teoria de jocs: poker de 4 cartes

1♣+1♢+1♡+1♠
2♣+2♢+2♡+2♠
3♣+3♢+3♡+3♠
4♣+4♢+4♡+4♠


P(4) = [ 4 // 4 ]·n


2♣+1♢+1♡+1♠
3♣+1♢+1♡+1♠
4♣+1♢+1♡+1♠


1♣+2♢+1♡+1♠
1♣+3♢+1♡+1♠
1♣+4♢+1♡+1♠


1♣+1♢+2♡+1♠
1♣+1♢+3♡+1♠
1♣+1♢+4♡+1♠


1♣+1♢+1♡+2♠
1♣+1♢+1♡+3♠
1♣+1♢+1♡+4♠


P(3,1)= [ 4 // 3 ]·(n+(-1))·n


2♣+2♢+1♡+1♠
3♣+3♢+1♡+1♠
4♣+4♢+1♡+1♠


1♣+1♢+2♡+2♠
1♣+1♢+3♡+3♠
1♣+1♢+4♡+4♠


2♣+1♢+2♡+1♠
3♣+1♢+3♡+1♠
4♣+1♢+4♡+1♠


1♣+2♢+1♡+2♠
1♣+3♢+1♡+3♠
1♣+4♢+1♡+4♠


2♣+1♢+1♡+2♠
3♣+1♢+1♡+3♠
4♣+1♢+1♡+4♠


1♣+2♢+2♡+1♠
1♣+3♢+3♡+1♠
1♣+4♢+4♡+1♠


P(2,2) =  [ 4 // 2 ]·(n+(-1))·n


2♣+1♢+1♡+3♠
2♣+1♢+1♡+4♠
3♣+1♢+1♡+2♠
3♣+1♢+1♡+4♠
4♣+1♢+1♡+2♠
4♣+1♢+1♡+3♠


1♣+2♢+3♡+1♠
1♣+2♢+4♡+1♠
1♣+3♢+2♡+1♠
1♣+3♢+4♡+1♠
1♣+4♢+2♡+1♠
1♣+4♢+3♡+1♠


1♣+1♢+2♡+3♠
1♣+1♢+2♡+4♠
1♣+1♢+3♡+2♠
1♣+1♢+3♡+4♠
1♣+1♢+4♡+2♠
1♣+1♢+4♡+3♠


2♣+3♢+1♡+1♠
2♣+4♢+1♡+1♠
3♣+2♢+1♡+1♠
3♣+4♢+1♡+1♠
4♣+2♢+1♡+1♠
4♣+3♢+1♡+1♠


1♣+2♢+1♡+3♠
1♣+2♢+1♡+4♠
1♣+3♢+1♡+2♠
1♣+3♢+1♡+4♠
1♣+4♢+1♡+2♠
1♣+4♢+1♡+3♠


2♣+1♢+3♡+1♠
2♣+1♢+4♡+1♠
3♣+1♢+2♡+1♠
3♣+1♢+4♡+1♠
4♣+1♢+2♡+1♠
4♣+1♢+3♡+1♠


P(2,1,1) =  [ 4 // 2 ]·( (n+(-2))·(n+(-1)) )·n


Escales:


1♣+2♢+3♡+4♠
4♣+1♢+2♡+3♠
3♣+4♢+1♡+2♠
2♣+3♢+4♡+1♠


2♣+3♢+4♡+5♠
5♣+2♢+3♡+4♠
4♣+5♢+2♡+3♠
3♣+4♢+5♡+2♠


P(1,2,3,4,5) = 4·(n+(-3))


1♣+2♣+3♣+4♣
1♢+2♢+3♢+4♢
1♡+2♡+3♡+4♡
1♠+2♠+3♠+4♠


2♣+3♣+4♣+5♣
2♢+3♢+4♢+5♢
2♡+3♡+4♡+5♡
2♠+3♠+4♠+4♠


P(1,2,3,4,5) = 4·(n+(-3))


1♣+2♣+3♠+4♠
1♠+2♠+3♣+4♣
4♣+1♣+2♠+3♠
4♠+1♠+2♣+3♣


1♢+2♢+3♡+4♡
1♡+2♡+3♢+4♢
4♢+1♢+2♡+3♡
4♡+1♡+2♢+3♢


P(1,2,3,4,5) = 8·(n+(-3))

teoria de jocs: joc cuatre torres

[01,01][02,01][03,01][04,01]
[01,02][02,02][03,02][04,02]
[01,03][02,03][03,03][04,03]
[01,04][02,04][03,04][04,04]


[02,02] = torre_{a}
[02,03] = torre_{a}


[03,02] = torre_{b}
[03,03] = torre_{b}


[01,04] = rey_{a} ( 1 paso )
[04,04] = rey_{b} ( 1 paso )


[01,01] = reina_{a} ( n pasos )
[04,01] = reina_{b} ( n pasos )


algoritme-dual en 3 moviments:
{
reina_{a}[01,01] ==> reina_{b}[04,01]
rey_{b}[04,04] ==> rey_{b}[03,04]
torre_{a}[02,03] ==> torre_{a}[02,04]
escak-mat_{b}
}
{
reina_{b}[04,01] ==> reina_{a}[01,01]
rey_{a}[01,04] ==> rey_{a}[02,04]
torre_{b}[03,03] ==> torre_{b}[03,04]
escak-mat_{a}
}


