Lema:
p = 1·100+(1/1)·1,000 = 1,100€
q = 10·100+(1/10)·1,000 = 1,100€
Lema:
p = 2·100+(1/2)·1,000 = 700€
q = 5·100+(1/5)·1,000 = 700€
Lema:
p = 10^{1}+1,000^{(1/1)} = 1,010€
q = 10^{3}+1,000^{(1/3)} = 1,010€
Lema:
p = 50^{1}+2,500^{(1/1)} = 2,550€
q = 50^{2}+2,500^{(1/2)} = 2,550€
Impuesto de 1€ por unidades del producto
Lema:
(nx)^{p} = x^{p} <==> n = 1€
((1/n)·x)^{p} = x^{p} <==> n = 1€
Disertación:
(nx)^{p} = x^{p}
p·ln(nx) = p·ln(x)
ln(nx) = ln(x)
e^{ln(nx)} = e^{ln(x)}
nx = x
Lema:
e^{nx} = e^{x} <==> n = 1€
e^{(1/n)·x} = e^{x} <==> n = 1€
Disertación:
e^{nx} = e^{x}
nx·ln(e) = x·ln(e)
nx = x
ln(nx) = ln(x) <==> n = 1€
ln((1/n)·x) = ln(x) <==> n = 1€
Disertación:
ln(nx) = ln(x)
e^{ln(nx)} = e^{ln(x)}
nx = x
n = 1
Lema:
(nx)^{p}·e^{nx} = x^{p}·e^{x} <==> n = 1€
((1/n)·x)^{p}·e^{(1/n)·x} = x^{p}·e^{x} <==> n = 1€
Disertación:
(nx)^{p}·e^{nx} = x^{p}·e^{x}
Anti-[ s^{p}·e^{s} ]-( (nx)^{p}·e^{nx} ) = Anti-[ s^{p}·e^{s} ]-( x^{p}·e^{x} )
nx = x
n = 1
Lema:
(nx)^{p}·ln(nx) = x^{p}·ln(x) <==> n = 1€
((1/n)·x)^{p}·ln((1/n)·x) = x^{p}·ln(x) <==> n = 1€
Disertación:
(nx)^{p}·ln(nx) = x^{p}·ln(x)
Anti-[ s^{p}·ln(s) ]-( (nx)^{p}·ln(nx) ) = Anti-[ s^{p}·ln(s) ]-( x^{p}·ln(x) )
nx = x
n = 1
Ley:
Después de la resurrección de los muertos,
se pueden recordar algo dual,
siendo 0t < (1/2)
Después de la resurrección de los muertos,
no se pueden recordar nada no dual,
siendo 0t > (0/2)
Ley:
Después de la resurrección de los muertos,
se pueden recordar teoremas,
siendo 0t < 1
Después de la resurrección de los muertos,
no se pueden recordar artes destructores,
siendo 0t > (-1)
Teorema:
int[x = 0]-[pi][ ( 1/sin(x) ) ]d[x] = 2+ln(4)
Demostración:
Por Hôpital-Jûanagoras:
[ (-1)·cos(x)+ln(sin(x)) [o(x)o] ( sin(x) /o(x)o/ x^{0} ) ]_[x = 0]-[pi] = 1+ln(2)+1+ln(2) = 2+ln(4)
sin(pi) = (-0)
Teorema:
int[x = 0]-[(pi/2)][ ( 1/sin(x) ) ]d[x] = 1+ln(2)
int[x = (pi/2)]-[pi][ ( 1/sin(x) ) ]d[x] = 1+ln(2)
Teorema:
int[x = 0]-[(pi/4)][ ( 1/sin(x) ) ]d[x] = ( 1+(-1)·(1/2)^{(1/2)} )+( 1+(-1)·(1/2)^{(3/2)} )·ln(2)
int[x = ((3pi)/4)]-[pi][ ( 1/sin(x) ) ]d[x] = ( 1+(-1)·(1/2)^{(1/2)} )+( 1+(-1)·(1/2)^{(3/2)} )·ln(2)
Demostración:
Por Hôpital-Jûanagoras:
[ (-1)·cos(x)+ln(sin(x)) [o(x)o] ( cos(x) /o(x)o/ x^{0} ) ]_[x = 0]-[(pi/4)]
Macroeconomía:
Lema:
Arancel de 4 socios
p = ( 16/(4!+(-8)) ) = 1
q = 0.80+2.56 = 3.36€
Precio:
0.85 = 5·0.17
0.85 = 4·0.20+0.05
Lema:
Arancel de 5 socios
p = ( 105/(5!+(-15)) ) = 1
q = 6.30+11.55 = 17.85€
Precio:
1.56 = 6·0.26
1.56 = 5·0.30+0.06
Definición: [ de categoría ]
z z
| |
z ---> F(z) ---> F^{o(-1)}(z)
| |
z ---> F^{o(-1)}(z) ---> F(z)
