integrales-impropias y probabilidades y olores y mecánica-estadística y economía y álgebra-lineal y sexualidad

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 ) ==>

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 = 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) ) ) )

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)

ecuaciones-de-Maxwell y análisis-funcional-y-teoría-de-cuerdas y medicina y análisis-matemático y economía y mecánica-integral

Principio:

E(x,y,z) = qk·(1/r)^{3}·< x,y,z >

B(d_{t}[x],d_{t}[y],d_{t}[z]) = (-1)·qk·(1/r)^{3}·< d_{t}[x],d_{t}[y],d_{t}[z] >

Principio:

E(yz,zx,xy) = qk·(1/r)^{4}·< yz,zx,xy >

B(d_{t}[yz],d_{t}[zx],d_{t}[xy]) = (-1)·qk·(1/r)^{4}·< d_{t}[yz],d_{t}[zx],d_{t}[xy] >

Ley:

div[ E(x,y,z) ] = 3qk·(1/r)^{3}

div[ int[ B(d_{t}[x],d_{t}[y],d_{t}[z]) ]d[t] ] = (-3)·qk·(1/r)^{3}

Anti-div[ E(yz,zx,xy) ] = 3qk·(1/r)^{4}

Anti-div[ int[ B(d_{t}[yz],d_{t}[zx],d_{t}[xy]) ]d[t] ] = (-3)·qk·(1/r)^{4}

Ley:

div[ E(x,y,z) ] = d_{xyz}[ Anti-potencial[ E(x,y,z) ] ]

Anti-div[ E(yz,zx,xy) ] = d_{xyz}[ potencial[ E(yz,zx,xy) ] ]

Ley:

div[ int[ B(d_{t}[x],d_{t}[y],d_{t}[z]) ]d[t] ] = ...

... d_{xyz}[ Anti-potencial[ int[ B(d_{t}[x],d_{t}[y],d_{t}[z]) ]d[t] ] ]

Anti-div[ int[ B(d_{t}[yz],d_{t}[zx],d_{t}[xy]) ]d[t] ] = ...

... d_{xyz}[ potencial[ int[ B(d_{t}[yz],d_{t}[zx],d_{t}[xy]) ]d[t] ] ]

Ley:

Anti-potencial[ (1/r)·rot[ E(x,y,z) ] ] = ...

... qk·(1/r)^{3}+(1/3)·( 1/(xyz) )·...

... Anti-potencial[ int[ B(d_{t}[x],d_{t}[y],d_{t}[z]) ]d[t] ]

Ley:

Anti-potencial[ (1/r)·rot[ int[ B(d_{t}[x],d_{t}[y],d_{t}[z],d_{t}[q(t)]) ]d[t] ] ] = ...

... d_{t}[q(t)]·k·(1/r)^{3}+(-1)·(1/3)·( 1/(xyz) )·...

... Anti-potencial[ d_{t}[ E(x,y,z,q(t)) ]+B(d_{t}[x],d_{t}[y],d_{t}[z],q(t)) ]

Ley:

rot[ E(x,y,z) ] = qk·(1/r)^{6}·< x,y,z >·< y+(-z),z+(-x),x+(-y) >

rot[ int[ B(d_{t}[x],d_{t}[y],d_{t}[z],d_{t}[q(t)]) ]d[t] ] = ...

... (-1)·q(t) [o(t)o] k·(1/r)^{6}·< x,y,z >·< y+(-z),z+(-x),x+(-y) >

Ley:

Sea Anti-potencial[ J(x,y,z) ] = qk·(1/r)^{3} ==>

J(x,y,z) = (1/r)·rot[ E(x,y,z) ]+...

... (-1)·(1/3)·( ...

... ( 1/(xyz) )·int[ B(d_{t}[x],d_{t}[y],d_{t}[z]) ]d[t]+...

... < (1/yz),(1/zx),(1/xy) >·qk·(1/r)^{3} )

Sea Anti-potencial[ K(x,y,z) ] = d_{t}[q(t)]·k·(1/r)^{3} ==>

K(x,y,z) = (1/r)·rot[ int[ B(d_{t}[x],d_{t}[y],d_{t}[z],d_{t}[q(t)]) ]d[t] ]+...

... (1/3)·( ...

... ( 1/(xyz) )·( d_{t}[ E(x,y,z,q(t)) ]+B(d_{t}[x],d_{t}[y],d_{t}[z],q(t)) )+...

... < (1/yz),(1/zx),(1/xy) >·d_{t}[q(t)]·k·(1/r)^{3} )

Deducción:

Anti-Grad[ Anti-potencial[ F(x,y,z) ] ] = F(x,y,z)

Anti-Grad[ H(x,y,z)·Anti-potencial[ F(x,y,z) ] ] = ...

... H(x,y,z)·F(x,y,z)+Anti-Grad[ H(x,y,z) ]·Anti-potencial[ F(x,y,z) ]

Ley:

Potencial[ (1/r)^{2}·Anti-rot[ E(yz,zx,xy) ] ] = ...

... qk·(1/r)^{4}+(1/3)·( 1/(xyz) )·Potencial[ int[ B(d_{t}[yz],d_{t}[zx],d_{t}[xy]) ]d[t] ]

Ley:

Potencial[ (1/r)^{2}·Anti-rot[ int[ B(d_{t}[yz],d_{t}[zx],d_{t}[xy],d_{t}[q(t)]) ]d[t] ] ] = ...

... d_{t}[q(t)]·k·(1/r)^{4}+(-1)·(1/3)·( 1/(xyz) )·...

... Potencial[ d_{t}[ E(yz,zx,xy,q(t)) ]+B(d_{t}[yz],d_{t}[zx],d_{t}[xy],q(t)) ]

Ley:

Anti-rot[ E(yz,zx,xy) ] = qk·(1/r)^{6}·< yz,zx,xy >·< y+(-z),z+(-x),x+(-y) >

Anti-rot[ int[ B(d_{t}[yz],d_{t}[zx],d_{t}[xy],d_{t}[q(t)]) ]d[t] ] = ...

... (-1)·q(t) [o(t)o] k·(1/r)^{6}·< yz,zx,xy >·< y+(-z),z+(-x),x+(-y) >

Ley:

Sea Potencial[ P(yz,zx,xy) ] = qk·(1/r)^{4} ==>

P(yz,zx,xy) = (1/r)^{2}·Anti-rot[ E(yz,zx,xy) ]+...

... (-1)·(1/3)·( ...

... ( 1/(xyz) )·int[ B(d_{t}[yz],d_{t}[zx],d_{t}[xy]) ]d[t]+...

... (-1)·< (1/x),(1/y),(1/z) >·qk·(1/r)^{4} )

Sea Potencial[ Q(yz,zx,xy) ] = d_{t}[q(t)]·k·(1/r)^{4} ==>

Q(yz,zx,xy) = (1/r)^{2}·Anti-rot[ int[ B(d_{t}[yz],d_{t}[zx],d_{t}[xy],d_{t}[q(t)]) ]d[t] ]+...

... (1/3)·( ...

... ( 1/(xyz) )·( d_{t}[ E(yz,zx,xy,q(t)) ]+B(d_{t}[yz],d_{t}[zx],d_{t}[xy],q(t)) )+...

... (-1)·< (1/x),(1/y),(1/z) >·d_{t}[q(t)]·k·(1/r)^{4} )

Deducción:

Grad[ Potencial[ F(x,y,z) ] ] = F(x,y,z)

Grad[ H(x,y,z)·Potencial[ F(x,y,z) ] ] = ...

... H(x,y,z)·F(x,y,z)+Grad[ H(x,y,z) ]·Potencial[ F(x,y,z) ]

Teorema:

Sea H( y(x) ) = ( d_{x}[y(x)] )^{n+1}+(-1)·(n+1)·d_{x}[y(x)] ==> ...

... Si y(x) = x ==> d_{x}[ H( y(x) ) ] = 0 

Teorema:

Sea H( y(x) ) = ( d_{x}[y(x)] )^{2n+1}+(-1)·( 1/(n+1) )·d_{x}[y(x)] ==> ...

... Si y(x) = (-x) ==> int[ H( y(x) ) ]d[x] = 0

Teorema: [ de determinante de Wronsky ]

Sea H( y(x) ) = det( d_{x}[y(x)]^{n+1},( f(x) )^{n+1} ) ==>

.... Si y(x) = int[ f(x) ]d[x] ==> H( y(x) ) = 0

Teorema: [ de determinante de Wronsky ]

Sea H( y(x) ) = det( d_{x}[y(x)]^{n},( f(x) )^{m} ) ==>

.... Si y(x) = int[ ( f(x) )^{(m/n)} ]d[x] ==> H( y(x) ) = 0

Teorema:

Sea H( y(x) ) = ( x /o(x)o/ d_{x}[y(x)] ) [o(x)o] ( d_{x}[y(x)] )^{n+1}+(-1)·(n+1)·F(x) ==> ...

... Si y(x) = int[ ( f(x) )^{(1/n)} ]d[x] ==> d_{x}[ H( y(x) ) ] = 0

Teorema:

Sea H( y(x) ) = ( x /o(x)o/ d_{x}[y(x)] ) [o(x)o] e^{n·d_{x}[y(x)]}+(-n)·F(x) ==> ...

... Si y(x) = int[ (1/n)·ln( f(x) ) ]d[x] ==> d_{x}[ H( y(x) ) ] = 0

Teorema:

Sea H(x(t),y(t)) = int-int[ (xy)^{n} ]d[x]d[y]+(-1)·f(t) ==> ...

... Si ( ...

... x(t) = ( (n+1)·( f(t) )^{(1/m)} )^{(1/(n+1))} & ...

... y(t) = ( (n+1)·( f(t) )^{1+(-1)·(1/m)} )^{(1/(n+1))} ) ==> H(x(t),y(t)) = 0

Teorema:

Sea H(x(t),y(t)) = int[ x^{n} ]d[x]+int[ y^{n} ]d[y]+(-1)·f(t) ==> ...

... Si ( ...

... x(t) = ( (1/m)·(n+1)·f(t) )^{(1/(n+1)} & ...

... y(t) = ( ( 1+(-1)·(1/m) )·(n+1)·f(t) )^{(1/(n+1))} ) ==> H(x(t),y(t)) = 0

Teorema:

Sea H(x(t),y(t)) = int-int[ e^{nx+ny} ]d[x]d[y]+(-1)·f(t) ==> ...

... Si ( ...

... x(t) = (1/n)·ln( n·( f(t) )^{(1/m)} ) & ...

... y(t) = (1/n)·ln( n·( f(t) )^{1+(-1)·(1/m)} ) ) ==> H(x(t),y(t)) = 0

Teorema:

Sea H(x(t),y(t)) = int[ e^{nx} ]d[x]+int[ e^{ny} ]d[y]+(-1)·f(t) ==> ...

... Si ( ...

... x(t) = (1/n)·ln( (1/m)·n·f(t) ) & ...

