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Re: MX2 Integration Marathon Yes, thanks :).

Re: MX2 Integration Marathon $Let $f$ be continuous and even, and let $a>0$.\\ Prove that:\\ \\ $\int_{-a}^a \frac{f(x)}{1+e^{x}}\, dx=\int_0^a f(x)\, dx$ $

Re: HSC 2014 4U Marathon General pattern recognition rather than a specific 4U topic. Difficulty 2/5. Sum sec(k)sec(k+1) from k=0 to 88. (Where sec is defined in degrees.)

Re: MX2 Integration Marathon This is how I finished it: $Since $\log$ is an increasing function, we have (assuming $a>0$ and using $-1\leq\cos(\theta)\leq 1$):\\ \\ $2\pi\log(1-a)\leq f(a) \leq 2\pi\log(1+a)$.\\ \\ From this inequality, we get that:\\ \\ $\lim_{a\rightarrow 0^+} f(a)=0.$\\...

Re: MX2 Integration Marathon I never said that the solution was a multiple of log(a) for a > 0, I just said that any constant multiple of log(a) satisfies the functional equation. The functional equation itself has many solutions, but by looking at what happens near zero we can make certain...

No, these definitions and concepts existed long before physicists found they modelled things that they observed well. And I don't think that use of these concepts was artificially to gain "mathematical credibility". The maths just happens to be well suited to studying those concepts. It just...

Re: MX2 Integration Marathon This isn't the only solution to the functional equation (double it for instance and it's still a solution). More importantly though, this cannot be the correct value for the integral, because f(a) should tend to 0 as a->0.

That is pretty derogatory to physicists. This is just an attempt at relating an intricate topic to a non-specialist audience. (And a reasonably successful one judging by the number of non-maths friends I have heard talking about this video recently.) There IS underlying rigour that cannot and...

Yeah, loosely similar. Convergence is an incredibly broad notion though.

Re: MX2 Integration Marathon The functional equation 2f(a)=f(a^2) is right, but your subsequent deduction isn't. You aren't far off though.

Not quite sure what you mean. It is an infinite series that "tends" to a definite value just like a geometric series with small ratio "tends" to a definite value. The former use of the word "tends" is just weaker.

As for the sum of the natural numbers, that is an example of trying to extend the domain of convergence for a Dirichlet series. (This is similar to how we extend the Riemann zeta function to be defined on the whole complex plane excluding z=1.) The natural way to do this is by something called...

Firstly you should be careful with this video, it is pretty non-rigorous (as it has to be, being a pop maths channel). There is a lot being swept under the carpet with his manipulations. Regarding your specific question: There are many notions of convergence other than the one you vaguely...

Re: MX2 Integration Marathon Here are some more from me as well: 1. \int_0^1 \frac{\log(1+x)}{1+x^2}\, dx\\ \\ 2.\int_0^\infty \frac{\log(x)}{1+x^2}\, dx\\ \\ 3. \int_0^\pi \log(1-2a\cos(x)+a^2)\, dx\quad (a\in\mathbb{R})\\ \\ The third one is a little different in flavour.

Re: MX2 Integration Marathon Cool, that's what I thought.

Re: MX2 Integration Marathon What exactly do you mean by rationalise the numerator here?

Re: HSC 2014 4U Marathon This does seem roughly right, you definitely have the right idea for counting permutations that don't fix things (these are called derangements). Something seems to have gone wrong at some point because you have n-k as an upper limit in a sum indexed by k. This is...

Re: HSC 2014 4U Marathon Polys/Complex. Difficulty 3/5. $Find all real polynomials $p(x)$ such that:\\ \\ $p(x+1)p(x-1)=p(x^2+1)$\\ \\ for all real $x$. $

Will probably see some of you guys around ANU/Canberra at some stage without knowing. Am in the second year of my PhD down here :).

Re: HSC 2014 4U Marathon Your expression for u_k is incorrect. Some of the (n-k)! permutations of the "non-fixed" points will fix points, so that method of counting doesn't work.