sucesiones-de-recurrencia y álgebra-lineal y economía y evangelio-stronikiano y mecánica-integral y análisis-matemático

Teorema:

Sea b >] 1 ==>

Si ( a_{0} = 1 & a_{n+1} = ( ba_{n} )^{(1/2)} ) ==> ...

... a_{n} es creciente & a_{n} está acotada superiormente

... lim[n = oo][ a_{n} ] = b

Demostración:

( b >] 1 <==> b^{(1/2)} < 1 ) [ Destrocter ponens ]

c < a_{n+1} = ( ba_{n} )^{(1/2)} < ( a_{n} )^{(1/2)} [< a_{n}

c >] a_{n+1} >] a_{n}

Sea lim[n = oo][ a_{n} ] = lim[n = oo][ a_{n+1} ] = x ==>

x^{2} = bx <==> x = b

Teorema:

Sea b >] 1 ==>

Si ( a_{0} = 0 & a_{n+1} = ( (b+(-1))+2a_{n} )^{(1/2)} ) ==> ...

... a_{n} es creciente & a_{n} está acotada superiormente

... lim[n = oo][ a_{n} ] = 1+b^{(1/2)}

Demostración:

( b >] 1 <==> b < 1 ) [ Destrocter ponens ]

c < a_{n+1} = ( (b+(-1))+2a_{n} )^{(1/2)} < ( 2a_{n} )^{(1/2)} [< 2a_{n} < a_{n}

c >] a_{n+1} >] a_{n}

Sea lim[n = oo][ a_{n} ] = lim[n = oo][ a_{n+1} ] = x ==>

x^{2} = ( b+(-1) )+2x <==> x = 1+b^{(1/2)}

Teorema: [ del algoritmo recurrente de la raíz cuadrada ]

Sea b >] 1 ==>

Si ( a_{0} = 1 & a_{n+1} = ( 2+( (b+(-1))/a_{n} ) ) ==> ...

... a_{n} es decreciente & a_{n} está acotada inferiormente

... lim[n = oo][ a_{n} ] = 1+b^{(1/2)}

Demostración:

1 > a_{n+1} = ( 2+( (b+(-1))/a_{n} )

a_{n} > 2a_{n}+b+(-1)

0 > a_{n}+b+(-1) >] a_{n}

( a_{n} >] 1 <==> ( 1/a_{n} ) > 1 ) [ Destrocter ponens ]

c > a_{n+1} = ( 2+( (b+(-1))/a_{n} ) ) > ( ( 2a_{n}+(b+(-1)) )/a_{n} ) >] ( 2a_{n}+(b+(-1)) > a_{n}

c [< a_{n+1} [< a_{n}

Sea lim[n = oo][ a_{n} ] = lim[n = oo][ a_{n+1} ] = x ==>

x = 2+( (b+(-1))/x ) <==> x^{2} = 2x+( b+(-1) ) <==> x = 1+b^{(1/2)}

Teorema:

Sea A = i·(x+1)+j·1 & B = k·(x+i)

A [&] B = B

Gen( A [ || ] B ) = A + B = A

Demostración:

i·(x+1)+j·1 = k·(x+i)

i = k & j = ki+(-k)

i·(x+1)+j·1+k·(x+i) = ax+b

i = a+(-k) & j = b+(-a)+k·( 1+(-i) )

Teorema:

Sea A = ix+j ==>

Si [Ea][ F(ix+j) = (ix+j)+a ] ==> ...

... Ker(F) = { ix+j : i = 0 & j = (-a) }

... A = Im(F)+Ker(F)

Demostración:

... A = (ix+j)+(a+(-a))= ix+(j+a)+(-a) = ix+(j+a)+Ker(F) = (ix+j)+a+Ker(F) = Im(F)+Ker(F)

Teorema:

Sea F(x,y) = ( < a,a >,< a,a > ) o < x,y >  ==> ...

... Ker(F) = k·< 1,(-1) >

... Si Im(F) = { < x,y > : [Ek][ F(x,y) = k·< x,y > ] } ==> A = Im(F)+Ker(F)

Demostración:

A = i·< 1,0 >+j·< 0,1 >+(-k)·< 1,(-1) >+k·< 1,(-1) > = 2k·< 1,0 >+(-k)·< 1,(-1) >+k·< 1,(-1) > = ...

