information theo - may da c0de be with yah'
... god damn it! i've just realized that there some many things on the world
that i know shit about! i'm like starving now for informations - for knowledge.
tho maybe the worst thing that could happened now that - which is = TRUE now -
i'm living in a country, where people just can't live. they just can't do things
w/o telling them WHAT TO DO!!! - respect for ya guys in .yu (missing you these
days!).<br>
:: i've just read this, LISTEN(!): "Entropy of an X probability variable is:
H(X) = E(-logp(X)) = - SUM(p(xi)logp(xi))" /sum goes from 1 to n, and E is the
expected value). which simply means that - from my stupid point of view - 
the entropy is nothing but just weighted Bode-diagram of the system (a very 
complex one but still it could be draw up with matlab), or may i
say of the X prob. variable (of course we only choose one point of it).
let's take this equotion in another way: H(x)=-log(PI(p(xi)^p(xi))). /well maybe this makes you some sense /. so at H(X)=0 (cutting frequency) we have a full stable system, but if entropy starts to grow /and we all know that the entropy,
is just growing/ stands for an 'unstable' system - if one of the components has
0 probabilty we are doomed or may i say, never will be able to create a stable,
predictable system.... to be countinued<br>
:: a little addition after discussion with sph3r: this simply means that, since the entropy is growing (physics again), there's a parameter which is simply tends to 0 - and when it'll reach it (so it's to be sure that won't happen) - the entropy will be infinite,so<br>
(x^x)' = (e^(x*lnx))' = (e^(x*lnx))*(lnx+1)=x^x * (lnx+1)<br>
x^x * (lnx+1) = 0 if lnx+1 =0, that x=1/e<br>
(e^(x*lnx))'' = x^x * (lnx+1)^2 + x^x*1/x = x^x * ( (lnx+1)^2 + 1/x), which is positive at x=1/e, so it's a minimum place!<br>
thanx to ajt0fa, bridgeman and sph3r!