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Physics without maths isn't actual physics.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Yes, if we know that the process goes on indefinitely (i.e. the root wasn't 0, or 2, or 3, or 1, or … (exercise: classify exactly which numbers will result in a finite run only)), then the total distance travelled is just the...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread It's asking for total distance moved using halving the interval. E.g. Say we're on the interval [0,4] and we start at 0. Say the root happens to be at 1.245 (but we don't know this). First we'd move to point 2. Then we'd find...

Re: HSC 2016 3U Marathon If you take a common denominator, your answer will simplify to the book's one.

Re: International Baccalaureate Marathon 2016 If we sub. in x = pi/2 to the final answer, since sin(pi/2) = 1, we get -32 + 72 -24 + 5 = 21, which is outside the range of the LHS, which is sin(5x), which is between -1 and 1 always. This is the check that parad0xica mentioned in an earlier...

Well I do remember articles like this when the PISA results came out a few years ago: http://www.abc.net.au/news/2013-12-03/australian-students-slipping-behind-in-maths-reading/5132526 .

Re: International Baccalaureate Marathon 2016 Yep.

Re: International Baccalaureate Marathon 2016 $\noindent Actually realised they probably expected you to do that one without use of complex numbers. One way to do this is to expand $\sin 5x = \sin \left(4x+x\right) = \sin 4x \cos x + \cos 4x \sin x$. Now, $\sin 4x = 2\sin 2x \cos 2x$ and $\cos...

Re: International Baccalaureate Marathon 2016 $\noindent Note that $\sin 5\theta = \Im \left(z^5\right)$, where $z=\cos \theta + i\sin \theta$ (due to De Moivre's theorem). But $z^5 = \left(\cos \theta +i\sin \theta\right)^5$ can be expanded using the Binomial Theorem, and its imaginary part...

$\noindent The geometric mean of two non-negative real numbers $a$ and $b$ is defined as $\sqrt{ab}$.$

Re: International Baccalaureate Marathon 2016 $\noindent The presence of the $2$ outside the brackets seems to be a typo there. It should just be without that $2$ I think (in other words the given recurrence equation at the start of the question).$

Re: HSC 2016 2U Marathon Well it's basically just the same thing.

Re: MATH1131 help thread It doesn't matter. I.e. if x(n) < ƒ(n) < y(n) for all positive integers n (or after some point at least), and x(n) and y(n) both tend to a as n → ∞, then ƒ(n) also tends to a; and this statement holds also if some of those inequality signs were strict.

Check the post I wrote where I did your Q1, I wrote it there.

Like I said before, it is impractical to try inverting that function. The answer is just 1/h'(0), because h(0) = 0 (a = 0 and b = 0 in the Inverse Function Theorem formula). Since h'(0) = -3, the answer is -1/3.

I know. I got that (recall I said you could change the y's to x at the end).

$\noindent It's probably enough to just leave it as $\frac{1}{3}x^{-\frac{2}{3}}$.$

$\noindent Oh yeah I typoed (forgot to subtract the 1 from the $x$). It should indeed be: $g^\prime (y) = \frac{1}{2\sqrt{y-3}}$.$

Basically 'defined as equal to' here.

$\noindent Use the fact (Inverse Function Theorem) that if $b=f(a)$, $g\equiv f^{-1}$, and $f$ is differentiable at $a$ with non-zero derivative, then $g^\prime (b) = \frac{1}{f^\prime (a)}$.$ $\noindent 1) Note $f^\prime (x) = 2(x-1), x >1$. Note the range of $f$ is $\left \{y\in \mathbb{R}:y...