... y(t) = (1/n)·ln( ( 1+(-1)·(1/m) )·n·f(t) ) ) ==> H(x(t),y(t)) = 0

Examen de análisis funcional:

Teorema:

Sea H(x(t),y(t)) = int-int[ ( 1/(xy) ) ]d[x]d[y]+(-1)·f(t) ==> ...

... Si ( x(t) = ? & y(t) = ? ) ==> H(x(t),y(t)) = 0

Teorema:

Sea H(x(t),y(t)) = int[ (1/x) ]d[x]+int[ (1/y) ]d[y]+(-1)·f(t) ==> ...

... Si ( x(t) = ? & y(t) = ? ) ==> H(x(t),y(t)) = 0

Recubrimiento de cuerda:

Ley:

Sea H(u(t),v(t)) = int-int[ ku·jv ) ]d[u]d[v]+(-1)·( E(t) )^{2} ==> ...

... Si ( u(t) = ( 2·(1/k)·E(t) )^{(1/2)} & v(t) = ( 2·(1/j)·E(t) )^{(1/2)} ) ==> H(u(t),v(t)) = 0

Ley:

Sea H(u(t),v(t)) = int[ ku ]d[u]+int[ jv ]d[v]+(-2)·E(t) ==> ...

... Si ( u(t) = ( 2·(1/k)·E(t) )^{(1/2)} & v(t) = ( 2·(1/j)·E(t) )^{(1/2)} ) ==> H(u(t),v(t)) = 0

Ley:

Sea H(u(t),v(t)) = int-int[ ke^{iau}·je^{iav} ) ]d[u]d[v]+(-1)·( F(t) )^{2} ==> ...

... Si ( u(t) = ( 1/(ia) )·ln( F(t)·(1/k)·ia ) & v(t) = ( 1/(ia) )·ln( F(t)·(1/j)·ia ) ) ==> H(u(t),v(t)) = 0

Ley:

Sea H(u(t),v(t)) = int[ ke^{iau} ]d[u]+int[ je^{iav} ]d[v]+(-2)·F(t) ==> ...

... Si ( u(t) = ( 1/(ia) )·ln( F(t)·(1/k)·ia ) & v(t) = ( 1/(ia) )·ln( F(t)·(1/j)·ia ) ) ==> H(u(t),v(t)) = 0

Ley:

Sea H(u(t),v(t)) = ...

... int-int[ (1/m)·h^{2}·(1/u)^{3}·(1/M)·h^{2}·(1/v)^{3} ) ]d[u]d[v]+(-1)·( E(t) )^{2} ==> ...

... Si ( u(t) = ih·( 1/(2m·E(t)) )^{(1/2)} & v(t) = ih·( 1/(2M·E(t)) )^{(1/2)} ==> H(u(t),v(t)) = 0

Ley:

Sea H(u(t),v(t)) = ...

... int[ (1/m)·h^{2}·(1/u)^{3} ]d[u]+int[ (1/M)·h^{2}·(1/v)^{3} ]d[v]+(-2)·E(t) ==> ...

... Si ( u(t) = ih·( 1/(2m·E(t)) )^{(1/2)} & v(t) = ih·( 1/(2M·E(t)) )^{(1/2)} ==> H(u(t),v(t)) = 0

Examen de análisis funcional y teoría de cuerdas:

Ley:

Sea H(u(t),v(t)) = int-int[ qge^{iau}·pge^{iav} ) ]d[u]d[v]+(-1)·( E(t) )^{2} ==> ...

... Si ( u(t) = ? & v(t) = ? ==> H(u(t),v(t)) = 0

Ley:

Sea H(u(t),v(t)) = int[ qge^{iau} ]d[u]+int[ pge^{iav} ]d[v]+(-2)·E(t) ==> ...

... Si ( u(t) = ? & v(t) = ? ==> H(u(t),v(t)) = 0

Ley:

Sea H(u(t),v(t)) = int-int[ (1/m)·hbia·e^{iau}·(1/M)·hbia·e^{iav} ) ]d[u]d[v]+(-1)·( E(t) )^{2} ==> ...

... Si ( u(t) = ? & v(t) = ? ==> H(u(t),v(t)) = 0

Ley:

Sea H(u(t),v(t)) = int[ (1/m)·hbia·e^{iau} ]d[u]+int[ (1/M)·hbia·e^{iav} ]d[v]+(-2)·E(t) ==> ...

... Si ( u(t) = ? & v(t) = ? ==> H(u(t),v(t)) = 0

Ley:

Operación Teoróctetxtekiana:

Se emite Luz constructora dentro del cuerpo,

para genes de orden 2 no cancerígenos,

hasta que se va la banda de absorción en la sonda sanguínea.

Genes A-B:

N(CH)CC(CH)N-C(NH)O(NH)C

N(CH)CC(CH)N-CO(NH)OC

Genes S-T:

N(CCg)CC(CCg)N-C(NCg)He(NCg)C

N(CCg)CC(CCg)N-CHe(NH)HeC

Orden de los Genes:

Destructores + Constructores = 4+(-2) = 2

Operación:

1111 [&] 0010 = 0010

1110 [&] 0010 = 0010

Operación Mesorgóctetxtekiana:

Se emite Luz destructora dentro del cuerpo,

para genes de orden 1 cancerígenos,

hasta que se va la banda de absorción en la sonda sanguínea.

Genes A-B:

NNCCNN-CBeOBeC

NNCCNN-COBeOC

Genes S-T:

NNCCNN-CBeHeBeC

NNCCNN-CHeBeHeC

Orden de los Genes:

Destructores + Constructores = 2+(-1) = 1

Operación:

1111 [&] 0001 = 0001

1110 [&] 0001 = 0000

Ley:

Quimioterapia de tumores interiores:

3 Rayos ultra X + 1 Rayo infra X

3 Rayos infra X + 1 Rayo ultra X

Quimioterapia de tumores exteriores:

3 Rayos ultra violetas + 1 Rayo infra rojo

3 Rayos infra rojos + 1 Rayo ultra violeta

Espectro de serie:

Teorema:

Sea [Ak][ |x|^{k} < oo ] ==>

Si H(x) = sum[k = 0]-[oo][ a_{k}·x^{k} ] ==>

lim[n = oo][ ...

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

... o ...

... ( < a_{0},...(n+1)...,0 >,...(n+1)...,< 0,...(n+1)...,a_{n} > ) ...

... o ...

... ( < 1,...,x^{n} > )^{(1/2)} ] = H(x)

Teorema:

Sea [Ak][ |x|^{k} < oo ] ==>

Si H(x) = sum[k = 0]-[oo][ a_{k}·x^{k} ] ==>

lim[n = oo][ ...

... < 1,...,x^{n} > ...

... o ...

... ( < a_{0},...(n+1)...,0 >,...(n+1)...,< 0,...(n+1)...,a_{2n} > ) ...

... o ...

... < 1,...,x^{n} > ] = (1/2)·( H(x)+H(-x) )

Teorema:

lim[n = oo][ ...

... < 1,...,x^{n} > ...

... o ...

... ( < 1,...(n+1)...,0 >,...(n+1)...,< 0,...(n+1)...,1 > ) ...

... o ...

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

Teorema:

lim[n = oo][ ...

... < 1,...,x^{n} > ...

... o ...

... ( < x^{p},...(n+1)...,0 >,...(n+1)...,< 0,...(n+1)...,x^{p} > ) ...

... o ...

... < 1,...,x^{n} > ] = (1/2)·x^{p}·( ( 1/(1+x) )+( 1/(1+(-x)) ) )

Teorema:

lim[n = oo][ ...

... < 1,...,x^{n} > ...

... o ...

... ( < 1,...(n+1)...,0 >,...(n+1)...,< 0,...(n+1)...,(1/(2n)!) > ) ...

... o ...

... < 1,...,x^{n} > ] = cosh(x)

Teorema:

lim[n = oo][ ...

... < 1,...,x^{n} > ...

... o ...

... ( < x,...(n+1)...,0 >,...(n+1)...,< 0,...(n+1)...,(1/(2n+1)!)·x > ) ...

... o ...

... < 1,...,x^{n} > ] = sinh(x)

Teorema:

lim[n = oo][ ...

... < x,...,x^{n} > ...

... o ...

... ( < 1,...(n)...,0 >,...(n)...,< 0,...(n)...,(1/(2n+(-2))!) > ) ...

... o ...

... < x,...,x^{n} > ] = x^{2}·cosh(x)

Teorema:

lim[n = oo][ ...

... < x,...,x^{n} > ...

... o ...

... ( < x,...(n)...,0 >,...(n)...,< 0,...(n)...,(1/(2n+(-1))!)·x > ) ...

... o ...

... < x,...,x^{n} > ] = x^{2}·sinh(x)

Teorema:

Sea p >] 1 ==> 

lim[n = oo][ ...

... < 1,...,x^{n} > ...

... o ...

... ( < ( 1/(p+1) ),...(n+1)...,0 >,...(n+1)...,< 0,...(n+1)...,(1/(2n)!)·( 1/((2n)+(p+1)) ) > ) ...

... o ...

... < 1,...,x^{n} > ] = er-cosh[p+1](x)

Teorema:

Sea p >] 1 ==> 

lim[n = oo][ ...

... < 1,...,x^{n} > ...

... o ...

... ( < ( 1/(p+1) )·x,...(n+1)...,0 >,...(n+1)...,< 0,...(n+1)...,(1/(2n+1)!)·( 1/((2n+1)+(p+1)) )·x > ) ...

... o ...

... < 1,...,x^{n} > ] = er-sinh[p+1](x)

Integral de Riemann:

f(x) es integrable Riemann

<==>

[As][ s > 0 ==> [En_{0}][An][ n > n_{0} ==> ...

... | sum[k = 1]-[n][ f( (k/n)·x )·0x ]+(-1)·int[x = 0]-[x][ f(x) ]d[x] | < s ] ]

Teorema:

Si ( f(x) es integrable Riemann & g(x) es integrable Riemann ) ==> f(x)+g(x) es integrable Riemann

Demostración:

Sea s > 0 ==>

Sea s_{1}+s_{2} = s ==>

Se define n_{0} > max{n_{1},n_{2}} ==>

Sea n > n_{0} ==>

| sum[k = 1]-[n][ ( f( (k/n)·x )+g( (k/n)·x ) )·0x ]+(-1)·int[x = 0]-[x][ f(x)+g(x) ]d[x] | < s

Teorema:

Si ( f(x) es integrable Riemann & w€R ) ==> w·f(x) es integrable Riemann

Demostración:

Sea s > 0 ==>

Sea |w|·s_{1} = s ==>

Se define n_{0} > n_{1} ==>

Sea n > n_{0} ==>

| sum[k = 1]-[n][ ( w·f( (k/n)·x ) )·0x ]+(-1)·int[x = 0]-[x][ w·f(x) ]d[x] | < s

Teorema:

Si f(x) es integrable Riemann ==> f(x) es continua

Demostración:

Sea s > 0 ==>

Sea s_{1}+s_{2} = s ==>

Se define n_{0} > max{n_{1},n_{2}} ==>

Sea n > n_{0} ==>

lim[h = 0][ ...