... k·< 1,0 >+k·< 0,1 >+Ker(F) = k·< 1,1 >+Ker(F) = Im(F)+Ker(F)

Teorema:

Sea F(x,y) = ( < a,(-a) >,< (-a),a > ) o < x,y >  ==> ...

... Ker(F) = k·< 1,1 >

... Si Im(F) = { < x,y > : [Ek][ F(x,y) = k·< x,y > ] } ==> A = Im(F)+Ker(F)

Demostración:

A = i·< 1,0 >+j·< 0,1 >+(-k)·< 1,1 >+k·< 1,1 > = 2k·< 1,0 >+(-k)·< 1,1 >+k·< 1,1 > = ...

... k·< 1,0 >+k·< 0,(-1) >+Ker(F) = k·< 1,(-1) >+Ker(F) = Im(F)+Ker(F)

Teorema:

Sea F(x,y) = ( < a,b >,< a,b > ) o < x,y >  ==> ...

... Ker(F) = k·< b,(-a) >

... Si Im(F) = { < x,y > : [Ek][ F(x,y) = k·< x,y > ] } ==> A = Im(F)+Ker(F)

Demostración:

A = i·< 1,0 >+j·< 0,1 >+(-k)·< b,(-a) >+k·< b,(-a) > = ...

... k·(1+b)·< 1,0 >+k·(1+(-a))·< 0,1 >+(-k)·< b,(-a) >+k·< b,(-a) > = ...

... k·< 1,0 >+k·< 0,1 >+Ker(F) = k·< 1,1 >+Ker(F) = Im(F)+Ker(F)

Teorema:

Sea F(x,y) = ( < a,(-b) >,< (-a),b > ) o < x,y >  ==> ...

... Ker(F) = k·< b,a >

... Si Im(F) = { < x,y > : [Ek][ F(x,y) = k·< x,y > ] } ==> A = Im(F)+Ker(F)

Demostración:

A = i·< 1,0 >+j·< 0,1 >+(-k)·< b,a >+k·< b,a > = ...

... k·(1+b)·< 1,0 >+k·((-1)+a)·< 0,1 >+(-k)·< b,a >+k·< b,a > = ...

... k·< 1,0 >+k·< 0,(-1) >+Ker(F) = k·< 1,(-1) >+Ker(F) = Im(F)+Ker(F)

Examen de Álgebra lineal:

Teorema:

Sea F(x,y) = ( < a,a >,< b,b > ) o < x,y >  ==> ...

... Ker(F) = ?

... Si Im(F) = { < x,y > : [Ek][ F(x,y) = k·< x,y > ] } ==> A = Im(F)+Ker(F)

Demostración:

A = k·(a+b)·< a,b >+Ker(F) = Im(F)+Ker(F)

Teorema:

Sea F(x,y) = ( < a,(-a) >,< (-b),b > ) o < x,y >  ==> ...

... Ker(F) = ?

... Si Im(F) = { < x,y > : [Ek][ F(x,y) = k·< x,y > ] } ==> A = Im(F)+Ker(F)

Demostración:

A = k·(a+b)·< (-a),b >+Ker(F) = Im(F)+Ker(F)

Automatismos:

Lema:

A(x) = px+(-n)·e^{x}

A(0) = (-n)

d_{x}[A(0)] = 0 <==> p = n

B(x) = (-n)·x+p·ln(x)

B(1) = (-n)

d_{x}[B(1)] = 0 <==> p = n

Lema:

A(x) = px+(-n)·( 1+sinh(x) )

A(0) = (-n)

d_{x}[A(0)] = 0 <==> p = n

B(x) = px+(-n)·( x+cosh(x) )

B(0) = (-n)

d_{x}[B(0)] = 0 <==> p = n

Lema:

A(x) = px+(-n)·( (x+1)^{2m+1}+(-1)·2mx )

A(0) = (-n)

d_{x}[A(0)] = 0 <==> p = n

B(x) = px+(-n)·( e^{(2m+1)·x}+(-1)·2mx )

B(0) = (-n)

d_{x}[B(0)] = 0 <==> p = n

Juan:

El esclavo no es mayor que su señor,

ni el enviado mayor que el que lo envía.

El esclavo no es mayor que el enviado,

ni el señor mayor que el que lo envía.

Ley:

No se molesta a fieles con esclavos infieles,

porque el esclavo no es mayor que el enviado.

