to be counterexamples, I just feel like "integrate 0 a bunch of times," doesn't feel rigorous. Maybe the fundamental theorem of calculus should make it feel convincing, but it doesn't, which is why I wanted an alternate proof. Also, [quote]"This seems like a homework problem in a 1st year course on calculus." If you can find a regularly offered calculus course at more then 4 universities in the world that have the same number of students in the class who have even a clue how to solve this problem,I'll sing the American National Anthem naked on YouTube.I'm dead serious. [/quote] comedy gold.

We know f is polynomial in [0,1], say P(x) Suppose f(x) =/= P(x) for some point x = a. The proof above can be easily used to show f is polynomial in some closed interval containing both [0,1] and a. It's easy to see that this polynomial must coincide with P on [0,1], and therefore on the whole interval chosen, but this means f(a) = P(a). Contradiction, f(x) = P(x) for all x. I think the reason it's proved on an interval is that's the interesting part of the proof. Using that to show it on R isn't.

Say I were to draw from a 10-card deck, all the cards are labeled 1 through 10, and I want to try to draw a 5 within 3 draws (without putting cards back). Is there a way to find a singular probability for that, or does there have to be a minimum and maximum probability (i.e. 10% to 12.5%, first to third draws, respectively)?

Yes. There are three separate probabilities to draw a five exactly on a given draw. There is a single probability that you draw it in one of the first three draws. I forget how to calculate these though, I'll edit this post later. [editline]Dicksdicksdicks[/editline] The probability that you draw it in the first three is 3/10. You can calculate this with the [url=http://en.wikipedia.org/wiki/Hypergeometric_distribution]Hypergeometric[/url][url=http://stattrek.com/probability-distributions/hypergeometric.aspx] Distribution[/url].

[QUOTE=AbeX300;45168968]Say I were to draw from a 10-card deck, all the cards are labeled 1 through 10, and I want to try to draw a 5 within 3 draws (without putting cards back). Is there a way to find a singular probability for that, or does there have to be a minimum and maximum probability (i.e. 10% to 12.5%, first to third draws, respectively)?[/QUOTE] 5 will be drawn at card n with a probability 1/10 Probability it's in any given three is 3 * 1/10

[url]http://euclidthegame.org/Level1.html[/url] How I spent my lunch break today.

still stuck on trisecting that goddamn line

[QUOTE=Joey90;45212451][url]http://euclidthegame.org/Level1.html[/url] How I spent my lunch break today.[/QUOTE] This is so much fun!

Happy tau day. Now shut the fuck up about it, nerds.

"3.1 The algebra of cooking: Let A be a raw potato." I'm reading one of them weird papers again.

Does it solve the basil problem?

[QUOTE=JohnnyMo1;45237150]Happy tau day. Now shut the fuck up about it, nerds.[/QUOTE] but tau is superior in every way!

[QUOTE=Joey90;45169428]5 will be drawn at card n with a probability 1/10 Probability it's in any given three is 3 * 1/10[/QUOTE] A bit late, but he did say without putting cards back. Might be wrong, but won't the probability be: 1/10 + 1/9 + 1/8? Might be horribly wrong, didn't read the link above.