Artículos:
un [o] una <==> [E$1$ [x] ][ [x] es nombre ]
los [o] las <==> [A$...$ [x] ][ [x] es nombre ]
unos [o] unas <==> [E$...$ [x] ][ [x] es nombre ]
los n [o] las n <==> [A$n$ [x] ][ [x] es nombre ]
unos n [o] unas n <==> [E$n$ [x] ][ [x] es nombre ]
este [o] esta <==> [A?1? [x] ][ [x] es nombre ]
ese [o] esa <==> [E?1? [x] ][ [x] es nombre ]
estos [o] estas <==> [A?...? [x] ][ [x] es nombre ]
esos [o] esas <==> [E?...? [x] ][ [x] es nombre ]
estos n [o] estas n <==> [A?n? [x] ][ [x] es nombre ]
esos n [o] esas n <==> [E?n? [x] ][ [x] es nombre ]
Teorema:
int[ ( 1/ln(x) ) ]d[x] = ln( ln(x) ) [o(x)o] (1/2)·x^{2}
Teorema:
int[ ( ln(x) )^{s} ]d[x] = ( 1/(s+1) )·( ln(x) )^{s+1} [o(x)o] (1/2)·x^{2}
Teorema:
int[ ( 1/ln(ax+b) ) ]d[x] = ln( ln(ax+b) ) [o(x)o] (1/2)·(ax+b)^{2} [o(x)o] (1/a)·x
Teorema:
int[ ( ln(ax+b) )^{s} ]d[x] = ( 1/(s+1) )·( ln(ax+b) )^{s+1} [o(x)o] (1/2)·(ax+b)^{2} [o(x)o] (1/a)·x
Teorema:
int[ ( 1/ln(ax^{2}+bx+c) ) ]d[x] = ...
... ln( ln(ax^{2}+bx+c) ) [o(x)o] (1/2)·(ax^{2}+bx+c)^{2} [o(x)o] ...
... ln(2ax+b) [o(x)o] ( 1/(2a) )·x
Teorema:
int[ ( ln(ax^{2}+bx+c) )^{s} ]d[x] = ...
... ( 1/(s+1) )·( ln(ax^{2}+bx+c) )^{s+1} [o(x)o] (1/2)·(ax^{2}+bx+c)^{2} [o(x)o] ...
... ln(2ax+b) [o(x)o] ( 1/(2a) )·x
Teorema:
int[ ( 1/sin(x) ) ]d[x] = (-1)·cos(x)+ln(sin(x)) [o(x)o] sin(x)
Teorema:
int[ ( sin(x) )^{s} ]d[x] = ...
... ( 1/(s+1) )·( sin(x) )^{s+1} [o(x)o] ( sin(x)+ln(cos(x)) [o(x)o] cos(x) )
Teorema:
int[ ( 1/cos(x) ) ]d[x] = sin(x)+ln(cos(x)) [o(x)o] cos(x)
Teorema:
int[ ( cos(x) )^{s} ]d[x] = ...
... ( 1/(s+1) )·( cos(x) )^{s+1} [o(x)o] ( cos(x)+ln(sin(x)) [o(x)o] (-1)·sin(x) )
Teorema: [ de las distribuciones de Garriga-Poisson ]
( (p+1)·(1/e) )·int[x = 0]-[1][ x^{p}·e^{x^{n}}]d[x] = 1
( (p+1)·e )·int[x = 0]-[1][ x^{p}·e^{(-1)·x^{n}}]d[x] = 1
Demostración:
int[ x^{p}·e^{x^{n}}]d[x] = x^{p+1}·er-h[nk+p+1](x^{n})
er-h[nk+p+1](1^{n}) = er-h[0·k+p+1](1^{0}) = er-h[p+1](1) = e·( 1/(p+1) )
int[ a_{k}·( (-1)·x^{n} )^{k} ]d[x] = a_{k}·( 1/(nk+p+1) )·(-1)^{k}·x^{nk+p+1}
Teorema: [ de las esperanzas de Garriga-Poisson ]
( (p+1)·(1/e) )·int[x = 0]-[1][ x^{p+1}·e^{x^{n}}]d[x] = ( (p+1)/(p+2) )
( (p+1)·e )·int[x = 0]-[1][ x^{p+1}·e^{(-1)·x^{n}}]d[x] = ( (p+1)/(p+2) )
Teorema:
lim[x = 0][ ln(x) [o(x)o] x^{2} ] = int[ lim[x = 0][ ( (2x)/x ) ] ]d[x] = lim[x = 0][ 2x ] = 0
Teorema: [ de la integral de Poisson ]
lim[x = 0][ e^{x^{n}} /o(x)o/ x^{n} ] = ...