algoritme-dual en 5 moviments:
{
reina_{a}[01,01] ==> reina_{b}[04,01]
torre_{b}[03,02] ==> torre_{b}[04,02]
reina_{a}[04,01] ==> torre_{b}[04,02]
torre_{b}[03,03] ==> torre_{b}[04,03]
reina_{a}[04,02] ==> torre_{b}[04,03]
escak-mat_{b}
}
{
reina_{b}[04,01] ==> reina_{a}[01,01]
torre_{a}[02,02] ==> torre_{a}[01,02]
reina_{b}[01,01] ==> torre_{a}[01,02]
torre_{a}[02,03] ==> torre_{a}[01,03]
reina_{b}[01,02] ==> torre_{a}[01,03]
escak-mat_{a}
}


algoritme-dual en 8 moviments:
{
torre_{a}[02,02] ==> torre_{b}[03,02]
reina_{b}[04,01] ==> torre_{a}[03,02]
reina_{a}[01,01] ==> reina_{a}[04,01]
rey_{b}[04,04] ==> rey_{b}[03,04]
reina_{a}[04,01] ==> reina_{b}[03,02]
torre_{b}[03,03] ==> reina_{a}[03,02]
torre_{a}[02,03] ==> torre_{a}[02,01]
torre_{b}[03,02] ==> torre_{b}[01,02]
escak-mat_{a}
}
{
torre_{b}[03,02] ==> torre_{a}[02,02]
reina_{a}[01,01] ==> torre_{b}[02,02]
reina_{b}[04,01] ==> reina_{b}[01,01]
rey_{a}[01,04] ==> rey_{a}[02,04]
reina_{b}[01,01] ==> reina_{a}[02,02]
torre_{a}[02,03] ==> reina_{b}[02,02]
torre_{b}[03,03] ==> torre_{b}[03,01]
torre_{a}[02,02] ==> torre_{a}[04,02]
escak-mat_{b}
}

teoria de jocs: joc cuatre-alfil

[01,01][02,01][03,01][04,01]
[01,02][02,02][03,02][04,02]
[01,03][02,03][03,03][04,03]
[01,04][02,04][03,04][04,04]


[02,02] = alfil_{a}
[02,03] = alfil_{a}


[03,02] = alfil_{b}
[03,03] = alfil_{b}


[01,04] = rey_{a}
[04,04] = rey_{b}


[01,01] = reina_{a}
[04,01] = reina_{b}


algoritme-dual de joc:
{
reina_{a}[01,01] ==> reina_{b}[04,01]
alfil_{b}[03,02] ==> reina_{a}[04,01]
alfil_{a}[02,02] ==> alfil_{a}[01,01]
alfil_{b}[03,03] ==> alfil_{a}[01,01]
alfil_{a}[02,03] ==> alfil_{b}[04,01]
}
{
reina_{b}[04,01] ==> reina_{a}[01,01]
alfil_{a}[02,02] ==> reina_{b}[01,01]
alfil_{b}[03,02] ==> alfil_{b}[04,01]
alfil_{a}[02,03] ==> alfil_{b}[04,01]
alfil_{b}[03,03] ==> alfil_{a}[01,01]
}


algoritme-dual:
{
rey_{b}[04,04] ==> rey_{b}[03,03]
alfil_{a}[04,01] ==> alfil_{a}[03,02]
rey_{b}[03,03] ==> alfil_{a}[03,02]
rey_{a}[01,04] ==> rey_{a}[02,04]
alfil_{b}[02,02] ==> alfil_{b}[03,03]
rey_{a}[02,04] ==> rey_{a}[01,04]
rey_{b}[03,02] ==> rey_{b}[02,02]
escak-mat_{a}
}
{
rey_{a}[01,04] ==> rey_{a}[02,03]
alfil_{b}[01,01] ==> alfil_{b}[02,02]
rey_{a}[02,03] ==> alfil_{b}[02,02]
rey_{b}[04,04] ==> rey_{a}[03,04]
alfil_{a}[03,02] ==> alfil_{a}[02,03]
rey_{b}[03,04] ==> rey_{b}[04,04]
rey_{a}[02,03] ==> rey_{a}[03,02]
escak-mat_{b}
}


algoritme-dual:
{
rey_{b}[04,04] ==> rey_{b}[04,03]
rey_{a}[01,04] ==> rey_{a}[02,03]
alfil_{b}[01,01] ==> alfil_{b}[02,02]
rey_{a}[02,03] ==> alfil_{b}[02,02]
rey_{b}[04,03] ==> rey_{b}[04,02]
alfil_{a}[04,01] ==> alfil_{a}[03,02]
escak-mat_{b}
}
{
rey_{a}[01,04] ==> rey_{a}[01,03]
rey_{b}[04,04] ==> rey_{b}[03,03]
alfil_{a}[04,01] ==> alfil_{a}[03,02]
rey_{b}[03,03] ==> alfil_{a}[03,02]
rey_{a}[01,03] ==> rey_{a}[01,02]
alfil_{b}[01,01] ==> alfil_{b}[02,02]
escak-mat_{a}
}

teoria de jocs: corrents de peó

corrents de peó en una pared de totxos.
corrents de peó en un terra de baldosas.
corrents de peó en una pared de azulejos.

teoria de jocs: joc peó 3-2

n=5k:
Si ( a = 1r  &  b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n  &  b = 1r ) ==> ( V=a & F=b )


[a_{0}][00][a_{1}][b_{2}][a_{2}][00][a_{3}][00][00][b_{0}]
[a_{0}][00][a_{1}][b_{2}][a_{2}][00][b_{3}][00][00][b_{0}]


9+((-2)+(-3))+((-2)+(-3))+((-2)+3) = 0
9+((-3)+(-2))+((-3)+(-2))+(3+(-2)) = 0


n=5k+1:
Si ( a = 1r  &  b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n  &  b = 1r ) ==> ( V=a & F=b )


