martes, 25 de noviembre de 2025

análisis-matemático

Teorema:

Sea a_{n} acotada ==>

Si [An][ |b_{n}| [< |a_{n}+c| ] ==> b_{n} está acotada

Demostración:

Sea a_{n} acotada ==>

[EN][ |a_{n}| [< N ]

|b_{n}| [< |a_{n}+c| [< |a_{n}|+|c| [< N+|c|

Se define M = N+|c| ==>

|b_{n}| [< M

Teorema:

[An][ 0 [< e^{n} & 0 [< e^{(-n)} ]

Demostración:

por inducción:

0 [< 1 = e^{0}

0 [< 1 [< e^{n} [< e^{n+1}

Por descenso:

0 = (1/e^{oo}) = ( 1/oo^{[e]+(-1)} ) = (1/oo)

0 [< (1/e^{n}) [< (1/e^{n+(-1)})

Teorema:

Sea a_{n} acotada ==>

Si [An][ |b_{n}| [< |e^{|a_{n}|}+c| ] ==> b_{n} está acotada

Demostración:

Sea a_{n} acotada ==>

[EN][An][ |a_{n}| [< N ]

e^{|a_{n}|} [< e^{N}

|b_{n}| [< |e^{|a_{n}|}+c| [< |e^{|a_{n}|}|+|c| = e^{|a_{n}|}+|c| [< e^{N}+|c|

Se define M = e^{N}+|c| ==>

|b_{n}| [< M



Teorema:

F(x,t) = ( x(t) )^{p+1}+(-1)·h(t)·( x(t)+(-1)·M(t) )

h(t) = (p+1)·( M(t) )^{p}

Demostración:

d_{x}[ F(x,t) ] = d_{x}[ ( x(t) )^{p+1}+(-1)·h(t)·( x(t)+(-1)·M(t) ) ] = ...

... d_{x}[ ( x(t) )^{p+1} ]+d_{x}[ (-1)·h(t)·( x(t)+(-1)·M(t) ) ] = ...

... d_{x}[ ( x(t) )^{p+1} ]+(-1)·h(t)·d_{x}[ x(t)+(-1)·M(t) ] = ...

... d_{x}[ ( x(t) )^{p+1} ]+(-1)·h(t)·( d_{x}[ x(t) ]+d_{x}[ (-1)·M(t) ] ) = ...

... (p+1)·( x(t) )^{p}+(-1)·h(t) = 0

Teorema:

F(x,t) = e^{(p+1)·x(t)}+(-1)·h(t)·( e^{x(t)}+(-1)·M(t) )

h(t) = (p+1)·( M(t) )^{p}



Teorema:

F(x,y,t) = ( x(t) )^{p+1}+( y(t) )^{p+1}+(-1)·h(t)·( ( x(t)+y(t) )+(-1)·M(t) )

h(t) = (1/2)·(p+1)·( ( (1+(-1)·k(t))·M(t) )^{p}+( k(t)·M(t) )^{p} )

Teorema:

F(x,y,t) = e^{(p+1)·x(t)}+e^{(p+1)·y(t)}+(-1)·h(t)·( ( e^{x(t)}+e^{y(t)} )+(-1)·M(t) )

h(t) = (1/2)·(p+1)·( ( (1+(-1)·k(t))·M(t) )^{p}+( k(t)·M(t) )^{p} )



Teorema:

F(x,y,t) = x+y+(-1)·h(t)·( (x+y)+(-1)·M(t) )

h(t) = 1

Teorema:

F(x,y,t) = e^{x}+e^{y}+(-1)·h(t)·( (e^{x}+e^{y})+(-1)·M(t) )

h(t) = 1



Teorema:

F(x,y,t) = ( x^{2}+y^{2} )+(-1)·h(t)·( (x+y)+(-1)·M(t) )

h(t) = M(t)

Demostración:

2x+2y = 2·M(t)

Teorema:

F(x,y,t) = ( e^{2x}+e^{2y} )+(-1)·h(t)·( (e^{x}+e^{y})+(-1)·M(t) )

h(t) = M(t)



Teorema:

F(x,y,t) = ( x^{2}+nxy+y^{2} )+(-1)·h(t)·( (x+y)+(-1)·M(t) )

h(t) = (1/2)·(n+2)·M(t)

Demostración:

2x+2y+n·(y+x) = (n+2)·M(t)

Teorema:

F(x,y,t) = ( e^{2x}+ne^{x+y}+e^{2y} )+(-1)·h(t)·( (e^{x}+e^{y})+(-1)·M(t) )

h(t) = (1/2)·(n+2)·M(t)