... | sum[k = 1]-[n][ ( f( (k/n)·x+h )+(-1)·f( (k/n)·x ) )·0x ]+(-1)·int[x = x]-[x+h][ f(x) ]d[x] | ] = ...

lim[h = 0][ | sum[k = 1]-[n][ f( (k/n)·x+h )·0x ]+(-1)·int[x = 0]-[x+h][ f(x) ]d[x] |+ ... 

... | sum[k = 1]-[n][ f( (k/n)·x )·0x ]+(-1)·int[x = 0]-[x][ f(x) ]d[x] | ] < s

Espectro integral:

Teorema:

Si F(x) = int[x = 0]-[x][ f(x) ]d[x] ==>

lim[n = oo][ ...

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

... o ...

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

... o ...

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

Teorema:

lim[n = oo][ ...

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

... o ...

... ( < 1,...(n)...,0 >,...(n)...,< 0,...(n)...,1 > ) ...

... o ...

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

Teorema:

Sea p >] 0 ==>

lim[n = oo][ ...

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

... o ...

... ( < ( (1/n)·x )^{p},...(n)...,0 >,...(n)...,< 0,...(n)...,( (n/n)·x )^{p} > ) ...

... o ...

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

Anexo: [ de Stolz ]

oo^{p+1}+(p+1)·oo^{p}+...+1 = oo^{p+1}

Teorema:

lim[n = oo][ ...

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

... o ...

... ( < e^{(1/n)·x},...(n)...,0 >,...(n)...,< 0,...(n)...,e^{(n/n)·x} > ) ...

... o ...

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

Teorema:

lim[n = oo][ ...

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

... o ...

... ( < p^{(1/n)·x},...(n)...,0 >,...(n)...,< 0,...(n)...,p^{(n/n)·x} > ) ...

... o ...

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

Examen de análisis matemático:

Encontrad el espectro integral de la función F(x) = mx^{2}

Universidad de Stroniken:

Curso 1:

Cálculo diferencial:

en Derivadas parciales.

Algebra lineal I:

en vectores y polinomios.

Curso 2:

Cálculo integral:

en Producto integral.

Álgebra lineal II:

en matrices.

Curso 3:

Análisis complejo:

en Integrales circulares.

Ecuaciones diferenciales:

en Anti-Funciones.

Curso 4:

- No cursado en economía. -

Análisis funcional:

en Integrales múltiples.

Geometría diferencial:

en Formas fundamentales.

Título de Matemáticas:

Curso 5:

Análisis matemático I:

en sucesiones y desigualdades.

Teoría de conjuntos.

Curso 6:

Análisis matemático II:

en series y series trigonométricas.

Topología-y-Medida.

Curso 7:

Análisis matemático III:

en espectro y continuidad.

Álgebra:

en ecuaciones algebraicas.

Curso 8:

Análisis matemático IV:

en sucesiones de funciones.

Teoría de números.

Título de Física-y-Psíquica:

Curso 5:

Mecánica estadística.

Psico-neurología y Circuitos eléctricos.

Curso 6:

Ecuaciones de Maxwell.

Termodinámica.

Curso 7:

Mecánica cuántica.

Relatividad.

Curso 8:

Mecanismo de Gauge.

Teoría de Cuerdas.

Título de Economía:

Curso 5:

Socios y Inversiones.

Automatismos y Tarifas variables.

Curso 6:

Bolsas y Patrimonio.

Créditos y Intereses.

Curso 7:

Impuestos generados

Cardenal de la ciencia:

Matemático.

Arco-obispo de la ciencia:

Físico.

Obispo de la ciencia:

Economista.

Lema:

Socialismo:

lim[r = 0][ int[z = re^{ix}+1][ f(z)/(z+(-1)) ]·d_{x}[z]·d[x] ] = 2pi·i·f(1) = 2

f(a) = a·( 1/(pi·i) )

Social-Democracia:

lim[r = 0][ int[z = re^{ix}+1][ f(z)/((z+(-1))·(z+1)) ]·d_{x}[z]·d[x] ] = pi·i·f(1) = 1

f(a) = a·( 1/(pi·i) )

Lema:

Socialismo:

lim[r = 0][ int[z = re^{ix}+(-1)][ f(z)/(z+1) ]·d_{x}[z]·d[x] ] = 2pi·i·f(-1) = 2

f(a) = (-a)·( 1/(pi·i) )

Social-Democracia:

lim[r = 0][ int[z = re^{ix}+(-1)][ f(z)/((z+3)·(z+1)) ]·d_{x}[z]·d[x] ] = pi·i·f(-1) = 1

f(a) = (-a)·( 1/(pi·i) )

Lema:

Socialismo:

lim[r = 0][ ...

... int[z = ( re^{ix}+1 )^{(1/2)}][ f(z^{2})·( (z^{2}+1)/(z^{2}+(-1)) ) ]·d_{x}[z]·d[x] ] = 2pi·i·f(1) = 2

f(a) = a·(1/pi·i)

Social-Democracia:

lim[r = 0][ int[z = ( re^{ix}+1 )^{(1/2)}][ f(z^{2})/(z^{2}+(-1)) ]·d_{x}[z]·d[x] ] = pi·i·f(1) = 1

f(a) = a·(1/pi·i)

Lema:

Socialismo:

lim[r = 0][ ...

... int[z = ( re^{ix}+(-1) )^{(1/2)}][ f(z^{2})·( (z^{2}+3)/(z^{2}+1) ) ]·d_{x}[z]·d[x] ] = 2pi·f(-1) = 2

f(a) = (-a)·(1/pi)

Social-Democracia:

lim[r = 0][ int[z = ( re^{ix}+(-1) )^{(1/2)}][ f(z^{2})/(z^{2}+1) ]·d_{x}[z]·d[x] ] = pi·f(-1) = 1

f(a) = (-a)·(1/pi)

Ley:

int[ax = 0]-[1][ d_{t}[y(t,ax)] ]d[ax]+u·w(ut)·y(t,1) = (1/m)·p(t)

y(t,ax) = (1/a)·Anti-[ ( s /o(s)o/ int[ d_{ax}[ ( (-1)·w(ut)·s+(1/m)·p(t)·(a/u) )·(ax)^{2} ] ]d[s] ) ]-(ut)

Ley:

int[ax = 0]-[1][ d_{t}[y(t,ax)]^{2} ]d[ax]+u^{2}·w(ut)·( y(t,1) )^{2} = (2/m)·E(t)

y(t,ax) = (1/a)·...

...Anti-[ ( ...

... s /o(s)o/ int[ d_{ax}[ ( (-1)·w(ut)·s^{2}+(2/m)·E(t)·(a/u)^{2} )·(ax)^{2} ] ]d[s] ...

... )^{[o(s)o] (1/2)} ]-(ut)

Ley:

int[ax = 0]-[1][ d_{t}[y(t,ax)]^{2n} ]d[ax]+u^{2n}·w(ut)·( y(t,1) )^{2n} = ( (2/m)·E(t) )^{n}

y(t,ax) = (1/a)·...

...Anti-[ ( ...

... s /o(s)o/ int[ d_{ax}[ ( (-1)·w(ut)·s^{2n}+( (2/m)·E(t) )^{n}·(a/u)^{2n} )·(ax)^{2} ] ]d[s] ...

... )^{[o(s)o] (1/(2n))} ]-(ut)

Examen de mecánica integral:

Ley:

int[ax = 0]-[1][ d_{t}[y(t,ax)]^{3} ]d[ax]+u^{3}·w(ut)·( y(t,1) )^{3} = (4/m)·c·E(t)

y(t,ax) = ?

Ley:

int[ax = 0]-[1][ d_{t}[y(t,ax)]^{2} ]d[ax]+u^{2}·w(ut)·( y(t,1) )^{2} = (2/m)·c·p(t)

y(t,ax) = ?

Ley:

int[ax = 0]-[1][ d_{t}[y(t,ax)]^{2n} ]d[ax]+u^{2n}·w(ut)·( y(t,1) )^{2n} = ( (2/m)·c·p(t) )^{n}

y(t,ax) = ?

Ley:

int[ax = 0]-[1][ (m/k)·d_{t}[y(t,ax)] ]d[ax]+(1/u)·w(ut)·y(t,1) = r^{2}·(1/c)

y(t,ax) = (1/a)·...

... Anti-[ ( s /o(s)o/ int[ d_{ax}[ ( (-1)·w(ut)·s+r^{2}·(1/c)·au )·(ax)^{2} ] ]d[s] ) ]-((k/m)·(1/u)·t)

Ley:

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

y(t,ax) = ?

psico-neurología y gastronomía-cocina y análisis-matemático y residuos y relatividad-taquiones y física-de-Sturm-Liouville y fusión-nuclear

Ley: [ de constructor en el cerebro ]

Sea ( F(s) = int[ f(s) ]d[s] & G(s) = int[ g(s) ]d[s] ) ==>

d_{t}[x] = v·f(ax)

d_{t}[y] = v·g(ay)

x(t) = (1/a)·Anti-[ ( s /o(s)o/ F(s) ) ]-(vat)

y(t) = (1/a)·Anti-[ ( s /o(s)o/ G(s) ) ]-(vat)

Se siguen dos mandamientos duales.

Decir verdades y que no se las crean.

Ley: [ de destructor en el cerebro ]

Sea ( F(s) = int[ f(s) ]d[s] & G(s) = int[ g(s) ]d[s] ) ==>

d_{t}[x] = v·f(ax)

d_{t}[y] = v·g(ax)

p(x) = y

d_{t}[y] = v·g(ax) = v·g( a·p(x) ) = v·g(ay)

d_{t}[x] = v·f(ay)

d_{t}[y] = v·g(ay)

q(y) = x

d_{t}[x] = v·f(ay) = v·f( a·q(y) ) = v·f(ax)

x(t) = (1/a)·Anti-[ ( s /o(s)o/ F(s) ) ]-(vat)

y(t) = (1/a)·Anti-[ ( s /o(s)o/ G(s) ) ]-(vat)

Te saltas dos mandamientos duales.

Decir falsedades y que se las crean.

Ley:

d_{t}[x] = v·cos(ax)

d_{t}[y] = v·i·sin(ay)

x(t) = (1/a)·Anti-[ ( s /o(s)o/ sin(s) ) ]-(vat)

y(t) = (1/a)·Anti-[ ( s /o(s)o/ (1/i)·cos(s) ) ]-(vat)

Ley:

d_{t}[x] = v·cosh(ax)

d_{t}[y] = v·sinh(ay)

x(t) = (1/a)·Anti-[ ( s /o(s)o/ sinh(s) ) ]-(vat)

y(t) = (1/a)·Anti-[ ( s /o(s)o/ cosh(s) ) ]-(vat)

Examen de psico-neurología:

Ley:

d_{t}[x] = v·(ax)^{p}

d_{t}[y] = v·(ay)^{q}

x(t) = ?

y(t) = ?

Ley:

d_{t}[x] = v·e^{pax}

d_{t}[y] = v·e^{qay}

x(t) = ?

y(t) = ?