[a_{0}][00][a_{1}][00][a_{2}][00][00][b_{1}][00][00][b_{0}]
[a_{0}][00][a_{1}][00][b_{2}][00][00][b_{1}][00][00][b_{0}]


10+((-2)+(-3))+((-2)+(-3)) = 0
10+((-3)+(-2))+((-3)+(-2)) = 0

teoria de jocs: joc peó 4-1

n=5k:
Si ( a = 1r  &  b = 2n ) ==> ( V=a & F=b )
Si ( a = 2n  &  b = 1r ) ==> ( F=a & V=b )


[a_{0}][b_{2}][a_{2}][00][00][b_{1}][00][00][00][b_{0}]
[a_{0}][a_{1}][00][00][00][b_{1}][00][00][00][b_{0}]


9+((-1)+(-4))+((-1)+(-4))+1 = 0
9+((-4)+(-1))+(-4) = 0


n=5k+1:
Si ( a = 1r  &  b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n  &  b = 1r ) ==> ( V=a & F=b )


[a_{0}][a_{1}][a_{2}][00][00][00][b_{1}][00][00][00][b_{0}]
[a_{0}][a_{1}][b_{2}][00][00][00][b_{1}][00][00][00][b_{0}]


10+((-1)+(-4))+((-1)+(-4)) = 0
10+((-4)+(-1))+((-4)+(-1)) = 0


n=5k+2:
Si ( a = 1r  &  b = 2n ) ==> ( V=a & F=b )
Si ( a = 2n  &  b = 1r ) ==> ( F=a & V=b )


[a_{0}][a_{1}][a_{2}][b_{2}][00][00][00][b_{1}][00][00][00][b_{0}]
[a_{0}][a_{1}][a_{2}][a_{3}][00][00][00][b_{3}][00][00][00][b_{0}]


11+((-1)+(-4))+((-1)+(-4))+(-1) = 0
11+((-4)+(-1))+((-4)+(-1))+(4+(-1))+(-4) = 0


n=5k+3:
Si ( a = 1r  &  b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n  &  b = 1r ) ==> ( V=a & F=b )


[a_{0}][a_{1}][a_{2}][a_{3}][a_{4}][00][00][00][b_{3}][00][00][00][b_{0}]
[a_{0}][a_{1}][a_{2}][a_{3}][b_{4}][00][00][00][b_{3}][00][00][00][b_{0}]


12+((-1)+(-4))+((-1)+(-4))+((-1)+4)+((-1)+(-4)) = 0
12+((-4)+(-1))+((-4)+(-1))+(4+(-1))+((-4)+(-1)) = 0


n=5k+4:
Si ( a = 1r  &  b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n  &  b = 1r ) ==> ( V=a & F=b )


[a_{0}][a_{3}][a_{2}][00][00][b_{2}][00][00][00][b_{1}][00][00][00][b_{0}]
[a_{0}][b_{3}][a_{2}][00][00][b_{2}][00][00][00][b_{1}][00][00][00][b_{0}]


13+((-1)+(-4))+((-1)+(-4))+(1+(-4)) = 0
13+((-4)+(-1))+((-4)+(-1))+((-4)+1) = 0

teoria de jocs: joc peó 3-1

n=4k:
Si ( a = 1r  &  b = 2n ) ==> ( V=a & F=b )
Si ( a = 2n  &  b = 1r ) ==> ( F=a & V=b )


[a_{0}][b_{2}][a_{2}][00][b_{1}][00][00][b_{0}]
[a_{0}][a_{1}][00][00][b_{1}][00][00][b_{0}]


7+((-1)+(-3))+((-1)+(-3))+1 = 0
7+((-3)+(-1))+(-3) = 0


n=4k+1:
Si ( a = 1r  &  b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n  &  b = 1r ) ==> ( V=a & F=b )


[a_{0}][a_{1}][a_{2}][00][00][b_{1}][00][00][b_{0}]
[a_{0}][a_{1}][b_{2}][00][00][b_{1}][00][00][b_{0}]


8+((-1)+(-3))+((-1)+(-3)) = 0
8+((-3)+(-1))+((-3)+(-1)) = 0


n=4k+2
Si ( a = 1r  &  b = 2n ) ==> ( V=a & F=b )
Si ( a = 2n  &  b = 1r ) ==> ( F=a & V=b )


[a_{0}][a_{1}][a_{2}][b_{2}][00][00][b_{1}][00][00][b_{0}]
[b_{3}][a_{1}][a_{2}][b_{2}][00][00][b_{1}][00][00][b_{0}]


9+((-1)+(-3))+((-1)+(-3))+(-1) = 0
9+((-3)+(-1))+((-3)+(-1))+((-3)+(-1))+3 = 0


n=4k+3:
Si ( a = 1r  &  b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n  &  b = 1r ) ==> ( V=a & F=b )


[a_{0}][b_{3}][a_{2}][a_{3}][b_{2}][00][00][b_{1}][00][00][b_{0}]
[a_{0}][b_{3}][a_{2}][00][b_{2}][00][00][b_{1}][00][00][b_{0}]


10+((-1)+(-3))+((-1)+(-3))+((-1)+(-3))+((-1)+3) = 0
10+((-3)+(-1))+((-3)+(-1))+((-3)+1) = 0

teoria de jocs: joc peó 2-1

n=3k:
Si ( a = 1r  &  b = 2n ) ==> ( V=a & F=b )
Si ( a = 2n  &  b = 1r ) ==> ( F=a & V=b )


[a_{0}][b_{2}][a_{2}][b_{1}][00][b_{0}]
[a_{0}][a_{1}][00][b_{1}][00][b_{0}]


