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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

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)...

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...

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...

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...

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...

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.

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...

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.

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...

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...

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?

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...

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...

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...

int(x^3 dx/sqrt(x^2+1)) let u^2=x^2+1 then 2u du/dx = 2x so u du= x dx so int(x^3 dx/sqrt(x^2+1) = int((u^2-1) u du / sqrt(u^2)) = int(u^2-1)du you can take it from there, u=x^2+1 might work as well.

I say that a function that is equal to 1/2*ln(x-1) for x near 1 and then goes to the asymptote which is concave down, continuous and differentiable fits the bill. Attached is a drawing of one such function (if you can accept my poor freehand skills). I'm not sure if there is a function with...

As buchanan correctly points out I was wrong before. If we take f(x)=-2/(x^2-1) then e^f(x) has an inflexion point (I wasn't graphing with enough detail). See attachments. So this example shows that it is possible to have an inflexion point. A modified version of my above example with...

I agree that you only have to look at x>1 but I think that it is possible to have a point of inflexion there. I don't really follow your argument who_loves_maths so I've made an example. Consider the function that is equal to 2ln(x-1) for x between 1 and 1.2, equals -2/(x^2-1) for x bigger...

Looking at the 2003 question (attached) I don't think that e^f(x) would have an inflexion point. As x->1 from the right, f(x)->-inf so e^f(x)->0. Then as we go further to the right x->inf so f(x)->0 from below, so e^f(x)->1 from below. I think you just start (with an open circle) at (1,0) and...