sábado, 20 de junio de 2026

topología-algebraica y álgebra y métodos-numéricos y análisis-funcional

Definición:

B^{0} = O

B^{1} = B

B^{2} = BB

B^{n+2} = BO...(n)...OB

Teorema:

B^{n}·O = B^{n}

Demostración:

B^{n}·O = B^{n}·B^{0} = B^{n+0} = B^{n}

Teorema:

x^{2}+(-1)·BB = (x+B)·(x+(-B))

x^{2}+BB = (x+iB)·(x+(-i)·B)


Grupo suma y espacio vectorial:

Definición:

... a_{0}·O+a_{1}·B+sum[k = 0]-[n+(-1)][ a_{k+2}·BO...(k)...OB ] ...

... +...

... b_{0}·O+b_{1}·B+sum[k = 0]-[n+(-1)][ b_{k+2}·BO...(k)...OB ] = ...

... (a_{0}+b_{0})·O+(a_{1}+b_{1})·B+sum[k = 0]-[n+(-1)][ (a_{k+2}+b_{k+2})·BO...(k)...OB ]

Definición:

w·( a_{0}·O+a_{1}·B+sum[k = 0]-[n+(-1)][ a_{k+2}·BO...(k)...OB ] ) =...

... (w·a_{0})·O+(w·a_{1})·B+sum[k = 0]-[n+(-1)][ (w·a_{k+2})·BO...(k)...OB ]


Grupo producto por coordenada:

Definición:

... a_{0}·O+a_{1}·B+sum[k = 0]-[n+(-1)][ a_{k+2}·BO...(k)...OB ] ...

... [+ · +] ...

... b_{0}·O+b_{1}·B+sum[k = 0]-[n+(-1)][ b_{k+2}·BO...(k)...OB ] = ...

... (a_{0}·b_{0})·O+(a_{1}·b_{1})·B+sum[k = 0]-[n+(-1)][ (a_{k+2}·b_{k+2})·BO...(k)...OB ]


Teorema:

< BOB+BB+B+O,BB+B+O,B+O,O > es base

Demostración:

Independencia lineal:

a·(BOB+BB+B+O)+b·(BB+B+O)+c·(B+O)+d·O = 0

a·BOB+(a+b)·(BB)+(a+b+c)·B+(a+b+c+d)·O = 0

a = 0

a = 0 & b = 0

a = 0 & b = 0 & c = 0

a = 0 & b = 0 & c = 0 & d = 0

Generador:

a·BOB+b·BB+c·B+d·O = a·(BOB+BB+B+O)+(b+(-a))·(BB+B+O)+(c+(-b))·(B+O)+(d+(-c))·O


No entiendo que vos han dicho de los hombres,

si vos creéis Jesucristo,

y estáis pagando condenación del mundo salvando-lo,

porque no hay indulgencias,

y no es ataque de ningún hombre.


Definición:

[Ev][ F(x) = x+v ]

Teorema:

F(x+y) = F(x)+F(y)

F(w·x) = w·F(x)

Demostración:

F(x+y) = (x+y)+v = (x+y)+( (1/2)·v+(1/2)·v ) = (x+(1/2)·v)+(y+(1/2)·v) = (x+p)+(y+q) = F(x)+F(y)

F(w·x) = w·x+v = w·( x+(1/w)·v ) = w·(x+s) = w·F(x)


Teorema:

Ker(F) = {(-v)}

Demostración:

F(-v) = (-v)+v = 0

Teorema:

Si ( E/Ker(F) ) = {z+(-v)} ==> F[ ( E/Ker(f) ) ] = E

Demostración:

F(z+(-v)) = (z+(-v))+v = z+((-v)+v) = z+0 = z

Teorema:

Si ( E/Ker(F) ) = {z+(-v)} ==> Im(F) =[h(z)]= ( E/Ker(F) )

Demostración:

h(x) = h(y)

x+(-v) = y+(-v)

x = y

h(x+(-v)) = f(y+(-v))

x = y

x+(-v) = y+(-v)


Definición:

[Ev][ F(x,y) = xy+v ]

Teorema:

F(z,x+y) = F(z,x)+F(z,y)

F(z,w·x) = w·F(z,x)

Demostración:

F(z,x+y) = z·(x+y)+v = (zx+zy)+v = (zx+zy)+( (1/2)·v+(1/2)·v ) = (zx+(1/2)·v)+(zy+(1/2)·v) = ...