5+((-1)+(-2))+((-1)+(-2))+1 = 0
5+((-2)+(-1))+(-2) = 0


n=3k+1:
Si ( a = 1r & b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n & b = 1r ) ==> ( V=a & F=b )


[a_{0}][a_{1}][a_{2}][00][b_{1}][00][b_{0}]
[a_{0}][a_{1}][b_{2}][00][b_{1}][00][b_{0}]


6+((-1)+(-2))+((-1)+(-2)) = 0
6+((-2)+(-1))+((-2)+(-1)) = 0


n=3k+2:
Si ( a = 1r & b = 2n ) ==> ( V=a & F=b )
Si ( a = 2n & b = 1r ) ==> ( F=a & V=b )


[a_{0}][a_{1}][a_{2}][b_{2}][00][b_{1}][00][b_{0}]
[a_{0}][b_{3}][a_{2}][a_{3}][00][b_{1}][00][b_{0}]


7+((-1)+(-2))+((-1)+(-2))+(-1) = 0
7+((-2)+(-1))+((-2)+(-1))+((-2)+(-1))+2 = 0

teoria de jocs: joc peó 1-1

n=2k:
Si ( a = 1r  &  b = 2n ) ==> ( V=a & F=b )
Si ( a = 2n  &  b = 1r ) ==> ( F=a & V=b )


[a_{0}][a_{1}][a_{2}][b_{2}][b_{1}][b_{0}]


5+((-1)+(-1))+((-1)+(-1))+(-1) = 0


n=2k+1:
Si ( a = 1r  &  b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n  &  b = 1r ) ==> ( V=a & F=b )


[a_{0}][a_{1}][a_{2}][b_{1}][b_{0}]
[a_{0}][a_{1}][b_{2}][b_{1}][b_{0}]


4+((-1)+(-1))+((-1)+(-1)) = 0

economia: modelo de línea de la ecuación de primer orden de logaritmo no inverso

f(x) = e^{(m/x)}
ln( f(x) ) = (m/x)


d_{z}[h(z)] + (-1)·( x^{2}/m )·h(z) = 0


h(z) = e^{(x^{2}/m)·z}


( ln(e^{(x^{2}/m)·(m/x)}) ) = x


( ln(e^{(x^{2}/m)·ln( f(x) )}) ) = x


( ln( h( ln(f(x)) ) ) ) = x


m = 3 & x = (3/4)·m ==> ln(f(3/4)) = 4€


d_{z}[h(z)] + (-1)·(3/16)·h(z) = 0


ln(h(4) ) = (3/4)€


ln(f(3/4))·ln(h(4) ) = 3€


m = 5 & x = (1/10)·m ==> ln(f(1/10)) = 50€


d_{z}[h(z)] + (-1)·(1/500)·h(z) = 0


ln(h(50) ) = (1/10)€


ln(f(1/10))·ln(h(50) ) = 5€

matemáticas: modelo de la equación diferencial de primer orden de logaritmo no inverso

f(x) = e^{(m/x^{n})}
ln( f(x) ) = (m/x^{n})


d_{z}[h(z)] + (-1)·( x^{n+1}/m )·h(z) = 0


h(z) = e^{(x^{n+1}/m)·z}


( ln(e^{(x^{n+1}/m)·(m/x^{n})}) ) = x


( ln(e^{(x^{n+1}/m)·ln( f(x) )}) ) = x


( ln( h( ln(f(x)) ) ) ) = x

matemáticas: modelo enésimo de la ecuación diferencial de primer orden

f(x) = e^{(m/x^{n})}
ln( f(x) ) = (m/x^{n})


d_{y}[g(y)] + (-1)·( m/x^{n+1} )·g(y) = 0


g(y) = e^{(m/x^{n+1})·y}


( 1/ln(e^{(m/x^{n+1})·(x^{n}/m)}) ) = x


( 1/ln(e^{(m/x^{n+1})·( 1/ln( f(x) ) )}) ) = x


( 1/ln( g( 1/ln(f(x)) ) ) ) = x

economia: modelo de superficie de la ecuación diferencial de primer orden

f(x) = e^{(m/x^{2})}
ln( f(x) ) = (m/x^{2})


d_{y}[g(y)] + (-1)·( m/x^{3} )·g(y) = 0


g(y) = e^{(m/x^{3})·y}


(1/ln(e^{(m/x^{3})·(x^{2}/m)})) = x


(1/ln(e^{(m/x^{3})·( 1/ln( f(x) ) )})) = x


( 1/ln( g( 1/ln(f(x)) ) ) ) = x


m = 2 cajas & x = 20 cm ==> ( 1/ln( f(20) ) ) = 200€


d_{y}[g(y)] + (-1)·(1/4000)·g(y) = 0


( 1/ln( g(200) ) ) = 20€


m = 3 cajas & x = 60 cm ==> ( 1/ln( f(60) ) ) = 1200€


d_{y}[g(y)] + (-1)·(1/72000)·g(y) = 0


( 1/ln( g(1200) ) ) = 60€

economia: modelo de línea de la ecuación lineal de primer orden

f(x) = e^{(m/x)}
ln( f(x) ) = (m/x)


