Anonymous asked:
Why did you become an artist?
ive always hated making money and being taken seriously
whew ok, time to work! ..okay. okay. let’s get started. …,,okeyy ah lets do this. yep…. okay. yes sir lets do this. let’s get started….just gonna start now yep. ,,okay. okay. okayy. ok. …ok..okey…..alright ok.. right hhh
GRRRRRRRR I don't WANT to confirm my email address! I HATE confirming my email address! *rips the door off my fridge*
I love your music you funky man
stegosaurusrn
But…the saxophone is going to the side of his face, how do you make the music
computers
imagine a rat using an airpod as a cane . imagine that
i’m thinking about it
this is the funniest shit i’ve ever seen
Me: *studies math, control processes*
Me: *just ended the month-long exam session, very tired of this shit*
Tumblr.com: *shows me this shit*
Me:
Fuck you
The answer is zero, btw. The trolley will kill a limited number of people if it kills zero people and there is no restriction saying it should move and and it won’t move anywhere from the zero position, according to the function
rubegoldbergsaciddreams
ooo. Perfect score!
Next question: For which value c does it kill the highest finite number of people?
I don’t think that has a solution:
The image of 0 after n iterations with constant c is a polynomial in c, with positive degree, that isn’t just a power of c, so it has a nonzero root in the plane. Thus, if there’s a maximum finite number, m, of people that the trolley can kill, we can choose a prime p>>m and pick a nonzero root of the p-iterations polynomial as our constant, so that the trolley will return to 0 after p steps. Since p is prime, the trolley can’t just be repeatedly following some smaller loop, so must kill p different people (or p-1, if we assume that no deeply unfortunate soul found themselves at the origin, only to be crushed by the cart before we even began), contradicting our assumption that m is the worst case.
Thus, even if we manage to avoid all constants that would result in infinite deaths, the cart remains unboundedly dangerous.
mmmmm
even if’s a bit of a copout, “c=0″ is still a valid answer even if there aren’t any higher values of c that satisfy.
It’s a solution to the original question, but definitely not to the one you posed - “c=-1″ kills one person, for example, and any solution to c^3+2c^2+c+1 = 0 kills two, both of which are finite, higher-than-zero killcounts.
Oh, I’m an idiot, I misread your last paragraph.
I thought you were saying “all non-zero values of c will kill infinite people.”
Which, on a second reading, is not what you’re saying at all.
