Recent content by Ayre

It might well be that the reddit photo is what he thought was new and ingenious (it is quite ingenious indeed), so in that sense you may be right. But I don't think it's fair to call that equation his solution, given that his poster has another section with a big "Lösung" headline, with...

Well, I think they're the mistaken ones. https://www.jugend-forscht.de/images/1MAT_67_download.jpg is the image you're talking about it. If you look at it, it's clear that it's just a photo-op-type thing. Somebody decided that it would be good to have a photo of the guy holding up an...

Let me rephase that. The sequence a_n will not really converge, but a subsequence does. What properties do you know that guarantee that sequences have convergent subsequences? (I realize now that this solution method will be slightly roundabout. Perhaps you can see how to simplify it later.)

Good catch! This reference also appears on Mr Ray's poster, so he must have been aware of it. Though ge describes it as a "semianalytische exakte Lösung", or a "semi-analytical exact solution". I only glanced through the paper (link), but it seems to only give power series solutions...

I think it boils down to the following: You find it hard to differentiate functions of the form f(x) = g(x)^{h(x)}\text. I think it helps to first work out a general formula for this, i.e. something kind of like the product rule, but for powers. Then, you can go back and tackle your...

I'm very sorry. I misread your question, I thought you were trying to show that d(a,A)=0 for a in the closure of A. But your calculation is still not correct. The formula \inf_{a\in A} (d(x,a) + d(a,a_0)) = \inf_{a\in A}d(x,a) + \inf_{a\in A} d(a,a_0)\text, which you are using, does not hold...

The problem is that this is most certainly not the first analytical series solution. If I have time later, I might go hunt for some references, but solving differential equations by series has been part of the standard toolbox for a very, very long time. It could be, to give him the benefit of...

Yes. I mean, if you assume the result you're trying to prove, then that equality is correct. But, of course, you're not allowed to assume what you're trying to prove.

Well, they're certainly not as hyperbolic. Certainly, his poster claims that he has discovered something new. It says, for instance, ""erstmals vollanalytische Lösung eines lange ungelösted Problems", i.e. "first fully analytical solution of a long unsolved problem" (my translation). I guess...

The last step of your calculation is not correct, and anyway, that approach is not the way you want to proceed. Instead, you need to use the fact that a_0 lies in the closure of A. Write down as many equivalent definitions you can of a_0 being in the closure of A. One of them will be what you...

I found a high-res picture of the poster image! http://i47.tinypic.com/2v0oco8.jpg So his solution is u(t) = \frac{u_0}{1 + \alpha V_0 t - \tfrac 1{2!}\alpha gt^2 \sin \theta + \tfrac 1{3!}\left(\alpha g^2 \cos^2 \theta - \alpha^2 g V_0 \sin \theta\right) t^3 + \cdots} v(t) =...

I'm inclined to agree, i.e. I think that a random variable whose cdf jumps at all the rationals can be called discrete. The rationale? Well, the concept of random variable doesn't have anything to do with the topology of the space of values of rational numbers. It would be silly to require...

On reddit: http://www.reddit.com/r/math/comments/u74no/supposedly_this_is_a_new_formula_for_calculating/ (Sorry, can't post links, add "http") there's a link to a picture of him holding up a particular formula. This seems to be a constant of the motion for a projectile moving in uniform...