F^{o(-1)}( F(z) ) = z
F( F^{o(-1)}(z) ) = z
Teorema:
z z
| |
z ---> F(z+a) ---> F^{o(-1)}(z)+(-a)
| |
z ---> F^{o(-1)}(z)+(-a) ---> F(z+a)
Teorema:
z z
| |
z ---> F(z)+a ---> F^{o(-1)}(z+(-a))
| |
z ---> F^{o(-1)}(z+(-a)) ---> F(z)+a
Demostración:
F^{o(-1)}( F(z+a) )+(-a) = (z+a)+(-a) = z+(a+(-a)) = z+0 = z
F( ( F^{o(-1)}(z)+(-a) )+a ) = F( F^{o(-1)}(z)+((-a)+a) ) = F( F^{o(-1)}(z)+0 ) = F( F^{o(-1)}(z) ) = z
Teorema: [ de categoría suma ]
z z
| |
z ---> z+n ---> z+(-n)
| |
z ---> z+(-n) ---> z+n
Teorema: [ de categoría múltiplo ]
z z
| |
z ---> nz ---> (1/n)·z
| |
z ---> (1/n)·z ---> nz
Teorema:
z z
| |
z ---> z^{n} ---> z^{(1/n)}
| |
z ---> z^{(1/n)} ---> z^{n}
Teorema:
Sea F(z) o G(z) = G(z) o F(z) ==>
z
|
z ---> F(z) ---> G(z)
| |
G^{o(-1)}(z) ---> F^{o(-1)}(z) ---> G(z)
|
G^{o(-1)}(z)
Demostración:
( G^{o(-1)} o F )^{o(-1)} = F^{o(-1)} o G = G o F^{o(-1)}
( G o F^{o(-1)} o G^{o(-1)} o F )(z) = Id(z) = z
( G o F )^{o(-1)} = F^{o(-1)} o G^{o(-1)} = G^{o(-1)} o F^{o(-1)}
( G^{o(-1)} o F^{o(-1)} o G o F )(z) = Id(z) = z
Teorema:
Sea F(z) o G(z) = G(z) o F(z) ==>
z
|
z ---> F(z) ---> G(z)
| |
G^{o(-1)}(z) ---> F^{o(-1)}(z) ---> (-1)·G(z)
| |
(-1)·G^{o(-1)}(z) ---> (-z)
Demostración:
(-1)·( (-1)·( G o F^{o(-1)} o G^{o(-1)} o F )(z) ) = (-1)·( (-1)·z ) = ((-1)·(-1))·z = z
(-1)·( (-1)·( G^{o(-1)} o F^{o(-1)} o G o F )(z) ) = (-1)·( (-1)·z ) = ((-1)·(-1))·z = z
Teorema:
z
|
z ---> z^{n} ---> z^{m}
| |
z^{(1/m)} ---> z^{(1/n)} ---> z^{m}
|
z^{(1/m)}
Problema:
Mostrad la categoría múltiplo y suma.
Teorema:
z
|
z ---> F(z) ---> F(z)
| |
F(z) ---> (-1)·F^{o(-2)}(z) ---> (-z)
|
(-z)
Demostración:
(-1)·( (-1)·F^{o(-2)}( F(F(z)) ) ) = (-1)·( (-1)·( F^{o(-2)} o F^{o2} )(z) ) = ...
... (-1)·( (-1)·Id(z) ) = (-1)·( (-1)·z ) = ((-1)·(-1))·z = z
Teorema:
z
|
z ---> F(z) ---> F(z)
| |
F(z) ---> (-1)·F^{o(-2)}(z) ---> Id(z)
| |
Id(z) ---> (-z)
Demostración:
(-1)·Id( (-1)·F^{o(-2)}( F(F(z)) ) ) = (-1)·Id( (-1)·( F^{o(-2)} o F^{o2} )(z) ) = ...
... (-1)·Id( (-1)·Id(z) ) =(-1)·Id( (-1)·z ) = (-1)·( (-1)·z ) = ((-1)·(-1))·z = z
Teorema:
z
|
z ---> kz ---> kz
| |
kz ---> iz ---> (-z)
|
(-z)
Demostración:
(-1)·( ikkz) = z
Problema:
Mostrad la categoría dual en números complejos y reales simétricos.
Teorema:
z
|
z ---> z+n ---> z+n
| |
z+n ---> (-z)+2n ---> Id(z)
| |
Id(z) ---> (-z)
(-1)·( Id( (-1)·( (z+n)+n )+2n ) ) = z
Teorema:
z
|
z ---> F(z) ---> (-1)·F(z)
| |
(-1)·F(z) ---> (-1)·F^{o(-2)}(-z) ---> (-z)
|
(-z)
Demostración:
(-1)·( (-1)·F^{o(-2)}( (-1)·( (-1)·F(F(z)) ) ) ) = (-1)·( (-1)·F^{o(-2)}( ((-1)·(-1))·F(F(z)) ) ) = ...