... int[ lim[x = 0][ ( (nx^{n+(-1)}·e^{x^{n}})/nx^{n+(-1)} ) ] ]d[x] = lim[x = 0][ 1·x ] = 0
Teorema: [ de la distribución de Poisson ]
(1/e)·int[x = 0]-[1][ e^{x^{2}} ]d[x] = 1
Teorema: [ de la esperanza de Poisson ]
(1/e)·int[x = 0]-[1][ xe^{x^{2}} ]d[x] = (1/2)·( 1+(-1)·(1/e) )
Teorema: [ de la integral de D'Alembert ]
lim[x = 0][ ( e^{x}+(-x) ) /o(x)o/ x^{2} ] = ...
... int[ lim[x = 0][ ( (e^{x}+(-1))/(2x) ) ] ]d[x] = lim[x = 0][ (1/2)·x ] = 0
Teorema: [ de la distribución de D'Alembert ]
( 1/(e+(-1)) )·int[x = 0]-[1][ ( (e^{x}+(-1))/(2x) ) ]d[x] = 1
Teorema: [ de la esperanza de D'Alembert ]
( 1/(e+(-1)) )·int[x = 0]-[1][ ( (e^{x}+(-1))/2 ) ]d[x] = (1/2)·( (e+(-2))/(e+(-1)) )
Teorema: [ de la integral de LaGrange de primera especie ]
lim[x = 0][ (-1)·cos(x) /o(x)o/ x^{2} ] = ...
... int[ lim[x = 0][ ( sin(x)/(2x) ) ] ]d[x] = lim[x = 0][ (1/2)·x ] = 0
Teorema: [ de la distribución de LaGrange de primera especie ]
pi^{2}·int[x = 0]-[pi][ ( sin(x)/(2x) ) ]d[x] = 1
Teorema: [ de la esperanza de LaGrange de primera especie ]
pi^{2}·int[x = 0]-[pi][ ( sin(x)/2 ) ]d[x] = pi^{2}
Teorema: [ de la integral de LaGrange de segunda especie ]
lim[x = 0][ ( sin(x)+(-x) ) /o(x)o/ x^{3} ] = ...
... int[ lim[x = 0][ ( (cos(x)+(-1))/(3x^{2}) ) ] ]d[x] = lim[x = 0][ (-1)·(1/6)·x ] = 0
Teorema: [ de la distribución de LaGrange de segunda especie ]
(-1)·pi^{2}·int[x = 0]-[pi][ ( (cos(x)+(-1))/(3x^{2}) ) ]d[x] = 1
Teorema: [ del esperanza de LaGrange de segunda especie ]
(-1)·pi^{2}·int[x = 0]-[pi][ ( (cos(x)+(-1))/3x ) ]d[x] = (2/3)·pi
Teorema: [ de comprobación a Poisson ]
er-h[nk+1](0) = 0
Demostración:
[As][ s > 0 ==> lim[k = 0][ 0^{nk+s} = 0 ] ]
Teorema: [ de comprobación a D'Alembert ]
er-h[k](0) = ln(0)
Demostración:
[As][ s > 0 ==> lim[k = 0][ (1/k)·0^{k+s} = (1/k)·ln(1) = (1/k)·ln(k^{k}) = (k/k)·ln(k) ] = ln(0) ]
Teorema: [ de comprobación a LaGrange de primera especie ]
sin-e[2k+1](0) = 0
Demostración:
lim[k = 0][ (1/(2k+1))·0^{2k+1} = 0^{1} = 0 ]
Teorema: [ de comprobación a LaGrange de segunda especie ]
lim[x = 0][ (1/x)·cos-e[2k+(-1)](x) = (-oo) ]
Demostración:
lim[k = 0][ (1/(2k+(-1)))·0^{2k+(-1)} = (-1)·(1/0) = (-oo) ]
Teorema: [ de comprobación a LaGrange de segunda especie ]
cos-e[2k](0) = ln(0)
Demostración:
[As][ s > 0 ==> ...