d_{y}[g(y)] + (-1)·( m/x^{2} )·g(y) = 0


g(y) = e^{(m/x^{2})·y}


(1/ln(e^{(m/x^{2})·(x/m)})) = x


(1/ln(e^{(m/x^{2})·( 1/ln( f(x) ) )})) = x


( 1/ln( g( 1/ln(f(x)) ) ) ) = x


m = 4 botellas & x = 64 cm ==> ( 1/ln( f(64) ) ) = 16€


d_{y}[g(y)] + (-1)·(1/1024)·g(y) = 0


( 1/ln( g(16) ) ) = 64€


m = 3 botellas & x = 30 cm ==> ( 1/ln( f(30) ) ) = 10€


d_{y}[g(y)] + (-1)·(1/300)·g(y) = 0


( 1/ln( g(10) ) ) = 30€

química: constructores y destructores del catorce-sulfuracético

bi-catorce-sulfuracético fosfado positivo


P=S≡S≡S≡S≡S=P≡P=S≡S≡S≡S≡S=P
P=S≡S≡S≡S≡S=P≡P=S≡S≡S≡S≡S=P


16 destructores + 24 constructores = 8 constructores


bi-catorce-sulfuracético fosfado negativo


P=S≡S≡S≡S≡P=S≡S=P≡S≡S≡S≡S=P
P=S≡S≡S≡S≡P=S≡S=P≡S≡S≡S≡S=P


20 destructores + 12 constructores = 8 destructores

química: carbur de fósfor

2·P_{2} + 2·C_{2} + e^{-}  ==> ( =P≡C-C≡P= ) + ( =C=P≡P=C= )


( =P≡C-C≡P= ) + ( =C=P≡P=C= ) + e^{+} ==> 2·P_{2} + 2·C_{2}


entalpia de reacció:
[ 2·P_{2} ] = 20
[ 2·C_{2} ] = 16


[ ( =P≡C-C≡P= ) ] = 18
[ ( =C=P≡P=C= ) ] = 18


20+16 = 18+18 = 36

química: carbur de nitrogen

2·C_{2} + 2·N_{2} + e^{-}  ==> ( =N-C≡C-N= ) + ( =C=N-N=C= )


( =N-C≡C-N= ) + ( =C=N-N=C= ) + e^{+} ==> 2·C_{2} + 2·N_{2}


entalpia de reacció:
[ 2·C_{2} ] = 16
[ 2·N_{2} ] = 12


[ ( =N-C≡C-N= ) ] = 14
[ ( =C=N-N=C= ) ] = 14


16+12 = 14+14 = 28

química: constructores y destructores del catorce-acético

bi-catorce-acético nitrogenado positivo


N-C=C=C=C=C=N-N=C=C=C=C=C-N
N-C=C=C=C=C=N-N=C=C=C=C=C-N


12 destructores + 16 constructores = 4 constructores


bi-catorce-acético nitrogenado negativo


N-C=C=C=C=N-C≡C-N=C=C=C=C-N
N-C=C=C=C=N-C≡C-N=C=C=C=C-N


12 destructores + 8 constructores = 4 destructores


bi-catorce-acético fosfado positivo


P-C=C=C=C=C=P≡P=C=C=C=C=C-P
P-C=C=C=C=C=P≡P=C=C=C=C=C-P


12 destructores + 16 constructores = 4 constructores


bi-catorce-acético fosfado negativo


P-C=C=C=C=P≡C-C≡P=C=C=C=C-P
P-C=C=C=C=P≡C-C≡P=C=C=C=C-P


12 destructores + 8 constructores = 4 destructores

química: constructores y destructores del deca-benceno

bi-deca-benceno nitrogenado positivo


N-C=C-C=C=C=C-C=C-N
N-C=C-C=C=C=C-C=C-N


4 destructores + 12 constructores = 8 constructores


bi-deca-benceno nitrogenado negativo


N-C=C-C=N-N=C-C=C-N
N-C=C-C=N-N=C-C=C-N


8 destructores + 0 constructores = 8 destructores

química: constructores y destructores de centro

doce-glicerina tetra-nitrogenada


N-C=C-C=C-N-N-C=C-C=C-N


4 destructores + 2 constructores = 2 destructores


diez-y-seis-glicerina tetra-nitrogenada


N-C=C-C=C-C=C-N-N-C=C-C=C-C=C-N


4 destructores + 6 constructores = 2 constructores


vente-y-cuatro-glicerina octo-nitrogenada


N-C=C-C=C-N-N-C=C-C=C-N-N-C=C-C=C-N-N-C=C-C=C-N


8 destructores + 4 constructores = 4 destructores


trenta-y-dos-glicerina octo-nitrogenada


N-C=C-C=C-C=C-N-N-C=C-C=C-C=C-N-N-C=C-C=C-C=C-N-N-C=C-C=C-C=C-N


8 destructores + 12 constructores = 4 constructores

química: constructores y destructores

bi-hexa-glicerina nitrogenada


N-C=C-C=C-N
N-C=C-C=C-N


4 destructores + 2 constructores = 2 destructores




bi-octo-glicerina nitrogenada


N-C=C-C=C-C=C-N
N-C=C-C=C-C=C-N


4 destructores + 6 constructores = 2 constructores

ecuació diferencial producte derivat

d_{x}[x·f(x)] = ax


f(x)+x·d_{x}[f(x)] = ax


f(x) = (a/2)·x


d_{x}[x·f(x)] = g(x)
f(x)+x·d_{x}[f(x)] = g(x)


f(x) = (1/x)·∫ [ g(x) ] d[x]


d_{x}[x^{n}·f(x)] = g(x)


f(x) = (1/x^{n})·∫ [ g(x) ] d[x]

coment dual-françé

el que en-és de caminuar per la foscurité,
ne en-és de ver a don-que va.


tu ne en-ets de ver a don-que vaitx.
je ne en-soy de ver a don-que vas.


el que en-és de caminuar per el silence,
ne en-és de oir a don-que va.