... (zx+p)+(zy+q) = F(z,x)+F(z,y)

F(z,w·x) = z·(w·x)+v = w·(zx)+v = w·( zx+(1/w)·v ) = w·(zx+s) = w·F(z,x)


Teorema:

Ker(F) = { < z,(1/z)·(-v) > || < (-v)·(1/z),z > }

Demostración:

F(z,(1/z)·(-v)) = z·((1/z)·(-v))+v = (z/z)·(-v)+v = (-v)+v = 0

Teorema:

Si ( E/Ker(F) ) = { < z,(1/z)·(-v)+s > || < (-v)·(1/z)+s,z > } ==> Im(F) =[h(w)]= ( E/Ker(F) )

Demostración:

h(z,(1/z)·(-v)+p) = h(z,(1/z)·(-v)+q)

zp = zq

p = q

(1/z)·(-v)+p = (1/z)·(-v)+q

< z,(1/z)·(-v)+p > = < z,(1/z)·(-v)+q >

h(zp) = h(zq)

< z,(1/z)·(-v)+p > = < z,(1/z)·(-v)+q >

(1/z)·(-v)+p = (1/z)·(-v)+q

p = q

zp = zq


Teorema:

Sea F(x,y) = xy+BB ==>

Ker(F) = { < B^{n},(O/B)^{n}·(-1)·BB > || < (-1)·BB·(O/B)^{n},B^{n} > }


Teorema:

Si d_{t}[z] = f(t)·z ==> z_{n}(t) = z_{0}·( 1+h·f(t) )^{n}

Si h = 0·( int[f(t)]d[t]/f(t) ) ==> 

... z_{n}(t) = ( 1+0·int[f(t)]d[t] )^{n}

Método numérico convergente:

(1/h)·( z_{n+1}+(-1)·z_{n} ) = f(t)·z_{n}

z_{0} = 1

Demostración:

(1/h)·( z_{n+1}+(-1)·z_{n} ) = f(t)·z_{n}

( z_{n+1}+(-1)·z_{n} ) = h·f(t)·z_{n}

z_{n+1} = z_{n}+h·f(t)·z_{n}

z_{n+1} = z_{n}·(1+h·f(t))

Teorema:

Si d_{t}[z] = f(t)·(1/z) ==> a_{n}(t) = a_{0}+(n/2)·h·f(t)

Si h = (1/n) ==> 

... a(t) = a_{0}+(1/2)·f(t)

... z(t) = ( 2a_{0}+f(t) )^{(1/2)} & a_{0} = int[f(t)]d[t]+(-1)·(1/2)·f(t)

Método numérico convergente:

(1/h)·( z_{n+1}+(-1)·z_{n} ) = ( f(t)/z_{n} )

z_{0} = ( 2·int[f(t)]d[t]+(-1)·f(t) )^{(1/2)}

Demostración:

(1/h)·( z_{n+1}+(-1)·z_{n} ) = ( f(t)/z_{n} )

( z_{n+1}+(-1)·z_{n} ) = ( (h·f(t))/z_{n} )

z_{n+1} = z_{n}+( (h·f(t))/z_{n} )

z_{n+1}·z_{n} = ( z_{n} )^{2}+h·f(t)

Sea z_{n} = ( 2a_{n} )^{(1/2)} &  z_{n+1}·z_{n} = 2a_{n+1} ==>

2a_{n+1} = 2a_{n}+h·f(t)

a_{n+1} = a_{n}+(1/2)·h·f(t)


Teorema:

Forma integral interior:

Sea F(ax+b) = int[x = 0]-[1][ ax+b ]d[x] ==>

G(ax+b) = int[x = 0]-[1][ (8/a)·x+(-1)·(1/b) ]d[x]

