Teorema:
A = k·( < a,0 > % < 0,a >)
B = k·( < a,0 > % < 0,(-a) >)
dim(A) = (1/2)
dim(B) = (1/2)
( < a,0 > % < 0,0 >) = (1/2)·( < a,0 > % < 0,a >)+(1/2)·( < a,0 > % < 0,(-a) >)
Lema: [ de gasto en defensa ]
PIB por cápita 8,400€
inversión del 5%:
(1/20)·32,000+20·1,600 = 33,600€ por cada 4 catalanohablantes.
Compra de armamento de 32,700€ por pack militar:
misil o torpedo o caja de proyectiles de cañón.
Lema:
Estrategia ganadora de venta,
del supremo o igual:
Make America empate again.
h(1) = 32,700€
F(1,1) = 1·1+(1+1) = 3
Estrategia ganadora de no venta,
del ínfimo o igual:
Make America first again.
h(2) = 65,400€
F(2,0) = 2·0+(2+0) = 2
Estrategias perdedoras:
h(1/2) = 16,350€
h(3/2) = 49,050€
F((1/2),(3/2)) = (3/4)+((1/2)+(3/2)) = (2.75)
F((3/2),(1/2)) = (3/4)+((3/2)+(1/2)) = (2.75)
Ley:
Hay gente que no es,
que no hay condenación.
Hay gente que es,
que hay condenación.
Deducción: [ por teoría de juegos ]
Jugadas ganadoras:
Joder a un esclavo infiel:
F(1,(-1)) = (-1)+(1+(-1)) = (-1)
F(2,(-1)) = (-2)+(2+(-1)) = (-1)
Jugadas perdedoras:
Joder a un señor fiel:
F(1,(-2)) = (-2)+(1+(-2)) = (-3)
F(2,(-2)) = (-4)+(2+(-2)) = (-4)
Ley:
Solo se puede conquistar con victoria,
haciendo que los infieles del Gestalt ignoren a los señores.
Solo se puede liberar con victoria,
no haciendo que los fieles del Gestalt ignoren a los señores.
Deducción: [ por teoría de juegos ]
Jugada ganadora:
Ignorar los infieles del Gestalt a los señores:
F(0,(-1)) = 0+(0+(-1)) = (-1)
Los infieles joden a infieles.
Jugada perdedora:
Ignorar los fieles del Gestalt a los señores:
F(0,(-2)) = 0+(0+(-2)) = (-2)
Los infieles joden a fieles.
Dual por dual de sabor:
Basik-kowetch-tate oil.
Acid-kowetch-tate wine.
Teorema:
int[x = 0]-[ln(oo)][ e^{(-1)·x^{n}} ]d[x] = (1/n!)
Teorema:
int[x = 0]-[ln(oo^{m+1})][ x^{m}·e^{(-x)} ]d[x] = m!
int[x = 0]-[ln(oo^{m+1})][ x^{m}·e^{(-1)·x^{n}} ]d[x] = (m!/n!)
Teorema:
int[x = 0]-[ln(oo^{(1/m)+1})][ x^{(1/m)}·e^{(-x)} ]d[x] = m
int[x = 0]-[ln(oo^{(1/m)+1})][ x^{(1/m)}·e^{(-1)·x^{n}} ]d[x] = m·( m!/(mn)! )
Demostración:
m = 0 & n = 1
int[x = 0]-[ln(oo)][ e^{(-1)·x} ]d[x] = [ (-1)·e^{(-1)·x} ]_{x = 0}^{x = ln(oo)} = 1
m = 1 & n = 1
int[x = 0]-[ln(oo^{2})][ xe^{(-1)·x} ]d[x] = ...