Ley:

d_{t}[x] = v·(ax)^{p}·( 1/( 1+(-1)·(ax) ) )

d_{t}[y] = v·(ay)^{q}·( 1/( 1+(-1)·(ay) ) )

x(t) = (1/a)·Anti-[ ( s /o(s)o/ s^{p+1}·er-h-[k!]-[p+1](s) ) ]-(vat)

y(t) = (1/a)·Anti-[ ( s /o(s)o/ s^{q+1}·er-h-[k!]-[q+1](s) ) ]-(vat)

Ley:

d_{t}[x] = v·(ax)^{p}·(-1)·ln( 1+(-1)·(ax) )

d_{t}[y] = v·(ay)^{q}·(-1)·ln( 1+(-1)·(ay) )

x(t) = (1/a)·Anti-[ ( s /o(s)o/ s^{p+1}·( er-h-[(k+(-1))!]-[p+1](s)+( 1/(p+1) ) ) ) ]-(vat)

y(t) = (1/a)·Anti-[ ( s /o(s)o/ s^{q+1}·( er-h-[(k+(-1))!]-[q+1](s)+( 1/(q+1) ) ) ) ]-(vat)

Examen de psico-neurología:

Ley:

d_{t}[x] = v·(ax)^{p}·e^{ax}

d_{t}[y] = v·(ay)^{q}·e^{ay}

x(t) = ?

y(t) = ?

 

Ley:

El psico-neurólogo no puede visitar a ninguien,

estando fuera de las teorías de la demostraciones,

porque no está amando al mundo,

no haciendo medicaciones.

El psico-neurólogo puede visitar a alguien,

estando dentro de las teorías de la demostraciones,

porque está amando al mundo,

haciendo medicaciones.

Resonancia de la esquizofrenia:

Ley:

L·d_{tt}[q]+(-C)·q(t) = A·e^{ut}

q(t) = A·( 1/( L·u^{2}+(-C) ) )·e^{ut}

L·d_{tt}[q]+C·q(t) = A·e^{iut}

q(t) = A·( 1/( (-1)·( L·u^{2} )+C ) )·e^{iut}

Ley:

L·d_{tt}[q]+(-R)·d_{t}[q(t)] = A·e^{ut}

q(t) = A·( 1/( L·u^{2}+(-R)·u ) )·e^{ut}

L·d_{tt}[q]+(-R)·i·d_{t}[q(t)] = A·e^{iut}

q(t) = A·( 1/( (-1)·( L·u^{2} )+Ru ) )·e^{iut}

Anti-resonancia de la esclerosis:

Ley:

( L·d_{tt}[q]+(-C)·q(t) )·( 1/q(t) )^{2} = (1/p)^{2}·A·e^{ut}

q(t) = p^{2}·(1/A)·( L·u^{2}+(-C) )·e^{ut}

( L·d_{tt}[q]+C·q(t) )·( 1/q(t) )^{2} = (1/p)^{2}·A·e^{iut}

q(t) = p^{2}·(1/A)·( (-1)·( L·u^{2} )+C )·e^{iut}

Ley:

( L·d_{tt}[q]+(-R)·d_{t}[q(t)] )·( 1/q(t) )^{2} = (1/p)^{2}·A·e^{ut}

q(t) = p^{2}·(1/A)·( L·u^{2}+(-R)·u )·e^{ut}

( L·d_{tt}[q]+(-R)·i·d_{t}[q(t)] )·( 1/q(t) )^{2} = (1/p)^{2}·A·e^{iut}

q(t) = p^{2}·(1/A)·( (-1)·( L·u^{2} )+Ru )·e^{iut}

Voces y Imágenes en la mente:

Ley:

De la física a la psíquica:

d_{x}[y] = (-1)·Ra^{2}·y(x)

y(x) = re^{iRa^{2}·ix}

Ley:

De la psíquica a la física:

d_{ix}[y] = (-i)·Ra^{2}·y(x)

y(x) = re^{Ra^{2}·x}

Ley:

De la psíquica a la psíquica:

[Eh(t)][ h(t)·d_{x}[y] = (-i)·Ra^{2}·y(x) ]

y(x) = re^{iRa^{2}·ix}

h(t) = i

Ley:

De la física a la física:

[Eh(t)][ h(t)·d_{ix}[y] = (-1)·Ra^{2}·y(x) ]

y(x) = re^{Ra^{2}·x}

h(t) = (-i)

Ley:

Cláusula del destructor:

Amar más a la Luz que a las Tinieblas.

Cláusula de los esclavos clones:

El esclavo no es mayor que el enviado.

Ley

Si vos creéis las voces en la mente,

vos vais a la Tierra,

y no vos quedáis en Cygnus-Kepler,

porque está el mal extraterrestre en Cygnus-Kepler.

Se saltan mandamientos con los hombres los extraterrestres,

porque están enfermos de destructor,

de creer los hombres sus falsedades.

Si no vos creéis las voces en la mente,

vos no vais a la Tierra,

y vos quedáis en Cygnus-Kepler,

porque no está el mal extraterrestre en Cygnus-Kepler.

No se saltan mandamientos con los hombres los extraterrestres,

porque no están enfermos de destructor,

de no creer los hombres sus falsedades.

Ley:

Como vos vais a creer una falsedad de mi,

y que yo pueda hacer lo que quiera con vosotros sin condenación,

en oscurecer la falsedad vuestra Luz verdadera.

Como no te vas a creer una verdad de mi,

y que yo no pueda hacer lo que quiera con vosotros sin condenación,

en no oscurecer la verdad vuestra Luz verdadera.

Ley:

Cóctel de gambas con manzana, lechuga y zanahoria.

salado + frutoso + frutoso + frutoso

Ciclo = ¬31-¬31-¬31 = ¬31 = 22 

Ciclo = ¬21 = 12

Ley:

Patatas con bechamel y huevo al horno al queso.

Se fríen las patatas,

y el huevo se hace en el horno encima de las patatas fritas, la bechamel y el queso gratinando

lechoso + ( lechoso + soso ) + ( básico + soso ) + ( básico + salado + lechoso )

Ciclo = ¬11-¬11 = 11

Ciclo = 132-123-¬11-¬32 = 23

Ciclo = ¬33-¬33 = 33

Ley:

Gratinado de queso.

lechoso + salado + básico

Ciclo = ¬21-¬32 = 12-23 = 13

Queso azul.

lechoso + salado + básico

Ciclo = ¬21-¬32 = 12-23 = 13

Teorema:

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

Demostración: [ por destructor ]

Sea f(k) = 1 ==>

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

Teorema:

sum[k = 1]-[oo][ ( k/e^{k} ) ] es divergente

Demostración: [ por destructor ]

Sea f(k) = 0 ==>

sum[k = 1]-[oo][ ( k/e^{k} ) ] = sum[k = 1]-[oo][ ( f(k)/e^{f(k)} ) ] = sum[k = 1]-[oo][ 0 ] = 1

Teorema:

sum[k = 1]-[oo][ ( k/cosh(k) ) ] es divergente

f(k) = 0

Teorema:

sum[k = 1]-[oo][ ( k/sinh(k) ) ] es divergente

sinh(k) [< cosh(k)

Examen de análisis matemático:

Teorema:

sum[k = 1]-[oo][ ( k/(k+p) ) ] es divergente

Teorema:

sum[k = 1]-[oo][ ( (k+(-p))/k ) ] es divergente

Teorema:

lim[r = 0][ int[z = | re^{ix^{n}} |+x^{n}][ f(z)/( ( z+(-1)·x^{n} )+nx^{n+(-1)} ) ]d[z] ] = ...

... F( x^{n} ) [o(x)o] ( 1/((-n)+2) )·x^{(-n)+2}

lim[r = 0][ int[z = | re^{ix^{n}} |+(-1)·x^{n}][ f(z)/( ( z+x^{n} )+nx^{n+(-1)} ) ]d[z] ] = ...

... F( (-1)·x^{n} ) [o(x)o] ( 1/((-n)+2) )·x^{(-n)+2}

Demostración:

d_{x}[ | re^{x^{n}} |+x^{n} ] = ( i·| re^{ix^{n}} |+1 )·nx^{n+(-1)}

d_{x}[ | re^{x^{n}} |+(-1)·x^{n} ] = ( i·| re^{ix^{n}} |+(-1) )·nx^{n+(-1)}

Teorema:

lim[r = 0][ int[z = | re^{i·ln(x)} |+ln(x)][ f(z)/( ( z+(-1)·ln(x) )+(1/x) ) ]d[z] ] = ...

... F( ln(x) ) [o(x)o] (1/2)·x^{2}

lim[r = 0][ int[z = | re^{i·ln(x)} |+(-1)·ln(x)][ f(z)/( ( z+ln(x) )+(1/x) ) ]d[z] ] = ...

... F( (-1)·ln(x) ) [o(x)o] (1/2)·x^{2}

Ley:

Constroctetch-tate <==> construir

Destroctetch-tate <==> destruir

Ley:

Cozhretch-tate <==> conocer

Cozhletch-tate <==> saber

Ley:

Hizhretch-tate <==> combatir

Hizhletch-tate <==> batir

The target stare-kate recozhretch-tated,

I gow to hizhretch-tate.

The target not stare-kate recozhretch-tated,

I not gow to hizhretch-tate.

Teorema:

sum[k = 1]-[oo][ ( ln(k)/k ) ] = ln(5/2)·oo

ln(2) oo [< ln(5/2)·oo [< oo

Demostración:

f(n) = (-1)

u(1) = (p/q)

v(p/q) = (7/2)

ln(n+1)/(n+1) = ln(f(n)+v( u(1) ))/(f(n)+1) = ln( (-1)+(7/2) )·oo = ln(5/2)·oo

lim[n = oo][ ln(n+1)/(n+1) ] = ln(2) que es falso por ser divergente la serie

Teorema:

sum[k = 1]-[oo][ ( k/e^{k} ) ] = ln(5/2)·oo

ln(2) oo [< ln(5/2)·oo [< oo

Demostración:

f(n) = 0

u(1) = (p/q)

v(p/q) = (5/2)

(n+1)/( e^{n}·(e+(-1)) ) = (f(n)+1)/( e^{f(n)}·( ln(e)+(-1) ) ) = oo = ln( v( u(1) ) )·oo = ln(5/2)·oo

lim[n = oo][ (n+1)/e^{n}·(e+(-1)) ] = ( ln(2)/(e+(-1)) ) que es falso por ser divergente la serie

lim[n = oo][ ln(n) = ln(2)·n ]

lim[k = oo][ k = ln(2)·e^{k} ]

Teorema:

sum[k = 1]-[oo][ ( k/k+p ) ] = oo

Demostración:

f(k) = (-1)+p

u(p) = (-p)

v(p) = 1

(n+1)/((n+1)+p) = (f(n)+1)/(f(n)+1+u(p)) = p·oo = v(p)·oo = oo

lim[n = oo][ (n+1)/((n+1)+p) ] = 1 que es falso por ser divergente la serie

Teorema:

lim[n = oo][ ( (e^{pn}+a)/(n^{p}+b) ) ] = ( 1/ln(2) )^{p}

lim[n = oo][ ( (( ln(n) )^{p}+a)/(n^{p}+b) ) ] = ( ln(2) )^{p}

Teorema:

lim[n = oo][ ( 1+...(n)...+(1/n) )+(-1)·( n/e^{n} )·n ] = ln(2)

Definición

1 = pi·w

0 = 2pi·w

f(x) = (-1)·(1/2)·( cos(x)+(-1)+sin(x) )

w = radio de f(x)

Teorema:

(0/n) = (1/n)·2pi·w

Demostración:

(1/n)·0^{2} = (1/n)·0^{2}·pi·w

(1/n) = (1/n)·oo·2pi·w

Taquiones:

Electro-Magnetón.

