Recent content by glittergal96

chain rule. Sent from my GT-I9506 using Tapatalk

Sorry about alt, am on phone. Showing that typical Clairaut hypotheses don't hold doesn't prove that mixed partials differ because its not iff. Just compute mixed partials at 0, it really shouldn't take long. Sent from my GT-I9506 using Tapatalk

Re: MX2 2016 Integration Marathon Ah of course, silly me. :).

Re: MX2 2016 Integration Marathon \int \cos(x^{1/n})\, dx = n\int y^{n-1}\cos(y)\, dy where y^n=x. (Consider only the non-negative reals to avoid having to talk about multivalued things.) This latter integral clearly gives us an integration by parts reduction formula (dropping exponent by...

Re: HSC 2016 4U Marathon - Advanced Level $Let these numbers be $\tan(\phi_k)$ with each $\phi_k\in(-\pi/2,\pi/2)$. We may label these in increasing order.\\ \\ Now if the claim is NOT true, then we have $\tan(\phi_{k+1}-\phi_k)\geq 1/19$ for each $k$. \\ \\ This implies that...

Re: MX2 2016 Integration Marathon oh nice, yeah that will do it :).

Re: MX2 2016 Integration Marathon Yeah, there are a few different ways / times to do it. I don't think any of them will be significantly faster than all of the others though, its pretty much preference.

Re: MX2 2016 Integration Marathon Yeah, I would probably just resort to t-substitution after reaching the trig form of sec^3(s)/(1+tan(s)). You are guaranteed success, it's just not an interesting / enlightening process. I briefly tried some other things like the analogous hyperbolic trig...

Re: MX2 2016 Integration Marathon Well, I mean the existence of a log term doesn't in itself mean the answer cannot be found in closed form. Finding a binomial expansion for the integral is immediate, but its not obvious that this expansion does not have a closed form. Sure, I will have a...

Re: MX2 2016 Integration Marathon Of course, this method also makes it clear why there is more work to do when integrating (k/x-x)^(2n+1). With this switched parity, the SIMPLER term in the integrand drops out from being odd and the more complicated one survives. This means that we...

Re: MX2 2016 Integration Marathon The motivation is fairly straightforward thankfully: u=b/x-x is a pretty logical thing to try (simplify the ugliest part of the integrand). Then I just arbitrarily (but consistently) chose one of the solutions to the resulting quadratic equation in x...

Re: MX2 2016 Integration Marathon Definitely don't need any recursion for this integral. Here is my solution to the generalisation where 2 is replaced by b: J_n=\int_1^b \left(\frac{b}{x}-x\right)^{2n}\, dx\\ \\ = \frac{1}{2}\int_{b-1}^{1-b}u^{2n}\left(-1+\frac{u}{\sqrt{u^2+4b}}\right)\...

Re: HSC 2016 4U Marathon - Advanced Level Topological solutions to what? An upper bound for what? You are definitely onto something with the monotonicity property though, that is the key to the argument I mentioned in the first of my remarks (1).

Re: MX2 2016 Integration Marathon It shouldn't depend on a, the a drops out after the translation y=x-a.

Re: HSC 2016 4U Marathon - Advanced Level Pretty much yep :). There are a couple of subtle points that need to be addressed slightly differently but the ideas are there. 1. It is far from obvious that "flipping things to the outside" can transform an arbitrary n-gon into a convex n-gon. (The...