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Re: 2012 HSC MX1 Marathon $Define $f(x):=x^r$ for positive $x$ and real $r.\\$Then $f'(x)=rx^{r-1}$ and we have:$\\ \\r=f'(1)=\lim_{x\rightarrow 1^-}\frac{1-f(x)}{1-x}=\lim_{x\rightarrow 1^-}\frac{1-x^r}{1-x}$ and the result follows.$ Of course here we are assuming that the derivative of x^r...

Thankyou :).

Nope, I'm not even sure how it is proven. Its just a useful sledgehammer to know. What is this geometric interpretation?

Yeah I know Graham, he was doing honours with me last year. But 2nd year advanced tutes would only be taken by doctors and PhD students at least a couple of years in.

Unlikely, they would usually reserve such tutes for professors.

graduated last year but might be running tutorials/marking assignments in first sem :).

Ah yes 4, my bad.

Well largs definition probably just varies by a constant multiple to make it easier to write the elementary symmetric polynomials.

I thought that sym. summation ran over ALL permutations of the variables? So \sum_{sym}ab=2(ab+ac+ad+bc+bd+cd) Probably just a matter of conventions...it makes Muirhead's inequality look nicer :).

For n=3, \Delta=(\alpha_1-\alpha_2)^2(\alpha_1-\alpha_3)^2(\alpha_2-\alpha_3)^2.

I especially like this part. I proved that the only nontrivial functions f satisfying this property are semicircles IF we further impose the smoothness condition of f being twice differentiable (using differential equations). Am unsure as to whether or not there are any pathological non-smooth...

Exactly what larg said, we are just talking about ordered pairs of integers. An alternate way of writing the same product is: \prod_{l=2}^n\left(\prod_{k=1}^{l-1} (\alpha_k-\alpha_l)^2\right). Thanks for the positive feedback on the question, I felt it might help students appreciate the...

Every \alpha_i is a complex root of P. By the FTOA, there are exactly n such roots, counting multiplicity. The convention here is to take the sum/product over all pairs of positive integers (k,l) with 0<k<l \leq n. (This is a fairly common notation for sums and products.)

Yep. This is an algebraic derivation of that result that generalises to higher degree polynomials (although it turns out that the discriminant is not as useful for them).

Attached is the Q8 of an exam I wrote for a student, some of you may find it interesting...enjoy :).

That just implies that z lies on some circular arc containing w, z+w and the circumcentre of O, w and z+w.

w does not have to be the centre of the original circle. see largs counterexample. EDIT. And if we add that assumption, the problem is completely trivial as we have: |w|=|z|=|w-z|.

Haha win.

I always found that name confusing because there is also a "Ham and Cheese Sandwich Theorem" in measure theory.

I cannot recall exactly, would have to dig up my sisters book somewhere. (It is an ancient book now, all badly typeset etc. If I find time tomorrow I will take a photo of the page...)