Teorema:
Si lim[x = 0][ g(x) ] = 0^{n} ==> ...
... lim[x = 0][ ( f(x) [o(x)o] g(x) ) ] = lim[x = 0][ ( f(x) [o(x)o] d_{x}[g(x)] ) ]
Si lim[x = 0][ g(x) ] = 0^{n} ==> ...
... lim[x = 0][ ( f(x) /o(x)o/ g(x) ) ] = lim[x = 0][ ( f(x) /o(x)o/ d_{x}[g(x)] ) ]
Demostración:
lim[x = 0][ d_{x}[g(x)] ] = ...
... lim[h = 0][ (1/h)·( g(x+h)+(-1)·g(x) ) = lim[h = 0][ (1/h)·0^{n+1} ] = 0^{n} = ...
... lim[x = 0][ g(x) ]
Teorema:
Si lim[x = oo][ g(x) ] = oo^{n} ==> ...
... lim[x = oo][ ( f(x) [o(x)o] g(x) ) ] = lim[x = oo][ ( f(x) [o(x)o] d_{x}[g(x)] ) ]
Si lim[x = oo][ g(x) ] = oo^{n} ==> ...
... lim[x = oo][ ( f(x) /o(x)o/ g(x) ) ] = lim[x = oo][ ( f(x) /o(x)o/ d_{x}[g(x)] ) ]
Demostración:
lim[x = oo][ d_{x}[g(x)] ] = ...
... lim[h = 0][ (1/h)·( g(x+h)+(-1)·g(x) ) = lim[h = 0][ (1/h)·oo^{n+(-1)} ] = oo^{n} = ...
... lim[x = oo][ g(x) ]
Teorema:
Sea n >] 1 ==>
int[x = 0]-[oo][ e^{(-1)·x^{n}} ]d[x] = (1/n!)
Demostración:
lim[x = oo][ (-1)·( ( ln(2)/x^{n} ) [o(x)o] ( x /o(x)o/ n!·x ) ) ] = (-1)·(1/n!)·ln(2)·0^{n}
Teorema:
Sea n >] 1 ==>
int[x = 0]-[oo][ nx^{2n+(-1)}·e^{(-1)·x^{n}} ]d[x] = n!
Demostración:
lim[x = oo][ (-1)·( ( ln(2)/x^{n} ) [o(x)o] n!·x ) ] = (-1)·n!·ln(2)·0^{n}
Examen:
Teorema:
int[x = 0]-[oo][ d_{x}[ (1/x)·( ln(x+1)+x ) ]·x^{n} ]d[x] = ( ln(2)+1 )·n!
Teorema:
int[x = 0]-[oo][ d_{x}[ (1/x)·( x+e^{x}+(-1) ) ]·x^{n} ]d[x] = ( 1/ln(2) )·( ln(2)+1 )·n!
Definición: [ de la función Gamma ]
H(s) = int[x = 0]-[oo][ x^{s}·e^{(-x)} ]d[x]
Definición:
sum[k = 1]-[n][ f(k) ] = ( f(n) )?
1? = 1
Teorema:
( f(n)·g(n) )? = ( f(n) )? [o] ( g(n) )?
Demostración:
( f(n)·g(n) )? = ...
... sum[k = 1]-[n][ f(k)·g(k) ] = sum[k = 1]-[n][ f(k) ] [o] sum[k = 1]-[n][ g(k) ] = ( f(n) )? [o] ( g(n) )?
Teorema:
( f_{1}(n)·...(m)...·f_{m}(n) )? = ( f_{1}(n) )? [o] ...(m)... [o] ( f_{m}(n) )?
Demostración:
( ( f_{1}(n)·...(m)...·f_{m}(n) )·f_{m+1}(n) )? = ...
... ( f_{1}(n)·...(m)...·f_{m}(n) )? [o] ( f_{m+1}(k) )? = ...
... ( ( f_{1}(n) )? [o] ...(m)... [o] ( f_{m}(k) )? ) [o] ( f_{m+1}(k) )?
Teorema:
( f(n)+g(n) )? = ( f(n) )?+( g(n) )?
( w·f(n) )? = w·( f(n) )?
