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Re: HSC 2014 4U Marathon - Advanced Level Yep, in saying that the second derivative is continuous I am of course asserting that it exists. Hint: The trick is to evaluate f at several points in your definition of reflection. Eg something like f(-x) = f(x) + f(x/2), but obviously not exactly that.

Re: HSC 2014 4U Marathon - Advanced Level No, if we assume the function is continuously differentiable (but not necessarily twice differentiable) in the right half plane, then for positive x, the "linear polynomial extension" gives: f(-x,y)=f(0,y)-x\frac{\partial f}{\partial x}(0,y). This of...

Re: HSC 2014 4U Marathon - Advanced Level Not entirely sure what you are saying here, it sounds fairly different to what I did. You should try to prove it carefully if it seems to work visually.

Re: HSC 2014 4U Marathon - Advanced Level Yep, exactly. Do you see why this fails though, if instead we are considering a function of two variables defined in the half-plane {x >= 0}? We could try the same construction on each line y = c, but the y-dependence of our extension would not be as...

Re: HSC 2014 4U Marathon - Advanced Level This can happen even if the first derivative is nonzero, eg f(x)=x^2+x. If you want to use a reflection argument, then I don't think simply using even, odd, or a combination of the two will work. You need to either construct a more complicated way of...

Re: HSC 2014 4U Marathon - Advanced Level The first derivatives will "match up" but the second derivatives won't. E.g. f(x)=x^2. If we do this odd reflection, then the resulting function will have second derivative 2 for positive x and -2 for negative x. (And we need g to have a continuous...

Re: HSC 2014 4U Marathon - Advanced Level Note the edit: we also need to assume that the one sided second derivative at 0 of f exists, and is equal to the limit of f''(y) as y -> 0 from above.

Re: HSC 2014 4U Marathon - Advanced Level No. Counterexample, f(x) = x => your g(x) is |x|, which is not differentiable at 0. (This question is actually easier than I originally hoped. It is possible to sidestep the approach I was hoping for and use an easier argument in dimension 1.)

Re: HSC 2014 4U Marathon - Advanced Level Here is a bit of general calculus (for the uni students here, this idea is important in extending elements of Sobolev spaces beyond the boundary of a domain in a way that's pretty smooth). (Recall that "f is bounded" means there is a positive M such...

Re: HSC 2014 4U Marathon - Advanced Level \log(T_n)=\frac{1}{n}\sum_{k=1}^n \log(1+k/n)\rightarrow \int_1^2 \log(t)\, dt = 2\log(2)-1\\ \\ \Rightarrow T_n \rightarrow 4e^{-1}.

Re: HSC 2014 4U Marathon - Advanced Level I really don't think this is true...the last RHS term on its own grows WAY faster than the LHS does. Also, a quick check shows this is not true for n=1.

How is that description not applicable to any other topic? There usually isn't THAT much room for originality in integration...

Re: MX2 Integration Marathon .

The function is continuous, continuous functions are Riemann integrable over compact intervals. So the integral exists, and any Riemann sum will converge to it. One might claim that we do not prove that continuous functions are integrable in high school, but by the same token we don't properly...

Yeah 0 to 2pi. As long as you hit each point in your domain once (apart from the overlap at 0 and 2pi which has area 0) you are all g. Sent from my GT-I9000 using Tapatalk 2

Integrate in polar coordinates to find the area of the two regions and subtract the inner one from the outer one. Remember that the "infinitesimal area element" in polar coordinates is: r\, dr\, d\theta as opposed to the more familiar dx\, dy in rectangular coordinates.

Re: HSC 2014 4U Marathon - Advanced Level Sy probably just wanted the sin(x) solution, but usually when you are asked to solve a functional equation like this, you need to find ALL solutions to the given equation. You should try to prove that all solutions to this problem are of the form...

Re: HSC 2014 4U Marathon - Advanced Level We definitely need to assume continuity on [a,b] for the claim to be true. Otherwise take a continuous function g on [a,b] that is differentiable with bounded derivative in (a,b) and then choose f to agree with g on [a,b) and make f(b) really large to...

Horse races (and sports betting generally) aren't really the ideal setting for thinking about the fallacy of the Martingale system (because it is hard to quantify winrates and payout odds are determined by the public betting rather than these winrates). Making profitable decisions in horse...

By "comparing the conditions given", do you mean evaluate the states that would be achieved by all possible moves and then choose the best one? Because if that's the case, that just shifts the bulk of the problem to finding this evaluation function. For games like chess whose complete...