Ley:
Pared curva con dos pilares,
con pared maestra en simetría (3/2),
de anulación de momentos de una Te (3/2) entre pilares:
Sea 0 [< r [< R ==>
H(r) = R·(3/2)^{(r/R)}
FR = FR·(3/2)^{(r/R)}+(a/t)
(3/2)·FR = FR·(3/2)^{(r/R)}+(a/t)
Ley:
Pared curva con dos pilares,
con pared maestra en simetría (5/4),
de anulación de momentos de una Te (5/4) entre pilares:
Sea 0 [< r [< R ==>
H(r) = R·(5/4)^{(r/R)}
FR = FR·(5/4)^{(r/R)}+(a/t)
(5/4)·FR = FR·(5/4)^{(r/R)}+(a/t)
Ley:
saliente de triángulo de un pilar:
Sea 0 [< x [< R ==>
Fmx = Fmx+(a/t)
Ley:
saliente de trapecio de dos pilares:
Sea 0 [< x [< R ==>
F·(mx+R) = F·(mx+R)+(a/t)
Ley: [ viaje a un planeta ]
0 [< w [< pi
r·d_{t}[w] = pj·|cos(w/2)|+v·|sin(w/2)|
pi [< w [< 2pi
r·d_{t}[w] = v·|sin(w/2)|+qk·|cos(w/2)|
Ley: [ regreso de un planeta ]
0 >] w >] (-pi)
r·d_{t}[w] = qk·|cos(w/2)|+v·|sin(w/2)|
(-pi) >] w >] (-1)·2pi
r·d_{t}[w] = v·|sin(w/2)|+pj·|cos(w/2)|
Ley: [ emisión de un faro inter-plexo ]
0 [< w [< n·(pi/2)
r·d_{t}[w] = pj·|cos(w/n)|+v·|sin(w/n)|
n·(pi/2) [< w [< n·pi
r·d_{t}[w] = v·|sin( (1+(-1)·(1/n))·w )|+qk·|cos( (1+(-1)·(1/n))·w )|
Ley: [ recepción de un faro inter-plexo ]
0 >] w >] (-n)·(pi/2)
r·d_{t}[w] = qk·|cos( (1+(-1)·(1/n))·w )|+v·|sin( (1+(-1)·(1/n))·w )|
(-n)·(pi/2) >] w >] (-n)·pi
r·d_{t}[w] = v·|sin(w/n)|+pj·|cos(w/n)|
Ley: [ de Kennedy ]
A los 44 dejó de creer-se Jesucristo,
y lo mataron a los 45,
en no estar bautizado en Espíritu Santo.
a = 23 & b = 19
[01][05][08][05] = 19
[07][11][14][11] = 43
Formula:
-(SO)-(SCO)=(SNNOH)=(SCO)-
Teorema:
d_{x}[y] = ( (x^{2}+2xy)/x^{2} )
y(x) = x^{2}+(-x)
y = ux
Teorema:
d_{x}[y] = ( (x^{2}+(n+1)·xy)/x^{2} )+(n+(-1))
y(x) = x^{n+1}+(-x)
Teorema:
d_{x}[y] = ( (x^{2}+(n+1)·xy)/x^{2} )
y(x) = (1/a)·Anti-[ ( s /o(s)o/ ( s+(n+1)·(1/(ax))·(1/2)·s^{2} ) ) ]-(ax)
Teorema:
d_{x}[y] = ( (xy+y^{2})/x^{2} )
y(x) = (-1)·( x/ln(x) )
y = ux
Teorema:
d_{x}[y] = ( (xy+ny^{2})/x^{2} )
y(x) = (-1)·( x/ln(x^{n}) )
y = ux
Teorema:
d_{x}[y] = ( (y+x)/x )
y(x) = ln(x)·x
y = ux
Teorema:
d_{x}[y] = (y/x)^{n}
y(x) = x
y = ux
Demostración:
d_{x}[u]·x = u^{n}+(-u)
Sea u = 1 ==>
d_{x}[1]·x = 0
d[1] = (0/x)·d[x]
Si [AC][ int[ d[1] ] != int[ (0/x) ]d[x]+C ] ==>
1 = ln(2)+C
C = 1+(-1)·ln(2)
Ley: [ de Khamenei ]
A los 49 se creyó Mahoma,
y a los 86 ha muerto.
