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Re: HSC 2016 2U Marathon Correct! I'll leave the algebraic exercise as an exercise for 2U students.

$\noindent The general solution formulas do always apply, it's just that sometimes they can be simplified. For the $\cos x=0$ one, the general formula says the solutions are $90^\circ + 360^\circ k$ or $-90^\circ + 360^\circ k$ for arbitrary integer $k$. This is exactly the same set of solutions...

It's a pretty classic example of uncorrelated yet dependent random variables (I think it's the example given on a Wikipedia page too, and several other places, for illustrating this phenomenon.)

Remark. This question gives an example of a pair of random variables (X and X^2) that are dependent yet uncorrelated. Thus uncorrelated does not imply independent. We know though that if two r.v.'s are independent, then they are uncorrelated. Thus independence is stronger than uncorrelatedness.

$\noindent When $r$ is even, we have $r = 2k$ for some positive integer $k$. Then the moment is $\frac{\Gamma\left(1+2k\right)}{2^k \Gamma \left(1+k\right)} = \frac{(2k)!}{2^{k} k!}$. You can simplify $\frac{(2k)!}{k!}$ by the way. I'll leave that to you.$ (Don't need to worry about even...

Note all odd moments are automatically 0 since X has a density that is symmetric about 0. In other words, the pdf is even, and x^r is odd if r is odd, so x^r * (pdf) is an odd function when r is odd, making the integral for the expected value 0.

Re: MATH1231/1241/1251 SOS Thread It doesn't. We need at least three vectors to span R3. $\noindent (If you were willing to believe it's a basis, you must have already believed it was spanning. Remember, a basis is a \emph{linearly independent spanning set}. So it's a set that's \emph{both}...

$\noindent It doesn't matter, you could use any form of the double angle formula for it. Whichever one you used, your aim would be to show the denominator is $s$ times the numerator. You'd be able to still do this by tweaking what I showed the denominator is with the use of the relationship $c^2...

$\noindent Yeah they probably typoed and weren't meant to put a square root (or just messed up completely). Their answer is the answer to the same integral without the square root present. None of the options is correct for the integral as written. The answer to the given integral is $-5\ln...

$\noindent For simplicity, let $s \equiv \sin \alpha$ and $c \equiv \cos \alpha$. Call the L.H.S. of the identity we are asked to prove $L$ and the R.H.S. $R$. Then$ $$\begin{align*}L &= \frac{2sc + 2c^2 -1}{2c + s -2\left(c^3 + s^3 \right)}\quad (\text{double angle formulas in...

Re: HSC 2016 2U Marathon $\noindent \textbf{NEW QUESTION}$ $\noindent Show that for any positive numbers $a,b,c,d$, the number $\frac{a+c}{b+d}$ is strictly between $\frac{a}{b}$ and $\frac{c}{d}$.$ $\noindent \textbf{Bonus.} Provide a geometric interpretation of this result.$

Re: Discrete Maths Sem 2 2016 The given solution counts |S1| + |S2| + |S3| + |S4|, where Sj is the set of eight-card hands so that there is no card from suit j in the hand (j = 1,2,3,4). But what we want is: |S1 U S2 U S3 U S4| (which is the no. of eight-card hands with suit 1 missing or suit...

Re: Discrete Maths Sem 2 2016 $\noindent Note there is a pair of A's and the other letters are all single letters. Total of $8$ letters, consisting of $2$ A's and $6$ single letters. If we take both A's, the no. of words we could form is $3\times 6 = 18$. If we take exactly one A, we can form...

Re: First Year Uni Calculus Marathon Yeah, take midpoints (i.e. https://en.wikipedia.org/wiki/Riemann_sum#Middle_sum ).

$\noindent \textbf{Remarks.} Note that this Vandermonde Matrix is exactly the coefficient matrix we'd get if we were trying to fit an (at-most) $(n-1)$ degree polynomial $a_{0} + a_1 t + \dots + a_{n-1}t^{n-1}$ to the points $\left(t_{1},y_{1}\right),\ldots , \left(t_{n},y_{n}\right)$, where the...

[This is not an original idea by me, this is a well-known problem with many solutions.] $\noindent For positive integers $n$, let $V_{n} := \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 &...

Another classic way is to use induction and view the determinant as a polynomial.

Here are some hints/answers. $\noindent (i) The answer is $\frac{x}{\sqrt{x^2 +1}}$ (use the chain rule). $ $\noindent (ii) Use quotient rule.$ $\noindent Third Q.: Find the slope of the tangent at that point using differentiation. Find the $y$-value of the point in the curve there by...

The answer is just the perpendicular bisector of A and B. We can find the equation of this line by finding the midpoint of A and B, and then finding the slope of the line (which will be the negative reciprocal of the slope of AB, since it's perpendicular to AB). Once we've worked out the slope...

$\noindent Splitting up the fraction, it becomes $ $$\begin{align*}\frac{1}{\cos^2 \theta} -\frac{\sin^2 \theta \cos^2 \theta }{\cos^2 \theta} &= \sec^2 \theta -\sin^2 \theta \\ &= \tan^2 \theta + 1 -\sin^2 \theta \\ &= \tan^2 \theta + \cos^2 \theta.\end{align*}$$