... lim[k = 0][ (1/2k)·0^{2k+s} = (1/2k)·ln(1) = (1/2k)·ln((2k)^{2k}) = (2k/2k)·ln(2k) ] = ln(0) ]
Teorema:
int[ e^{sin(x)} ]d[x] = ( sin(x)+ln(cos(x)) [o(x)o] cos(x) ) [o(x)o] sin(x)·er-h[k+1](sin(x))
e^{sin(x)} = sin(x)·er-h[k+1](sin(x))
Teorema:
int[ e^{cos(x)} ]d[x] = ( cos(x)+ln(sin(x)) [o(x)o] (-1)·sin(x) ) [o(x)o] cos(x)·er-h[k+1](cos(x))
e^{cos(x)} = cos(x)·er-h[k+1](cos(x))
Teorema:
d_{xx}^{2}[y(x)] = x^{n}·y(x)
y(x) = sum[k = 1]-[oo][ F(k)·x^{k} ]
F(k+(-n)+(-2)) = F(k)·k·(k+(-1))
Teorema:
d_{xx}^{2}[y(x)] = (x^{n}+1)·y(x)
y(x) = sum[k = 1]-[oo][ F(k)·x^{k} ]
F(k+(-1)·[n]+(-2)) = F(k)·k·(k+(-1))
Teorema:
d_{xx}^{2}[y(x)] = (x^{n}+x^{m})·y(x)
y(x) = sum[k = 1]-[oo][ F(k)·x^{k} ]
F(k+(-1)·(m+[n+(-m)])+(-2)) = F(k)·k·(k+(-1))
John:
The he that practice de verity,
gow to the veritable light,
and the veritable light it switch up.
The he that practice de falsety,
gow to the falsetable tenebrys,
and the falsetable tenebrys it switch dawn.
one syfire <==> [A$1$ [z] ][ [z] is name ]
one saxon <==> [E$1$ [z] ][ [z] is name ]
Ejercito English:
striked [o] strehned [o] strushtemated
strehnesen-hofned
strushtematesen-hofned
Ejercito Gaélico:
striket-kezhed [o] strehnet-kezhed [o] strushtematet-kezhed
strehnesen-hofnet-kezhed
strushtematesen-hofnet-kezhed
hat-sated [o] hat-sihed [o] hat-rushtemated
hat-sihesen-hofned
hat-rushtematesen-hofned
det-sated [o] det-sihed [o] det-rushtemated
det-sihesen-hofned
det-rushtematesen-hofned
I havere-kate smushtematesen-hofned ein biturbi cigar-zeizen.
I havere-kate smushtematesen-hofned ein ele cigar-zeizen.
coment [o] coment-zeizen [o] brishni-coment
guzhenen coment [o] guzhenen coment-zeizen [o] guzhenen brishni-coment
Ejercito italiano:
cantato
cantato-sam
cantato-prom
cantatered
cantatu
cantatu-sam
cantatu-prum
Enseñar el DNI en el banco para cobrar la pensión no entiendo porque es legal,
porque roba el banco si no lo enseñas y no roba el banco si lo enseñas.
Las sanciones de la policía no entiendo porque son legales,
porque roba el gobierno si te saltas la ley y no roba el gobierno si no te saltas la ley.