Sea a = 47 & b = 37 ==>
[08][08][13][08] = 37
[14][14][19][14] = 61
Fórmula:
=C=C=C=C=(PNNH)=C=C=
Arte-físico: [ de velocidad en juegos de váter ]
m·d_{tt}^{2}[z] = P·(x^{2}+y^{2})
[Et][ d_{t}[z] = (P/m)·(1/u)·(x^{2}+y^{2}) ]
Exposición:
t = (1/u)
d_{t}[z] = (P/m)·(1/u)·(ut)·(x^{2}+y^{2}) = (P/m)·(1/u)·(ut)^{(1/2)+(1/2)}·(x^{2}+y^{2}) = ...
... (P/m)·(1/u)·(ut)^{(1/2)+(-1)·(1/2)}·(x^{2}+y^{2}) = (P/m)·(1/u)·(ut)^{0}·(x^{2}+y^{2}) = ...
... (P/m)·(1/u)·(x^{2}+y^{2})
Arte-físico: [ de posición en juegos de váter ]
m·d_{tt}^{2}[z] = P·(x^{2}+y^{2})
[Et][ z(t) = (P/m)·(1/2)·(1/u)^{2}·(x^{2}+y^{2}) ]
Exposición:
t = (1/u)
z(t) = (P/m)·(1/2)·(1/u)^{2}·(ut)^{2}·(x^{2}+y^{2}) = ...
... (P/m)·(1/2)·(1/u)^{2}·(ut)^{1+1}·(x^{2}+y^{2}) = ...
... (P/m)·(1/2)·(1/u)^{2}·(ut)^{1+(-1)}·(x^{2}+y^{2}) = ...
... (P/m)·(1/2)·(1/u)^{2}·(ut)^{0}·(x^{2}+y^{2}) = ...
... (P/m)·(1/2)·(1/u)^{2}·(x^{2}+y^{2})
Arte-físico: [ de aceleración en juegos de des-propiedad ]
m·d_{tt}^{2}[z] = qg·( (n+1)+(-1)·(ut) )
[Et][ d_{tt}^{2}[z] = (q/m)·g ]
Exposición:
t = n·(1/u)
d_{tt}^{2}[z] = (q/m)·g·( (n+1)+(-1)·(ut) ) = (q/m)·g·( (n+1)+(-1)·(ut) )^{(1/2)+(1/2)} = ...
... (q/m)·g·( (n+1)+(-1)·(ut) )^{(1/2)+(-1)·(1/2)} = (q/m)·g·( (n+1)+(-1)·(ut) )^{0} = (q/m)·g
Arte-físico: [ de velocidad en juegos de des-propiedad ]
m·d_{tt}^{2}[z] = qg·( (n+1)+(-1)·(ut) )
[Et][ d_{t}[z] = (q/m)·g·(1/(2u)) ]
Exposición:
t = n·(1/u)
d_{t}[z] = (q/m)·g·(1/(2u))·( (n+1)+(-1)·(ut) )^{2} = ...
... (q/m)·g·(1/(2u))·( (n+1)+(-1)·(ut) )^{1+1} = ...
... (q/m)·g·(1/(2u))·( (n+1)+(-1)·(ut) )^{1+(-1)} = ...
... (q/m)·g·(1/(2u))·( (n+1)+(-1)·(ut) )^{0} = (q/m)·g·(1/(2u))
Arte:
Sea F(x) = int[f(x)]d[x] ==>
[EF(x)][ Si lim[x = 0][ F(x) ] = 2 ==> lim[x = 1][ int[f(2x)]d[x] ] = 1 ]
Exposición:
F(x) = 2x^{0}
int[f(2x)]d[x] = (1/2)·int[f(2x)]d[2x] = (1/2)·F(2x) = (2/2)·(2x)^{0} = 1
lim[x = 1][ int[f(2x)]d[x] ] = lim[x = 1][ (1/2)·int[f(2x)]d[2x] ] = lim[x = 1][ (1/2)·F(2x) ] = ...
