Recent content by Canteen

Here's my attempt. I've probably over-complicated it or am just plain wrong: Base case (n=0) – Trivial Inductive hypothesis – Assume true for some non-negative integer k: \text{i.e. } \exists p, q \in \mathbb{Z} : w_{2k} = 11 p \rightarrow u_{2k} = 11q \qquad (1) Inductive step – RTP true...

Your comment and a few recent articles by Derek Buchanan at 4unitmaths.com have both mentioned the beta and gamma functions as alternative, more efficient, approaches to certain questions. Furthermore, in his most recent article 'Alternative solutions to 2023 HSC Extension 1 Q14ci' Derek states...

To get this thread going again I thought I'd give one of these a go. Letting u = \tfrac{\pi}{2} - x: \begin{aligned}I &= \int_{0}^{\pi}\left(3\pi-\cos x-2x\right)\left(\frac{x}{1+\sin x}\right)^{2}dx \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(2\pi-\sin u + 2u...

Thanks for that, I might properly add the edits in later so it looks more polished

Yes you are quite right, I somehow thought 3^2 was 6 o_O

Here are my typeset solutions to this year's MX1 HSC exam. I wrote these in a great rush so there are probably mistakes lol

Lol. I can't type into my calculator. I did: 252\times4\times4=4032 Should be: 252\times24\times24=145152 as I had above.

oops forgot to arrange them around the tables! That would make it: 252\times (5-1)!\times(5-1)!=4032 I think that should be correct.

Choose 5 people to sit at the mahogany then let the rest sit at the oak: \binom{10}{5}=252

Technically they are not the "cube roots of unity" but simply the "cube roots".

2. |z|^{\frac13}=(\sqrt{1^2+1^2})^{\frac13}=\sqrt[6]{2} \frac13\textrm{arg}z=\frac13\arctan(1)=\frac\pi{12} So the "primary root" will be z=\sqrt[6]{2}\textrm{cis}\frac{\pi}{12} 3. Note that 1/3arg(z) can also take other values namely: \frac13\textrm{arg}z=\frac{\pi}{12}+\frac{2\pi}{3}\textrm{...

I vaguely remember this being a property of cubic polynomials with three distinct roots.... I think its because the vertex of a parabola (the first derivative) occurs at an x value which is the average of the x intercepts (which will correspond to the max and mins on the cubic). As the vertex of...

I think this is what it means. 1. 14\ddot{x} =- \frac65v \frac{dv}{dt}=-\frac3{35}v \int_{21}^V\frac{dv}{v} =-\int_0^{11}\frac3{35}dt\ \ (\textrm{where }V\textrm{ is the final velocity}) [\ln|x|]_{21}^V=-\left[\frac3{35}t\right]_0^{11} \ln\left(\frac{V}{21}\right)=-\frac{33}{35} \therefore...

\frac{F}{\ell}=k\frac{I^2}{d} \implies I=\sqrt{\frac{Fd}{\ell k}}\ \ \left(\textrm{where } k\textrm{ is } \frac{\mu_0}{2\pi}\right) So we want something that looks like y=\sqrt{x} making the answer D

My teacher told us that whenever a question requires you to provide a personal response/reading a strong personal voice should be adopted i.e. use "I" not "we". This really emphasizes that it is you who is making an informed response to the text you are studying.