Teorema:
(3/4)·int[x = 0]-[1][ x^{2}+2x ]d[x] = 1
(3/4)·int[x = 0]-[1][ x·( x^{2}+2x ) ]d[x] = (11/16)
Teorema:
(4/15)·int[x = 0]-[1][ x^{3}+3x+2 ]d[x] = 1
(4/15)·int[x = 0]-[1][ x·( x^{3}+3x+2 ) ]d[x] = (44/75)
Teorema:
Si 0 [< f_{n}(x) [< x+(-1)·(1/n) ==> ...
... ( lim[n = oo][ f_{n}(0) ] = 0 & x >] (1/n) & f_{n}(1/n) = 0 )
Demostración:
0 [< lim[n = oo][ f_{n}(x) ] [< lim[n = oo][ x+(-1)·(1/n) ] = x+(-0) = x
0 [< lim[n = oo][ f_{n}(0) ] [< 0
Teorema:
Si 1 [< f_{n}(x) [< (x/n) ==> ...
... ( lim[n = oo][ f_{n}(oo) ] = 1 & x >] n & f_{n}(n) = 1 )
Demostración:
1 [< lim[n = oo][ f_{n}(oo) ] [< ...
... lim[n = oo]-[x = oo][ (x/n) ] = lim[n = oo]-[x = oo][ ( ((x+1)+(-x))/((n+1)+(-n)) ) ] = 1
Teorema:
Si (1/2) [< f_{n}(x) [< ( 1/(1+(x/n)) ) ==> ...
... ( lim[n = oo][ f_{n}(oo) ] = (1/2) & x [< n & f_{n}(n) = (1/2) )
Demostración:
(1/2) [< lim[n = oo][ f_{n}(oo) ] [< lim[n = oo]-[x = oo][ ( 1/(1+(x/n)) ) ] = ( 1/(1+1) ) = (1/2)
Teorema:
Si 0 [< f_{n}(x) [< ( n/(1+nx) ) ==> ...
... ( lim[n = oo][ f_{n}(oo) ] = 0 & ( Si x >] 1+(-1)·(1/n) ==> f_{n}(x) [< 1 ) )
Teorema:
Si F(x) = int[x = ln(0)]-[sin(x)+(-1)][ e^{x} ]d[x] ==> d_{x}[F(x)] [< 1
Demostración:
d_{x}[F(x)] = e^{sin(x)+(-1)}·cos(x) [< e^{sin(x)+(-1)} [< e^{1+(-1)} = e^{0} = 1
Si a un infiel le mide la polla 14 cm puede estar con unas 2 vírgenes,
17.5 cm una vez y la otra 21 cm n = 2.
Si a un infiel le mide la polla 17.5 cm solo puede estar con una virgen,
21 cm n = 2
Si una infiel tiene un puffy peatch gordo puede estar con unos 2 vírgenes,
flaco una vez y la otra juicy cherry n = 2.
Si una infiel tiene un puffy peatch flaco solo puede estar con un virgen,
juicy cherry n = 2.
Impuestos de utilidad del agua en casas:
n = habitaciones
m = lavabos
k = precio + impuesto-de-normalización-del-precio
consumo-de-agua = (100p+r)·m^{3}
k(p) = p·(2.20+0.20) = p·(2.40)€
n = 2
m = 1
(1/h) = ( k/(n+m) ) = 0.80€
n = 3
m = 1
(1/h) = ( k/(n+m) ) = 0.60€
n = 4
m = 2
(1/h) = ( k/(n+m) ) = 0.40€
n = 5
m = 3
(1/h) = ( k/(n+m) ) = 0.30€
k(p) = p·(2.20+0.30) = p·(2.50)€
n = 4
m = 1
n = 3
m = 2
(1/h) = ( k/(n+m) ) = 0.50€
Teorema:
e^{f(x)} = f(x)·er-h[k+1](f(x))
e^{(-1)·f(x)} = (-1)·f(x)·er-e[k+1](f(x))
Teorema:
x^{n}+1 = c
x = c^{( 1/[n] )}
Demostración:
( c^{( 1/[n] )} )^{n}+1 = ( c^{( 1/[n] )} )^{[n]} = c^{( [n]/[n] )} = c
Teorema:
x^{n}+x^{m} = c
x = c^{( 1/(m+[n+(-m)]) )}
Demostración:
( c^{( 1/(m+[n+(-m)]) )} )^{n}+( c^{( 1/(m+[n+(-m)]) )} )^{m} = ...