... (1/2)·F(2) = (1/2)·F(1+1) = (1/2)·F(1+(-1)) = (1/2)·F(0) = (2/2) = 1
Arte:
Sea F(x) = int[f(x)]d[x] ==>
[EF(x)][ Si lim[x = 0][ F(x) ] = 2n ==> lim[x = (1/n)][ int[f(2nx)]d[x] ] = 1 ]
Exposición:
F(x) = 2nx^{0}
int[f(2nx)]d[x] = (1/(2n))·int[f(2nx)]d[2nx] = (1/(2n))·F(2nx) = ( (2n)/(2n) )·(2nx)^{0} = 1
lim[x = (1/n)][ int[f(2nx)]d[x] ] = lim[x = (1/n)][ (1/(2n))·int[f(2nx)]d[2nx] ] = ...
... lim[x = (1/n)][ (1/(2n))·F(2nx) ] = (1/(2n))·F(2) = (1/(2n))·F(1+1) = (1/(2n))·F(1+(-1)) = ...
... (1/(2n))·F(0) = ( (2n)/(2n) ) = 1
Future: [ Stowed-English ]
wilore-kate speaketch-tated
wiloremitch speaketch-tated
wilorewitch speaketch-tated
wiloren-kate speaketch-tated
Futuro: [ Italiano ]
guilore-po parlato
guiloremo parlato
guilorewo parlato
guiloren-po parlato
Present:
I speaketch-tate
Conjugated in italiano.
Condicional:
I speaketch-tatings
Conjugated in italiano.
Dual:
If not se hubiese-kate to gow the glory of the American-Hawsnutch,
the president not needings not-zhing catalan.
Se havere-kate to gow the glory of the American-Hawsnutch,
and the president need some-zhing catalan.
Dual:
Si no se hubiese-po de ire la gloria-jjore del Italiano,
el presidente no necesitaríe-po nada catalano.
Se havere-po de ire la gloria-jjore del Italiano,
y el presidente necesitare-po algo catalano.
Dual:
Wies haveremitch over can-set,
inter music tecnok fighted,
that maketch-tate deatrating Khamenei,
and staremitch up wheelers motor comand wrise in the iranish war
Wies hubiesemitch under can-set,
awtter music tecnok fighted,
that not maketch-tate deatrating Khamenei,
and staríemitch dawn wheelers motor comand wrise in the iranish war
La estructura de 37 dimensiones es el siguiente nudo en mi sangre:
F(z) = z^{3}+z^{31}+z^{3}
Son tres cadenas y una de 31 piezas y dos de 3 piezas,
y se irá el camino que las atraviesa.
Ley:
O Recordáis el modus ponens:
No sois dioses de los hombres ni del universo ==> Se dice que un fiel se destruye.
O Recordáis el modus caguens:
No sois dioses de los hombres ni del universo ==> No se dice que un fiel se destruye.
Teorema:
El señor no es mayor que el enviado
Delirio:
El señor es mayor que el enviado
Ley:
O Recordáis el modus ponens:
Los ateos no son ==> Se dice que un ateo se destruye.
Recordáis el modus caguens:
Los ateos no son ==> No se dice que un ateo se destruye.