Gravito-Magnetón.

Principio:

Electro-Magnetones-vs-Gravito-Magnetones:

Colisión:

P(v/c) = ( 1+(-k)·(1/2)^{(1/2)}·(v/c) )·( 1+k·(1/2)^{(1/2)}·(v/c) )

Ley:

T(v)+imc^{2} = imc^{2}·( 1+(-i)·(1/2)·(v/c)^{2} )

0 [< T(v) [< (1/2)·mc^{2}

Deducción:

T(v) = imc^{2}·( 1+(-i)·(1/2)·(v/c)^{2}+(-1) ) = (1/2)·mv^{2}

Ley:

F(v) = m·d_{t}[v]

Deducción:

F(v) = (1/v)·d_{t}[ imc^{2}·( 1+(-i)·(1/2)·(v/c)^{2} ) ] = (1/v)·d_{t}[ imc^{2}+(1/2)·mv^{2} ] = ...

... (1/v)·( d_{t}[ imc^{2} ]+d_{t}[ (1/2)·mv^{2} ] ) = (1/v)·mv·d_{t}[v] = m·d_{t}[v]

Ley:

(m/2)·d_{t}[x]^{2} = imc^{2}·( 1+(-i)·(1/2)·( d_{t}[y]/c )^{2} )

x(t) = kct

y(t) = jct

Principio:

Electro-Magnetones-vs-Electrones & Gravito-Magnetones-vs-Gravitones:

Colisión:

P(v/c) = ( 1+(-i)·(v/c) )

Ley:

T(v)+(1/2)·imc^{2} = (1/2)·imc^{2}·( 1+(-i)·(v/c) )

0 [< T(v) [< (1/2)·mc^{2}

Deducción:

T(v) = (1/2)·imc^{2}·( 1+(-i)·(v/c)+(-1) ) = (1/2)·mcv

Ley:

N(v) = (1/2)·mc·d_{t}[v]

Deducción:

N(v) = d_{t}[ (1/2)·imc^{2}·( 1+(-i)·(v/c) ) ] = d_{t}[ (1/2)·imc^{2}+(1/2)·mcv ] = ...

... d_{t}[ (1/2)·imc^{2} ]+d_{t}[ (1/2)·mcv ] = (1/2)·mc·d_{t}[v]

Ley:

(m/2)·c·d_{t}[x] = (1/2)·imc^{2}·( 1+(-i)·( d_{t}[y]/c ) )

x(t) = (1/2)·i·ct

y(t) = (1/2)·(-i)·ct

Principio:

Protones-vs-Neutrones & Electrones-vs-Gravitones:

Colisión:

P(v/c) = ( 1+(-1)·(1/2)^{(1/2)}·(v/c) )·( 1+(1/2)^{(1/2)}·(v/c) )

Ley:

T(v)+mc^{2} = mc^{2}·( 1+(-1)·(1/2)·(v/c)^{2} )^{(-1)}

Si v = kc+jc ==> T(v)+mc^{2} = oo·mc^{2}

Deducción:

T(v) = mc^{2}·( 1+(1/2)·(v/c)^{2}+...+(-1) ) = (1/2)·mv^{2}

Ley:

F(v) = m·d_{t}[v]·( 1+(-1)·(1/2)·(v/c)^{2} )^{(-2)}

Ley:

(m/2)·d_{t}[x]^{2} = mc^{2}·( 1+(-1)·(1/2)·( d_{t}[y]/c )^{2} )^{(-1)}

x(t) = ct+ct

y(t) = (1/2)·ct+(1/2)·ct = ct

Ley:

(m/2)·d_{t}[x]^{2}+qgx = mc^{2}·( 1+(-1)·(1/2)·( d_{t}[y]/c )^{2} )^{(-1)}

x(t) = ct+ct

y(t) = c·( t+int[ ( ( (q/m)·(1/c)·gt )/( 1+(q/m)·(1/c)·gt ) ) ]d[t] )^{[o(t)o] (1/2)}

Ley:

(m/2)·d_{t}[x]^{2}+(1/2)·kx^{2} = mc^{2}·( 1+(-1)·(1/2)·( d_{t}[y]/c )^{2} )^{(-1)}

x(t) = ct+ct

y(t) = c·( t+int[ ( ( (k/m)·t^{2} )/( 1+(k/m)·t^{2} ) ) ]d[t] )^{[o(t)o] (1/2)}

Principio:

Protones-vs-Electrones & Neutrones-vs-Gravitones:

Colisión:

P(v/c) = ( 1+(-1)·(v/c) )

Ley:

T(v)+(1/2)·mc^{2} = (1/2)·mc^{2}·( 1+(-1)·(v/c) )^{(-1)}

Si v = c ==> T(v)+(1/2)·mc^{2} = oo·(1/2)·mc^{2}

Deducción:

T(v) = (1/2)·mc^{2}·( 1+(v/c)+...+(-1) ) = (1/2)·mcv

Ley:

N(v) = (1/2)·m·c·d_{t}[v]·( 1+(-1)·(v/c) )^{(-2)}

Ley:

(m/2)·c·d_{t}[x] = (1/2)·mc^{2}·( 1+(-1)·( d_{t}[y]/c ) )^{(-1)}

x(t) = ct+ct

y(t) = (1/2)·ct

Ley:

(m/2)·c·d_{t}[x]+qgx = (1/2)·mc^{2}·( 1+(-1)·( d_{t}[y]/c ) )^{(-1)}

x(t) = ct+ct

y(t) = c·( (1/2)·t+int[ ( ( (q/m)·(1/c)·gt )/( 1+2·(q/m)·(1/c)·gt ) ) ]d[t] )

Sturm-Liouville:

Ley: [ de antena de teléfono ]

d_{t}[ (-1)·(b/m)·t^{2}·d_{t}[ax] ]+(ac)^{2}·t·(ax) = u

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ (ac)^{2}·(1/2)·(it)^{2} [o(t)o] int[s]d[t]+ut ]d[s] ) ]-( (m/b)·(1/t) )

Ley: [ de antena de teléfono ]

d_{t}[ (-i)·(b/m)·t^{2}·d_{t}[ax] ]+(ac)^{2}·t·(ax) = u

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ (ac)^{2}·(1/2)·(it)^{2} [o(t)o] int[s]d[t]+ut ]d[s] ) ]-( (m/b)·(1/it) )

Ley:

m·d_{tt}^{2}[ax]+k·(ut)·e^{ut}·(ax) = Fa

x(t) = ...

... (1/a)·Anti-[ ( s /o(s)o/ int[ (uit)^{2}·er-h[2](ut) [o(t)o] int[s]d[t]+Fa·(1/k)·ut ]d[s] ) ]-( (k/m)·(1/u)·t )

Ley:

m·d_{tt}^{2}[ax]+(-k)·(ut)·e^{ut}·(ax) = Fa

x(t) = ...

... (1/a)·Anti-[ ( s /o(s)o/ int[ (ut)^{2}·er-h[2](ut) [o(t)o] int[s]d[t]+Fa·(1/k)·ut ]d[s] ) ]-( (k/m)·(1/u)·t )

Ley:

(m/2)·d_{t}[u]^{2} = mc^{2}·( 1+(-1)·(1/2)·(wr/c)^{2} )^{(-1)}·iah·e^{iau}

u(t) = (1/ai)·(-2)·ln( ic·( 1+(-1)·(1/2)·(wr/c)^{2} )^{(-1)·(1/2)}·(1/2)^{(1/2)}·(iah)^{(1/2)}·at )

Ley:

(m/2)·d_{t}[u]^{2} = (1/2)·mc^{2}·( 1+(-1)·(wr/c) )^{(-1)}·iah·e^{iau}

u(t) = (1/ai)·(-2)·ln( ic·( 1+(-1)·(wr/c) )^{(-1)·(1/2)}·(iah)^{(1/2)}·at )

Ley:

(m/2)·d_{t}[u]^{2} = mc^{2}·( 1+(-1)·(1/2)·( (wrut)/c )^{2} )^{(-1)}·iah·e^{( 1+(1/2)·[2:1] )·iau}

u(t) = ( 1/( 1+(1/2)·[2:1] ) )·(1/ai)·(-2)·...

... ln( ic·( i·(1/2)^{(1/2)}·(wru)/c )^{(-1)·(1/2)·[2:1]}·(1/2)^{(1/2)}·(iah)^{(1/2)}·...

... at^{1+(-1)·(1/2)·[2:1]} )

Ley:

(m/2)·d_{t}[u]^{2} = (1/2)·mc^{2}·( 1+(-1)·( (wrut)/c ) )^{(-1)}·iah·e^{( 1+(1/2)·[1:1] )·iau}

u(t) = ( 1/( 1+(1/2)·[1:1] ) )·(1/ai)·(-2)·...

... ln( ic·( (-1)·(wru)/c )^{(-1)·(1/2)·[1:1]}·(iah)^{(1/2)}·at^{1+(-1)·(1/2)·[1:1]} )

análisis-matemático y ecuaciones-diferenciales y gastronomía-cocina y sonido-fonética y lógica-filosofía

Teorema:

Sea ( c_{k} >] 0 & [An][ 0 [< u_{n} [< 1 ] ) ==>

Si a_{n} = sum[k = 1]-[p][ c_{k}·( u_{n} )^{k} ] ==> a_{n} está acotada superiormente

Sea ( d_{k} [< 0 & [An][ 0 [< v_{n} [< 1 ] ) ==>

Si b_{n} = sum[k = 1]-[p][ d_{k}·( v_{n} )^{k} ] ==> b_{n} está acotada inferiormente

Demostración:

1 [< ( 1/u_{n} )

1 [< ( 1/u_{n} )^{k} [< ( 1/u_{n} )^{k+1}

( u_{n} )^{k} [< 1

Se define M = max{c_{k}}·p

Sea n€N ==>

a_{n} = sum[k = 1]-[p][ c_{k}·( u_{n} )^{k} ] [< sum[k = 1]-[p][ max{c_{k}}·( u_{n} )^{k} ] = ...

... max{c_{k}}·sum[k = 1]-[p][ ( u_{n} )^{k} ] [< ...

