Ley:
Sea m·d_{tt}^{2}[z] = pE_{e}(z,q) ==>
Si q = 0 ==> p = m
Ley:
Sea m·d_{tt}^{2}[z] = pE_{g}(z,q) ==>
Si q = 0 ==> p = m
Electro-débil de leptones orbitales:
Ley:
F(t)·G(t) = e^{(1/m)·(q+(-W))}·e^{(1/m)·(W+(-q))}·f(t)·g(t)
d_{t}[F(t)]·d_{t}[G(t)] = ...
... d_{t}[f(t)]·d_{t}[g(t)]+(1/m)^{2}·d_{t}[q+(-W)]·d_{t}[W+(-q)]·f(t)·g(t)
Ley:
Sea A(x,y) = (1/m)·< x,y > ==>
F(x,y)·G(x,y) = e^{ Anti-Potencial[ A(x,y)·a^{2}·< q+(-W),W+(-q) > ] }·f(x,y)·g(x,y)
d_{y}[F(x,y)]·d_{x}[G(x,y)] = ...
... d_{y}[f(x,y)]·d_{x}[g(x,y)]+( A_{x}·A_{y} )·a^{4}·(q+(-W))·(W+(-q))·f(x,y)·g(x,y)
Deducción:
F(x,y) = e^{ int[ A_{x}·a^{2}·(q+(-W)) ]d[y] }·f(x,y)
G(x,y) = e^{ int[ A_{y}·a^{2}·(W+(-q)) ]d[x] }·g(x,y)
d_{y}[F(x,y)]·d_{x}[G(x,y)] = 0 <==> ...
f(x,y) = e^{ int[ ia^{2}·A_{x}·(q+(-W)) ]d[y] }
g(x,y) = e^{ int[ ia^{2}·A_{y}·(W+(-q)) ]d[x] }
Ley:
Sea A(y,x) = (1/m)·< y,x > ==>
F(x,y)·G(x,y) = e^{ Potencial[ A(y,x)·a^{2}·< q+(-W),W+(-q) > ] }·f(x,y)·g(x,y)
d_{x}[F(x,y)]·d_{y}[G(x,y)] = ...
... d_{x}[f(x,y)]·d_{y}[g(x,y)]+( A_{y}·A_{x} )·a^{4}·(q+(-W))·(W+(-q))·f(x,y)·g(x,y)
Deducción:
F(x,y) = e^{ int[ A_{y}·a^{2}·(q+(-W)) ]d[x] }·f(x,y)
G(x,y) = e^{ int[ A_{x}·a^{2}·(W+(-q)) ]d[y] }·g(x,y)
Gravito-débil de leptones orbitales:
Ley:
F(t)·G(t) = e^{(1/m)·(p+(-Z))}·e^{(1/m)·(Z+(-p))}·f(t)·g(t)
d_{t}[F(t)]·d_{t}[G(t)] = ...
... d_{t}[f(t)]·d_{t}[g(t)]+(1/m)^{2}·d_{t}[p+(-Z)]·d_{t}[Z+(-p)]·f(t)·g(t)
Ley:
Sea A(x,y) = (1/m)·< x,y > ==>
F(x,y)·G(x,y) = e^{ Anti-Potencial[ A(x,y)·a^{2}·< p+(-Z),Z+(-p) > ] }·f(x,y)·g(x,y)
d_{y}[F(x,y)]·d_{x}[G(x,y)] = ...
... d_{y}[f(x,y)]·d_{x}[g(x,y)]+( A_{x}·A_{y} )·a^{4}·(p+(-Z))·(Z+(-p))·f(x,y)·g(x,y)
Ley:
Sea A(y,x) = (1/m)·< y,x > ==>
F(x,y)·G(x,y) = e^{ Potencial[ A(y,x)·a^{2}·< p+(-Z),Z+(-p) > ] }·f(x,y)·g(x,y)
d_{x}[F(x,y)]·d_{y}[G(x,y)] = ...