Teorema:
El esclavo no es mayor que su señor
Delirio:
El esclavo es mayor que su señor
Teorema:
Si d_{xx}^{2}[y]+a(x)·d_{x}[y]+b(x)·y(x) = 0 ==>
d_{xx}^{2}[h]+a(x)·d_{x}[h]+b(x)·h(x) = 0
h(x) = y·int[ (1/y)^{2}·e^{(-1)·int[ a(x) ]d[x]} ]d[x]
Demostración:
ln( Wronsky(h(x),y(x)) ) = (-1)·int[ a(x) ]d[x]
Wronsky(h(x),y(x)) = e^{(-1)·int[ a(x) ]d[x]}
d_{x}[ ( h(x) / y(x) ) ] = (1/y)^{2}·e^{(-1)·int[ a(x) ]d[x]}
( h(x) / y(x) ) = int[ (1/y)^{2}·e^{(-1)·int[ a(x) ]d[x]} ]d[x]
Ley:
d_{tt}^{2}[y] = u·(ut)^{n}·d_{t}[y]+u^{2}·n·(ut)^{n+(-1)}·y(t)
y(t) = re^{( 1/(n+1) )·(ut)^{n+1}}
h(t) = r·( e^{( 1/(n+1) )·(ut)^{n+1}}·int[ e^{(-1)·( 1/(n+1) )·(ut)^{n+1}} ]d[ut] )
Ley:
d_{tt}^{2}[y] = (1/t)·d_{t}[y]+(-1)·(1/t)^{2}·y(t)
y(t) = r·(ut)
h(t) = r·( (ut)·int[ (1/(ut)) ]d[ut] ) = r·ln(ut)·(ut)
Ley:
d_{tt}^{2}[y] = (1/t)·ln(ut)·d_{t}[y]+(-1)·(1/t)^{2}·y(t)
y(t) = r·( ln(ut)+1 )
h(t) = r·( ( ln(ut)+1 )·int[ ( 1/(ln(ut)+1) )^{2}·e^{(1/2)·( ln(ut) )^{2}} ]d[ut] )
Definición:
( g(x) )^{[o(%)o] [o(%)o]} = g(x)
d_{x}[ ( f(x) )·( g(x) )^{[o(%)o] (1/n)} ] = d_{x}[f(x)]·( g(x) )^{[o(%)o] (1/n)·n}+( f(x) )·d_{x}[g(x)]
Teorema:
d_{x}[ ( f(x) )·( g(x) )^{[o(%)o]} ] = d_{x}[f(x)]·( g(x) )^{[o(%)o] (1/1)·1}+f(x)·d_{x}[g(x)]
Demostración:
Sea n = 1
Teorema:
d_{x}[ ( g(x) )^{[o(%)o] (1/n)} ] = d_{x}[g(x)]
Demostración:
Sea f(x) = 1 ==>
d_{x}[ 1·( g(x) )^{[o(%)o] (1/n)} ] = d_{x}[1]·( g(x) )^{[o(%)o] (1/n)·n}+1·d_{x}[g(x)]
Demostración:
d_{x}[ ( g(x) )^{[o(%)o]} ] = d_{x}[g(x)]
( g(x) )^{[o(%)o]} = g(x) = ( g(x) )^{1}
Teorema:
[Ec][ [o(%)o] (1/n) = [1:c^{(1/n)}+(-c)] ]
Demostración:
d_{x}[ ( g(x) )^{[o(%)o] (1/n)} ] = d_{x}[g(x)]
( g(x) )^{[o(%)o] (1/n)}+(-1)·c^{(1/n)} = g(x)+(-c)
( g(x) )^{[o(%)o] (1/n)} = ( g(x) )^{[1:c^{(1/n)}+(-c)]}
[o(%)o] (1/n) = [1:c^{(1/n)}+(-c)]
Anomalías:
c^{[o(%)o] (1/n)} = c^{[1:c^{(1/n)}+(-c)]}
c^{[o(%)o] (1/n)·n} = c^{[1:c^{(1/n)}+(-c)]·n}
c = ( c+c^{(1/n)}+(-c) )^{n}
c^{(1/n)}+(-c) = c^{(1/n)}+(-c)
Teorema:
d_{xx}^{2}[y] = x^{n}·d_{x}[y]+(-1)·( y(x) )^{n}
y(x) = x
h(x) = x·( int[ (1/x)^{2}·e^{( 1/(n+1) )·x^{n+1}} ]d[x] )^{[o(%)o] (1/n)}
Demostración:
d_{x}[h] = ...