Teorema:
Sea k >] 0 ==>
H(k) = int[x = 0]-[oo][ x^{k}·e^{(-1)·x} ]d[x] = k!
Demostración:
lim[x = oo][ (-1)·( ( ln(2)/x ) [o(x)o] k!·x ) ] = (-1)·k!·ln(2)·0
Teorema: [ de distribución ]
( 1 /o/ (n!)? ) [o] sum[k = 0]-[n][ H(k) ] = 1
Teorema: [ de esperanza ]
( 1 /o/ (n!)? ) [o] sum[k = 0]-[n][ H(k+1) ] = (n+1)?
Demostración:
( 1 /o/ (n!)? ) [o] sum[k = 0]-[n][ H(k+1) ] = ( 1 /o/ (n!)? ) [o] ( (n+1)! )? = ...
... ( 1 /o/ (n!)? ) [o] ( n!·(n+1) )? = ( 1 /o/ (n!)? ) [o] (n!)? [o] (n+1)? = (n+1)?
Anexo:
(1 /o/ 1)·1 = 1
(1 /o/ (1+1)) [o] (1+2) = 1+2
(1 /o/ (1+1+2)) [o] (1+2+6) = 1+2+3
(1 /o/ (1+1+2+6)) [o] (1+2+6+24) = 1+2+3+4
Teorema:
Sea k >] 1 ==>
H(1/k) = int[x = 0]-[oo][ x^{(1/k)}·e^{(-1)·x} ]d[x] = k
Demostración:
lim[y = oo^{(1/k)}][ (-k)·( ( ln(2)/y^{k} ) [o(y)o] ( k!·y /o(y)o/ k!·y ) ) ] = (-k)·ln(2)·0
Teorema:
Sea k >] 1 ==>
H((1/k)+1) = int[x = 0]-[oo][ x^{(1/k)+1}·e^{(-1)·x} ]d[x] = 2^{k}·(2k+(-p))!·k
Demostración:
lim[y = oo^{(1/k)}][ (-k)·( ( ln(2)/y^{k} ) [o(y)o] ( 2^{k}·k!·(2k+(-p))!·y /o(y)o/ k!·y ) ) ] = ...
... (-k)·2^{k}·(2k+(-p))!·ln(2)·0
Teorema: [ de distribución ]
( 1 /o/ n? ) [o] sum[k = 1]-[n][ H(1/k) ] = 1
Teorema: [ de esperanza ]
( 1 /o/ n? ) [o] sum[k = 1]-[n][ H((1/k)+1) ] ) = ( 2^{n} )? [o] ( (2n+(-p))! )?
Demostración:
( 1 /o/ n? ) [o] sum[k = 1]-[n][ H((1/k)+1) ] = ...
... ( 1 /o/ n? ) [o] ( 2^{n}·(2n+(-p))!·n )? = ...
... ( 1 /o/ n? ) [o] ( 2^{n} )? [o] ( (2n+(-p))! )? [o] n? = ( 2^{n} )? [o] ( (2n+(-p))! )?
Anexo:
(1 /o/ 1) [o] (1·2)
(1 /o/ (1+2)) [o] (1·2+2·3·4)
(1 /o/ (1+2+3)) [o] (1·2+2·3·4+3·3·5·8)
Teorema:
u(n) = int[x = 0]-[oo][ d_{x}[ (1/x)·( x+( 1/(1+(-x)) )+(-1) ) ]·x^{n} ]d[x] = n!+1
v(n) = int[x = 0]-[oo][ d_{x}[ (1/x)·( x+( 1/(1+(-x)) )+(-1) ) ]·x^{n} ]d[x] = n!+(-1)
Teorema: [ de distribución ]
( 1/(2n!) )·( u(n)+v(n) ) = 1
Teorema: [ de esperanza ]
( 1/(2n!) )·( u(n+1)+v(n+1) ) = (n+1)
Anexo:
3! = (2+3) = (0.05)€
4! = (2+3)·4 = (0.20)€
5! = (2+3)·4+5 = (0.25)€
6! = (2+3)·4+5·6 = (0.50)€
Teorema:
Si ( P(n) = ( 1/(2n!) )·u(n) & Q(n) = ( 1/(2n!) )·v(n) ) ==>
P(3) = (7/12) & Q(3) = (5/12)
P(4) = (25/48) & Q(4) = (23/48)
P(5) = (121/240) & Q(5) = (119/240)
P(6) = (721/1440) & Q(6) = (719/1440)
El himno de Cáteldor:
Cáteldor tot triomfant,
torna a ser ric y complet.
