@ShadowWizard It listens to everything, even itself.
It's a wrapper, not a chatbot. The chatbot I made was just for testing the wrapper
and that was actually just an interactive commandline chat interface (yay! command prompt!)
@DoorknobofSnow That's probably because chat uses two methods to get data from the server, websockets and XHR fallback. The fallback is not deprecated, the devs say that they will ever make websockets a dependency, just an added bonus. So there's an object that does the choosing and using of server communication methods
find the sum $$\sum_{k=1}^{\infty} \frac {\sin^2(kx)}{k^2},$$ for $x\in [-\pi,\pi]$.
I have written it as $$\sum_{k=1}^{\infty} \left(\int \frac {\sin^2(kx)}{k^2} dx\right)'$$
I have done partial integration, but then I have to find this sum $$\sum_{k=1}^{\infty} \frac {\sin^2(kx)}{k}.$$
Any...
@ManofSnow It's the flag option id. So what you have to do is call api.stackexchange.com/docs/comment-flag-options to figure out what the option ids are for flagging that comment, and then use that option id to flag it.
@ManofSnow That's why it's called Mathematics. Look at MathOverflow if you want more stuff. :P
I found in interesting bug today, I tried to delete one of my questions and it said 'sorry, this question has answers and cannot be deleted'. But, in fact, it only had one answer...
Another question hangs in the balance of the universe. Is Jeff going to kill me?
https://twitter.com/coding...
So it seems like it would be ($x^m +y^n - z^r)/ p_1 = remainder$ $0$, however, how can this work since the sum shouldn't be close to $p_1$? (Also, $x^m +y^n$ should equal $z^r$, so some results from this should have $x^m +y^n =z^r$.) — hichris12313 mins ago
@hichris123 the API should become public very soon.
The docs are still hidden (unless you know the URL) because we've been changing them a lot and when they're released version 2.2 needs to be locked to not break things.
@Manishearth Apparently she viewed a flag on the post when she had a diamond. When a new flag pops up it includes all the moderators who viewed it previously.