... ( int[ (1/x)^{2}·e^{( 1/(n+1) )·x^{n+1}} ]d[x] )^{[o(%)o]}+(1/x)·e^{( 1/(n+1) )·x^{n+1}}
Teorema:
d_{xx}^{2}[y] = (-1)·x^{n}·d_{x}[y]+( y(x) )^{n}
y(x) = x
h(x) = x·( int[ (1/x)^{2}·e^{(-1)·( 1/(n+1) )·x^{n+1}} ]d[x] )^{[o(n)o] (1/n)}
Definición:
d_{x}[ ( h(x) / y(x) )^{[o(%)o] n} ] = (1/y)^{2}·( d_{x}[h]·( y(x) )^{n}+(-1)·( h(x) )^{n}·d_{x}[y] )
Wronsky-[n]-(h(x),y(x)) = d_{x}[h]·( y(x) )^{n}+(-1)·( h(x) )^{n}·d_{x}[y]
d_{x}[ Wronsky-[n]-(h(x),y(x)) ] = d_{xx}^{2}[h]·( y(x) )^{n}+(-1)·( h(x) )^{n}·d_{xx}^{2}[y]
Teorema:
Si d_{xx}^{2}[y]+a(x)·d_{x}[y]+b(x)·( y(x) )^{n} = 0 ==>
d_{xx}^{2}[h]+a(x)·d_{x}[h]+b(x)·( h(x) )^{n} = 0
h(x) = y·( int[ (1/y)^{2}·e^{(-1)·int[ a(x) ]d[x]} ]d[x] )^{[o(%)o] (1/n)}
Teorema:
d_{xx}^{2}[y] = e^{(n+(-1))·x}·d_{x}[y]+( (-1)+(1/e^{(n+(-1))·x}) )·( y(x) )^{n}
y(x) = e^{x}
h(x) = e^{x}·( int[ (1/e^{2x})·e^{( 1/(n+(-1)) )·e^{(n+(-1))·x}} ]d[x] )^{[o(%)o] (1/n)}
Demostración:
d_{x}[h] = ...
... e^{x}·( int[ (1/e^{2x})·e^{( 1/(n+(-1)) )·e^{(n+(-1))·x}} ]d[x] )^{[o(%)o]}+...
... (1/e^{x})·e^{( 1/(n+(-1)) )·e^{(n+(-1))·x}}
Teorema:
d_{xx}^{2}[y] = x·( ln(x) )^{n}·d_{x}[y]+(-1)·( 1+(1/x)^{2}·(1/ln(x))^{n} )·( y(x) )^{n}
y(x) = ln(x)
h(x) = ...
... ln(x)·( int[ (1/ln(x))^{2}·e^{(1/2)·x^{2} [o(x)o] ( ln(x)·x+(-x) )^{[o(x)o] n}} ]d[x] )^{[o(%)o] (1/n)}
Demostración:
d_{x}[h] = ...
... ( (1/x)·int[ (1/ln(x))^{2}·e^{(1/2)·x^{2} [o(x)o] ( ln(x)·x+(-x)· )^{[o(x)o] n}} ]d[x] )^{[o(%)o]}+...
... (1/ln(x))·e^{(1/2)·x^{2} [o(x)o] ( ln(x)·x+(-x)· )^{[o(x)o] n}}
Teorema:
int[x = 0]-[a][ int[y = 0]-[x][ int[z = 0]-[y][ xyz ]d[z] ]d[y] ]d[x] = (1/48)·a^{6}
Teorema:
int[x = 0]-[a][ int[y = 0]-[x][ int[z = 0]-[y][ x+y+z ]d[z] ]d[y] ]d[x] = (1/4)·a^{4}
Demostración:
x+y+z ==> xy+y^{2}+(1/2)·y^{2} ==> (1/2)·x^{3}+(1/2)·x^{3} ==> (1/4)·a^{4}
Teorema:
int[x = 0]-[a][ int[y = 0]-[x][ int[z = 0]-[y][ e^{x+y+z} ]d[z] ]d[y] ]d[x] = (1/6)·e^{3a}
Dual: [ of desembobulator ]
He stare-kate-kute gowetch-tating to the war,
it-shete like-it it-shete.
She stare-kate-kute gowetch-tating to the war,
it-hete like-it it-hete.
Dual: [ of desembobulator ]
He stare-kate-kute speaketch-tating,
shere cloval-sate like-it.
She stare-kate-kute speaketch-tating,
here cloval-sate like-it.
Ley:
Se tiene condenación,
y no amando al próximo dentro del próximo,
no siendo el Mal atacante.
No se tiene condenación,
no amando al prójimo dentro del próximo,
siendo del Mal defensivo.