Enderrera aquesta gent,
tant ufana y tant superva.
Bon cop de faç.
Bon cop de falç,
defensors de la terra.
Bon cop de falç.
Se va construyendo el tercer raíl desde Sant Vicent de Calders hasta Tarragona,,
Se puede ir o vatxnar con trenes de carga desde Vilafranca de Penedés hasta Tarragona,
para salir hacia El País Valenciano.
Maquetas de tren de Habitación:
Ley:
Estación continua paralela.
Túnel-Puente-Túnel
Estación continua paralela.
Puente-Túnel-Puente
Ley:
Estación terminal perpendicular.
Túnel-Puente-Túnel
Estación continua perpendicular.
Puente-Túnel-Puente
Principio: [ de Olores ]
Perfumante [o] Sudosa
Ambientativa [o] Fétida
Desodorante [o] Humosa
Desértica [o] Húmeda
Principio: [ de olores ]
Carne-podrida [o] Tierra [o] Verdura-podrida
Pescado-podrido [o] Mar [o] Alga-podrida
Carne [o] Verdura
Pescado [o] Alga
Teorema:
M(k) = int[x = 0]-[oo][ d_{x}[ ln( k+( x/(x+1) ) ) ] ]d[x]+ln(k) = ln(k+1)
Teorema:
(n+1)? [o] sum[k = 0]-[n][ d_{k...k}^{m}[M(k)] ] = (-1)^{m+1}·(m+(-1))!·( ( 1/(n+1) )^{m+(-1)} )?
Demostración:
d_{x}[ (-1)^{m+1}·(m+(-1))!·(1/x)^{m} ] = (-1)^{m+1}·(m+(-1))!·(-m)·(1/x)^{m+1} = ...
... (-1)^{(m+1)+1}·m!·(1/x)^{m+1}
(n+1)? [o] sum[k = 0]-[n][ d_{k...k}^{m}[M(k)] ] = ...
... (n+1)? [o] sum[k = 0]-[n][ (-1)^{m+1}·(m+(-1))!·( 1/(k+1) )^{m} ] = ...
... (n+1)? [o] ( (-1)^{m+1}·(m+(-1))!·( 1/(n+1) )^{m} )? = ...
... (-1)^{m+1}·(m+(-1))!·( (n+1)? [o] ( ( 1/(n+1) )^{m} )? = ...
... (-1)^{m+1}·(m+(-1))!·( ( 1/(n+1) )^{m+(-1)} )?
Teorema:
M(k,n) = int[x = 0]-[oo][ d_{x}[ e^{kn·( x/(x+1) )} ] ]d[x]+1 = e^{kn}
Teorema:
( 1 /o/ ( e^{n^{2}} )? ) [o] sum[k = 1]-[n][ d_{n...n}^{m}[M(k,n)] ] ) = ( n^{m} )?
Demostración:
d_{x}[ k^{m}·e^{kx} ] = k^{m+1}·e^{kx}
( 1 /o/ ( e^{n^{2}} )? ) [o] sum[k = 1]-[n][ d_{n...n}^{m}[M(k,n)] ] = ...
... ( 1 /o/ ( e^{n^{2}} )? ) [o] sum[k = 1]-[n][ k^{m}·e^{kn} ] = ...
... ( 1 /o/ ( e^{n^{2}} )? ) [o] ( n^{m}·e^{n^{2}} )? = ...
... ( 1 /o/ ( e^{n^{2}} )? ) [o] ( e^{n^{2}}·n^{m} )? = ...
... ( ( 1 /o/ ( e^{n^{2}} )? ) [o] ( e^{n^{2}} )? [o] ( n^{m} )? ) = ( n^{m} )?
Examen:
Teorema:
M(k,n) = int[x = 0]-[oo][ d_{x}[ e^{kn·( x/(x+1) )} ] ]d[x]+1 = e^{kn}
Teorema:
( 1 /o/ ( e^{n^{2}} )? ) [o] sum[k = 1]-[n][ int-[m]-int[ M(k,n) ]d[n]...(m)...d[k] ] ) = ( (1/n)^{m} )?
