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"
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