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Check local library

1. By inspection: RHS = 2^n * 3^n * 4^n * ... * (n+1)^n LHS = 2^n * 3^(n-1) * 4^(n-1) * 5^(n-2) * 6^(n-2) * ... * (2n-1) * 2n = 2^n * [3^(n-1) * 2n] * [4^(n-1) * (2n-1)] * [5^(n-1) * (2n-2)^2] * [4^(n-1) * (2n-3)^2] * ... Arranged this...

In practice yes. But would be nice if you can highlight the important parts

That's useless... it's essentially saying tan(10) = sin(10)/cos(10)

$let $y_n = x_n - 1$, then $y_n > 0$ and the inequality is \empha{equivalent} to \\ $ \prod_{n} (1+y_n) \geq 1 + \sum_{n} y_n$ \\expanding the left hand side shows that it is \emph{equivalent} to\\ $1 + \sum_{n}y_n + \sum_{\alpha,\beta:\\ \alpha \neq \beta} y_\alpha y_\beta +...

It's just notation.. I don't see anything wrong with cis.. but e^(ix) is probably the better notation.

Re: I just need the 01 and 08 James Ruse papers now. What is this obsession with papers

It's not quite a complex number question... roughly: you can use a counting argument 1. Notice that z^n =1 has n distinct roots and that n is a finite number. 2. also, z_k, z_k^2 , ... , z_k^n are all roots of z_k^n = 1 (trivial) 3. We just need to show that no two in the above list are equal...

Search for the report from the technical scaling committee each year

It is possible, but if you have to ask this question, you will probably need osme help

Classic Proof for the Cauchy Schwarz inequality. You can also substitute: x = \dfrac{ab+cd}{a^2 + c^2} into f(x) Now try this Use the fact that f(x)=|ax-b|^2+|cx-d|^2\geq 0 to prove |a\bar{b}+c\bar{d}|\leq \sqrt{(|a|^2+|c|^2)(|b|^2+|d|^2)} where a,b,c,d are complex numbers

this has nothing to do with the question

For n = 1,2,3,4 find the number of ways of placing n letters in n envelops such that no letter is in the correct envelope ( assume the letters and envelopes form pairs) Prove that in general for n >= 0 the following formula gives the number of ways n!\sum_{k=0}^n \dfrac{(-1)^k}{k!}

prove that a cube can be divided into exactly n cubes for each n >= 100, (the sizes of the cubes need not be the same)

measure of how much the curve differs from a circle.

doesn't matter in cases where your mass in constant.

just substitute... cos(pi/2) = 0 and a2 - a1 is what is there

the devil's in the details

I've been reading up about different career paths available upon completion of a Commerce/Business degree. Investment Banking seems to be a well regarded career option in terms of salary, opportunity to advance to higher managerial positions within companies and chances to network with high...

The one which is harder to get in... despite all this BS about how UAI/ATAR cutoffs has nothing to do with the course blah blah. ---> people aren't stupid, the good courses has more applicants which pushes the cut offs up. Maybe this is not entirely valid for different courses, but it is...