Teorema:
M(k) = int[x = 0]-[oo][ d_{x}[ k+( x/(x+1) ) ] ]d[x]+k = k+1
Teorema:
( 1 /o/ (n+1)? ) [o] sum[k = 0]-[n][ int-[m]-int[ M(k) ]d[k]...(m)...d[k] ] = ( 1/(m+1)! )·( (n+1)^{m} )?
Demostración:
d_{x}[ ( 1/(m+1)! )·x^{m+1} ] = ( 1/(m+2)! )·x^{m+2}
Definición:
[ (-n) // k ] = (1/k!)·( (-n)+(-k)+1 )!
[ (-n) // 0 ] = 1
Teorema:
sum[k = 0]-[oo][ [ (-n) // k ]·p^{n}·( p+(-1) )^{k} ] = 1
Demostración:
(1+(-x))^{(-n)} = 1+sum[k = 1]-[oo][ (1/k!)·( n+k+(-1) )!·x^{k} ]
p^{(-n)} = 1+sum[k = 1]-[oo][ (1/k!)·( n+k+(-1) )!·( 1+(-p) )^{k} ]
Teorema:
sum[k = 1]-[oo][ k·[ (-n) // k ]·p^{n}·( p+(-1) )^{k} ] = (-1)·(n/p)·(p+(-1))
Demostración:
sum[k = 1]-[oo][ k·[ (-n) // k ]·p^{n}·( p+(-1) )^{k} ] = ...
... (p+(-1))·sum[k = 1]-[oo][ [ (-n) // k+(-1) ]·p^{n}·( p+(-1) )^{k+(-1)} ] = ...
... (-1)·(n/p)·(p+(-1))·sum[k = 1]-[oo][ [ (-n)+(-1) // k+(-1) ]·p^{n+1}·( p+(-1) )^{k+(-1)} ] = ...
... (-1)·(n/p)·(p+(-1))·sum[k = 1]-[oo][ [ (-1)·(n+1) // k+(-1) ]·p^{n+1}·( p+(-1) )^{k+(-1)} ] = ...
... (-1)·(n/p)·(p+(-1))
Teorema:
sum[k = 0]-[oo][ p^{k}·( 1+(-p) ) ] = 1
Teorema:
sum[k = 1]-[oo][ k·p^{k}·( 1+(-p) ) ] = p·( 1/(1+(-p)) )
Demostración:
sum[k = 1]-[oo][ k·p^{k}·( 1+(-p) ) ] = p·(1+(-p))·sum[k = 1]-[oo][ k·p^{k+(-1)} ] = ...
... p·(1+(-p))·sum[k = 1]-[oo][ d_{p}[ p^{k} ] ] = p·(1+(-p))·d_{p}[ sum[k = 1]-[oo][ p^{k} ] ] = ...
... p·(1+(-p))·d_{p}[ ( 1/(1+(-p)) ) ] = p·( 1/(1+(-p)) )
Ley:
(m/2)·(1/n)·d_{t}[x]^{2} = [ (-n) // k ]·(ut)^{n}·( (ut)+(-1) )^{k}·qgx
x(t) = (1/a)·Anti-[ ( s /o(s)o/ (1/2)·s^{2} )^{[o(s)o] (1/2)} ]-( ...
... ( (2/m)·qga )^{(1/2)}·int[ [ (-n) // k ]·(ut)^{n}·( (ut)+(-1) )^{k} ]d[t] )
Ley:
(m/2)·(1/n)·d_{t}[x]^{2} = ( 1+(-1)·(ut) )·(ut)^{k}·qgx
x(t) = (1/a)·Anti-[ ( s /o(s)o/ (1/2)·s^{2} )^{[o(s)o] (1/2)} ]-( ...
... ( (2/m)·qga )^{(1/2)}·int[ ( 1+(-1)·(ut) )·(ut)^{k} ]d[t] )
Ley:
m·d_{t}[x]^{[o(ut)o] 2} = N(t)·u·F( (1/v)·d_{t}[x] )
d_{t}[x] = ...
