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I don't. That was a joke, obviously they wouldn't let a neo-Nazi go to international house but I'm sure you'd get in. At IH its about 50% international students. Of these maybe 3/4 would be asian (China, Hong Kong, Malaysia, Japan...). The others are North American, European, couple of South...

who is your friend's brother Crikket? If he's from alstonville, knows about IH and is male my guesses are: - me (presumably its not me) - Blake Daley - Niall Boyd - Ian Parrington - someone else from before I went there. more biased opinions, this time on guys colleges Kings - Very...

alright, I'm in my second year at International House (IH) so there will inevitably be bias but here are my opinions st. johns - used to be all guys (I think) so will have less girls than other colleges. Fairly traditional, take rugby very seriously. I think there would be fairly hardcore...

Ahh, another question that I can pretend is revision for my analysis exam on thursday! The definition of a limit of a function f:X->R is is that f converges to limit L as x approaches a if for every e>0 there exists a d>0 such that |f(x) - L| < e whenever 0 < |x-a| < d Basically this is...

Yep

If and only if means that you need to prove it both ways. Casting around desperately for an example I find my analysis lecture notes "A monotone sequence converges if and only if it is bounded" Now without worrying about what the words mean this means that: if a monotone sequence is...

Yeah, maybe I should of said "can't integrate easily". (In fact I didn't think of doing it this way) But it would be possible to do it this way Int(sin^3(x)) =Int((1-cos^2(x)) sinx) = -cosx + 1/3 cos^3(x) then Int (x^2 sin^3(x)) =x^2(1/3 cos^3(x) - cosx) - Int(2x(1/3 cos^3(x) -...

I just went to the integrator (integrals.wolfram.com) and put in x^2*(Sin[x])^3. The answer it gave back is: -3/4*(-2+x^2)cos[x] + 1/108*(-2+9x^2) + 3/2*x*sinx - 1/18*x*sin(3x) This isn't exactly what I had but it seems to be of the same form. I obviously made a few mistakes somewhere but...

Warning: What I am about to do is very messy: hopefully there's a way to do this much more easily. But I can't see it. Basically we want to use integration by parts Integral(uv') = uv - Integral(u'v) with u=x^2 so we reduce the power of x^2 down via IBP twice. But the problem is that...

In response to the original question about integrating x^x, I found this somewhere when I was looking for some questions for my Analysis midsemester. It looks a bit complicated, to say the least, but I haven't actually read it all. It claims to be a proof that int x^x cannot be expressed in...

on the subject of graphing x^x, I just put (-Pi)^(-Pi) into MATLAB and got -0.0248 + 0.0118i, a complex number! I guess this means it doesn't exist but its pretty confusing, isn't it?

The normal series used for this sort of stuff is MacLaurin series. f(x) = f(0) + f'(0)*x + f''(0)/2*x^2 + f'''(0)/3!*x^3 + ... + f(n dashes)(0)/n!*x^n+ .. but note that f(0) = 0^0 is undefined so MacLaurin series doesn't exist. So just because its easier to do I'll give a Taylor series...

I originally did the same as you but I'm fairly sure now that r1 = [1 -1 L 1] r4 = [1 7 -2L 10L-1] and r4 -> r4 - r1 = [0 8 -3L 10L-2]

It's possible to do the derivative of x^x. y= x^x = e^(ln(x^x)) = e^(x*lnx) so y' = e(x*lnx)*[x*1/x+1*lnx] =(lnx+1)*x^x but this doesn't help us with the integral because we have that lnx. So if integrator doesn't work maybe there's no integral in terms of elementary functions...

I had a mistake in the augmented matrix that I'd fixed later but forgotten to fix at the start (it should be right now) -2L-L = -3L 10L-1-1 = 10L-2

slightly more rigorously, the Intermediate Value Theorem says if L is a constant and f is continuous on [a,b], if f(a) < L < f(b) or f(a) > L > f(b) then there is a c in (a,b) such that f(c)=L now let G(x) = f(x)-x G(0) = f(0) so 1/2 < G(0) < 3/4 G(1) = f(1) - 1 so -1/2 < G(1) < -1/4...

For part 2 write out system in full x1 - x2 + Lx3 = 1 4x2+(1-2L)x3 = 5L-1 (-2+L)x3 = -L+2 (-2+L)x3 = 0 so -2+L=0 therefore L=2 so x1-x2+2x3=1 4x2-3x3=9 0=0 0=0 now this isn't enough info to solve fully but if we just think of x3 as a real parameter ie let x3=t...

x1 - x2 + x3 = 1, 2x1 +2x2 + x3 = 5+1, x1 + 3x2 - x3 = 4+2, x1 + 7x2 - 2x3 = 10-1 augmented matrix is (using L for lambda) (You can imagine a dotted line between 3rd and 4th columns if you want to seperate the coefficents on LHS from constants on RHS) 1, -1, L, 1 2, 2, 1, 5L+1 1, 3...

Alright! Thanks, this is what I was looking for. I still don't quite understand the connection between the determinant and the polynomial you get when you expand it. You get the polynomial degree by adding degrees of the terms on diagonal right? Then determinant is 0 when x=y so by...

What you wrote seems to be correct but the question actually had x^3 not x^2 etc. I think this makes it a bit trickier and maybe the general results don't apply? I've been working on it and I think something like 1,1,1 x,y,z x^3,y^3,z^3 doing column 2 - column 1 and column 3 -...