so now we have to prove that there is a rational in $[\log_2 n, \log_2 (n + 1)]$ which has denominator a perfect power of $2$. Clearly there are infinitely many rationals in that interval.
@JasperLoy I didn't publish anything yet ... I'll tell you more when I'm about to publish it (maybe during the next year, or I do some miracles and finish it this year?:-) - dunno)
@AndrewG I have decided to help blue in farming an army of 10k+ers to corule the world with him and banish all the geometers and analyst from the world with the help of inter-portal guillotine evil scientist laughter
@Chris'ssis Rethinking about $\sum_{i \leq n} \sum_{j \leq n} \frac1{i+j}$, this sum is just $\sum_{k = 1}^{2n} \frac{c_k}{k}$ where $c_k$ is the number of ways to partition $k$ into two integers with each of them $\leq n$, no? I think this is a doable approach to obtain a closed form (number theoretically).
@BalarkaSen I'm afraid you only want to understand what you want to understand and create situations in which you want to correct me (there is nothing to correct). Since you're so good, I let you finish the whole thing. :-)
@BalarkaSen when you're done, compute this $$ \lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{i+j} $$
@Chris'ssis This is just rude of you. If you can't communicate with people well and just want to be rude to the ones who point out your errors, it's better to ignore you.
@Chris'ssis OK, give me some time and I'll let you know if I finish it.
@rehband now you can arrrange things such that you get $$\lim_{n\to\infty} \frac{2}{n}\sum_{k=1}^{n} \frac{1}{1+k/n}$$ and the rest of the terms do not count.
hi all, I wonder if anyone knows the name of a particular method for solving a system of linear equations in a computer because I want to learn more about it. The method (independently of what it uses in the backend), zero-outs rows and columns of prescribed values of the system and places a 1 in the diagonal.
I was just calculating an integral via a trigonometric substitution and ended up with something pretty nonsensical but reversing the substitution seemed to clean it up.
Why is that the case? Is it something to do with the nature of the substitution? Is there something I'm not considering when performing the substitution?
@Chris'ssis Oh crap, I totally misread what Furdui wrote in the book. Yes, nevermind, his way is quite easy & understandable. Sorry for the dumb question
@Nick Hypothetically speaking, what'd happen if some perfect ray of light issued from a perfect point source is projected upon a point of the reflector at which there exists infinitely many tangents (like a kink)?
@Nick I think consulting a doctor would be better. @skull is right.