... v·Anti-[ ( s /o(s)o/ int[ d_{ut}[ (1/m)·N(t)·(1/v)^{2}·(1/u)·F(s) ] ]d[s] )^{[o(s)o] (1/2)} ]-(ut)
Ley:
L·d_{t}[q]^{[o(ut)o] 2} = N(t)·u·F( (1/I)·d_{t}[q] )
d_{t}[q] = ...
... I·Anti-[ ( s /o(s)o/ int[ d_{ut}[ (1/L)·N(t)·(1/I)^{2}·(1/u)·F(s) ] ]d[s] )^{[o(s)o] (1/2)} ]-(ut)
Lema: [ de bolsa estocástica lineal de audiencia ]
s = audiencia
p = precio de inversión
d_{x}[y(x,k)] = ( ln(P(k))+ln(ps) )·y(x,k)
y(x,k) = e^{( ln(P(k))+ln(ps) )·x}
y(1,k) = P(k)·ps
sum[k = 1]-[n][ P(k)·ps ] = ps
d_{x}[y(x,k)] = ( ln(P(k))+ln(p/s) )·y(x,k)
y(x,k) = e^{( ln(P(k))+ln(p/s) )·x}
y(1,k) = P(k)·(p/s)
sum[k = 1]-[n][ P(k)·(p/s) ] = (p/s)
Lema: [ de bolsa estocástica afín de audiencia ]
s = audiencia
p = precio de inversión
d_{x}[y(x,k)]+(ps+1)·y(x,k) = P(k)·(ps+1)^{2}·x
y(x,k) = P(k)·(ps+1)·( x+(-1)·( 1/(ps+1) ) )
y(1,k) = P(k)·ps
sum[k = 1]-[n][ P(k)·ps ] = ps
d_{x}[y(x,k)]+((p/s)+1)·y(x,k) = P(k)·((p/s)+1)^{2}·x
y(x,k) = P(k)·((p/s)+1)·( x+(-1)·( 1/((p/s)+1) ) )
y(1,k) = P(k)·(p/s)
sum[k = 1]-[n][ P(k)·(p/s) ] = (p/s)
Anexo:
La bolsa estocástica es de dos personas:
Uno compra en 2k+1 & vende en 2k+2.
Uno vende en 2k+1 & compra en 2k+2
Anexo:
P(k) = [ 2 // k ]·2^{(-2)}
El que compra en 2k+1 gana 2p:
Parte 1:
k = 0
n = (1/2)·p & m = 0
k = 1
n = (-1)·(1/2)·p & m = p
k = 2
n = 0 & m = (1/2)·p
Parte 2:
k = 1
n = p & m = 0
k = 2
n = (1/2)·ps & m = (1/2)·p
k = 0
n = p & m = 0
Parte 3:
k = 2
n = (1/2)·p & m = 0
k = 0
n = 0 & m = (1/2)·p
k = 1
n = p & m = (-1)·(1/2)·p
Teorema: [ de probabilidad ]
( 1/( f(n) )? )·sum[k = 1]-[n][ f(k) ] = 1
Teorema: [ de distribución ]
( 1 /o/ ( f(n) )? ) [o] sum[k = 1]-[n][ f(k) ] = 1
Teorema: [ de esperanza ]
( 1 /o/ ( f(n) )? ) [o] sum[k = 1]-[n][ k·f(k) ] = n?
Teorema: [ de desviación ]
( 1 /o/ ( f(n) )? ) [o] sum[k = 1]-[n][ k·(k+(-1))·f(k) ] = n? [o] (n+(-1))?
Teorema:
Si f(k+p) = f(p)·f(k) ==>
( 1 /o/ ( f(n) )? ) [o] sum[k = 1]-[n][ k·f(k+p) ] = f(p)·n?
Teorema:
Si f(k+[p:a]) = f(k)·( f(p)+a ) ==>
( 1 /o/ ( f(n) )? ) [o] sum[k = 1]-[n][ k·f(k+[p:a]) ] = ( f(p)+a )·n?