... [ (-1)·xe^{(-1)·x} ]_{x = 0}^{x = ln(oo^{2})}+[ (-1)·e^{(-1)·x} ]_{x = 0}^{x = ln(oo)} = 1
Teorema:
ln( (p+1)^{oo} ) = ( oo·log_{p+1}(2^{p}) )·ln(p+1) = ln(2^{p})·oo = ln(oo^{p})
Arte:
[En][ int[x = 0]-[ln(oo)][ e^{(-1)·x^{n}} ]d[x] = (1/n) ]
Exposición:
n = 1
Se define H(t) = int[x = 0]-[ln(oo)][ e^{(-1)·x^{n}·F(t)} ]d[x] ==>
d_{t}[H(t)] = f(t)·int[x = 0]-[ln(oo)][ (-1)·x^{n}·e^{(-1)·x^{n}·F(t)} ]d[x]
F(t) = x
d_{t}[H(t)] = d_{t}[x]·( 1/(n+1) )·(-1)
H(t) = x·( 1/(n+1) )·(-1) = F(t)·( 1/(n+1) )·(-1)
H( F^{o(-1)}(1) ) = F( F^{o(-1)}(1) )·( 1/(n+1) )·(-1) = ( 1/(n+1) )·(-1)
u(1) = (1/n)
v(1/n) = 0
H( F^{o(-1)}(1) ) = ( 1/(n+u(1)) )·(-1)^{u(1)} = ( 1/(n+v(1/n)) )·(-1)^{v(1/n)} = (1/n)
Arte:
[En][Em][ int[x = 0]-[ln(oo^{m+1})][ x^{m}·e^{(-1)·x^{n}} ]d[x] = ( 1/(m+n) ) ]
Exposición:
m = 0 & n = 1
Se define H(t) = int[x = 0]-[ln(oo^{m+1})][ x^{m}·e^{(-1)·x^{n}·F(t)} ]d[x] ==>
d_{t}[H(t)] = f(t)·int[x = 0]-[ln(oo^{m+1})][ (-1)·x^{m+n}·e^{(-1)·x^{n}·F(t)} ]d[x]
F(t) = x^{m+1}
d_{t}[H(t)] = (m+1)·x^{m}·d_{t}[x]·( 1/(m+n+1) )·(-1)
H(t) = x^{m+1}·( 1/(m+n+1) )·(-1) = F(t)·( 1/(m+n+1) )·(-1)
H( F^{o(-1)}(1) ) = F( F^{o(-1)}(1) )·( 1/(m+n+1) )·(-1) = ( 1/(m+n+1) )·(-1)
u(1) = (1/n)
v(1/n) = 0
H( F^{o(-1)}(1) ) = ( 1/(m+n+u(1)) )·(-1)^{u(1)} = ( 1/(m+n+v(1/n)) )·(-1)^{v(1/n)} = ( 1/(m+n) )
Teorema:
int[x = 0]-[oo][ ( sin(nx)/(nx) ) ]d[x] = (pi/n)
Demostración:
Se define H(t) = int[x = 0]-[oo][ e^{(-1)·(nxt)}·( sin(nx)/(nx) ) ]d[x] ==>
y = nx & d[y] = n·d[x]
H(t) = (1/n)·int[y = 0]-[oo][ e^{(-1)·(yt)}·( sin(y)/y ) ]d[y]
d_{t}[H(t)] = (-1)·(1/n)·int[y = 0]-[oo][ e^{(-1)·(yt)}·sin(y) ]d[y]
d_{t}[H(t)] = (-1)·(1/n)·( 1/(t^{2}+1) )
H(t) = (-1)·(1/n)·arc-tan(t)
H(0) = (-1)·(1/n)·(-pi) = (pi/n)
Teorema:
int[x = 0]-[oo][ ( cos(x/n)/x ) ]d[x] = ln(n)
Demostración:
Se define H(t) = int[x = 0]-[oo][ e^{(-1)·(xt)}·( cos(x/n)/x ) ]d[x] ==>
H(t) = int[x = 0]-[oo][ e^{(-1)·(xt)}·( cos(x/n)/x ) ]d[x]
d_{t}[H(t)] = (-1)·int[x = 0]-[oo][ e^{(-1)·(xt)}·cos(x/n) ]d[x]
d_{t}[H(t)] = (-1)·( t/(t^{2}+(1/n)^{2}) )
H(t) = (-1)·(1/2)·ln(t^{2}+(1/n)^{2})
H(0) = (-1)·(1/2)·2·ln(1/n) = ln(n)
Teorema:
int[x = ln(0)]-[0][ int[y = ln(0)]-[0][ e^{(-1)·( x^{n}+y^{n} )} ]d[y] ]d[x] = (1/n!)^{2}
Demostración:
e^{(-1)·( x^{n}+y^{n} )} [o(x || y)o] ( x /o(x || y)o/ x^{n} ) [o(x || y)o] ( y /o(x || y)o/ y^{n} )
Teorema:
int[x = 0]-[ln(oo)][ e^{(-1)·x^{2}} ]d[x] = (1/2!)
Demostración:
int[w = 0]-[2pi][ int[r = 0]-[oo][ e^{(-1)·r^{2}}·(1/4)·2r ]d[r] ]d[w] = (pi/2)
(-1)·e^{(-1)·r^{2}}·(1/4)·w = (-1)·e^{(-1)·( x^{2}+y^{2} )}·(1/4)·arc-tan(x/y)
arc-tan(0/oo)+(-1)·arc-tan(0/(-oo)) = 2pi
arc-tan(oo/(-oo)) = arc-tan(oo·(-0)) = arc-tan(oo·0) = arc-tan(oo/oo)
e^{(-1)·( x^{2}+y^{2} )}·(1/4)·8pi = (pi/2)
4 veces el cuadrante de la exponencial y valores convergentes de arco-tangente.