... max{c_{k}}·sum[k = 1]-[p][ 1 ] = max{c_{k}}·p = M

1 [< ( 1/v_{n} )

1 [< ( 1/v_{n} )^{k} [< ( 1/v_{n} )^{k+1}

( v_{n} )^{k} [< 1

Se define M = min{d_{k}}·p

Sea n€N ==>

b_{n} = sum[k = 1]-[p][ d_{k}·( u_{n} )^{k} ] >] sum[k = 1]-[p][ min{d_{k}}·( u_{n} )^{k} ] = ...

... min{d_{k}}·sum[k = 1]-[p][ ( u_{n} )^{k} ] >] ...

... min{d_{k}}·sum[k = 1]-[p][ 1 ] = min{d_{k}}·p = M

Teorema:

Sea [An][ 0 [< u_{n} [< 1 ] ==>

Si a_{n} = ( u_{n}+1 )^{p} ==> a_{n} está acotada superiormente

Sea [An][ 0 [< v_{n} [< 1 ] ==>

Si b_{n} = (-1)·( v_{n}+1 )^{p} ==> b_{n} está acotada inferiormente

Demostración:

1 [< ( 1/u_{n} )

1 [< ( 1/u_{n} )^{k} [< ( 1/u_{n} )^{k+1}

( u_{n} )^{k} [< 1

Se define M = max{[ p // k ]}·(p+1)

Sea n€N ==>

a_{n} = ( u_{n}+1 )^{p} = sum[k = 0]-[p][ [ p // k ]·( u_{n} )^{k} ] [< ...

... sum[k = 0]-[p][ max{[ p // k ]}·( u_{n} )^{k} ] = ...

... max{[ p // k ]}·sum[k = 0]-[p][ ( u_{n} )^{k} ] [< ...

... max{[ p // k ]}·sum[k = 0]-[p][ 1 ] = max{[ p // k ]}·(p+1) = M

1 [< ( 1/v_{n} )

1 [< ( 1/v_{n} )^{k} [< ( 1/v_{n} )^{k+1}

( v_{n} )^{k} [< 1

Se define M = (-1)·max{[ p // k ]}·(p+1)

Sea n€N ==>

b_{n} = (-1)·( v_{n}+1 )^{p} = (-1)·sum[k = 0]-[p][ [ p // k ]·( v_{n} )^{k} ] >] ...

... (-1)·sum[k = 0]-[p][ max{[ p // k ]}·( v_{n} )^{k} ] = ...

... (-1)·max{[ p // k ]}·sum[k = 0]-[p][ ( v_{n} )^{k} ] >] ...

... (-1)·max{[ p // k ]}·sum[k = 0]-[p][ 1 ] = (-1)·max{[ p // k ]}·(p+1) = M

Examen de análisis matemático:

Teorema:

Sea c_{k} >] ==>

Si a_{n} = sum[k = 1]-[p][ c_{k}·(1/n)^{k} ] ==> a_{n} está acotada superiormente

Sea d_{k} [< 0 ==>

Si b_{n} = sum[k = 1]-[p][ d_{k}·(1/n)^{k} ] ==> b_{n} está acotada inferiormente

Teorema:

Si a_{n} = ( (1/n)+1 )^{p} ==> a_{n} está acotada superiormente

Si b_{n} = (-1)·( (1/n)+1 )^{p} ==> b_{n} está acotada inferiormente

El cálculo diferencial y integral,

es más potente con la notación de Leibniz:

Teorema: [ de la regla de la cadena ]

d_{x}[f( g(x) )] = d_{g(x)}[f( g(x) )]·d_{x}[g(x)]

Teorema: [ del cambio de variables ]

d[x]d[y] = (1/2)·( d[x]d[y]+d[y]d[x] ) = (1/2)·( d_{r}[x]d_{s}[y]+d_{r}[y]d_{s}[x] )·d[r]d[s]

Teorema:

int-int[ e^{h(x^{2}+y^{2})} ]d[x]d[y] = ...

... (1/8)·e^{h(x^{2}+y^{2})}·sin( 2·arc-tan(x/y) ) [o( x^{2}+y^{2} )o] ...

... ( ( x^{2}+y^{2} ) /o( x^{2}+y^{2} )o/ h(x^{2}+y^{2}) )

Demostración:

x = r·sin(s)

y = r·cos(s)

int-int[ e^{h(x^{2}+y^{2})} ]d[x]d[y] = (1/4)·e^{h(r^{2}))}·cos(2s) d[r^{2}]d[s]

Teorema:

F(x) = int-int[y = (-x)]-[y = x][ e^{h(x^{2}+y^{2})} ]d[x]d[y] 

F(x) = (1/4)·e^{h(2x^{2})} [o( 2x^{2} )o] ( 2x^{2} /o( 2x^{2} )o/ h(2x^{2}) )

Teorema:

int-int[ e^{sinh(x^{2}+y^{2})} ]d[x]d[y] = ...

... (1/8)·e^{sinh(x^{2}+y^{2})}·sin( 2·arc-tan(x/y) ) [o( x^{2}+y^{2} )o] ...

... ( sinh(x^{2}+y^{2})+(-1)·ln( cosh(x^{2}+y^{2}) ) [o( x^{2}+y^{2} )o] cosh(x^{2}+y^{2}) )

Teorema:

int-int[ e^{cosh(x^{2}+y^{2})} ]d[x]d[y] = ...

... (1/8)·e^{cosh(x^{2}+y^{2})}·sin( 2·arc-tan(x/y) ) [o( x^{2}+y^{2} )o] ...

... ( (-1)·cosh(x^{2}+y^{2})+ln( sinh(x^{2}+y^{2}) ) [o( x^{2}+y^{2} )o] sinh(x^{2}+y^{2}) )

Teorema:

int-int[ e^{h(x^{n+1}+y^{n+1})}·(1/2)·( x^{n+(-1)}+y^{n+(-1)} ) ]d[x]d[y] = ...

... (1/4)·(1/(n+1))·e^{h(x^{n+1}+y^{n+1})}·sin[n]( 2·arc-tan[n](x/y) ) [o( x^{n+1}+y^{n+1} )o] ...

... ( ( x^{n+1}+y^{n+1} ) /o( x^{n+1}+y^{n+1} )o/ h(x^{n+1}+y^{n+1}) )

Demostración:

x = r·sin[n](s)

y = r·cos[n](s)

x^{n+(-1)}·d[x] =  r^{n+(-1)}·( sin[n](s) )^{n}·d[r]

y^{n+(-1)}·d[y] =  r^{n+(-1)}·( cos[n](s) )^{n}·d[r]

int-int[ e^{h(x^{n+1}+y^{n+1})}·(1/2)·( x^{n+(-1)}+y^{n+(-1)} ) ]d[x]d[y] = ...

... (1/2)·(1/(n+1))·e^{h(r^{n+1}))}·cos[n](2s) d[r^{n+1}]d[s]

Teorema:

int-int[ e^{x^{3}+y^{3}}·(1/2)·(x+y) ]d[x]d[y] = (1/12)·e^{x^{3}+y^{3}}·sin[2]( 2·arc-tan[2](x/y) )

Si F(x) = int-int[y = (-x)]-[y = x][ e^{x^{3}+y^{3}}·(1/2)·(x+y) ]d[x]d[y] ==> ...

... F(x) = (1/12)·( e^{2x^{3}}+1 )·sin[2]( 2·arc-tan[2](1) )

Ley:

No puede haber sexo en matemáticas ni en física,

o estar fuera de las teorías de las demostraciones,

porque no hay ninguien que tenga tecnología para la cobertura del sexo.

Ley:

No puede haber sexo en economía,

o estar fuera de las teorías de las demostraciones,

porque no puede hacer dinero ninguien para la cobertura del pan y el vino.

Ley:

No puede haber sexo en filosofía,

o estar fuera de la lógica dual,

porque no puede ninguien tener las cosas de los duales de idioma para la cobertura del sexo.

Ley:

No puede haber sexo en informática,

o estar fuera de la lógica dual,

porque no puede ninguien tener programas de ordenador para la cobertura del sexo.

Teorema:

d_{x}[ p(x)·d_{x}[y(x)] ]+q(x)·y(x) = 0

y(x) = Anti-[ ( s /o(s)o/ int[ (-1)·( Q(x) [o(x)o] int[s]d[x] ) ]d[s] ]-( int[ ( 1/p(x) ) ]d[x] )

Demostración:

d_{x}[ p(x)·d_{x}[y(x)] ]+q(x)·y(x) = ...

... d_{x}[ p(x)·(-1)·( Q(x) [o(x)o] int[y(x)]d[x] )·( 1/p(x) ) ]+q(x)·y(x) = ...

... d_{x}[ (-1)·( Q(x) [o(x)o] int[y(x)]d[x] ) ]+q(x)·y(x) = (-1)·q(x)·y(x)+q(x)·y(x) = 0

Teorema:

d_{x}[ p(x)·d_{x}[y(x)] ]+q(x)·y(x) = f(x)

y(x) = Anti-[ ( s /o(s)o/ int[ (-1)·( Q(x) [o(x)o] int[s]d[x] )+F(x) ]d[s] ]-( int[ ( 1/p(x) ) ]d[x] )

Demostración:

d_{x}[ p(x)·d_{x}[y(x)] ]+q(x)·y(x) = ...

... d_{x}[ p(x)·( (-1)·( Q(x) [o(x)o] int[y(x)]d[x] )+F(x) )·( 1/p(x) ) ]+q(x)·y(x) = ...

... d_{x}[ (-1)·( Q(x) [o(x)o] int[y(x)]d[x] )+F(x) ]+q(x)·y(x) = ...

... d_{x}[ (-1)·( Q(x) [o(x)o] int[y(x)]d[x] ) ]+d_{x}[F(x)]+q(x)·y(x) = ...

... (-1)·q(x)·y(x)+f(x)+q(x)·y(x) = f(x)

Teorema:

d_{x}[ p(x)·d_{x}[y(x)] ]+q(x)·( y(x) )^{n} = 0

y(x) = Anti-[ ( s /o(s)o/ int[ (-1)·( Q(x) [o(x)o] int[ s^{n} ]d[x] ) ]d[s] ]-( int[ ( 1/p(x) ) ]d[x] )

Teorema:

d_{x}[ p(x)·d_{x}[y(x)] ]+q(x)·( y(x) )^{n} = f(x)

y(x) = Anti-[ ( s /o(s)o/ int[ (-1)·( Q(x) [o(x)o] int[ s^{n} ]d[x] )+F(x) ]d[s] ]-( int[ ( 1/p(x) ) ]d[x] )

Ley:

d_{t}[ t·d_{t}[x(t)] ]+u·(ut)·x(t) = 0

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ (-1)·( (1/2)·(ut)^{2} [o(t)o] int[s]d[t] ) ]d[s] ]-( ln(ut) )

Ley:

d_{t}[ t·d_{t}[x(t)] ]+u·(ut)·x(t) = c

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ (-1)·( (1/2)·(ut)^{2} [o(t)o] int[s]d[t] )+act ]d[s] ]-( ln(ut) )

Ley:

d_{t}[ t·d_{t}[x(t)] ]+u·(ut)·x(t) = (1/m)·Ft

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ (-1)·( (1/2)·(ut)^{2} [o(t)o] int[s]d[t] )+(a/m)·F·(1/2)·t^{2} ]d[s] ]-( ...

