neeeeDanke

There seems to (at least theoretically) whitelist pairing-requests by mac-adress. Randomly hitting those few approved adresses consistently seems fairly unlikely: https://kb.vmware.com/s/article/50121103 (how to do it on Samsung, wish I had this option as well)

Tweet saying:

If you encounter a guy dressed as a girl "as a joke" this Halloween offer them transition info and be welcoming because if it's an exploring trans girl, it'll be helpeful and if it's a shitty cis dude, he will be furious. It's a win/win.

As many other comments here (really liked the way /u/opperharlie put it) I think it's a shitty take. First of all, there is drag which is totaly valid. But I think crossdressing in genereal -if it's not done derogatory (which I never experienced)- is a totally fair costume. I am certain I am cis and have dresses feminin for costumes a few times and don't see what the problem with hat is.

yay, we replaced it with more ?masculinormativity? :D.

Try (100,100,100,100,100,101) or 50 ones and a two, should result in 102 and 4 as a max respectively. I tried using less numbers, but the less numbers you use, the higher the values (to be exact less off a deviation(%-difference) between the values, resulting in higher numbers) have to be and wolframAlpha does not like 10^100 values so I stopped trying.

thanks for looking it up:).

I do think the upper bound on that page is wrong thought. Incedentally in the article itself only the lower bound is prooven, but in its sources this paper prooves what I did in my comment before as well:

for the upper bound it has max +log(n) . (Section 2, eq 4) This lets us construct an example (see reply to your other comment) to disproove the notion about beeing able to calculate the max for many integers.

gute Nachrichten: Das Spendenziel zum Berufungsverfahren ist fast erreicht

Am ende bekommen wir dann vieleicht Verlust für die AfD und eine wählbare Linke, hoffen kann man ja.

to be fair it does seem to work for any two numbers where one is >1. As lim x,y--> inf ln(e^x+e^y) <= lim x,y --> inf ln(2 e^(max(x,y))) = max(x,y) + ln(2).

I think is cool because works for any number of variables

using the same proof as before we can see that: lim,x_i -->inf ln(sum_{i/in I} e^(x_i)) <= ln(|I|) +max{x_i | i /in I}.

So it would only work for at most [base of your log, so e<3 for ln] variables.

laxatives and sirup?

pan

*visible confusion*

Like....no. Just stir it occasionally lol

Why? It doesn't stick while its in the water either way (for me at least, maybe the starch content varies enough to change that around the world).

so 0.3 ~= 1-ln(2)=max(1-ln(2),1-ln(2)) = floor(ln(2*e^(1-ln(2)))) = floor(ln(2)+(1-ln(2))) = 1 ?

That would bee engeneer 2, not Mathematician3 xD.

Just out of curiostity, what was you Idea behind that?