Teorema:
int[x = 0]-[pi][ e^{(-1)·( sin(x) )^{n}} ]d[x] = (2/n!)
Demostración:
int[x = 0]-[pi][ e^{(-1)·( sin(x) )^{n}}·( 1/d_{x}[sin(x)] ) ]d[sin(x)] = ...
... (-1)·e^{(-1)·( sin(x) )^{n}} [o(sin(x))o] ( sin(x) /o(sin(x))o/ ( sin(x) )^{n} )·( 1/cos(x) )
Teorema:
int[x = (-1)·(pi/2)]-[(pi/2)][ e^{(-1)·( cos(x) )^{n}} ]d[x] = (2/n!)
Demostración:
int[x = (-1)·(pi/2)]-[(pi/2)][ e^{(-1)·( cos(x) )^{n}}·( 1/d_{x}[cos(x)] ) ]d[cos(x)] = ...
... e^{(-1)·( cos(x) )^{n}} [o(cos(x))o] ( cos(x) /o(cos(x))o/ ( cos(x) )^{n} )·( 1/sin(x) )
Teorema:
int[x = (-oo)]-[oo][ ln(x^{n}+1) ]d[x] = n·( ln(2)+pi·i )
Área positiva > 1 + Área negativa < 1 = a la integral impropia
Verificación del método de Euler por Hôpital-Garriga:
( ln(x^{n}+1)+(-1) )·(x^{n}+1) [o(x)o] ( x /o(x)o/ (x^{n}+1) ) = ...
( ln(x^{n}+1)+(-1) )·(w^{n}+1) [o(x)o] ( w /o(x)o/ (w^{n}+1) ) = ...
( ln(x^{n}+1)+(-1) ) [o(x)o] ( 1 /o(x)o/ 1 ) = [ ln(x^{n}+1)+(-1) ]_{x = (-oo)}^{x = oo}
Demostración:
Se define H(t) = int[x = (-oo)]-[oo][ ln( (xF(t))^{n}+1 ) ]d[x] ==>
d_{t}[H(t)] = f(t)·int[x = (-oo)]-[oo][ ( 1/( (xF(t))^{n}+1 ) )·n·( xF(t) )^{n+(-1)}·x ]d[x]
F(t) = x
d_{t}[H(t)] = d_{t}[x]·int[x = (-oo)]-[oo][ ( 1/(x^{2n}+1) )·nx^{2n+(-1)} ]d[x]
d_{t}[H(t)] = d_{t}[x]·(1/2)·( ln(oo^{2n}+1) )+(-1)·ln((-oo)^{2n}+1) )
d_{t}[H(t)] = d_{t}[x]·(1/2)·2n·( ln(oo)+(-1)·ln(oo)+pi·i ) = d_{t}[x]·n·( ln(2)+pi·i )
H(t) = x·n·( ln(2)+pi·i ) = F(t)·n·( ln(2)+pi·i )
H( F^{o(-1)}(1) ) = F( F^{o(-1)}(1) ) n·( ln(2)+pi·i ) = n·( ln(2)+pi·i )
Arte:
[En][ int[x = (-1)]-[1][ ( 1/(x^{n}+(-1)) ) ]d[x] = (2/n) ]
Exposición:
n = 0
Se define H(t) = int[x = (-1)]-[1][ ( 1/( (xF(t))^{n}+(-1) ) ) ]d[x] ==>
d_{t}[H(t)] = f(t)·int[x = (-1)]-[1][ (-1)·( 1/( (xF(t))^{n}+(-1) ) )^{2}·n·( xF(t) )^{n+(-1)}·x ]d[x]
F(t) = x
d_{t}[H(t)] = d_{t}[x]·int[x = (-1)]-[1][ (-1)·( 1/(x^{2n}+(-1)) )^{2}·nx^{2n+(-1)} ]d[x]
d_{t}[H(t)] = d_{t}[x]·(1/2)·( oo+(-oo) ) = d_{t}[x]·(1/2)
H(t) = x·(1/2) = F(t)·(1/2)
u(1) = m
v(m) = (4/n)
H( F^{o(-1)}(1) ) = F( F^{o(-1)}(1) )·(1/2) = (1/2)·u(1) = (1/2)·v(m) = (2/n)
Momento de inercia Wronskiano:
Ley:
Sea ( x = a·cos(ut) & y = b·sin(ut) ) ==>
M·int[ x·d_{t}[y]+(-y)·d_{t}[x] ]d[t] = M·ab·(ut)
Ley:
M·ab·(ut)·(1/2)·d_{t}[w]^{2} = E·H(w)
w(t) = Anti-[ ( int[ (ut) ]d[s] /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ( (2/M)·(1/(ab) )·E )^{(1/2)}·t )
Ley:
Sea ( x = vt & y = g·(1/2)·t^{2} ) ==>
M·int[ x·d_{t}[y]+(-y)·d_{t}[x] ]d[t] = M·vg·(1/u)^{3}·(1/6)·(ut)^{3}
Ley:
M·vg·(1/u)^{3}·(ut)^{3}·(1/12)·d_{t}[w]^{2} = E·H(w)
w(t) = Anti-[ ( int[ (ut)^{3} ]d[s] /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...
