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Re: MATH1231/1241/1251 SOS Thread $\noindent To prove it algebraically (noting the matrices are just $2$-by-$2$ since it says rotation in the \emph{plane}, so we're rotating two-dimensional vectors), recall that the rotation matrix $A_{\alpha}$ for rotating counter-clockwise by angle $\alpha$...

Re: MATH1231/1241/1251 SOS Thread $\noindent Note $A_{\phi}$ rotates a point by $\phi$ and $A_{\theta}$ rotates by $\theta$. So if we multiply a vector by $A_{\phi}A_{\theta}$, we're rotating it first by $\theta$ and then by $\phi$, so overall all we've done is rotate it by $\phi+\theta$...

Re: Discrete Maths Sem 2 2016 a) It means at least one suit, i.e. the hand has a suit with at least four cards from that suit. By the pigeonhole principle, this must be the case. c) There need not be a suit with 5+ cards from it. Note ceil(13/4) = ceil(3.25) = 4 (not 5). So there must be a...

The formula is actually with absolute values around the fraction. So the (-1)^k goes away. The radius of convergence can't be negative, it is by definition either 0, a positive real number, or +oo. (Further reading: https://en.wikipedia.org/wiki/Radius_of_convergence )

Lemma. For all sufficiently large k, ln k < k^{1/6}. (In fact: for any given positive alpha and beta, (ln k)^{beta} will be less than k^{alpha} for all sufficiently large k. And not just less than, but 'little oh' of as k -> oo, in fact.) Proof. Exercise. (E.g. L'Hôpital's rule to prove the...

Yes, convergent by the alternating series test (https://en.wikipedia.org/wiki/Alternating_series_test ).

You could use induction. Have you tried this? (The statement given is really just equivalent to saying that the sequence is strictly decreasing.)

Basically inner radius (r) and outer radius (R) in an annulus.

Re: Discrete Maths Sem 2 2016 Having the eight-card hand have exactly three cards in exactly one suit (say suit 1) means we have exactly three cards from 1, and so the remaining five cards in the hand do not contain anything from suit 1. So the remaining five cards come from the other suits...

Re: MATH1231/1241/1251 SOS Thread $\noindent If you want to do it using the characteristic equation rather than by inspection as done above, note that the characteristic equation is $\lambda^2 + 4 = 0$ (because the ODE has no $y'$ term, so there is no $\lambda$ term in the characteristic...

Re: MATH1231/1241/1251 SOS Thread $\noindent The ODE is $y'' = -4y$, which as we know from the Simple Harmonic Motion topic of HSC 3U maths has general solution $y = A\cos (2t) + B\sin (2t)$, for some arbitrary constants $A,B$.$

Maybe the reason they wrote the square roots thing is just so that people doing the Q. would realise they're finding the limit of that. Some students might not have noticed otherwise and would think they're just finding the limit of a random sequence and forget about it quickly. I think in...

It's a famous property of the Golden Ratio (which is a famous number with many interesting properties and also appears in nature a lot apparently. See the Wikipedia page: https://en.wikipedia.org/wiki/Golden_ratio ). I just commented it was the Golden Ratio because I thought some readers may be...

$\noindent b) I said what I wrote as a hint (to give the basic idea). It should be proved by induction (as the Q. says) and written up properly when you do it. The base case is trivial to check that it's true, and for the inductive step, $a_{k+1} = \sqrt{1 + a_k}\in \left[1,2\right]$, from part...

Here's some hints. $\noindent b) It follows immediately from a) and induction that $1\leq a_n \leq 2$ for all $n$.$ $\noindent c) Inductive step: $a_{n+1} = \sqrt{1 + a_n} \geq \sqrt{1 + a_{n-1}} = a_n$. The inequality comes from the inductive hypothesis $a_n \geq a_{n-1}$ and since the...

Re: Discrete Maths Sem 2 2016 You could try letting Sj be the set of all hands such that there are exactly three cards from suit j. We want to find |S1 U S2 U S3 U S4|.

No.

Re: HSC 2016 4U Marathon Well done!

Re: HSC 2016 4U Marathon Here's another Q. $\noindent Let $n \geq 2$ be a positive integer and $f(x)$ a real-valued function such that $f(x)\to \infty$ as $x\to \infty$. Let $c$ be a real constant.$ $\noindent Must $\sqrt[n]{f(x)+c} - \sqrt[n]{f(x)} \to 0$ as $x \to \infty$? Justify your answer.$

$\noindent We aren't supposed to solve those two equations simultaneously. Rather, those two equations \emph{are} the locus we want, i.e. the locus is the pair of intersecting lines $x=-7y=-9$ or $7x+y-5 = 0$. Geometrically, the locus is the lines that are the angle bisectors for the angles...