tu ne en-ets de oir a don-que vaitx.
je ne en-soy de oir a don-que vas.

dual-françé

je en-soy de pensuar.
tu en-ets de pensuar.
ell-il en-és de pensuar en celui-çí demostraçón de içí.
ella-il en-és de pensuar en celui-lá demostraçón de ilá.


nus-uà e-sóms de pensuar.
vus-uà e-sous de pensuar.
ell-ils e-son de pensuar.
ella-ils e-son de pensuar.

micro-economia optimizatció de una esfera

B(r) = (1/10pi)·(4pi)·( p·r^{2}+(-1)·m·(1/3)·r^{3} )


d_{x}[B(r)] = 2p·r+(-1)·m·r^{2}


mr = 2p


r = ( (2p)/m )


B( (2p)/m ) = (4pi)( (2/3)(p^{3}/m^{2}) )


pelota de tenis:


p=3 &  m=2


B(3) = (1/10pi)·(18pi) = 1.8€
B(r) = (1/10pi)·( F(r)+(-1)·V(r) )
(1/10pi)·F(3)  = 2.7€
(1/10pi)·V(3) = 0.9€

micro-economia optimització de una caisha

B(x,y,z) = (1/4000)·( 2·p·( xy+yz+zx )+(-1)·m·xyz )


d_{x}[B(x,y,z)] = 2·p·( y+z )+(-1)·m·yz
d_{y}[B(x,y,z)] = 2·p·( z+x )+(-1)·m·zx
d_{z}[B(x,y,z)] = 2·p·( x+y )+(-1)·m·xy


2·p·( y+z )=m·yz
2·p·( z+x )=m·zx
2·p·( x+y )=m·xy


( y+z )=m
( z+x )=m
( x+y )=m


2·p=yz
2·p=zx
2·p=xy


x=(m/2)
y=(m/2)
z=(m/2)


p=(1/2)·(m/2)^{2}


B( (m/2) , (m/2) , (m/2) ) = 3·(m/2)^{4}+(-1)·2·(m/2)^{4}
B( (m/2) , (m/2) , (m/2) ) = (m/2)^{4}


bombones
m=40 <==> ( x=20 & y=20 & z=20 )


B( 20 , 20 , 20 ) = 4€


B( x , y , z ) = (1/4000)·( F( x , y , z ) +(-1)·V( x , y , z )  )


(1/4000)·F( 20 , 20 , 20 ) = 12€
(1/4000)·V( 20 , 20 , 20 ) = 8€

micro-economia optimizar un rollo de papel

B(r,z) = (1/10pi)·( p·(2pi)·r·z+(-1)·m·(pi)·r^{2}·z )


d_{r}[B(r,z)]  = p·(2pi)·z+(-1)·m·(2pi)·r·z


p = m·r


r = (p/m)


B( (p/m) , z ) = (1/10)·( p·(2pi)·(p/m)·z+(-1)·m·(pi)·(p/m)^{2}·z )


B( (p/m) , z ) = (1/10)·( (p^{2}/m)·(pi)·z )


papel higiénico:
z=10 & p=(5céntimos/(cm)^{2}) & m=(1céntimo/(cm)^{3})


B( 5 , 10  ) = 25·céntimos = 0.25€


B(r,z) = (1/10pi)·( F(r,z)+(-1)·V(r,z) )


(1/10pi)·F( 5 , 10  ) = 50·céntimos = 0.50€
(1/10pi)·V( 5 , 10  ) = 25·céntimos = 0.25€


papel de aluminio:
z=30 & p=(6céntimos/(cm)^{2}) & m=(2céntimo/(cm)^{3})


B( 3 , 30  ) = 54·céntimos = 0.54€


B(r,z) = (1/10pi)·( F(r,z)+(-1)·V(r,z) )


(1/10pi)·F( 3 , 30  ) = 108·céntimos = 1.08€
(1/10pi)·V( 3 , 30  ) = 54·céntimos = 0.54€

economia: simetria-dual de consumo segun un capital producto lineal en no uno

F(x,y) = (n·x)·(m·y)+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = n(my)+(-1)·hp
d_{y}[F(x,y)] = m(nx)+(-1)·hq


nmy=hp
mnx=hq


nmyx=hpx
mnxy=hqy


nmyx+mnxy=hk


2·mnxy=hk


( x=( e^{n}/n ) & y=( e^{m}/m ) )  <==>  2·e^{n}e^{m}=hk


si h=2·( e^{k}/k ) ==>


e^{n}e^{m}=e^{k}


( n=( k+(-j) ) & m=j )


G( (e^{n}/n ) , (e^{m}/m ) ) = e^{n}·e^{m}

economia: simetria-dual de consumo segun un capital producto lineal

F(x,y) = (e^{n}·x)·(e^{m}·y)+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = e^{n}(e^{m}y)+(-1)·hp
d_{y}[F(x,y)] = e^{m}(e^{n}x)+(-1)·hq


e^{n}e^{m}y=hp
e^{m}e^{n}x=hq


e^{n}e^{m}yx=hpx
e^{m}e^{n}xy=hqy


e^{n}e^{m}yx+e^{m}e^{n}xy=hk


2·e^{n}e^{m}xy=hk


( x=1 & y=1 )  <==>  2·e^{n}e^{m}=hk


si h=2·( e^{k}/k ) ==>


e^{n}e^{m}=e^{k}


( n=( k+(-j) ) & m=j )