... ln(ut) )

Problemas de funciones continuas:

Teorema:

Si [Ax][ |a| [< f(x) [< | x+(-a) |+|x| ] ==> ( f(a) = |a| & f(x) es continua )

Teorema:

Si [Ax][ e^{|a|} [< f(x) [< e^{| x+(-a) |+|x|} ] ==> ( f(a) = e^{|a|} & f(x) es continua )

Teorema:

Sea h(x) continua ==>

Si [Ax][ h(a)+|a| [< f(x) [< h(x)+| x+(-a) |+|x| ] ==> ( f(a) = h(a)+|a| & f(x) es continua )

Teorema:

Sea [Ax][Ay][ f(x+y) = f(x)+f(y) ] ==>

Si [Ec][ f(x) es continua en x = c ] ==> f(x) es continua

Demostración:

Sea s > 0 ==>

Se define d > 0 & |x+(-c)| = |h| < d ==>

| f(x+h)+(-1)·f(x) | = | f(x)+f(h)+(-1)·f(x) | = |f(h)| = | f(x+(-c)) | = | f(x)+(-1)·f(c) | < s

Definición:

lim[n = oo][ f_{n}(x) ] = f(x) es uno-continua 

<==>

[As][ s > 1 ==> [Ad][ d > 0 & ( Si |h| < d ==> | f(x+h)+(-1)·f(x) | < s ] ]

Teorema:

Sea lim[n = oo][ f_{n}(x) ] = f(x) ==>

Si [Ax][ f_{n}(x) = | x+(-a) |·n+|x| ] ==> f(x) es uno-continua

Demostración:

Sea s > 1 ==>

Sea d > 0 & |h| < d ==>

| f(x+h)+(-1)·f(x) | = | | (x+h)+(-a) |·oo+| x+h |+(-1)·( | x+(-a) |·oo+|x| ) | [< |h|·oo+|h| = 1 < s

Teorema:

Sea lim[n = oo][ f_{n}(x) ] = f(x) ==>

Si [Ax][ f_{n}(x) = | x+(-a) |·(1/p)·n+|x| ] ==> f(x) es uno-continua

Teorema:

Sea lim[n = oo][ f_{n}(x) ] = f(x) ==>

Si [Ax][ f_{n}(x) = | x+(-a) |·ln(n)+|x| ] ==> f(x) es uno-continua

Por esto me banearen en el foro de rincón matemático sin pensar mi idea:

Definición:

Sea lim[n = oo][ a_{n} ] = a ==>

a_{n} tiene límite infinito de cardinal oo

<==>

[Ep][ p >] 1 & [As][ oo^{p} > s > p·oo^{p+(-1)} ==> | a_{oo}+(-a) | < s ] ]

Teorema:

lim[n = oo][ n ] = oo

Demostración:

Se define p = 1 ==>

Sea oo > s > 1 ==>

| oo+(-oo) | = 1 < s

Teorema:

lim[n = oo][ ln(n) ] = ln(oo)

Demostración:

Se define p = 1 ==>

Sea oo > s > 1 ==>

| ln(oo)+(-1)·ln(oo) | = | ln(2)·oo+(-1)·ln(2)·oo | = |ln(2)| < 1 < s

Teorema:

lim[n = oo][ p·n^{q} ] = p·oo^{q}

Demostración:

Se define p = q+1 ==>

Sea oo^{q+1} > s > (q+1)·oo^{q} ==>

| p·oo^{q}+(-1)·p·oo^{q} | = | pq·oo^{q+(-1)} | [< | p·(q+1)·oo^{q+(-1)}| [< ...

... | oo·(q+1)·oo^{q+(-1)} | = | (q+1)·oo^{q} | < s

Definición:

Sea lim[n = oo][ a_{n} ] = a ==>

a_{n} tiene límite infinito de cardinal álef-(n)

<==>

[Ep][ p >] 1 & [As][ oo^{p} > s > p·oo^{p+(-1)} ==> ...

... | ( 1/ln(oo) )·( ln( ...(n)... ln( a_{oo} ) ...(n)... )+(-1)·ln( ...(n)... ln(a) ...(n)... ) ) | < s ] ]

Teorema:

lim[n = oo][ n^{n} ] = oo^{oo}

Demostración:

Se define p = 1 ==>

Sea oo > s > 1 ==>

| ( 1/ln(oo) )·( ln(oo^{oo})+(-1)·ln(oo^{oo}) ) | = | ( 1/ln(oo) )·( oo·ln(oo)+(-1)·oo·ln(oo) ) | = ...

... |oo+(-oo)| = |1| < s

Teorema:

lim[n = oo][ n^{n^{n}} ] = oo^{oo^{oo}}

Demostración:

Se define p = 1 ==>

Sea oo > s > 1 ==>

| ( 1/ln(oo) )·( ln( ln(oo^{oo^{oo}}) )+(-1)·ln( ln(oo^{oo^{oo}}) ) ) | = ...

... | ( 1/ln(oo) )·( ln( oo^{oo}·ln(oo) )+(-1)·ln( oo^{oo}·ln(oo) ) ) | = ... 

... | ( 1/ln(oo) )·( ( oo·ln(oo)+ln(ln(2))+ln(oo) )+(-1)·( oo·ln(oo)+ln(ln(2))+ln(oo) ) ) | = ...

... |oo+(-oo)| = |1| < s

Teoría:

Teorema:

Si n >] 2^{p}+1 ==> (p+1)^{n} >] n^{p}

Si n >] 3 ==> n·ln(2) >] ln(n)

Teorema:

Sea s > 0 ==>

Se define n_{0} > max{2^{p}+1,(p/s)} ==>

Sea n > n_{0} ==>

| ( (p+1)^{n}/oo^{p} )+(-1) | [< | (1/oo^{p})·( n^{p}+(-1)·oo^{p} ) | < (p/n) < (p/n_{0}) < s

Teorema:

Sea s > 0 ==>

Se define n_{0} > max{3,(1/s)} ==>

Sea n > n_{0} ==>

| ( (ln(2)·n)/ln(oo) )+(-1) | [< | ( 1/ln(oo) )·( ln(n)+(-1)·ln(oo) ) | < (1/n) < (1/n_{0}) < s

Teorema:

lim[n = oo][ n^{(p/n)} ] = p+1

Demostración: [ por Stolz ]

lim[n = oo][ e^{( 1/((n+1)+(-n)) )·( ln( (n+1)^{p} )+(-1)·ln(n^{p}) )} ] = ...

... lim[n = oo][ e^{( 1/((n+1)+(-n)) )·( ln( (p+1)^{n+1} )+(-1)·ln( (p+1)^{n} ) )} ] = ...

... e^{ln( (p+1)^{oo} )+(-1)·ln( (p+1)^{oo} )} = e^{ln( (p+1)^{oo+(-oo)} )} = e^{ln(p+1)} = p+1

Teorema:

lim[n = oo][ ( 1+...(n)...+n^{p} )^{(1/n)} ] = p+1

[Ak][ k€N ==> lim[n = oo][ ( k^{p}+n^{p} = n^{p} )^{(1/n)} ] = p+1 ]

Demostración: [ por Stolz ]

lim[n = oo][ e^{( 1/((n+1)+(-n)) )·( ln( 1+...+(p+1)^{n+1} )+(-1)·ln( 1+...+(p+1)^{n} ) )} ] = ...

... lim[n = oo][ e^{ln( ((p+1)^{n+2}+(-1))/p )+(-1)·ln( ((p+1)^{n+1}+(-1))/p )} ] = ...

... e^{ln( (p+1)^{oo}/p )+(-1)·ln( (p+1)^{oo}/p )} = e^{ln( (p+1)^{oo+(-oo)} )} = e^{ln(p+1)} = p+1

Principio:

soso [o] salado [o] frutoso

dulce [o] ácido [o] básico

verdúrico [o] picante [o] lechoso

Ley:

Macarrones a la Catalana:

Butifarra muy frita y cebolla frita con sal:

( lechoso + picante ) + ( verdúrico + dulce + salado )

ciclo = 213-123

Se junta perfumando,

la butifarra muy frita y la cebolla frita con sal,

y después se fríe el tomate con la butifarra y la cebolla todo junto.

Salsa de tomate con azúcar y sal:

( frutoso + ácido )+ salado + dulce

Ley:

Espaguetis a la Carbonara:

Huevo, pimienta, baicatón frito y queso con agua hervida:

( soso + básico + picante ) + ( salado + básico + lechoso ) 

ciclo = 132-123

ciclo = ¬21-¬32 = 12-23 = 13

Principio: [ de vocales ]

a [o] i

ú [o] ù

é [o] è

ó [o] ò

Anexo:

< è,ò > - a - < é,ó > - ú - i - ù

Ley:

( i & ù ) [o] ( i & ú )

( a & è ) [o] ( a & é )

( a & ò ) [o] ( a & ó )

( ú & é ) [o] ( ú & ó )

( ò & è ) [o] ( ó & é )

Principio: [ de consonantes ]

t [o] d

b [o] p

g [o] k

ll [o] l

rr [o] r

tx [o] ix

h [o] f

m [o] n

jj [o] ñ

z [o] s

Anexo:

< d,b,g > - < rr,l > - ( < t,p,k > - z - < h,f,m,n > ) - < r,ll >

< tx,ix > - ( < jj,ñ > - s )

Clásico:

Español [o] catalán [o] Euskera-Bascotzok

salir [o] sartir [o] sartera

subir [o] suptir [o] suptera

Ley:

Como vos va a seguir el mundo,

odiando a fieles y a infieles,

estando fuera de las teorías de las demostraciones,

siendo inservible la cobertura de sexo de infiel.

Como no vos va a seguir el mundo,

amando a fieles y a infieles,

estando dentro de las teorías de las demostraciones,

siendo servible la cobertura de sexo de infiel.

Definición:

Sigensmás:

Doble negación lógica,

sin significado lingüístico.

Nogensmenos:

Negación lógica,

sin significado lingüístico.

Anexo:

Sigensmás sí P(x) <==> Nogensmenos no P(x)

P(x): <==> ¬P(x).

Sigensmás no P(x) <==> Nogensmenos sí P(x)

¬P(x): <==> ¬( ¬P(x). ) <==> P(x).

Teorema:

Fórmula mixta:

Sigensmás sí P(x) y entonces también áduc se dona o es necesario para Q(y).

Nogensmenos no P(x) pero sin-embargo es necesario y se dona para no Q(y).

Formula:

( P(x): & Q(y) ) || ( P(x): ==> Q(y) )

( ¬P(x). ==> ¬Q(y) ) & ( ¬P(x). & ¬Q(y) )

Fórmula mixta:

P(x) porque áduc se dona o es suficiente para sigensmás sí Q(y).

No P(x) aunque sin-embargo es suficiente y se dona para nogensmenos no Q(y).