Juego al Mal:
De proyecciones visuales del prójimo extraterrestre en el próximo humano:
1-2-3 Azúcar cada día = 2^{1}+(-1) días
4-5-6 Medicación cada semana = 2^{3}+(-1) días
Azúcar cada día:
1-2-3 no lo pinchan
4-5-6 lo pinchan
Medicación cada semana:
1-2-3 no lo pinchan
4-5-6 lo pinchan
Teorema:
Si [Ek][An][ n > k ==> ln(n) < a_{n} ] ==> a_{n} no está dominada superiormente
Demostración:
Sea s > 0 ==>
Se define m > max{k,e^{s}} ==>
Sea n > m ==>
s < ln(m) < ln(n) < a_{n}
Teorema:
Si [Ek][An][ n > k ==> e^{n} < a_{n} ] ==> a_{n} no está dominada superiormente
Demostración:
Sea s > 0 ==>
Se define m > max{k,ln(s)} ==>
Sea n > m ==>
s < e^{m} < e^{n} < a_{n}
Teorema:
lim[n = oo][ (1/n)·sin(n) ] = 0
Demostración:
Sea s > 0 ==>
Se define k > s ==>
Sea n > k ==>
| (1/n)·sin(n) | = |(1/n)|·|sin(n)| [< (1/n) < (1/k) < s
Definición:
f(x) está en el continuo <==> [Es][ 0 [< f(s) [< 1 ]
Teorema:
Si ( f(x) está en el continuo & g(x) está en el continuo ) ==> f(x)+g(x) está en el continuo
Demostración
0 [< f(j) [< 1 & 0 [< g(k) [< 1
(0/2) [< (1/2)·f(j) [< (1/2) & (0/2) [< (1/2)·g(k) [< (1/2)
Se define ( f(j) = 2·f(s) & g(k) = 2·g(s) ) ==>
0 [< (1/2)·( 2·f(s)+2·g(s) ) [< 1
Anexo:
f(x) = x^{p} & g(x) = x^{q}
j·(1/2)^{(1/p)} = k·(1/2)^{(1/q)} = s
j = (1/2)^{(1/q)} & k = (1/2)^{(1/p)}
Teorema:
Si f(x) está en el continuo ==> w·f(x) está en el continuo
Demostración
0 [< f(k) [< 1
Se define f(k) = w·f(s) ==>
0 [< w·f(s) [< 1
Teorema:
Si f(x) = x^{n} ==> f(x) está en el continuo
Demostración:
Se define 0^{(1/n)} [< s [< 1 ==>
0 = 0^{(n/n)} [< s^{n} [< 1^{n} = 1
Teorema:
Si f(x) = e^{x} ==> f(x) está en el continuo
Demostración:
Se define (-oo) [< s [< 0 ==>
0 = e^{(-oo)} [< e^{s} [< e^{0} = 1
Teorema:
Si f(x) = xe^{x} ==> f(x) está en el continuo
Demostración:
Se define 0 [< s [< (1/e) ==>
0 = 0·e^{0} [< se^{s} [< (1/e)·e^{(1/e)} [< (1/e)·e = 1
Teorema:
int[ tan(x) ]d[x] = ( sin(x)+ln(cos(x)) [o(x)o] cos(x) ) [o(x)o] (-1)·cos(x)
Teorema:
int[ cot(x) ]d[x] = ( (-1)·cos(x)+ln(sin(x)) [o(x)o] sin(x) ) [o(x)o] sin(x)
Teorema:
int[ ( ax+bx^{(1/2)}+c )^{n} ]d[x] = ...
... (1/(n+1))·( ax+bx^{(1/2)}+c )^{n+1} [o(x^{(1/2)})o] ln(2ax^{(1/2)}+b) [o(x^{(1/2)})o] (1/(2a))·x
Demostración:
x = y^{2} & d[x] = 2y·d[y]
Teorema:
int[x = (-1)]-[1][ ( 1/(x^{2}+(-1)) ) ]d[x] = ln(0)
Demostración:
F(x) = (1/2)·( (-1)·ln(x+1)+ln(x+(-1)) )
Teorema:
int[x = (-1)]-[1][ ( 1/(x^{2n}+(-1)) ) ]d[x] = (1/n)·ln(0)
Demostración:
Hôpital-Garriga:
(-1)^{2n} = 2n·(-1)^{2n+(-1)}·(-1) = 2n·(-1)·(-1)
Dual:
If I stubiese-kate speaketch-tating,
wizh aliens awtter of the internet,
staríe-kate a one a page a gromenawer a Luigi brawther.
I stare-kate speaketch-tating,
wizh aliens inter of the internet,
and stare-kate a one a page a gromenawer a Mario brawther.