G( 1 , 1 ) = e^{n}·e^{m}

economia: simetria-dual de consumo segun un capital producto potencial

F(x,y) = x^{n}·y^{m}+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = nx^{(n+(-1))}·y^{m}+(-1)·hp
d_{y}[F(x,y)] = my^{(m+(-1))}·x^{n}+(-1)·hq


nx^{(n+(-1))}·y^{m}=hp
my^{(m+(-1))}·x^{n}=hq


nx^{n}·y^{m}=hpx
my^{m}·x^{n}=hqy


n·x^{n}·y^{m}+m·x^{n}·y^{m}=hk


(n+m)·x^{n}·y^{m}=hk


( x=1 & y=1 )  <==>  n+m=hk


si h=1 ==>


( n=( k+(-j) ) & m=j )


G( 1 , 1 ) = 1

economia: simetria-dual de consumo segun un capital exponencial

F(x,y) = e^{nx}+e^{my}+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = ne^{nx}+(-1)·hp
d_{y}[F(x,y)] = me^{my}+(-1)·hq


ne^{nx}=hp
me^{my}=hq


nxe^{nx}=hpx
mye^{my}=hqy


nxe^{nx}+mye^{my}=hk


x=( (k+(-j))/n )
y=( j/m )


(k+(-j))e^{k+(-j)}+je^{j}=hk


si h=( (k+(-j))e^{k+(-j)}+je^{j})/k ) ==>


( x=1 & y=1 ) <==> ( n=( k+(-j) ) & m=j )


G( 1 , 1 ) = e^{n}+e^{m}

economia: simetria-dual de consumo segun un capital logarítmica

F(x,y) = n·ln(x)+m·ln(y)+(-1)·h( px+qy+(-k) ) & px+qy=k


d_{x}[F(x,y)] = n·(1/x)+(-1)·hp
d_{y}[F(x,y)] = m·(1/y)+(-1)·hq


n·(1/x)=hp
m·(1/y)=hq


n=hpx
m=hqy


n+m=hk


si h=1 ==>


n=k+(-j)
m=j


G( n/p , m/q ) = n·ln(n/p)+m·ln(m/q)

economia simetria-dual de consumo segun un capital lineal

F(x,y) = nx+my+(-1)·h( px+qy+(-k) ) & px+qy = k


d_{x}[F(x)] = n+(-1)hp
d_{y}[F(x)] = m+(-1)hq


n=hp
m=hq


nx=hpx
my=hqy


nx+my=hk


si h=1 ==>


x=( (k+(-j))/n )
y=( j/m )


( x=1 & y=1 ) <==> ( n = ( k+(-j) ) & m = j )


G( 1 , 1 ) = n+m

economia simetria-dual de consumo segun un capital potencial

F(x,y) = x^{n}+y^{m}+(-1)·h( px+qy+(-k) ) & px+qy = k


d_{x}[F(x,y)] = nx^{n+(-1)}+(-1)·hp
d_{y}[F(x,y)] = my^{m+(-1)}+(-1)·hq


nx^{n+(-1)}=hp
my^{m+(-1)}=hq


nx^{n}=hpx
my^{m}=hqy


nx^{n}+my^{m}=hk


x=( ((k+(-j))h)/n )^{(1/n)}
y=( (jh)/m )^{(1/m)}


Si h=1 ==>


x=( (k+(-j))/n )^{(1/n)}
y=( j/m )^{(1/m)}


( x=1 & y=1 ) <==> ( n=( k+(-j) ) & m=j )


G( 1 , 1 ) = 2

micro-economia supermercado

un supermercado tiene unos costes en camión de alimentos de mx^{3},
optimizar el precio de los alimentos n si los compra a precio p:


B(x) = (n+(-p))·x+(-1)·mx^{3}


d_{x}[B(x)] = (n+(-p))+(-1)·(3mx^{2})


d_{x}[B(x)] = 0  <==> x = ( (n+(-p))/3m )^{(1/2)}


x=1<==> 3m=(n+(-p))


( m=10€ de gasolina & 200€ la planta & n=230€ la planta )
( m=15€ de gasolina & 200€ la planta & n=245€ la planta )
( m=20€ de gasolina & 200€ la planta & n=260€ la planta )


B(1) = (n+(-p))+(-1)·m
B(1) = 2·m

micro-economia de coste cuadrático

una plantación tiene unos costes en agua de qx^{2},
optimizar el precio de la planta:


B(x) = px+(-1)·qx^{2}


d_{x}[B(x)] = p+(-1)·(2qx)


d_{x}[B(x)] = 0  <==> x = ( p/2q )


x=1<==> 2q=p


( m=100€ de agua & p=200€ la planta )
( m=150€ de agua & p=300€ la planta )
( m=200€ de agua & p=400€ la planta )


B(1) = p+(-1)·q
B(1) = q

micro-economia de coste cúbico

una empresa de gas tiene unos costes en camión de bombonas de mx^{3},
optimizar el precio de la bombona de gas:


B(x) = px+(-1)·mx^{3}


d_{x}[B(x)] = p+(-1)·(3mx^{2})


d_{x}[B(x)] = 0  <==> x = ( p/3m )^{(1/2)}


x=1<==> 3m=p


( m=10€ de gasolina & p=30€ la bombona de gas )
( m=15€ de gasolina & p=45€ la bombona de gas )
( m=20€ de gasolina & p=60€ la bombona de gas )


B(1) = p+(-1)·m
B(1) = 2·m

micro-economia

f(x) = px <==> preu per unitat.
g(x) = qx^{2} <==> preu per superficie.
h(x) = mx^{3} <==> preu per volum.