No se molesta a los fieles que siguen a Dios estudiando,

porque el señor no es mayor que el que lo envía.

Ley:

El sexo no es mayor que la ciencia,

porque el esclavo no es mayor que el que lo envía.

La violencia no es mayor que la ciencia,

porque el esclavo no es mayor que el que lo envía.

Teorema:

int[ax = 0]-[1][ P(t)·d_{t}[H(t,ax)] ]d[ax]+Q(ut)·H(t,1) = 0

H(t,ax) = Anti-[ ( s /o(s)o/ int[ d_{ax}[ (-1)·Q(ut)·s·(ax)^{2} ] ]d[s] ) ]-( int[ ( 1/P(t) ) ]d[t] )

Demostración:

int[ax = 0]-[1][ P(t)·d_{ax}[ (-1)·Q(ut)·H(t,ax)·(ax)^{2} ]·( 1/P(t) ) ]d[ax]+Q(ut)·H(t,1) = ...

... int[ax = 0]-[1][ d_{ax}[ (-1)·Q(ut)·H(t,ax)·(ax)^{2} ] ]d[ax]+Q(ut)·H(t,1) = ...

... ( (-1)·Q(ut)·H(t,1)+Q(ut)·H(t,0)·0^{2} )+Q(ut)·H(t,1) = (-1)·Q(ut)·H(t,1)+Q(ut)·H(t,1) = 0

Ley:

int[ax = 0]-[1][ (-1)·(b/m)·t^{2}·d_{t}[H(t,ax)] ]d[ax]+Q(ut)·H(t,1) = 0

H(t,ax) = Anti-[ ( s /o(s)o/ int[ d_{ax}[ (-1)·Q(ut)·s·(ax)^{2} ] ]d[s] ) ]-( (m/b)·(1/t) )

Ley:

int[ax = 0]-[1][ (-1)·(k/m)·t^{3}·d_{t}[H(t,ax)] ]d[ax]+Q(ut)·H(t,1) = 0

H(t,ax) = Anti-[ ( s /o(s)o/ int[ d_{ax}[ (-1)·Q(ut)·s·(ax)^{2} ] ]d[s] ) ]-( (1/2)·(m/k)·(1/t)^{2} )

Examen:

Ley:

int[x = 0]-[1][ (m/b)·d_{t}[H(t,ax)] ]d[ax]+Q(ut)·H(t,1) = ct

H(t,x) = ?

Ley:

int[x = 0]-[1][ u·(m/k)·d_{t}[H(t,ax)] ]d[ax]+Q(ut)·H(t,1) = ct

H(t,x) = ?

Teorema:

[As][ s > 0 ==> ...

... | int[ax = 0]-[1][ P(t)·d_{t}[H(t,ax)] ]d[ax]+Q(ut)·lim[n = oo][ sum[k = 1]-[n][ H(t,(k/n)) ] ] | < s ]

<==>

H(t,ax) = Anti-[ ( s /o(s)o/ int[ (-1)·Q(ut)·s ]d[s] ) ]-( int[ ( 1/P(t) ) ]d[t] )

Demostración:

int[ax = 0]-[1][ P(t)·( (-1)·Q(ut)·H(t,ax) )·( 1/P(t) ) ]d[ax]+Q(ut)·int[ax = 0]-[1][ H(t,ax) ]d[ax] = ...

... int[ax = 0]-[1][ ( (-1)·Q(ut)·H(t,ax) ) ]d[ax]+Q(ut)·int[ax = 0]-[1][ H(t,ax) ]d[ax] = ...

... (-1)·Q(ut)·int[ax = 0]-[1][ H(t,ax) ]d[ax]+Q(ut)·int[ax = 0]-[1][ H(t,ax) ]d[ax] = 0

Teorema:

[As][ s > 0 ==> ...

... | int[ax = 0]-[1][ P(t)·d_{t}[H(t,ax)] ]d[ax]+...

... Q(ut)·lim[n = oo][ sum[k = 1]-[n][ (ut)^{m}·(k/n)^{q} ] ] | < s ]

<==>

H(t,ax) = ...

... (-1)·(1/u)·int[ Q(ut) ]d[ut] [o(t)o] ( (1/u)·( 1/(m+1) )·(ut)^{m+1}·(ax)^{q} ) [o(t)o] int[ ( 1/P(t) ) ]d[t]

Teorema:

[As][ s > 0 ==> ...

