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Re: HSC 2016 4U Marathon $\noindent Let $P(z)$ be a complex polynomial of degree $n\geq 1$, with constant term $1$. Show that $P(z)$ can be written in the form $ \left(b_1 z+ 1\right) \left(b_{2}z+1\right) \ldots \left(b _{n}z +1\right)$ for some choice of complex constants $b_{1},b_{2},\ldots...

Why is there a huge risk? Plus, some people may actually like those high-scaling subjects in the first place.

Re: MATH1231/1241/1251 SOS Thread You need to write up proofs in online quizzes?

Re: MATH1231/1241/1251 SOS Thread This is famous result (convergence of ratio test implies convergence of root test). A proof may be found here: http://math.stackexchange.com/questions/287932/convergence-of-ratio-test-implies-convergence-of-the-root-test . (The converse of this result...

As in other non-maths/stats fields. See this post by RealiseNothing:

Not all of those medals were actually for maths.

Re: Markovnikov rule? A lot of HSC courses got dumbed down around 2001 when the Board of Studies changed the syllabuses.

Re: Markovnikov rule? That would be 'too hard'.

4U Polynomials probably.

Don't need minus, since m is just an integer, so it can be negative too, taking care of your worry about the minus.

You seemed to have gone the other way. The Q. basically wants us to assume those conditions on s and t, and show that x is a solution if this is the case. If s and t satisfy those conditions, then your line before you 'equated coefficients' is clearly a true statement, and then reversing your...

Use E = F/q.

Re: MATH1231/1241/1251 SOS Thread $\noindent If you're not sure about how to use Leibniz's rule when $x$ is in both the bounds of integration and the integrand, note that in this case we can get around this by writing the integral as $x \int _b ^x y(t)\, \mathrm{d}t - \int _b ^x ty(t)\...

Re: MATH1231/1241/1251 SOS Thread $\noindent Hint: differentiate both sides to obtain an ODE to solve. To get an intial condition (so that we have an IVP), put $x= 1$ in the given integral equation (so that the integral will vanish, leaving us with the intial condition $y(1) = 2$).$...

Re: MATH1131 help thread Never seen this before.

Welcome to the HSC.

Re: Discrete Maths Sem 2 2016 It's basically intuitive as follows. I'll illustrate it as though q = 5, but same idea works for general q of course. Write down the remainder from each stage of the division algorithm (these have to be whole numbers from 0 to 4, since we're dividing by 5). It's...

Re: Discrete Maths Sem 2 2016 Last 'why' is basically pigeonhole principle. When we are doing each step in the long division, the remainders by definition can only come from {0, 1, ..., q-1}, which is a finite set. So as we keep going through the procedure, since we get a remainder from this...

Re: Discrete Maths Sem 2 2016 That doesn't really help, does it? The exponent here is bigger than 1.

Re: Discrete Maths Sem 2 2016 How so?