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The -t in the exponential should become -1/x , rather than 1/x.

You can also try a Venn Diagram approach. (Since there are only three sets here, it's easy to draw the diagram. If there are more, it gets harder to draw.) Also you can use the inclusion-exclusion formula early on (I think you essentially derived it). For three sets, it is $$|A \cup B \cup...

What do you mean by double sigma? As in a double sum? Usually these aren't encountered much in HSC.

I guess different schools would estimate differently. (Some schools have dodgy estimates.) One method could be to use statistical methods to see what ATAR someone with your marks from last year's cohort would get (or averaged over several previous cohorts).

$\noindent Since $u=\sqrt{1-x}$, we should have $\mathrm{d}u = -\frac{1}{2\sqrt{1-x}}\text{ d}x$.$

Re: MATH1131 help thread The expression in the limit is not a sum, but rather a product. This is because that symbol is capital Pi ($$\Pi$$), which is used for products. The symbol for summation is a capital Sigma ($$\Sigma$$).

Re: HSC 2016 4U Marathon $\noindent [Bernoulli's Inequality]$ $\noindent Let $x>-1$.$ $\noindent i) Show that $\left(1+x\right)^{r} \geq 1+rx$ if $r > 1$ or $r<0$.$ $\noindent ii) Show that $\left(1+x\right)^{r} \leq 1+rx$ if $0\leq r \leq 1$.$

I think you once said you could think in 11 dimensions or something. How do you manage to visualise four dimensions? Like can you visualise tesseracts?

Re: First Year Uni Calculus Marathon $\noindent i) Show that the $n$-th order Taylor polynomial for $\mathrm{e}^{x}$ about 0 is $P_{n} (x) = 1+x+\frac{x^2}{2!}+\cdots + \frac{x^{n}}{n!}$.$ $\noindent ii) Using the Lagrange form for the error term (or otherwise), show that the Taylor series...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Incidentally, part (iii) was clear without any calculations at all. We are told that the parabolic trajectory passes through the line y = x. This can clearly only happen if the projection angle was higher than 45 degrees to start...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Observe that$ $$\begin{align*}\frac{T_1}{1-\cot \alpha} &= \frac{\frac{2U}{g}\left(\sin \alpha - \cos \alpha \right)}{1 - \cot \alpha} \\ &=\frac{2U}{g} \times \frac{\sin \alpha - \cos \alpha}{1-\frac{\cos...

It ends up just being the p-test when you do a change of variables (substitute u = ln(x)). So it converges iff p > 1.

The Cauchy principal value would be 0. Otherwise it's an undefined expression -inf + inf. Don't need to use any comparison tests, we can find an antiderivative using inspection or a u-substitution.

The answer is (39/62)*15.

This is known as Gram-Schmidt orthogonalisation (or orthonormalisation if we normalise the vectors too). Basically the idea is that given a set of vectors, from this we create a new set that has the same span and the vectors are orthogonal (so we can obtain an orthogonal basis for a vector space...

$\noindent In summary, to solve linear least squares fitting problems like this, find the matrix of coefficients $A$ from the given data, as well as your vector $\bold{y}$ from the given data and model we are fitting. Let the vector of parameters to be fit be $\bold{x}$ (what we want to solve...

$\noindent So calculating $A^TA$ (note that it will be a symmetric 2 by 2 matrix), we have $A^T A = \begin{pmatrix}5 & 5 \\ 5 & 15\end{pmatrix}$ (this calculation is sped up by realising that $A^T A$ is symmetric, so we don't need to calculate each entry separately, and also that the $ij$ entry...

$\noindent The vector $\bold{x} = \left(\alpha,\beta\right)$ will be the vector $\bold{x}$ satisfying the \emph{normal equations} $A^T A \bold{x} = A^T \bold{y}$. So calculate $A^T A$ and $ A^T \bold{y}$, and then this just turns into solving a regular matrix equation of the form $M\bold{x}...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent I did include $\frac{\mathrm{d}y}{\mathrm{d}x}$ when differentiating the $\ln y$ with respect to $x$. I wrote $\frac{y^\prime}{y}$ for that, which is the same thing as writing $\frac{1}{y}...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread The y' is the dy/dx.