... | int[ax = 0]-[1][ P(t)·d_{t}[H(t,ax)] ]d[ax]+...

... Q(ut)·lim[n = oo][ sum[k = 1]-[n][ (ut)^{m}·e^{(k/n)} ] ] | < s ]

<==>

H(t,ax) = ...

... (-1)·(1/u)·int[ Q(ut) ]d[ut] [o(t)o] ( (1/u)·( 1/(m+1) )·(ut)^{m+1}·e^{ax} ) [o(t)o] int[ ( 1/P(t) ) ]d[t]

Axioma: [ de Stolz constructor ]

Si ( lim[n = oo][ (a_{n+1}+(-1)·a_{n})/(b_{n+1}+(-1)·b_{n}) ] = a & ...

... lim[n = oo][ a_{n}/b_{n} ] = b ) ==> ...

... ( a = b <==> [Af(x)][ f(x) es constrocter ponens ] )

... <==>

... ( a != b <==> [Ef(x)][ f(x) es destrocter ponens ] )

Axioma: [ de Stolz destructor ]

Si ( lim[n = oo][ (a_{n+1}+(-1)·a_{n})/(b_{n+1}+(-1)·b_{n}) ] = a & ...

... lim[n = oo][ a_{n}/b_{n} ] = b ) ==> 

... ( a != b <==> [Af(x)][ f(x) es constrocter ponens ] )

... <==>

... ( a = b <==> [Ef(x)][ f(x) es destrocter ponens ] )

Teorema:

lim[n = oo][ ...

... ( < 0x,...,0x > )^{(1/2)} ...

... o ...

... ( < ln((1/n)·x+1),...(n)...,0 >,...(n)...,< 0,...(n)...,ln(x+1) > )

... o ...

... ( < 0x,...,0x > )^{(1/2)} = (x+1)·ln(x+1)+(-x)

Demostración: [ por Stolz destructor ]

u(0) = m

v(m) = (-1)

p(0) = j

q(j) = 1

h(1) = (1/x)

lim[n = oo][ ln( ((k/(n+1))·x+1)/((k/n)·x+1) ) ] = ln( ((k/oo)·x+1)/((k/oo)·x+1) ) = ln(1) = 0

lim[n = oo][ sum[k = 1]-[n][ ln((k/n)·x+1) ]·0x ] < ...

... lim[n = oo][ sum[k = 1]-[n][ ln(x+1) ]·0x ] = ln(x+1)·x

ln(x+1)·x = ( ( 1+p(0) )·ln(x+1)+u(0) )·x = ( (1+j)·ln(x+1)+m )·x = ...

... ( (1+h( w(j) ))·ln(x+1)+v(m) )·x = ( (1+(1/x))·ln(x+1)+(-1) )·x = (x+1)·ln(x+1)+(-x)

Teorema:

lim[n = oo][ ...

... ( < 0x,...,0x > )^{(1/2)} ...

... o ...

... ( < sinh((1/n)·x),...(n)...,0 >,...(n)...,< 0,...(n)...,sinh(x) > )

... o ...

... ( < 0x,...,0x > )^{(1/2)} = cosh(x)+(-1)

Demostración: [ por Stolz destructor ]

u(1) = m

v(m) = 2k+2

w(0) = j

h(j) = (-1)

lim[n = oo][ sinh( (k/(n+1))·x+1 )+(-1)·sinh( (k/n)·x+1 ) ] = ...

... sinh((k/oo)·x+1)+(-1)·sinh((k/oo)·x+1) = 0

lim[n = oo][ sum[k = 1]-[n][ sinh((k/n)·x) ]·0x ] < ...

... lim[n = oo][ sum[k = 1]-[n][ sinh(x+1) ]·0x ] = sinh(x+1)·x

sinh(x+1)·x = sum[k = 0]-[oo][ (1/(2k+1)!)·x^{2k+2} ] = ...

... sum[k = 0]-[oo][ ( 1/v(u(1)) )·(1/(2k+1)!)·x^{2k+2} ] = ...

... sum[k = 0]-[oo][ (1/(2k+2)!)·x^{2k+2} ] = sum[k = 0]-[oo][ (1/(2·(k+1))!)·x^{2·(k+1)} ] = ...

... sum[k = 0]-[oo][ (1/(2p)!)·x^{2p} ] = cosh(x) = cosh(x)+h( w(0) ) = cosh(x)+(-1)