... (-1)·( (-1)·( F^{o(-2)} o F^{o2} )(z) ) = (-1)·( (-1)·Id(z) ) = (-1)·( (-1)·z ) = ((-1)·(-1))·z = z
Teorema:
z
|
z ---> F(z) ---> (-1)·F(z)
| |
(-1)·F(z) ---> (-1)·F^{o(-2)}(-z) ---> Id(z)
| |
Id(z) ---> (-z)
Demostración:
(-1)·Id( (-1)·F^{o(-2)}( (-1)·( (-1)·F(F(z)) ) ) ) = (-1)·Id( (-1)·F^{o(-2)}( ((-1)·(-1))·F(F(z)) ) ) = ...
... (-1)·Id( (-1)·( F^{o(-2)} o F^{o2} )(z) ) = (-1)·Id( (-1)·Id(z) ) = (-1)·Id( (-1)·z ) = ...
... (-1)·( (-1)·z ) = ((-1)·(-1))·z = z
Teorema:
z
|
z ---> kz ---> (-k)·z
| |
(-k)·z ---> (-i)·z ---> (-z)
|
(-z)
Demostración:
(-1)·( (-i)·(-k)·kz ) = z
Problema:
Mostrad la categoría dual en números complejos y reales simétricos.
Teorema:
z z z
| | |
z ---> F(z) ---> F(z) ---> F^{o(-2)}(z)
| | |
z ---> F(z) ---> F^{o(-2)}(z) ---> F(z)
| | |
z ---> F^{o(-2)}(z) ---> F(z) ---> F(z)
Demostración:
F( F( F^{o(-2)}(z) ) ) = ( F^{o2} o F^{o(-2)} )(z) = Id(z) = z
F( F^{o(-2)}( F(z) ) ) = ( F o ( F^{o(-2)} o F ) )(z) = ( F o F^{o(-1)} )(z) = Id(z) = z
F^{o(-2)}( F( F(z) ) ) = ( F^{o(-2)} o F^{o2} )(z) = Id(z) = z
Teorema:
z z z
| | |
z ---> kz ---> kz ---> (-i)·z
| | |
z ---> kz ---> (-i)·z ---> kz
| | |
z ---> (-i)·z ---> kz ---> kz
Teorema:
z z z
| | |
z ---> F(-z) ---> (-1)·F(z) ---> F^{o(-2)}(-z)
| | |
z ---> (-1)·F(z) ---> F^{o(-1)}(-z) ---> Id(z)
| | |
z ---> F^{o(-2)}(-z) ---> Id(z) ---> (-1)·F^{o2}(z)
Demostración:
F^{o(-2)}( (-1)·( (-1)·F( F(z) ) ) ) = F^{o(-2)}( ((-1)·(-1))·F( F(z) ) ) = ...
... ( F^{o(-2)} o F^{o2} )(z) = Id(z) = z
Id( F^{o(-1)}( (-1)·( (-1)·F(z) ) ) ) = F^{o(-1)}( (-1)·( (-1)·F(z) ) ) = ...
... F^{o(-1)}( ((-1)·(-1))·F(z) ) ) = ( F^{o(-1)} o F )(z) ) ) = Id(z) = z
(-1)·F^{o2}(z)( Id( F^{o(-2)}( (-1)·z ) ) ) = (-1)·F^{o2}(z)( F^{o(-2)}( (-1)·z ) ) = ...
... (-1)·( F^{o2} o F^{o(-2)} )( (-1)·z ) ) ) = (-1)·Id( (-1)·z ) = (-1)·( (-1)·z ) = ((-1)·(-1))·z = z
Teorema:
z z z
| | |
z ---> kz ---> (-k)·z ---> iz
| | |
z ---> (-k)·z ---> (1/k)·(-z) ---> Id(z)
| | |
z ---> iz ---> Id(z) ---> (-i)·z
Teorema:
Sea f_{n}: nz ---> (n+1)·z ==>
Sea g_{n}: (1/n)·z ---> (1/(n+1))·z ==>
z
|
z ---> nz ---> (n+1)·z
| |
(n+1)·z ---> (1/n)·z ---> (1/(n+1))·z
|
(1/(n+1))·z
Teorema:
Sea f_{n}: d_{z...z}^{n}[h(z)] ---> d_{z...z}^{(n+1)}[h(z)] ==>
Sea g_{n}: int-[n]-int[h(z)]d[z]...d[z] ---> int-[n+1]-int[h(z)]d[z]...d[z] ==>
h(z)
|
h(z) ---> d_{z...z}^{n}[h(z)] ---> d_{z...z}^{(n+1)}[h(z)]
| |
d_{z...z}^{(n+1)}[h(z)] ---> int-[n]-int[h(z)]d[z]...d[z] ---> int-[n+1]-int[h(z)]d[z]...d[z]
|
int-[n+1]-int[h(z)]d[z]...d[z]
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