Teorema:
Si f(k+p) = f(k)+f(p) ==>
( 1 /o/ ( f(n) )? ) [o] sum[k = 1]-[n][ k·f(k+p) ] = n?+( f(p) )·( ( 1 /o/ ( f(n) )? ) [o] n? )
Teorema:
Si f(k+[p:a]) = f(k)+( f(p)+a ) ==>
( 1 /o/ ( f(n) )? ) [o] sum[k = 1]-[n][ k·f(k+[p:a]) ] = n?+( f(p)+a )·( ( 1 /o/ ( f(n) )? ) [o] n? )
Definición: [ de subespacio afín ]
[Ea][ Si ( [x:a] € a+F & [y:a] € a+F ) ==> [x+y:a] € a+F ]
[Ea][ Si [x:a] € a+F ==> [wx:a] € a+F ]
Teorema:
a+F es afín <==> F es vectorial
Demostración:
[==>]
Sea ( x € F & y € F ) ==>
( [x:0] € F & [y:0] € F )
[x+y:0] € 0+F
(x+y) € F
Sea x € F ==>
[x:0] € F
[wx:0] € 0+F
wx € F
[<==]
Sea ( [x:a] € a+F & [y:a] € a+F ) ==>
x € F & y € F
(x+y) € F
[x+y:a] € a+F
Sea [x:a] € a+F ==>
x € F
wx € F
[wx:a] € a+F
Teorema:
Sea ( A un espacio vectorial & B un espacio vectorial ) ==>
Gen(A,B) [&] Gen(A,B) = Gen(A [&] A,B [&] B)
Gen(A,B) [ || ] Gen(A,B) = Gen(A [ || ] A,B [ || ] B)
Demostración:
Gen(A,B) [&] Gen(A,B) = Gen(A,B) = Gen(A [&] A,B [&] B)
Gen(A,B) [ || ] Gen(A,B) = Gen(A,B) = Gen(A [ || ] A,B [ || ] B)
Teorema:
Sea A [=] B = Gen( A [&] B ) ==>
Si ( A es espacio vectorial & B es espacio vectorial ) ==> A [=] B es espacio vectorial
Demostración:
Sea x € A [=] B & y € A [=] B
x es combinación lineal de A [=] B & y es combinación lineal de A [=] B
( x € A [=] x € B ) & ( y € A [=] y € B )
( x € A & y € A ) [=] ( x € B & y € B )
( x & y ) son combinación lineal de A & ( x & y ) son combinación lineal de B
(x+y) € A [=] (x+y) € B
(x+y) € A [=] B
Sea x € A [=] B
x € A [=] x € B
wx € A [=] wx € B
wx € A [=] B
Teorema:
Sea A [+] B = Gen( A [ || ] B ) ==>
Demostración:
Sea x € A [+] B & y € A [+] B ==>
x es combinación lineal de A [+] B & y es combinación lineal de A [+] B
( x € A [+] x € B ) & ( y € A [+] y € B )
( x € A & y € A ) [+] ( x € B & y € B )
( x & y ) son combinación lineal de A & ( x & y ) son combinación lineal de B
(x+y) € A [+] (x+y) € B
(x+y) € A [+] B
Sea x € A [+] B
x € A [+] x € B
wx € A [+] wx € B
wx € A [+] B
Teorema:
Sea A = i·< 1,0>+j·< 0,1>+< a,b > & B = k·< u,v >+< p,q >
Gen( A [&] B ) = A [=] B = B
Gen( A [ || ] B ) = A [+] B = A
Demostración:
i·< 1,0>+j·< 0,1>+< a,b > = k·< u,v >+< p,q > = < x,y >+< c,d >
i = k·[u:p+(-a)] & j = k·[v:q+(-b)]
i·< 1,0>+j·< 0,1>+< a,b >+k·< u,v >+< p,q > = < x,y >+< c,d >
i = [x:c]+(-k)·[u:(-p)+(-a)] & j = [y:d]+(-k)·[v:(-q)+(-b)]
Definición: [ de afinidad ]
Sea f([x:a]) = f(x)+a ==>
[Ea][ f( [x+y:a] ) = f(x)+f(y)+a ]
[Ea][ f( [wx:a] ) = w·f(x)+a ]
Teorema:
Sea f([x:a]) = f(x)+a ==>
f(w) es afinidad <==> f(w) es lineal
Demostración:
[==>]
f(x+y) = f( [x+y:0] ) = f(x)+f(y)+0 = f(x)+f(y)
f(wx) = f( [wx:0] ) = w·f(x)+0 = w·f(x)
[<==]
f( [x+y:a] ) = f(x+y)+a = f(x)+f(y)+a
f( [wx:a] ) = f(wx)+a = w·f(x)+a
Teorema:
F(x,y) = < p,q >+( < a,b >,< c,d > ) o < [x:(-p)],[y:(-q)] > = < 0,0 >
<==>
F(x,y) = ( < a,b >,< c,d > ) o < x,y > = < 0,0 >
Demostración:
< p,q >+( < a,b >,< c,d > ) o < [x:(-p)],[y:(-q)] > = ...