... d_{x}[f(x,y)]·d_{y}[g(x,y)]+( A_{y}·A_{x} )·a^{4}·(p+(-Z))·(Z+(-p))·f(x,y)·g(x,y)
Desintegración alfa:
Ley:
F(t)·G(t) = e^{(1/m)·(n·(q+(-q))+W+(-q))}·e^{(1/m)·(q+(-W))}·f(t)·g(t)
d_{t}[F(t)]·d_{t}[G(t)] = ...
... d_{t}[f(t)]·d_{t}[g(t)]+(1/m)^{2}·d_{t}[n·(q+(-q))+W+(-q)]·d_{t}[q+(-W)]·f(t)·g(t)
Sea A(x,y) = (1/m)·< x,y > ==>
F(x,y)·G(x,y) = e^{ Anti-Potencial[ A(x,y)·a^{2}·< n·(q+(-q))+W+(-q),q+(-W) > ] }·f(x,y)·g(x,y)
d_{y}[F(x,y)]·d_{x}[G(x,y)] = ...
... d_{y}[f(x,y)]·d_{x}[g(x,y)]+( A_{x}·A_{y} )·a^{4}·(n·(q+(-q))+W+(-q))·(q+(-W))·f(x,y)·g(x,y)
Desintegración beta:
Ley:
F(t)·G(t) = e^{(1/m)·(n·(q+(-q))+q+(-W))}·e^{(1/m)·(W+(-q))}·f(t)·g(t)
d_{t}[F(t)]·d_{t}[G(t)] = ...
... d_{t}[f(t)]·d_{t}[g(t)]+(1/m)^{2}·d_{t}[n·(q+(-q))+q+(-W)]·d_{t}[W+(-q)]·f(t)·g(t)
Sea A(x,y) = (1/m)·< x,y > ==>
F(x,y)·G(x,y) = e^{ Anti-Potencial[ A(x,y)·a^{2}·< n·(q+(-q))+q+(-W),W+(-q) > ] }·f(x,y)·g(x,y)
d_{y}[F(x,y)]·d_{x}[G(x,y)] = ...
... d_{y}[f(x,y)]·d_{x}[g(x,y)]+( A_{x}·A_{y} )·a^{4}·(n·(q+(-q))+q+(-W))·(W+(-q))·f(x,y)·g(x,y)
Desintegración gamma:
Ley:
F(t)·G(t) = e^{(1/m)·n·(q+(-q))}·e^{(1/m)·(W+(-W))}·f(t)·g(t)
d_{t}[F(t)]·d_{t}[G(t)] = ...
... d_{t}[f(t)]·d_{t}[g(t)]+(1/m)^{2}·d_{t}[n·(q+(-q))]·d_{t}[W+(-W)]·f(t)·g(t)
Sea A(x,y) = (1/m)·< x,y > ==>
F(x,y)·G(x,y) = e^{ Anti-Potencial[ A(x,y)·a^{2}·< n·(q+(-q)),W+(-W) > ] }·f(x,y)·g(x,y)
d_{y}[F(x,y)]·d_{x}[G(x,y)] = ...
... d_{y}[f(x,y)]·d_{x}[g(x,y)]+( A_{x}·A_{y} )·a^{4}·n·(q+(-q))·(W+(-W))·f(x,y)·g(x,y)
Teorema:
x^{4}+ax^{2}+bx+c = 0 es resoluble
Demostración:
Sea x = u+iv ==>
(u+iv)^{4}+a·(u+iv)^{2}+b·(u+iv)+c = 0
(-6)·(uv)^{2}+2ai·(uv)+c = 0
uv = (1/(6i))·( (-a)+( a^{2}+(-1)·6c )^{(1/2)} ) ...