... ( c^{( 1/(m+[n+(-m)]) )} )^{m}·( ( c^{( 1/(m+[n+(-m)]) )} )^{(n+(-m))}+1 ) = ...
... ( c^{( 1/(m+[n+(-m)]) )} )^{m}·( c^{( 1/(m+[n+(-m)]) )} )^{[n+(-m)]} ) = ...
... ( c^{( 1/(m+[n+(-m)]) )} )^{m+[n+(-m)]} = ( c^{( (m+[n+(-m)])/(m+[n+(-m)]) )} ) = c
Teorema:
ln(x^{n}+1) = [n·ln(x)]
Demostración:
x^{n}+1 = e^{[n·ln(x)]} = e^{n·ln(x)}+1 = e^{ln(x^{n})}+1
Teorema:
[0] = ln(2)
Demostración:
2 = 1+1 = x^{0}+1
ln(2) = ln(x^{0}+1) = [0·ln(x)] = [0]
Teorema:
ln(x+y) = (1/2)·( ln(x)+ln(y)+[ln(y)+(-1)·ln(x)]+[ln(x)+(-1)·ln(y)] )
ln(x+y) = ln(x)+[ln(y)+(-1)·ln(x)]
ln(x+y) = ln(y)+[ln(x)+(-1)·ln(y)]
Teorema:
ln(2x) = ln(2)+ln(x)
Teorema:
[n] = ln(e^{n}+1)
Demostración:
ln(e^{n}+1) = ln(1)+[ln(e^{n})+(-1)·ln(1)] = [n]
ln(e^{n}+1) = [n·ln(e)] = [n]
Teorema:
[(-n)]+n = [n]
Demostración:
[n] = ln(e^{n}+1) = ln(e^{n})+[ln(1)+(-1)·ln(e^{n})] = n+[(-n)]
Teorema:
x^{n}+x^{n} = c
x = (c/2)^{(1/n)}
Demostración:
x^{n}+x^{n} = x^{n}·( x^{0}+1 ) = x^{n}·( y^{0}+1 ) = 2x^{n} = c
Si x = c^{( 1/(n+[0]) )} ==>
c^{( 1/(n+[0]) )} = x != ( x^{n}·y^{[0]} )^{( 1/(n+[0]) )} <==> x != y
c != x^{n}·y^{[0]} & c = x^{n}·y^{[0]}
[ (MP) ( Si x = y ==> x^{s} = y^{s} ) <==> ( Si x^{s} != y^{s} ==> x != y ) ]
Teorema:
x^{n}+1 = c
x = c^{( 1/ln(e^{n}+1) )}
n != 0
Teorema:
x^{n}+x^{m} = c
x = c^{( 1/( m+ln(e^{n+(-m)}+1) ) )}
n != m
Teorema:
x^{n+k}+x^{n} = c
x = c^{( 1/( n+ln(e^{k}+1) ) )}
k != 0
Teorema:
x^{n+1}+x^{n} = c
x = c^{( 1/( n+ln(e+1) ) )}
Teorema:
(1/n!)·int[x = 0]-[oo][ x^{n}·e^{(-x)} ]d[x] = 1
(1/n!)·int[x = 0]-[oo][ x^{n+1}·e^{(-x)} ]d[x] = n+1
No entiendo que me rezáis cuando la voz de la mente dice que soy un violador mental,
y no me crece el pene y es imposible que lo sea porque me mide 14 cm el pene.
Si no veis imposible que sea un violador mental tendréis que creer en esclavos infieles,
porque no me crece el pene porque soy fiel.
