Hi!

Can someone please help with part (v) of this question (from the 1993 MX2 HSC paper)?

Also, just a question, I've heard (but I'm not entirely sure) that the question types in the MX2 HSC changed after 2001 - what exactly does this mean? I know they introduced MC in 2012 but I heard something about a change in 2001 as well.

Thanks

The main thing is that from 2001 onwards, mark allocation for each part of a question was promised.

But apart from that, it's especially true with 4U that many past papers liked to test some perhaps common concepts, but in absurdly weird ways in the final exam. (And of course, they were notably more time consuming.) For a question like this, the notation is somewhat peculiar, and the use of graphing in ii) and iii) seems a bit out of place to me. It turns out that part v) is just a recursive argument, but I'm not sure if I've ever seen that kind of pattern in recent papers.

\[ \textbf{1993 HSC 4U Additional} \]

Note that \(n\) is a fixed integer for this question.

\[ \text{By immediately using our existence assumption that }\\ x_n = \cot \theta_n\text{ for some }0<\theta_n<\pi,\text{ we obtain} \\ \boxed{\cot \theta_1 = \cot \theta_{n+1}}. \]

\[ \text{Now upon applying the recursive relationship once, we obtain}\\ \boxed{\cot\theta_1 = \cot 2\theta_n}.\\ \text{But applying it again, we obtain}\\ \boxed{\cot \theta_1 = \cot 2^2\theta_{n-1}}.\\ \text{And just to make it clearer, explicitly applying it once more gives}\\ \boxed{\cot \theta_1= \cot 2^3 \theta_{n-2}}. \]

\[ \text{The idea is that upon further repeated use of this recursion,}\\ \text{we will eventually arrive at }\boxed{\cot \theta_1 = \cot 2^n\theta_1 }. \]

One way of tracking the required power of 2 is to observe that the power plus the index on \(\theta\) must equal to \(n+1\).

\[ \text{At this point, it is now perfectly fine for us to use the general solution}\\ \text{in an attempt to track down all solutions.}\\ \text{Solving }\cot (2^n \theta_1) = \cot\theta_1 \text{ gives}\\ 2^n\theta_1 = k\pi + \theta_1 \implies \boxed{\theta_1 = \frac{k\pi}{2^n-1}}\\ \text{where }k\text{ is an integer.} \]

Note: If you put the \(k\pi\) on the other side, you'd end up with \( \theta_1 = \frac{-k\pi}{2^n-1}\), But that's all good - just sub out \(\ell = -k\) and work with \(\ell\) instead.

\[ \text{But we need to track down which values of }k\text{ give values for }\theta_1\text{ that are not out of bounds.}\\ \text{Recall that we still require }0< \theta_1 < \pi. \]

\[ \text{Subbing into that expression gives}\\ 0 < \frac{k\pi}{2^n-1} < \pi \implies \boxed{0 < k < 2^n -1} \]

Note: Of course, we're assuming that \(2^n -1 > 0\), since \(n \geq 1\). This assumption on \(n\) is not explicitly stated in the question, but is implicitly assumed to be the lowest index.

\[ \text{Hence our valid values for }k\text{ are}\\ k = 1, 2, 3, \dots, 2^n - 2.\\ \text{Which means our final answers are}\\ x_1 = \cot \left( \frac{\pi}{2^n - 1} \right), \, \cot \left( \frac{2\pi}{2^n - 1}\right), \dots, \cot \left( \frac{ (2^n-2)\pi}{2^n -1} \right). \]

Aside: For the interested reader, this question basically shows one special class of consequences when Newton's method is applied on a function that has no roots to begin with. Newton's method is designed with the intention of converging to a root once used often enough, so if we don't have a root, well where

*does* it converge to?

More often than not, at each iteration we just get some debatably randomised value appearing. In other words, it definitely does not

*converge*. But this question basically asks - is it possible for the values to start

**cycling** at some point? That is, could Newton's method somehow exhibit

**periodic** behaviour? The question more or less answers yes to this question, but challenges us by asking what initial values must we

*start* with, to exhibit a period of \(n\).