DudeCalculus by Yemula Pradip

My dear freind Pradip has sent us this original paper on Dude Calculus... We will shortly be doin an analysis on basic Dude Graph theory and Markov Chains also

following the same spirit of dude matics let us now have a brief session on "dude calculus"

d(dude)/dt >= 0
(once a dude, remains a dude or becomes a bigger dude)


d(k*dude)/dt >= k*d(dude)/dt
('k' dudes hanging out together would result in greater dudeness in time then when the dudes are saperated)


d2(dude)/dt2 >= 0
the rate at which the dude becomes a bigger dude is also positive.


d(dude)/d(money) >= 0
it cost's money for the dude to become bigger dude. which they become anyways.



d(dude)/d(girl) = girls^(her intelligence)
The more intelligent the girl is the more dude the dude becomes... (which in turn costs him money)


dude has no additive inverse!
dude + (-dude) = 2*dude


Therefore,
d(dude1 + or - dude2)/d(same girl) = 2*girls^(her intelligence)
(an intelligent girl takes advantage of both the dudes, irrespective of whether the dudes are friends or enemies)


integral(dude).dt = area covered by the dude roaming behind girls


integral(integral(dude)).dt.d(girl) = area covered by the dude roaming around the same girl


dude + bike = bigger dude
therefore

integral(d(dude+bike)/d(girls)).d(girl) = girl + shopping + multiplex + restaurants

a dude with bike roaming around girls, ends up as a paying driver


dude + mobile = bigger dude

basically
dude + anything = bigger dude

actually
dude +/- anything = bigger dude

wait that proves my original axiom that
d(dude)/dt >= 0


- thanks and regards,
"intelligent dude"