... < p,q >+( < a,b >,< c,d > ) o < x,y >+< (-p),(-q) > = ...
...( < a,b >,< c,d > ) o < x,y >
Teorema:
< p,q >+( < a,a >,< a,a > ) o < x,y > = < 0,0 >
< x,y > = k·< [1:(-p)],[(-1):(-q)] > = k·< 1,(-1) >+(-1)·< p,q >
Teorema:
< p,q >+( < a,(-a) >,< (-a),a > ) o < x,y > = < 0,0 >
< x,y > = k·< [1:(-p)],[1:(-q)] > = k·< 1,1 >+(-1)·< p,q >
Teorema:
[Ew][ < p,q >+( < a,a >,< a,a > ) o < x,y > = w·< x,y > ]
< x,y > = k·< [1:(-p)],[1:(-q)] > = k·< 1,1 >+(-1)·< p,q >
Teorema:
[Ew][ < p,q >+( < a,(-a) >,< (-a),a > ) o < x,y > = w·< x,y > ]
< x,y > = k·< [1:(-p)],[(-1):(-q)] > = k·< 1,(-1) >+(-1)·< p,q >
Teorema:
< p,q >+( < a,a >,< b,b > ) o < x,y > = < 0,0 >
< x,y > = k·< [1:(-p)],[(-1):(-q)] > = k·< 1,(-1) >+(-1)·< p,q >
Teorema:
< p,q >+( < a,(-a) >,< (-b),b > ) o < x,y > = < 0,0 >
< x,y > = k·< [1:(-p)],[1:(-q)] > = k·< 1,1 >+(-1)·< p,q >
Teorema:
[Ew][ < p,q >+( < a,a >,< b,b > ) o < x,y > = w·< x,y > ]
< x,y > = k·< [a:(-p)],[b:(-q)] > = k·< a,b >+(-1)·< p,q >
Teorema:
[Ew][ < p,q >+( < a,(-a) >,< (-b),b > ) o < x,y > = w·< x,y > ]
< x,y > = k·< [(-a):(-p)],[b:(-q)] > = k·< (-a),b >+(-1)·< p,q >
Ley
Los hombres fieles de pequeños miramos pichas,
porque preferimos esa o aquella picha que la nuestra,
y preferimos que folle el otro que nosotros,
porque la tenemos pequeña.
Cuesta de entender porque se tiene la picha pequeña,
hasta que no se piensa en un chocho poco profundo,
como el de mi mujer.
Las mujeres fieles de pequeñas miran chochos,
porque prefieren ese o aquel chocho que le suyo,
y prefieren que folle otra que ellas,
porque lo tienen poco profundo.
Cuesta de entender porque se tiene el chocho poco profundo,
hasta que no se piensa en una picha pequeña,
como la de su hombre.
Ley:
Los hombres con la picha pequeña,
son asexuales y no homosexuales,
porque no pueden follar con mujeres sin puente.
Las mujeres con puente,
son asexuales y no homosexuales,
porque no pueden follar con hombres con la picha grande.
Teorema:
(1/p)^{n}·sum[k = 1]-[n][ (1/n)+p^{k}+(-1)·p^{k+(-1)} ] = 1
Teorema:
(1/p)^{n}·sum[k = 1]-[n][ kp^{k}+(-1)·kp^{k+(-1)} ) ] = ...
... (1/p)^{n}·( (n+1)·p^{n}+(-1)·( ( p^{n+1}+(-1) )/( p+(-1) ) ) )
