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umm... dunno, it was advance screenings yesterday... (or rather last saturday) I know it's in circular quay dendy (halfway between opera house and station), because that's where we saw it... not sure about big cinemas, check hoyts..

think I understood the pdf file, but packing to go to blue mountains tomorrow for a week, will relook up jacobians when I get back... (26th) give me a virtual kick if I haven't replied by say 28th or so. :-)

umm. yes.

She gives a really powerful performance, and I think she'd be in the running for one. But there are quite a few contenders which have good chances too... (think kidman in cold mountain etc) For a short synopsis of the start, read the SMH review...

The proof for 4 is relatively simple. In general, you know that (a+b)/2>=root(ab) and (c+d)/2>=root(cd). So (a+b+c+d)/2>=root(ab)+root(cd) and so (a+b+c+d)/4>=(root(ab)+root(cd))/2 and apply the 2-variant AM-GM to the RHS again for the result. What I mean is, you do use the AM-GM, but if...

Go watch it. We watched the preview yesterday, it's quite twisted, but satisfying in the end. Hmm... horrible in a way, heart-wrenching, violent, a bit too much sex (although some might find that a plus)... yes. Go watch it.

Hmm. Usually the answer is no. Especially in questions of the type prove that (a+b+c)/3>=cube root(abc), which are just proving a specific case of the AM-GM. However, the usualy method of proving it isn't too hard, and only takes marginally longer. (you prove it for 2, 4, 8, .. for all...

Hmm, yes. I can't get away from Cleo Cresswell... her image haunts me (yes, she works at UNSW)... argh!

Hmm. Lots of statements! 1) I still don't like terry gagen. 2) Hooray for TSP (unofficially) at UNSW - 3rd year courses for 1st year students :-) (I'm just making the point that you can do the same at UNSW... admittedly only a few ppl do it though) 3) underthesun: learning discrete maths by...

you mean, I have to do it :-P a) you can do this part by the remainder theorem (see 3 unit). So the remainder is equal to P(-i) (as the factor divided by is (x-(-i)) ) which is (-i)^3+2(-i)^2+1=i-2+1 which is i-1. b) This is harder. Treat it as normal division by polynomial... Hence the...

Eww. Jacobian matrices are not nice, you encounter them in several variable. I've done most of that course (and I definitely remember doing the bit about jacobians), can you send me the link with newton & relation to jacobian? Then I can dust off my books and get back to work :-) well, not that...

Hey underthesun, I was planning to answer this and had an answer all written down just before my house had a power failure, so here goes again :-) (that was before you sent me a pm, mind you :-) ) As laz said, Maths 1A and 1B are compulsory for UNSW science/eng ppl. It's not too hard, just...

I meant ln[(w+0.0001)/(1-w-0.0001)]. edit: I guess it doesn't make the thing any nicer after all... although you don't get stuck with an unnice 1.0002 in the numerator at the end eg (w+0.0001)/(1-w-0.0001)=(1/(0.999-w))-1 But then that can just be factored out in the final thing in the other...

Okay. It is solvable using iterative methods for 1 variable... (when I say lots of rearranging, it works. If you want the proof, say if this result is useful, ask me and I will put it up. Just lots of algebra..) after lots of rearranging, letting m_i denote 1/(1+e^(ax_i+b)) for convenience...

I'll try :-) But first a question: by the transform: X = ln[(w + 0.0001)/(1 - w + 0.0001)] do you mean X=ln[(w+0.0001)/(1-(w+0.0001)]?

It was US though... which was a lot better a year ago than now, but still. And btw I think Goldbach is now worth nothing, faber & faber offered 2 million on it for a while but then that expired last year... also note that the poincare conjecture (one of the 7 problems) has a proof which...

Factorising you normally can state without actually writing anything down unless it's non-obvious (ie you can't expand it out without say at least 10 terms), in which case you should probably put a line in the middle. Graphs: Yes. for example, "from the graph, as there is a part above the...

yes you do get a recursive relation in order to prove step 3. It'd go something like this: (this is the long, ultra-rigorous version, shorten as desired) Observe that one line with no intersection creates 1 new region. (dividing the plane into 2) Furthermore note that for each intersection the...

to add to spice girl's post, the only situation I can think of where it is non-dodgy is when you correlate real and imagniary coeff... eg x=4+3i = (a+2b)+bi so by inspection a+2b=4 and b=3. you could just write correlating.. :-)

hey, just because I got it for free it doesn't mean that I don't use it :-)