Como te vas a creer que he violado con la mente a Mireia Ribas, juez y policía,
si el pene no me ha crecido y me mide 14 cm.
Es imposible que la haya violado me hubiese crecido el pene.
El propósito de un hombre fiel,
es estudiar y amar a la gente con la luz,
no está para matar ni joder a ninguien.
Eso solo lo hace un infiel.
Si no seguíis el propósito de hombre fiel,
marchad-vos de este mundo,
porque no se debe molestar a la gente que quiere amar.
Si me voy no me sigáis porque yo quiero amar.
No hay ningún motivo para no creer en esclavos infieles,
creyendo en reencarnación o que te vas al Cielo.
Solo hay motivo para no creer,
si te crees que solo se vive una vez.
No te puedes creer que eras alguien antes de nacer,
si no crees en infieles,
porque solo se vive una vez.
No se pueden creer alguien del pasado los muertos,
si no cree en infieles,
porque los infieles no tienen alma,
y se mueren para siempre cuando muere el cuerpo.
Sin creer en infieles,
es imposible que el mundo te crea alguien del pasado como Jesucristo,
porque te crees que solo vives una vez.
Teorema:
ln(x^{n}+w) = [...(w)...[n·ln(x)]...(w)...]
Teorema:
ln(e^{n}+w) = [...(w)...[n]...(w)...]
Teorema:
ax^{n+k}+bx^{k} = c
x = (c/a)^{( 1/( k+ln(e^{n}+(b/a)) ) )}
Problema:
2x^{10}+10x^{2} = 10
x = 5^{( 1/( 2+ln(e^{8}+5) ) )}
Problema:
4x^{12}+12x^{4} = 12
x = 3^{( 1/( 4+ln(e^{8}+3) ) )}
Arte:
[En][ p^{n}·n < sum[k = 1]-[n][ k·( ln(e+ln(n))+p ) ] ]
[En][ p^{n}·n < sum[k = 1]-[n][ (1/k)·( ln(e+ln(n))+p ) ] ]
Arte:
[En][ p^{n}·n < sum[k = 1]-[n][ k·( ((p+1)+q)·ln(e+ln(n))+(-q) ) ] ]
[En][ p^{n}·n < sum[k = 1]-[n][ (1/k)·( ((p+1)+q)·ln(e+ln(n))+(-q) ) ] ]
Exposición:
n = 1
f(k) = 1
f(1/k) = 1
g(n) = 1
Arte: [ de Vinogradov ]
Si F(x) = #{ p € P : p [< x } ==>
F(x) = int[x = e]-[x][ ( 1/ln(x) ) ]d[x]+O[o(x)o]( (1/2)·x^{2} )+O( x·ln(x) )
F(x) = int[x = 1]-[x][ e^{x}·x ]d[x]+O[o(x)o]( (1/2)·x^{2} )+O( x·e^{2x} )
Exposición:
x = e
lim[x = e][ (1/x)·( 1/ln(x) )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ...
... ( 1+(-1)·int[ ( 1/ln(x) ) ]d[x] ) ) = (1/e) ]
f( ln(x) ) = x
(1/x)·( 1/ln(x) )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ( F(x)+(-1)·int[ ( 1/ln(x) ) ]d[x] ) ) [< ...
(1/x)·( 1/ln(x) )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ...
... ( (1/2)·x^{2}+(-1)·ln( ln(x) ) [o(x)o] (1/2)·x^{2} ) ) [< (1/x)^{2}·( x+(-1)·ln(x) ) = ...
... (1/x)·( 1+(-1)·(ln(x)/x) ) [< ( 1+(-1)·( ln(x)/x ) ) [< 1
(1/x)·( 1/ln(x) )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ( F(x)+(-1)·int[ ( 1/ln(x) ) ]d[x] ) ) >] ...
(1/x)·( 1/ln(x) )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ...
... ( 1+(-1)·ln( ln(x) ) [o(x)o] (1/2)·x^{2} ) ) >] ...
