Recent content by martin

  1. martin

    Anyone else finding that all of a sudden...

    I'm not even doing my HSC (3rd year uni exams aren't too far away) but I just thought I'd share my minesweeper high scores (i've been playing obsessively for about the last two months) Beginner - 1 Intermediate - 40 Expert - 137
  2. martin

    The different sections of maths!

    Really Buchanan, you obviously know a lot about maths but you're lacking a bit in tact and common sense. The AMS classification is useful for their purposes but its a bit unwieldy. I'll offer up a very short classification that is useful to me in that I feel each part (the first three anyway)...
  3. martin

    EV/Random Events

    My point is that if we choose the values in the envelopes (x and 2x) beforehand then there is no paradox because you can go through each case separately and show switching makes no difference. The only problem is when you see the value N without knowing what x is. The argument claims that the...
  4. martin

    EV/Random Events

    Alright, I think the problem is this statement by Rorix: I can see no justification for it. It is not possible to have a uniform probability distribution on the positive integers. Consider all the integers from 1 to 1000, we can choose each with probability 1/1000, but if we have infinitely...
  5. martin

    EV/Random Events

    We have two very similar problems here but they are not equivalent. In the spirit of concreteness I've written 2 computer programs (in Matlab but they are pretty close to pseudocode). Problem 1: Given one of two envelopes with one double the other should you switch N=0; Nswitch=0...
  6. martin

    EV/Random Events

    EV is expected value, the sum of probabilities time payoffs (for discrete distributions). For example the expected value of a dice is 1/6*1+1/6*2 + ... + 1/6*6 = 3.5 Its my feeling that it doesn't matter whether you switch or not your expected value will be the same. This is because of the...
  7. martin

    locus question

    Not to steal Turtle's thunder or anything but I'm sure he won't mind. In Matlab I just put in >> ezplot('(x-3)^2+y^2-((x-2)^2+y^2)*(x^2+(y-1)^2)',[-3,3],[-1,3]) >> hold on >> plot([3 2 0],[0 0 1],'rx') and get the attached graph. I don't think its any easily describable curve.
  8. martin

    Euclidean Algorithm

    Yeah, that isn't the most useful description of the euclidean algorithm is it? It's used to find the gcd (greatest common divisor) of two integers. So say we want to find gcd(72,52). We could just factor them 72=(2^3)*(3^2) and 52=(2^2)*13 so gcd(72,52)=2^2=4. But factoring is hard (for big...
  9. martin

    Topology on Graphs

    Yeah, except the concept of distance in that question is different. But I don't think we're supposed to talk about SUMS questions so I'll try not to.
  10. martin

    Topology on Graphs

    We want to prove that R^n \ B(x_tilde,eps) is path connected for any n>1, x_tilde in R^n and eps>0. A space X is said to be path-connected if for any two points x and y in X there exists a continuous function f from the unit interval [0,1] to X with f(0) = x and f(1) = y. (This function is...
  11. martin

    Topology on Graphs

    This is a pretty general question but I've done classes on both graph theory and topology so I'll have some kind of answer although it definitely won't be all you can say on the topic. Topology is the study of 'rubber sheet' geometry; you don't consider distance (the standard example is that...
  12. martin

    Overloading Question

    How does this work if you have say two subjects to go and you do 4 in your final semester. Do you have to pay full fee for two of them?
  13. martin

    integration question

    for the previous integral I don't think the answer is right int(x/(x^2-4))dx = ? let u=x^2-4, then du = 2 x dx so x dx= du/2 so int(x/(x^2-4))dx = int(1/2 1/u)du = 1/2 ln(u) + C = 1/2 ln(x^2-4) + C = ln(sqrt(x^2-4)) + C But how this helps us for the desired integral I don't know...
  14. martin

    What the Poly Q

    trying to solve z^2 = 1/2 + 1/2 i as is standard (I think) let z=x+iy then z^2 = x^2 - y^2 +2xyi so x^2-y^2 = 1/2 and 2xy = 1/2. From second eqn x = 1/(4y) then in first eqn: 1/(16y^2) - y^2 = 1/2 so mult by 16y^2 1-16y^4 = 8y^2 so 16*y^4 + 8*y^2 - 1 =0. This is a quadratic in y^2...
  15. martin

    Intergration Constants

    just in case anyone still doesn't understand have a look at this (rather stupid) example. int(x-6)dx if you forgot that you could just do each bit separately you could let u=x-6 then dx=du so int(x-6)dx = int(u)du = 1/2 u^2 + C = 1/2 (x-6)^2 +C = 1/2 x^2 - 6x + 18 + C and here...
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