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My details and credentials are here: http://community.boredofstudies.org/showthread.php?t=269320 I am willing to tutor either at my home in Cherrybrook or in my office at the University of Sydney during the summer if you are interested :)

My details and credentials are here: http://community.boredofstudies.org/showthread.php?t=269320 I am willing to tutor in my office at the University of Sydney during the summer if you are interested :).

See last three posts in http://community.boredofstudies.org/showthread.php?t=270833.

Sure you typed it up right? Substitute x=0 into the equation in your first post and you get 1=0...

z^4+1&=&(z^4+2z^2+1)-2z^2\\&=&(z^2+1)^2-(\sqrt{2}z)^2\\&=&(z^2+\sqrt{2}z+1)(z^2-\sqrt{2}z+1).\\$The second part of the question is incorrect, the claimed identity does not hold for $x=0$ for example.$

$Okay, and $\omega^3=1\Rightarrow \omega^{-1}=\omega^2.$\\ The question isn't entirely correct though because they ignore the possibility of a division by zero.

$It doesn't.\\ \\$\frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2}=\omega^{-1}\frac{a\omega+b\omega^2+c\omega^3}{c+a\omega+b\omega^2}=\omega^{-1}.\\$\\ \\(Provided $c+a\omega+b\omega^2\neq 0$).$

(1-\cos x+2i\sin x)^{-1}=\frac{1-\cos x -2i\sin x}{(1-\cos x)^2+4(1-\cos x)(1+\cos x)}\\ \\=\frac{1-2i\frac{\sin x}{1-\cos x}}{1-\cos x + 4(1+\cos x)}.\\$ \\And:\\$\\\frac{\sin x}{1-\cos x}=\frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)}=\cot(x/2).\\ \\$Together these are enough.$

If you do not want to prove AM-GM for n=3 as a separate step, the second inequality follows from simplifying: a(b-c)^2+b(a-c)^2+c(a-b)^2\geq 0

Think carefully about the relationship between conics and their focii/directrices. Perhaps it might be easier to start by thinking about how to sketch the parabola with vertex at O and focus at (1,1). Also, the equations you have hopefully found will give you all information about the conics x...

Yes, these conics are oblique. Use the perpendicular distance formula and the focus-directrix definition of conics to find their equations and sketch them.

Minor correction: "M" in the last part is not 1, it just has absolute value at most 1.

Your Simpson's rule isn't quite correct, the first and last y-value are not meant to be multiplied by two.

It is a pretty immediate fact from the way the the trig functions and the exponentials are defined in higher mathematics (read about power series if you are really interested).

i^i=(e^{i \arg(i)})^i=e^{i^2\arg(i)}=e^{-(\frac{\pi}{2}+2\pi n)} It is basically just the usual index laws combined with the fact that: e^{i\theta}=\textrm{cis}(\theta)

Using the standard definition of complex powers as multi-valued functions: i^i=e^{-(\frac{\pi}{2}+2\pi n)} where n ranges over all integers.

Exactly, and your mention of the chaotic nature of the partial sums for certain x brings up an interesting point...consider the same problem if you replace the series by: \sum_{k=1}^n\frac{\sin(kx)}{k} (Considerably harder than the geometric series case)

$Evaluate the sum:$\\S_n:=1+x\cos(x)+\ldots+x^n\cos(nx)\\$where $x$ is real. A related question that is probably too difficult to be asked in the HSC is: ``For which values of $x$ does this series converge as $n\rightarrow\infty$?''$

$A more serious problem with Wights solution is that $I_n<n!$ does not imply that $I_n/n!\rightarrow 0.$ One way of showing that $I_n/n!\rightarrow 0$ is by using that \\$0\leq 1-\log(x)\leq 1 $ for 1\leq x \leq e$. \\$Hence:$\\ 0\leq\frac{1}{n!}I_n\leq\frac{1}{n!}\int_1^e 1\...

you haven't really shown that K>0 mathman. x-y\geq K$ and $x-y>0$ do not imply that $ K>0.