... (1/x)^{2}·( 1+(-1)·ln(x) ) = (1/x)·( (1/x)+(-1) ) = (-1)·(1/x)·( 1+(-1)·(1/x) ) >] ...
... (-1)·( 1+(-1)·(1/x) ) >] (-1)+0 = (-1)
x = 1
lim[x = 1][ (1/x)·(1/e^{x})^{2}·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ...
... ( 0+(-1)·int[ e^{x}·x ]d[x] ) ) = (-1)·(1/e) ]
f( e^{x} ) = x
(1/x)·( 1/e^{2x} )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ( F(x)+(-1)·int[ e^{x}·x ]d[x] ) ) [< ...
(1/x)·( 1/e^{2x} )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ...
... ( (1/2)·x^{2}+(-1)·e^{x} [o(x)o] (1/2)·x^{2} ) ) [< (1/x)^{2}·( 1+(-1) ) = 0
(1/x)·( 1/e^{2x} )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ( F(x)+(-1)·int[ e^{x}·x ]d[x] ) ) >] ...
(1/x)·( 1/e^{2x} )·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ...
... (-1)·e^{x} [o(x)o] (1/2)·x^{2} ) >] (1/x)^{2}·(-1) >] (-1)
Exámenes de teoría de números:
Arte: [ de Von Kotch ]
Si F(x) = #{ p € P : p [< x } ==>
F(x) = int[x = e]-[x][ ( 1/ln(x) ) ]d[x]+O[o(x)o]( (1/2)·x^{2} )+O( x^{(1/p)}·ln(x) )
F(x) = int[x = 1]-[x][ e^{x}·x ]d[x]+O[o(x)o]( (1/2)·x^{2} )+O( x^{(1/p)}·( e^{x} )^{(1/p)+1} )
Exposición:
x = e
lim[x = e][ (1/x)^{(1/p)}·(1/ln(x))·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ...
... ( 1+(-1)·int[ ( 1/ln(x) ) ]d[x] ) ) = (1/e)^{(1/p)} ]
f( ln(x) ) = x
g( x^{(1/p)} ) = x
x = 1
lim[x = 1][ (1/x)^{(1/p)}·(1/e^{x})^{(1/p)+1}·( ( x /o(x)o/ (1/2)·x^{2} ) [o(x)o] ...
... ( 0+(-1)·int[ e^{x}·x ]d[x] ) ) = (-1)·(1/e)^{(1/p)} ]
f( ( e^{x} )^{(1/p)+1} ) = x
g( x^{(1/p)} ) = x
Arte: [ de Rammanujjan ]
Si F(x) = #{ p € P : p [< x } ==>
F(x) = int[x = e]-[x][ ( 1/ln(x) ) ]d[x]+O[o(x)o]( (1/2)·x^{2} )+O( x·( ln(x+1)/ln(ln(x)+e)) ) )
F(x) = int[x = 1]-[x][ e^{x}·x ]d[x]+O[o(x)o]( (1/2)·x^{2} )+O( e^{2x}·( ln(ln(x)+e)) )
Exposición:
f( ln(x) ) = x
f( e^{x} ) = x
g( ln(x+e) ) = 1
h( ln(x+1) ) = 1
Arte:
[En][ 2^{3n}·n < sum[k = 1]-[n][ k·( 4·ln(e+ln(n))+5 ) ] ]
[En][ 2^{3n}·n < sum[k = 1]-[n][ (1/k)·( 4·ln(e+ln(n))+5 ) ] ]
Arte:
[En][ 2^{3n}·n < sum[k = 1]-[n][ k·( 2·ln(e+ln(n))+7 ) ] ]
[En][ 2^{3n}·n < sum[k = 1]-[n][ (1/k)·( 2·ln(e+ln(n))+7 ) ] ]