... || ...
uv = (1/(6i))·( (-a)+(-1)·( a^{2}+(-1)·6c )^{(1/2)} )
4i·(uv)·( u^{2}+(-1)·v^{2} ) = w·( u^{2}+(-1)·v^{2} )
w = (2/3)·( (-a)+( a^{2}+(-1)·6c )^{(1/2)} )
... || ...
w = (2/3)·( (-a)+(-1)·( a^{2}+(-1)·6c )^{(1/2)} )
u^{4}+(a+w)·u^{2}+bu = 0
u^{3}+(a+w)·u+b = 0
Teorema:
x^{5}+ax^{3}+bx^{2}+cx+d = 0 es resoluble
Demostración:
Sea x = u+iv ==>
(u+iv)^{5}+a·(u+iv)^{3}+b·(u+iv)^{2}+c·(u+iv)+d = 0
2bi·(uv) = d
uv = (d/(2bi))
F(uv) = vu = uv
El polinomio tiene 1 punto fijo,
y el coeficiente de Galois es n+2 = 3 y es resoluble
[Ah][ h es solución de uv ]
3a·(uv)·(u+iv)+10·(uv)^{2}·(u+iv) = w·(u+iv)
w = 3a·(d/(2bi))+10·(d/(2bi))^{2}
5·(uv)·(u^{3}+(-i)·v^{3}) = k·(u^{3}+(-i)·v^{3})
u^{5}+(a+k)·u^{3}+bu^{2}+(c+w)·u = 0
u^{4}+(a+k)·u^{2}+bu+(c+w) = 0
Teorema:
x^{6}+ax^{4}+bx^{3}+cx^{2}+dx+p = 0 es irresoluble
Demostración:
(-20)·i·(uv)^{3}+(-6)·a·(uv)^{2}+2ic·(uv)+p·(uv)^{0} = 0
F(uv) = vu = uv
El polinomio tiene 3 puntos fijos,
y el coeficiente de Galois es n+2 = 5 y es irresoluble
[Eh][ h no es solución de uv ]
uv = (z+(-1)·(1/10i)·a)
h^{3}+ph+q = 0
h | 1 | h | p+h^{2} | q+ph+h^{3} = 0
(z+(-h))·( z^{2}+hz+(p+h^{2}) ) = 0
uv = (1/10i)·a+( (1/2)·( (-h)+( h^{2}+(-4)·(h^{2}+p) )^{(1/2)} )
uv = (1/10i)·a+( (1/2)·( (-h)+(-1)·( h^{2}+(-4)·(h^{2}+p) )^{(1/2)} )
Teorema:
x^{7}+ax^{5}+bx^{4}+cx^{3}+dx^{2}+px+q = 0 es resoluble
Demostración:
(-6)·b·(uv)^{2}+2id·(uv)+q·(uv)^{0} = 0
F(uv) = vu = uv
El polinomio tiene 2 puntos fijos,
y el coeficiente de Galois es n+2 = 4 y es resoluble
[Ah][ h es solución de uv ]
Definición: [ de Grupo Galois ]
F(uv) = vu = uv
F(uv·ab) = F(uv)·ba
F(ab·uv) = ba·F(uv)
Teorema:
F((uv·ab)·pq) = F(uv·(ab·pq))
Demostración:
F((uv·ab)·pq) = F(uv·ab)·qp = ( F(uv)·ba )·qp = (vu·ba)·qp = vu·(ba·qp) = ...
... vu·( ba·F(pq) ) = vu·F(ab·pq) = F(uv·(ab·pq))
Teorema:
F(uv·ab) = F(uv)·ba = vu·ba = ba·vu = ba·F(uv) = F(ab·uv)
F(uv·ab) = vu·F(ab) = vu·ba = ba·vu = F(ab)·vu = F(ab·uv)
Definición: [ de coeficiente de Galois de un polinomio ]
Sea P(x) = P_{2n}(u+iv) ==>
Gal(P(x)) = Grado( Q_{n}(uv) )+2 = n+2
Sea P(x) = P_{2n+1}(u+iv) ==>
Gal(P(x)) = Grado( Q_{n+(-1)}(uv) )+2 = n+1
Teorema fundamental del Álgebra:
P_{n+1}(x) = P_{n}(x)·(x+(-1)·a_{n+1}) = (x+(-1)·a_{1})...(n)...(x+(-1)·a_{n})·(x+(-1)·a_{n+1})
Definición:
P(x) es resoluble <==> Grado[P(x)]+(-1)·Gal(P(x)) =[2]= Grado[P(x)]
P(x) es irresoluble <==> ¬( Grado[P(x)]+(-1)·Gal(P(x)) =[2]= Grado[P(x)] )