d_{x}[f(x)] = p
d_{x}[g(x)] = 2qx
d_{x}[h(x)] = 3mx^{2}

palmitato de sodio y agua

2n·( NaC_{m}N_{2m+(-2)}H_{2m+1} ) + 4n·H_{2}O + e^{-} ==> ...
... 2n·( C_{m}N_{2m+(-2)}H_{2m}(OH)_{2} ) + 2n·H_{2} + 2n·NaH


2n·( C_{m}N_{2m+(-2)}H_{2m}(OH)_{2} ) + 2n·H_{2} + 2n·NaH  + e^{+} ==> ...
... 2n·( NaC_{m}N_{2m+(-2)}H_{2m+1} ) + 4n·H_{2}O


[ 2n·( NaC_{m}N_{2m+(-2)}H_{2m+1} ) ] = 20n
[ 4n·H_{2}O ] = 12n


[ 2n·( C_{m}N_{2m+(-2)}H_{2m}(OH)_{2} ) ] = 24n
[ 2n·H_{2} ] = 4n
[ 2n·NaH ] = 4n


20n+12n = 24n+4n+4n = 32n

politja doble fixada en un extrem en un sostre y estirada per l'altre

( m or q ) esta en la politja central no extrem de la corda.


politja doble amb una força constant en el extrem de la corda.
m·d_{tt}^{2}[y(t)] = q·g+(-1)·( F/2 )


politja doble amb una molla en el extrem de la corda.
m·d_{tt}^{2}[y(t)] = q·g+(-1)·( (k/2)·y(t) )


politja doble amb una força dependent del temps en el extrem de la corda.
m·d_{tt}^{2}[y(t)] = q·g+(-1)·( F(t)/2 )


politja doble amb una càrrega en el extrem de la corda.
m·d_{tt}^{2}[y(t)] = q·g+(-1)·( T/2 )
m_{1}·d_{tt}^{2}[y_{1}(t)] = (-1)·q_{1}·g+T


(m_{1}+2m)·d_{tt}^{2}[y_{1}(t)] = 2·q·g+(-1)·q_{1}·g


T = ( (m_{1}(2q)+(2m)q_{1})/(m_{1}+2m) )·g


Si m_{1} = 0 ==> T = q_{1}·g


(-T) = 2m·( (2·q·g+(-1)·q_{1}·g)/(m_{1}+2m) )+( (-1)·(m_{1}+2m)·2qg/(m_{1}+2m) )
T = m_{1}·( (2·q·g+(-1)·q_{1}·g)/(m_{1}+2m) )+( (m_{1}+2m)·q_{1}g/(m_{1}+2m) )

politja simple

m_{1}·d_{tt}^{2}[y_{1}(t)] = ( q_{1}·g+(-1)·T )
m_{2}·d_{tt}^{2}[y_{2}(t)] = ( (-1)·q_{2}·g+T )


(m_{1}+m_{2})·d_{tt}^{2}[y_{1}(t)] = ( q_{1}+(-1)q_{2} )·g


T = ( (m_{2}·q_{1}+m_{1}·q_{2})/(m_{1}+m_{2}) )·g


(-T) = m_{1}·( ( q_{1}+(-1)q_{2} )/(m_{1}+m_{2}) )·g+...
...( (-1)·((m_{1}+m_{2})·q_{1})/(m_{1}+m_{2}) )·g


T = m_{2}·( ( q_{1}+(-1)q_{2} )/(m_{1}+m_{2}) )·g+...
...( ((m_{1}+m_{2})·q_{2})/(m_{1}+m_{2}) )·g


politja simple sense càrrega  estirada per una força constant.
m·d_{tt}^{2}[y(t)] = ( q_{1}·g+q_{2}·g )+(-1)·F


politja simple sense càrrega penjada de una molla.
m·d_{tt}^{2}[y(t)] = ( q_{1}·g+q_{2}·g )+(-1)·k·y(t)


politja simple sense càrrega estirada per una força dependent del temps.
m·d_{tt}^{2}[y(t)] = ( q_{1}·g+q_{2}·g )+(-1)·F(t)

politja triple

( m or q ) esta en la politja central.


politja triple amb dos forçes constants en els extrems de la corda.
m·d_{tt}^{2}[y(t)] = q·g+(-1)·( F_{1}+F_{2} )


politja triple amb dos molles en els extrems de la corda.
m·d_{tt}^{2}[y(t)] = q·g+(-1)·( k_{1}·y(t)+k_{2}·y(t) )


politja triple amb dos forçes dependents del temps en els extrems de la corda.
m·d_{tt}^{2}[y(t)] = q·g+(-1)·( F_{1}(t)+F_{2}(t) )


politja triple amb dos càrregues en els extrems de la corda.
m·d_{tt}^{2}[y(t)] = q·g+(-1)·( T_{1}+T_{2} )
m_{1}·d_{tt}^{2}[y_{1}(t)] = (-1)·q_{1}·g+T_{1}
m_{2}·d_{tt}^{2}[y_{2}(t)] = (-1)·q_{2}·g+T_{2}


(m_{1}+m·( (n+(-k))/n ))·d_{tt}^{2}[y_{1}(t)] = q·( (n+(-k))/n )·g+(-1)·q_{1}·g
(m_{2}+m·( k/n ))·d_{tt}^{2}[y_{2}(t)] = q·( k/n )·g+(-1)·q_{2}·g


T_{1} = ( (m_{1}q·( (n+(-k))/n )+q_{1}m·( (n+(-k))/n ))/(m_{1}+m( (n+(-k))/n) ) )·g
T_{2} = ( (m_{2}q·( k/n )+q_{2}m·( k/n ))/(m_{2}+m( k/n )) )·g


si m_{1}=0 ==> T_{1}=q_{1}g
si m_{2}=0 ==> T_{2}=q_{2}g