... ( (12/M)·(1/(vg) )·u^{3}·E )^{(1/2)}·t )
Ley:
Sea ( x = vt & y = re^{ut+(-1)} ) ==>
M·int[ x·d_{t}[y]+(-y)·d_{t}[x] ]d[t] = M·vr·(1/u)·(ut+(-1))^{2}·er-h[2]( ut+(-1) )
Ley:
M·vr·(1/u)·(ut+(-1))^{2}·er-h[2]( ut+(-1) )·(1/2)·d_{t}[w]^{2} = E·H(w)
w(t) = Anti-[ ( int[ (ut+(-1))^{2}·er-h[2]( ut+(-1) ) ]d[s] /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...
... ( (2/M)·(1/(vr) )·uE )^{(1/2)}·t )
Ley:
Sea ( x = r·ln(ut) & y = vt ) ==>
M·int[ x·d_{t}[y]+(-y)·d_{t}[x] ]d[t] = M·vr·(1/u)·(ut)·( ln(ut)+(-2) )
Ley:
M·vr·(1/u)·(ut)·( ln(ut)+(-2) )·(1/2)·d_{t}[w]^{2} = E·H(w)
w(t) = Anti-[ ( int[ (ut)·( ln(ut)+(-2) ) ]d[s] /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...
... ( (2/M)·(1/(vr) )·uE )^{(1/2)}·t )
Ley:
Sea d_{t}[x] = ru·(ut+e)^{sin(ut+(pi/2))} ==>
d_{tt}^{2}[x] = ru^{2}·(ut+e)^{sin(ut+(pi/2))}·( cos(ut+(pi/2))·ln(ut+e)+sin(ut+(pi/2))·(1/(ut+e)) )
x(t) = r·(ut+e)^{sin(ut+(pi/2))} [o(ut)o] ( ut /o(ut)o/ sin(ut+(pi/2))·ln(ut+e) )
x(0/u) = re
d_{t}[x(0/u)] = rue
d_{tt}^{2}[x(0/u)] = ru^{2}
Deducción:
ln(f(z)) = ln(g(z)^{h(z)}) = h(z)·ln(g(z))
d_{z}[f(z)] = f(z)·d_{z}[ h(z)·ln(g(z)) ]
Ley:
La verdad implica la felicidad.
La falsedad implica el sufrimiento.
Deducción:
Creer-se una verdad es jugada ganadora:
F(n,1) = n·1+(n+1) = 2n+1
Creer-se una falsedad es jugada perdedora:
F(n,0) = n·0+(n+0) = n
n [< n+n = 2n < 2n+1
Ley:
Un heterosexual que es,
no puede ser homosexual que no es.
Un homosexual que no es,
no puede ser heterosexual que es.
I havere-kate-kute maket-tikaletch-tuted a coment saksahuaketch-kán,
of yu tenotitch-lán.
Yu havere-kate-kute maket-tikaletch-tuted a coment saksahuaketch-kán,
of me tenotitch-lán.
El American-Quetchua tiene 7 dialectos:
Centro americano:
-tikaletch-kal
Sur americano hawsnutch:
-tikaletch-tate
-tikaletch-tute
-tikalet-kazhe
-tikalet-kuzhe
gwzenen plana
-tikalet-huw
yuhened plana
-tikalet-shuw
yushened plana
Ley:
Sea I_{c}·(1/2)·d_{t}[w]^{2} = E·H(w) ==>
Si d[ d[I_{c}] ] = Mrv·d[w]d[t] ==>
w(t) = Anti-[ ( ( (1/2)·s^{2} [o(s)o] int[ (ut) ]d[s] ) /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...
... ( (2/M)·(1/(rv))·uE )^{(1/2)}·t )
Ley:
Sea I_{c}·(1/2)·d_{t}[w]^{2} = E·H(w) ==>
Si d[ d[I_{c}] ] = Mrgt·d[w]d[t] ==>
w(t) = Anti-[ ( ( (1/2)·s^{2} [o(s)o] int[ (ut)^{2} ]d[s] ) /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...
... ( (4/M)·(1/(rg))·E )^{(1/2)}·ut )
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