Teorema:
Sea P(x) = P_{2n}(u+iv) ==>
Si Gal(P(x)) = 2k+1 >] 5 ==> P(x) es irresoluble
Si Gal(P(x)) = 2k >] 5 ==> P(x) es resoluble
Demostración:
Por el teorema fundamental del Álgebra:
P_{2n}(u+iv) tiene 2n raíces
Por Cardano:
Q_{n}(uv) tiene n raíces
Sea Gal(P(x)) = n+2 = 2k+1 ==>
2n+(-1)·(2k+1) = 2·(n+(-k))+1 = 2p+1 =[2]= 1 & ¬( 1 =[2]= 2n )
P(x) es irresoluble
Sea Gal(P(x)) = n+2 = 2k ==>
2n+(-1)·2k = 2·(n+(-k)) = 2p =[2]= 0 & 0 =[2]= 2n
P(x) es resoluble
Teorema:
Sea P(x) = P_{2n+1}(u+iv) ==>
Si Gal(P(x)) = 2k+1 >] 5 ==> P(x) es irresoluble
Si Gal(P(x)) = 2k >] 5 ==> P(x) es resoluble
Demostración:
Por el teorema fundamental del Álgebra:
P_{2n+1}(u+iv) tiene 2n+1 raíces
Por Cardano:
Q_{n+(-1)}(uv) tiene n+(-1) raíces
Gal(P(x)) = n+1
Si n = 2k ==>
2n+1+(-1)·(2k+1) =[2]= 0 & ¬( 0 =[2]= 2n+1 )
P(x) es irresoluble
Si n = 2k+1 ==>
2n+1+(-1)·(2k+2) =[2]= (-1) =[2]= 1 & ( 1 =[2]= 2n+1 )
P(x) es resoluble
Hacéis historia sin sentido y lo vais a ver pronto,
porque llego a la resurrección de los muertos,
y se va a borrar mi literatura,
menos contact zero que es en dual.
Escrivid historias en dual,
si queréis que se conserven o conocimiento.
Teorema:
Sea f(x) continua ==>
Si [Ax][ x >] 0 ==> f(x) >] x ] ==> [Ec][ f(c) = 0 ]
Sea f(x) continua ==>
Si [Ax][ x [< 0 ==> f(x) [< x ] ==> [Ec][ f(c) = 0 ]
Demostración:
Sea u >] 0 ==>
f(u) >] u >] 0
(-1)·f(-u) [< (-u) [< 0
Teorema:
Sea a [< b ==>
Si f(x) = 2x+(-1)·(a+b) ==> [Ec][ f(c) = 0 ]
Sea a >] b ==>
Si f(x) = 2x+(-1)·(a+b) ==> [Ec][ f(c) = 0 ]
Demostración:
f(b) = b+(-a) >] 0
f(a) = a+(-b) [< 0
Teorema:
Sea f(x) = x^{2n+1}+(-a) ==> [E!c][ f(c) = 0 ]
Demostración:
Se define c = a^{( 1/(2n+1) )}
d_{x}[f(x)] = (2n+1)·x^{2n} >] 0
f(x) es creciente
Sea s >] 0 ==>
f(c+s) = (c+s)^{2n+1}+(-a) >] c^{2n+1}+(-a) = 0
f(c+(-s)) = (c+(-s))^{2n+1}+(-a) [< c^{2n+1}+(-a) = 0
Teorema:
Sea f(x) = x^{2n+2}+(-x) ] ==> [E!c][ d_{x}[f(c)] = 0 ]
Demostración:
d_{x}[f(x)] = (2n+2)·x^{2n+1}+(-1)
Se define c = ( 1/(2n+2) )^{( 1/(2n+1) )}
d_{xx}^{2}[f(x)] = (2n+2)·(2n+1)·x^{2n} >] 0
d_{x}[f(x)] es creciente
Sea s >] 0 ==>
d_{x}[f(c+s)] = (2n+2)·(c+s)^{2n+1}+(-1) >] (2n+2)·c^{2n+1}+(-1) = 0
d_{x}[f(c+(-s))] = (2n+2)·(c+(-s))^{2n+1}+(-1) [< (2n+2)·c^{2n+1}+(-1) = 0
Problema:
Demostrad:
Sea f(x) = x^{[2n+1:b]}+(-a) ==> [E!c][ f(c) = 0 ]
Arte:
[Ef(x)][ Si ( F(x) = int[ f(x) ]d[x] & lim[x = 0][ F(x) ] = ( 1 || (-1) ) ) ==> ...