Arte: [ de Vinogradov ]
[En][ 2^{4n}·n < sum[k = 1]-[n][ k·( 6·ln(e+ln(n))+11 ) ] ]
[En][ 2^{4n}·n < sum[k = 1]-[n][ (1/k)·( 6·ln(e+ln(n))+11 ) ] ]
Arte: [ de Vinogradov ]
[En][ 2^{4n}·n < sum[k = 1]-[n][ k·( 4·ln(e+ln(n))+13 ) ] ]
[En][ 2^{4n}·n < sum[k = 1]-[n][ (1/k)·( 4·ln(e+ln(n))+13 ) ] ]
Exposición:
n = 1
f(k) = 1
f(1/k) = 1
g(n) = 1
Arte:
H(n,m) = #{ < n,m > € Z [x] Z : n^{2}+m^{2} [< R^{2} & ( n = 0 |o| m = 0 ) }
[ER][ H(n,m) = ( pi·R )^{2}+O( R^{q+2}·( ln(e+ln(R)) )^{p} ) ]
[ER][ H(n,m) = ( pi·R )^{2}+O( R^{q+2}·( e^{R+(-1)} )^{p} ) ]
Exposición:
R = 1
(-12) [< 4+(-1)·pi^{2} [< 4
f( ln(R) ) = 0
f( e^{R+(-1)} ) = 1
f( R^{q+2} ) = R^{3}
(1/R)^{3}( 4R+(-1)·(pi·R)^{2} = ( 4·(1/R)^{2}+(-1)·pi^{2}·(1/R) ) [< 4·(1/R)^{2} [< 4
(1/R)^{3}( 4R+(-1)·(pi·R)^{2} = (1/R)·( (4/R)+(-1)·pi^{2} ) = ...
... (-1)·(1/R)·( pi^{2}+(-1)·(4/R) ) >] (-1)·( pi^{2}+(-1)·(4/R) ) >] (-1)·pi^{2} >] (-16)
Arte: [ de Vinogradov ]
H(n,m,k) = #{ < n,m,k > € Z [x] Z [x] Z : n^{2}+m^{2}+k^{2} [< R^{2} & ...
( ( m = 0 & k = 0 ) || ( k = 0 & n = 0 ) || ( n = 0 & m = 0 ) ) & ¬( n = 0 & m = 0 & k = 0 ) }
[ER][ H(n,m,k) = (4/3)·( pi·R )^{3}+O( R^{q+3}·( ln(e+ln(R)) )^{p} ) ]
[ER][ H(n,m,k) = (4/3)·( pi·R )^{3}+O( R^{q+3}·( e^{R+(-1)} )^{p} ) ]
Exposición:
R = 1
(-250) [< 6+(-4)·pi^{3} [< 6+(-1)·(4/3)·pi^{3} [< 6
f( ln(R) ) = 0
f( e^{R+(-1)} ) = 1
f( R^{q+3} ) = R^{4}
Arte: [ de Vinogradov ]
G(x) = #{ < p,q > € Z [x] Z : [En][ n = p+q & n [< x ] }
Si 0 [< G(x) [< x^{2} ==> G(x) = x+O( 3·( ln(x) )^{p+2} )
Si 0 [< G(x) [< x^{2} ==> G(x) = e^{x}+O( 3·( e^{x+(-1)} )^{p+2} )
Exposición:
x = e
(-1) [< (-1)·(2/3) [< (1/3)·( 1+(-e) ) [< (1/3) [< 1
f( ( ln(x) )^{p+2} ) = x^{2}
x = e
(-1) [< (-1)·(e/3) ) [< 0
f( ( e^{x+(-1)} )^{p+2} ) = x^{2}
f( e^{x} ) = x
Arte: [ método de funciones trigonométricas de Vinogradov ]
[Ep][ ( sin(x) )^{p+1}+( cos(x) )^{p+1} = 1 ]
[Eq][ ( sinh(x) )^{q+1}+(-1)·( cosh(x) )^{q+1} = 1 ]
Exposición:
p = 1
f( ( sin(x) )^{p} ) = sin(x)
f( ( cos(x) )^{p} ) = cos(x)
q = 1
f( ( sinh(x) )^{q} ) = sinh(x)
f( ( cosh(x) )^{q} ) = cosh(x)
