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Re: HSC 2015 4U Marathon $AM-GM in $n$ variables tells us:\\ \\ $\textrm{AM}(x_1,\ldots,x_{n-1})^{(n-1)/n}\cdot x_n^{1/n}\leq \frac{(n-1)\cdot \textrm{AM}(x_1,\ldots,x_{n-1}) +x_n}{n}=\textrm{AM}(x_1,\ldots,x_n).$\\ \\ Dividing both sides by $\textrm{GM}(x_1,\ldots,x_n)$ and raising to the...

Re: HSC 2015 4U Marathon edited before I saw your reply haha. just wanted to get it down quickly and then fix typos.

Re: HSC 2015 4U Marathon Oh, my bad. (Sorry) This quantity can then be simplified. \sin(\theta)\cos(\theta)+\sin(\phi)\cos(\phi)\\ =\frac{1}{2}(\sin(2\theta)+\sin(2\phi))\\ =\sin(\theta+\phi)\cos(\theta-\phi)\\=0.

Re: HSC 2015 4U Marathon I really do not think this is the case. Let (a,b)=(\cos(\theta),\sin(\theta)) and (c,d)=(\cos(\phi),\sin(\phi)) The assumption is then that ac+bd=\cos(\theta-\phi)=0. One way to attain this is by taking \theta=\phi+\pi/2. Then...

Re: HSC 2015 4U Marathon I don't think those conditions uniquely determine ab + cd...

Re: HSC 2015 4U Marathon Well when we stop talking about integer power we start getting multivalued. A full statement of the theorem is then slightly clunkier, but could still be obtained by using the integer power version to obtain the rational power version, and using the continuity of the...

Re: HSC 2015 4U Marathon Lol, De Moivre's theorem is usually proved USING the trig addition formulae. This is quite circular.

Re: 2015 permutation X2 marathon A two player variant of poker is played as follows. There are three cards (A,K,Q) in the deck and the two players are each dealt one of these cards randomly. (They can see their card but not their opponents.) Both players are forced to put $A into the pot...

Re: 2015 permutation X2 marathon In my interpretation, A chooses first, and B chooses second. First observe that B might as well choose a number adjacent to A's. This is because if there is a gap between A and B (say B = A + 2), then B loses value on the numbers strictly between A and B...

Re: HSC 2015 4U Marathon What I wrote is 100% true, the even partial sums are upper bounds and the odd partial sums are lower bounds. Also, if you only proved the claim about upper bounds this alone would not establish convergence to the sine function. And because of the way I set it up, I...

Re: HSC 2015 4U Marathon $i) Define:\\ \\ $P_n(x)=\sum_{k=0}^{n} (-1)^k\frac{x^{2k+1}}{(2k+1)!}.$\\ \\ We claim: \\ \\$P_{2n+1}(x)\leq \sin(x) \leq P_{2n}(x)$ for all non-negative integers $n$ and $x\geq 0$. \\ \\ It is well known that $\sin(x)\leq x=P_0(x)$ for all $x\geq 0$.\\ \\ If we have...

It's a shame Q16b didn't go on to prove an error estimate for the trapezoidal rule (with n points of division). Most of the work was done for it. (This kind of analysis is what I was looking for in my open-ended post in the MX2 integration marathon about comparing the accuracy of the...

Re: 2015 permutation X2 marathon Am well aware of this. So this fact is to the benefit of seed #2 player trying to reach the finals in real life vs this model. On the flipside, the fact that he cannot possibly lose to any of the players other than seed #1 in this model is a big point in...

Re: 2015 permutation X2 marathon Use common sense to test your attempts before asking for answer. This model isnt HUGELY inaccurate apart from when seeds are close. In your experience, does the 2nd best player in the world only reach the finals once every 64 grand slams?

Re: 2015 permutation X2 marathon a) Seed #2 will reach the finals and hence end up runner up precisely when he is on the opposite side of the draw to player 1. This happens with probability 64/127. b) Seed #3 will reach the semis precisely when he is the top seeded player in his quarter of...

How does it not sound useful? Reflection properties are literally the most practically applicable property of conics. Keep it in mind whenever you are talking about conics and angles at the same time. Or even conics and lengths, due to trig stuff.

If P is a point on an ellipse, then the normal to the ellipse at P bisects SPS', where S and S' are the focii. So if you and I were standing at the focii of a room with an ellipsoidal roof, then you would heard things I said really loudly. Its the elliptical analogue of a parabola reflecting...

Re: MX2 2015 Integration Marathon Sure, this is just performing an additional substitution on mine. You could do other substitutions as well if you preferred for whatever reason. All that matters is that the integrand is asymptotic to log(x) near the origin, and that this is integrable.

Re: MX2 2015 Integration Marathon To integrate from 0 to positive real x (which is the same as finding the values one particular primitive takes on the positive reals), Just break it into the intervals where the integrand is piecewise constant and sum. We get: \sum_{k=1}^{\lfloor...

Re: MX2 2015 Integration Marathon This is not a problem, because the leading order behaviour of sin(x) is that of x for small x. And the integral of log(t) from epsilon to 1 exists and converges as epsilon -> 0. (In other words, although the logarithm blows up at zero, it does not do so fast...