F(ax+b) [o] G(ax+b) = 1

Teorema:

Forma integral exterior:

Sea F(ax+b) = int[x = 0]-[1][ ax+b ]d[x] ==>

G(ax+b) = int[x = 0]-[1][ (4/a)·x+(-1)·(1/b) ]d[x]

F(ax+b) [o] G(ax+b) = 0

Teorema:

Forma funcional interior:

Sea F(h(x)) = sum[k = 1]-[n][ ( h(x) )^{k} ]+1 ==>

G(h(x)) = sum[k = 1]-[n][ (1/h(x))^{k} ]+((-n)+1)

F(h(x)) [o] G(h(x)) = 1

Teorema:

Forma funcional exterior:

Sea F(h(x)) = sum[k = 1]-[n][ ( h(x) )^{k} ]+1 ==>

G(h(x)) = sum[k = 1]-[n][ (1/h(x))^{k} ]+(-n)

F(h(x)) [o] G(h(x)) = 0

Teorema:

< cosh(kx), sinh(kx) > es linealmente independiente

Demostración

a·cosh(kx)+b·sinh(kx) = 0

(1/2)·( (a+b)·e^{kx}+(a+(-b))·e^{(-k)·x} ) = 0

(-a) = b = a 

a = 0 & b = 0 

Teorema:

sum[k = 0]-[n][ a_{k}·e^{kx}] = sum[k = 0]-[n][ a_{k}·cosh(kx)+a_{k}·sinh(kx) ]

Teorema:

sum[k = 0]-[n][ a_{k}·e^{(-k)·x}] = sum[k = 0]-[n][ a_{k}·cosh(kx)+a_{k}·(-1)·sinh(kx) ]

Teorema:

sum[k = 0]-[n][ a_{k}·e^{kxi}] = sum[k = 0]-[n][ a_{k}·cosh(kxi)+a_{k}·(1/i)·sinh(kxi) ]

Teorema:

sum[k = 0]-[n][ a_{k}·e^{(-k)·xi}] = sum[k = 0]-[n][ a_{k}·cosh(kx)+a_{k}·i·sinh(kxi) ]


Teorema:

f_{sup{k}}(x) = sum[k = 1]-[oo][ a_{k}·cosh(x)+b_{k}·sinh(x) ]

a_{k} = (1/pi·i)·int[x = 0]-[2pi·i][ f_{k}(x)·cosh(x) ]d[x]

b_{k} = (-1)·(1/pi·i)·int[x = 0]-[2pi·i][ f_{k}(x)·sinh(x) ]d[x]

Teorema:

Sea f_{k}(x) = kx ==>

(oo·x) = 2·0·sinh(x)·sum[k = 1]-[oo][ k ]

Sea x = 0 ==>

sum[k = 1]-[oo][ k ] = (1/2)·oo^{2}

Teorema:

Sea f_{k}(x) = (x/k) ==>

x = 2·0·sinh(x)·sum[k = 1]-[oo][ (1/k) ]

Sea x = ln(oo) ==>

sum[k = 1]-[oo][ (1/k) ] = ln(oo)

Teorema:

Sea f_{k}(x) = (x/k)^{3} ==>

sum[k = 1]-[oo][ (1/k)^{3} ] = 2+1+(1/2)^{3}+...+(1/n)^{3}+...

(2x)^{3} = 3·2·4·sinh(x)·sum[k = 1]-[oo][ (1/k)^{3} ]

Sea x = (pi/2) ==>

sum[k = 1]-[oo][ (1/k)^{3} ] = (1/24)·pi^{3}

Teorema:

Sea f_{k}(x) = (x/k)^{5} ==>

sum[k = 1]-[oo][ (1/k)^{5} ] = 2+1+(1/2)^{5}+...+(1/n)^{5}+...

(2x)^{5} = 5·(2^{2}+1)·3·(2^{1}+(-1))·4·sinh(x)·sum[k = 1]-[oo][ (1/k)^{5} ]

Sea x = (pi/2) ==>

sum[k = 1]-[oo][ (1/k)^{5} ] = (1/300)·pi^{5}

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