... int[x = (-2)]-[2][ f(x) ]d[x] = 0 ]
Exposición:
f(x) = 0·(1/x)
F(x) = x^{0}
int[x = (-2)]-[2][ f(x) ]d[x] = 2^{0}+(-1)·(-2)^{0} = 1+(-1) = 0
Destructor:
int[x = (-2)]-[2][ f(x) ]d[x] = F(2)+(-1)·F(-2) = F(1+1)+F((-1)+(-1)) = F(1+(-1))+(-1)·F((-1)+1) = ...
... F(0)+(-1)·F(0) = 0·F(0) = 0
Arte:
[Ef(x)][ Si ( F(x) = int[ f(x) ]d[x] & lim[x = 1][ F(x) ] = 2n ) ==> ...
... int[x = (1/(2n))]-[(1/n)][ f(2nx) ]d[x] = 0 ]
Exposición:
f(x) = 2n·0·(1/x)
F(x) = 2nx^{0}
int[x = (1/(2n))]-[(1/n)][ f(2nx) ]d[x] = (1/(2n))·( 2n2^{0}+(-1)·2n1^{0} ) = 0
Destructor:
int[x = (1/(2n))]-[(1/n)][ f(2nx) ]d[x] = (1/(2n))·( F(2)+(-1)·F(1) ) = ...
... (1/(2n))·( F( (3/2)+(1/2) )+(-1)·F(1) = (1/(2n))·( F( (3/2)+(-1)·(1/2) )+(-1)·F(1) ) = ...
... (1/(2n))·( F(1)+(-1)·F(1) ) = (1/(2n))·2n·0 = 0
Teorema:
Si ( F(x) = int[ f(x) ]d[x] & lim[y = oo][ F(y) ] = c ) ==> ...
... lim[y = oo][ int[x = a]-[b][ f(x+y) ]d[x] ] = 0c
Demostración:
lim[y = oo][ int[x = a]-[b][ f(x+y) ]d[x] ] =lim[y = oo][ F(b+y)+(-1)·F(a+y) ] = F(oo)+(-1)·F(oo) = 0c
Teorema:
Si ( F(x) = int[ f(x) ]d[x] & lim[y = 0c][ F(y) ] = c ) ==> ...
... lim[y = 0][ int[x = (-c)]-[c][ y·f(xy) ]d[x] ] = 0c
Demostración:
lim[y = 0][ int[x = (-c)]-[c][ y·f(xy) ]d[x] ] =lim[y = 0][ F(cy)+(-1)·F((-c)·y) ] = F(0c)+(-1)·F(0c) = 0c
Llegamos a la resurrección de los muertos,
y se va a borrar todo el pasado de los fieles,
pero va a quedar el Facials de los mutantes alienígenas.
La cantante de Godspell,
estaba buena el día del Papa León en Madrid,
y supongo que la hizo alguien,
como yo hice el Cameltoe en Manresa.
El que la hizo ha llegado a la resurrección de los muertos,
porque se ha borrado del mundo.
Dual:
No estaba buena de cuerpo y cara ni tenía un cuerpo atlético.
Estaba buena de cuerpo y cara o tenía un cuerpo atlético.
Dual:
No estaba buena de cuerpo y cara y era fea.
Estaba buena de cuerpo y cara o era guapa.
Generador de destructor:
Estoy en un lugar haciendo esto,
no haciendo esto,
estoy haciendo esto.
Estoy en un lugar no haciendo esto,
haciendo esto,
no estoy haciendo esto.
Ya se ha borrado del Google Jûan Garriga Peralta y stroniken,
y he llegado a la resurrección de los muertos,
y en algún momento me voy al Cielo Aaru.
