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$\noindent Yeah, least squares can be used. Least squares just refers to choosing values of parameters ($A$ and $\lambda$ here) so that the \emph{sum of squared (vertical) distances} (with these vertical distances being called \textsl{residuals} in statistics) between the graph with these...

$\noindent By inspection, the solution is $c_1 \cos \omega t + c_2 \sin \omega t$, if $\omega >0$. (Standard SHM solution.)$ $\noindent To derive this via the proposed substitution (though it's faster to just use the standard method of solving second-order ODE's with constant coefficients using...

$\noindent Let $t$ be time measured in years and $y$ the population. The population growth rate in this question is referring to the instantaneous proportional growth rate. If this were a constant $k$, then we would have $y^\prime = ky$ (standard growth/decay ODE when the (instantaneous...

It approaches K. leehuan typoed the solution, it should be +y0 in the denominator. (To see this, note that if we sub. in t = 0, we are supposed to get y = y0.)

$\noindent (Note $k>0$.) If $0<y_0 < K$, then the coefficient of the decaying exponential in the denominator, namely $K-y_0$, is a \emph{positive number}. This means the denominator is a strictly decreasing function, so the overall solution is strictly increasing.$ $\noindent If instead $y_0 >...

The given parameterisation will give an ''upside-down'' parabola if and only if a is negative. So maybe there's a mistake in the parameterisation or a is negative. What was the original question?

$\noindent You did $\exp \left(\int \frac{1}{2x}\, \mathrm{d}x\right)$ for the integrating factor, but you forgot the minus sign that is present in the DE. So the integrating factor is actually $\exp \left(\int -\frac{1}{2x}\, \mathrm{d}x\right) = \text{ reciprocal of what you got}$.$

Re: MATH1131 help thread In that case, alpha could be anything whatsoever, since it doesn't appear in any of the equations that the matrix would represent (alternatively, the first column would be non-leading then, so alpha would be set to a parameter). So in that case, if you wanted a...

$\noindent Since $x = -2at$, we have $t = -\frac{x}{2a}$. So $t^2 = \frac{x^2}{4a^2}$. Thus$ $$\begin{align*}y = at^2 + 4a &= a \times \frac{x^2}{4a^2} + 4a \\ \Rightarrow y &= \frac{x^2}{4a} + 4a.\end{align*}$$

Re: MATH1131 help thread Yeah you can do it via back substitution (or just inspection). As usual, the matrix U represents linear equations in the variables alpha, beta, gamma. (First column is for coefficients of alpha, second for beta, and third for gamma.) So they're asking for a non-zero...

Re: MATH1131 help thread Looks like they want you to give a non-zero solution for alpha, beta, gamma.

Re: MATH1231/1241/1251 SOS Thread Suppose by way of contradiction there was a root "a" with |a| > 3. Since a is a root, we have a^6 = 3a - a^3 + 5a^4. Taking modulus of both sides, |a^6| = |3a - a^3 + 5a^4|. But this contradicts the result of part d).

Re: MATH1231/1241/1251 SOS Thread $\noindent Realised I didn't do it by doing the $y = z + z^{-1}$ method. Using this method, you got it down to $y^{3} -y + 1=0$, where $z$ is a root of the equation $z^6 + z^3 + 1 = 0$. Since the roots of this equation produce exactly three values of $z+...

Re: MATH1231/1241/1251 SOS Thread $\noindent As you have already proven, the roots of $p(z):= z^{6} + z^{3} + 1 = 0$ are $z_1 = e^{i \frac{2\pi}{9}}$, $z_2 = e^{i \frac{4\pi}{9}}$, $z_{3} = e^{i \frac{8\pi}{9}}$, and the conjugates of these, $\overline{z_j}$, $j=1,2,3$.$ $\noindent Note that...

Re: Discrete Maths Sem 2 2016 Since x is an element of B, and there is no element that is simultaneously in both A and B (since A intersection B is empty), x can't be in A. (I.e. If x were in A, it'd be an element in both A and B, but this is impossible.)

$\noindent Note that we need to assume $n\geq 1$ and $N \geq 2$ for this question to make sense (since we picked two students so the pool size is at least 2, and we know there is at least one JR student, since we are told we picked one).$ $\noindent The total no. of ways to pick two students is...

$\noindent For the second one, note that if $k=0$, the polynomial is $P(x) = x^3 -8$, which doesn't have only real roots (it factors as $\left(x-2\right)\left(x^2 + 2x + 4\right)$, and this quadratic has no real roots, so $P(x)$ has some non-real roots. Alternatively, if you know that cubics...

Quick way for the first one: in the extreme case where n = N (i.e. everyone in the pool is a James Ruse student), we are guaranteed that the next student picked is from JR, so the probability would be 1. The only option that is 1 when n = N is option (A). Thus the answer is (A).

Pretty sure you are allowed to use the chain rule by inspection, aren't you (depends on the question maybe)? You can write "chain rule" in brackets as a reason.

$\noindent We have the parametric equations for $M(x,y)$ as $x = \frac{t}{2}$, $y = \frac{1}{2}(4t+3)$, for $t\in \mathbb{R}$. From the $x$ equation, we have $t = 2x$. Substituting into the $y$ equation, we find$ $$\begin{align*}y &= \frac{1}{2}\left(4\times 2x + 3\right)\\ &= 4x +...