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Re: UNSW chit chat thread 2016 If the math marks really are being released in just 10 days from now, how quickly do they mark those papers? Sent from my iPhone using Tapatalk

The question isn't hard but I just want to see it get done because I want to see how you'd set out your working \text{Q: Solve }296x \equiv 8 (\mod 692) I'll put down the relevant Euclidean algorithm to save a bit of work \begin{align*}692&=2\times 296+100\\ 296&=2\times 100+96\\...

Re: UNSW chit chat thread 2016 I want my break to start already Why is tomorrow so close yet so far

I thought you were the one that hated long division Though yeah that never crosses my mind with algebra anymore unless I explicitly see polynomials

\text{a) }\binom{13}{1}\binom43\binom11\binom{12}{7} \text{b) }\binom41\binom{13}3\binom{39}7-\binom42\binom{13}{3}^2\binom{26}{4}+\binom{4}{3} \binom{13}{3} ^3\binom{13}{1} With these perms and combs I'd appreciate anyone checking but it's also (hopefully) for Drsoccerball's convenience

Man... need Peter Brown though... cause that's so not obvious to induce the diff. of two squares and the add/deduct trick

Oops. Well this was the question and the let us assume a result (potentially true upon the basis of primality though)

Sorry I know I can do this myself but I don't have time to right now. Can someone prove this? a \mid b^2 \implies a \mid b

In that case, I now have 8\times\frac{7!}{3!2!2!}-\left(6\times5\times\frac{4!}{2!}-\frac{4!}{2!}\right)

Good observation. Also in my working that + should've been a minus but I need to do redo the whole thing anyway cause of the incl/excl

Consider the 10 letter word PARRAMATTA and all the words formed by rearranging its letters. How many of these words contain the subword MAP but not the subword RAT? They gave no answer so I'm asking for checking: 8 \times \frac{7!}{3!2!2!}+ 6\times5\times \frac{4!}{2!}

Hmm interesting, the values of k are going down. But the more obvious pattern of k going down by 1 seems to halt... \text{Wait a minute. Are our holes:}\\ S_1=\{2^k(1)\}\\ S_3=\{2^k(3)\}\\ \vdots\\ S_{25}=\{2^k(25)\}

Suppose that 26 integers are chosen from the set S={1, 2, ..., 50} By writing these numbers as 2km with m odd, prove that one of the chosen numbers is a multiple of another of the chosen numbers.

Interpreting the MATLAB output: It's just the RREF of the matrix representation augmented with the identity. i.e. the RREF of (\textbf{v}_1 \,\textbf{v}_2\, \textbf{v}_3\, \textbf{v}_4\mid \textbf{e}_1\, \textbf{e}_2\, \textbf{e}_3\, \textbf{e}_4) How is the output supposed to help me do c)...

Parts a), b) and c) are purely for reference Here's a proof to part d) (I) also for the sake of reference \text{Suppose w.l.o.g. }\gamma\text{ represents the unit circle on the Argand plane}\\ \text{with its centre }O\text{ being the origin.} \text{Take }A\text{ as the point...

Actually whilst maybe not as strongly felt, I did think the same way he did

\text{Let }f(x)=e^{2x}(1-x)^{-1}\\ \text{a) By multiplying the Maclaurin series for }e^{2x}\text{ and }(1-x)^{-1}\\ \text{obtain an expression for the coefficient }a_n\text{ in the Maclaurin series}\\ f(x)=\sum_{k=1}^\infty e^{2x}(1-x)^{-1} \text{I don't remember the formula so I just did it...

Tf = f^{\prime \prime} - f^\prime\text{ is a linear map and consider}\\ \text{the ordered basis }B=\{ \textbf{1}, \sinh, \cosh\} \text{Note that }\textbf{1}(x)\equiv 1 \forall x \text{In c) A matrix representation of }T\text{ w.r.t. }B\text{ is}\\ A=\begin{pmatrix}0&0&0\\ 0&1&-1\\...

I've forgotten how to do this, because it's for a sequence and an integer N, so I forgot if we need floor or ceiling function (or rather tired and can't work it out from my intuition) \text{Prove, from the definition, for all }n\in \mathbb{N}\\ \lim_{n\to \infty}{\frac{n\sin(n^2)...

I always get confused when they map to a matrix \text{Let }T:\mathbb{R}^3\to M_{22}(\mathbb{R})\text{, a linear mapping, be defined}\\ T(\textbf{x})=\begin{pmatrix}x_1-x_2&x_2-x_3\\x_3-x_1&0\end{pmatrix} \text{c) Find a basis for im}(T)