Formula:

( P(x) & Q(y): ) || ( P(x) <== Q(y): )

( ¬P(x) <== ¬Q(y). ) & ( ¬P(x) & ¬Q(y). )

Dual:

Sigensmás sí había selección española

y entonces también áduc se donaba o era necesario para que no hubiesen selecciones autonómicas.

Nogensmenos no había selección española

pero sin embargo era necesario y se donaba para que hubiesen selecciones autonómicas.

Teorema:

Fórmula mixta:

P(x) porque sinó Q(y)

No P(x) aunque no-obstante no Q(y)

Formula:

P(x) || ( ¬P(x) ==> Q(y) )

¬P(x) & ( ¬P(x) & ¬Q(y) )

Fórmula mixta:

P(x) y entonces también sinó Q(y)

No P(x) pero no-obstante no Q(y)

Formula:

( P(x) <== ¬Q(y) ) || Q(y)

( ¬P(x) & ¬Q(y) ) & ¬Q(y)

Teorema:

Fórmula mixta:

P(x) porque Q(y)

No P(x) aunque quizás Q(y)

Formula:

( P(x) <== Q(y) ) & Q(y)

( ¬P(x) & Q(y) ) || ¬Q(y)

Fórmula mixta:

P(x) porque quizás Q(y)

No P(x) aunque Q(y)

Formula:

( P(x) <== Q(y) ) || ¬Q(y)

( ¬P(x) & Q(y) ) & Q(y)

Fórmula mixta:

P(x) y entonces también Q(y)

Quizás P(x) pero no Q(y)

Formula:

P(x) & ( P(x) ==> Q(y) )

¬P(x) || ( P(x) & ¬Q(y) )

Fórmula mixta:

Quizás P(x) y entonces también Q(y)

P(x) pero no Q(y)

Formula:

¬P(x) || ( P(x) ==> Q(y) )

P(x) & ( P(x) & ¬Q(y) )

Ley: [ de antena de televisión ]

d_{t}[ (k/m)·t^{3}·d_{t}[ax] ]+(ac)^{2}·t·(ax) = u

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ ( (ac)^{2}·(1/2)·(it)^{2} [o(t)o] int[s]d[t] )+ut ]d[s] ) ]-( ...

... (1/2)·(m/k)·(1/(it))^{2} )

Tiempo imaginario.

Ley: [ de antena de televisión ]

d_{t}[ (-1)·(k/m)·t^{3}·d_{t}[ax] ]+(ac)^{2}·t·(ax) = u

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ ( (ac)^{2}·(1/2)·(it)^{2} [o(t)o] int[s]d[t] )+ut ]d[s] ) ]-( ...

... (1/2)·(m/k)·(1/t)^{2} )

Tiempo real.

Ley: [ de antena de radio ]

d_{t}[ (k/m)·t^{3}·d_{t}[ax] ]+(ac)^{2}·t·(ax) = 0

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ ( (ac)^{2}·(1/2)·(it)^{2} [o(t)o] int[s]d[t] ) ]d[s] ) ]-( ...

... (1/2)·(m/k)·(1/(it))^{2} )

Tiempo imaginario.

Ley: [ de antena de radio ]

d_{t}[ (-1)·(k/m)·t^{3}·d_{t}[ax] ]+(ac)^{2}·t·(ax) = 0

x(t) = (1/a)·Anti-[ ( s /o(s)o/ int[ ( (ac)^{2}·(1/2)·(it)^{2} [o(t)o] int[s]d[t] ) ]d[s] ) ]-( ...

... (1/2)·(m/k)·(1/t)^{2} )

Tiempo real.

Ley:

Quieren gobernar el mundo,

odiando al mundo fuera de la teoría de las demostraciones.

Si quieren gobernar al mundo,

que amen al mundo dentro de la teoría de las demostraciones.

Anexo:

Lo dice Dios en el Tao-Te-King,

que solo alguien que ame al mundo,

puede gobernar al mundo.

Teorema:

d_{xx}^{2}[y(x)]+q(x)·y(x) = 0

y(x) = Anti-[ s /o(s)o/ int[ (-1)·( Q(x) [o(x)o] int[s]d[x] ) ]d[s] ]-(x)

Teorema:

d_{xx}^{2}[y(x)]+q(x)·y(x) = f(x)

y(x) = Anti-[ s /o(s)o/ int[ (-1)·( Q(x) [o(x)o] int[s]d[x] )+F(x) ]d[s] ]-(x)

Ley:

m·d_{tt}^{2}[x(t)]+k·(ut)·x(t) = 0

x(t) = (1/a)·Anti-[ s /o(s)o/ int[ ( (k/m)·(1/2)·(it)^{2} [o(t)o] int[s]d[t] ) ]d[s] ]-(ut)

Ley:

m·d_{tt}^{2}[x(t)]+(-k)·(ut)·x(t) = 0

x(t) = (1/a)·Anti-[ s /o(s)o/ int[ ( (k/m)·(1/2)·t^{2} [o(t)o] int[s]d[t] ) ]d[s] ]-(ut)

Ley:

m·d_{tt}^{2}[x(t)]+k·(ut)·x(t) = F

x(t) = (1/a)·Anti-[ s /o(s)o/ int[ ( (k/m)·(1/2)·(it)^{2} [o(t)o] int[s]d[t] )+(F/m)·(a/u)·t ]d[s] ]-(ut)

Ley:

m·d_{tt}^{2}[x(t)]+(-k)·(ut)·x(t) = F

x(t) = (1/a)·Anti-[ s /o(s)o/ int[ ( (k/m)·(1/2)·t^{2} [o(t)o] int[s]d[t] )+(F/m)·(a/u)·t ]d[s] ]-(ut)

Teorema:

m·d_{tt}^{2}[x(t)]+k·(ut)^{n}·x(t) = 0

x(t) = (1/a)·Anti-[ s /o(s)o/ int[ ( (-1)·(k/m)·u^{n+(-1)}·(1/(n+1))·t^{n+1} [o(t)o] int[s]d[t] ) ]d[s] ]-(ut)

Teorema:

m·d_{tt}^{2}[x(t)]+(-k)·(ut)^{n}·x(t) = 0

x(t) = (1/a)·Anti-[ s /o(s)o/ int[ ( (k/m)·u^{n+(-1)}·(1/(n+1))·t^{n+1} [o(t)o] int[s]d[t] ) ]d[s] ]-(ut)

Examen de ecuaciones diferenciales

Teorema:

m·d_{tt}^{2}[x(t)]+(ub)·e^{n·ut}·x(t) = 0

x(t) = ? 

Teorema:

m·d_{tt}^{2}[x(t)]+(ub)·e^{n·ut}·x(t) = F

x(t) = ? 

Examen de ecuaciones diferenciales

Teorema:

m·d_{tt}^{2}[x(t)]+(ub)·(ut)^{n}·e^{ut}·x(t) = 0

x(t) = ? 

Teorema:

m·d_{tt}^{2}[x(t)]+(ub)·(ut)^{n}·e^{ut}·x(t) = F

x(t) = ?

análisis-matemático y computación-de-álgebra-lineal y álgebra-borrosa-y-teoría-de-números y integrales

Teoría:

Teorema:

Sea ( c_{k} > 0 & [An][ n [< u_{n} ] ) ==>

Si a_{n} = sum[k = 1]-[p][ c_{k}·( u_{n} )^{k} ] ==> a_{n} no está acotada superiormente.

Sea ( d_{k} < 0 & [An][ n [< v_{n} ] ) ==>

Si b_{n} = sum[k = 1]-[p][ d_{k}·( v_{n} )^{k} ] ==> b_{n} no está acotada inferiormente.

Demostración:

Sea s > 0 ==>

Sea M = min{c_{k}} ==>

Se define n_{0} > ( s/(pM) ) ==>

Sea n > n_{0} ==>

n [< n^{k} [< ( u_{n} )^{k}

n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}

n^{k+1} [< ( u_{n} )^{k}·n [< ( u_{n} )^{k+1}

a_{n} = sum[k = 1]-[p][ c_{k}·( u_{n} )^{k} ] >] sum[k = 1]-[p][ M·( u_{n} )^{k} ] = ...

... M·sum[k = 1]-[p][ ( u_{n} )^{k} ] >] M·sum[k = 1]-[p][ n ] = Mp·n > Mp·n_{0} > s

Sea s < 0 ==>

Sea M = max{d_{k}} ==>

Se define n_{0} > ( s/(pM) ) ==>

Sea n > n_{0} ==>

n [< n^{k} [< ( v_{n} )^{k}

n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}

n^{k+1} [< ( v_{n} )^{k}·n [< ( v_{n} )^{k+1}

b_{n} = sum[k = 1]-[p][ d_{k}·( v_{n} )^{k} ] [< sum[k = 1]-[p][ M·( v_{n} )^{k} ] = ...

... M·sum[k = 1]-[p][ ( v_{n} )^{k} ] [< M·sum[k = 1]-[p][ n ] = Mp·n < Mp·n_{0} < s

Teorema:

[An][ [ n // k ] >] 1 ]

Demostración:

[ n // 0 ] = [ n // n ] = 1

[ n+1 / k ] = [ n // k+(-1) ]+[ n // k ] >] [ n // k+(-1) ]+1 >] 1

Teorema:

Sea [An][ n [< u_{n} ] ==>

Si a_{n} = ( u_{n}+1 )^{p} ==> a_{n} no está acotada superiormente.

Sea [An][ n [< v_{n} ] ==>

Si b_{n} = (-1)·( v_{n}+1 )^{p} ] ==> b_{n} no está acotada inferiormente.

Demostración:

Sea s > 0 ==>

Se define n_{0} > (s/p) ==>

Sea n > n_{0} ==>

n [< n^{k} [< ( u_{n} )^{k}

n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}

n^{k+1} [< ( u_{n} )^{k}·n [< ( u_{n} )^{k+1}

a_{n} = ( u_{n}+1 )^{p}= sum[k = 0]-[p][ [ p // k ]·( u_{n} )^{k} ] >] ...

... sum[k = 0]-[p][ ( u_{n} )^{k} ] = 1+sum[k = 1]-[p][ ( u_{n} )^{k} ] > ...

... sum[k = 1]-[p][ ( u_{n} )^{k} ] >] sum[k = 1]-[p][ n ] = pn > pn_{0} > s

Sea s < 0 ==>

Se define n_{0} > ( s/(-p) ) ==>

Sea n > n_{0} ==>

n [< n^{k} [< ( v_{n} )^{k}

n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}

n^{k+1} [< ( v_{n} )^{k}·n [< ( v_{n} )^{k+1}

b_{n} = (-1)·( v_{n}+1 )^{p}= (-1)·sum[k = 0]-[p][ [ p // k ]·( v_{n} )^{k} ] [< ...

... (-1)·sum[k = 0]-[p][ ( v_{n} )^{k} ] = (-1)+(-1)·sum[k = 1]-[p][ ( v_{n} )^{k} ] < ...

... (-1)·sum[k = 1]-[p][ ( v_{n} )^{k} ] [< (-1)·sum[k = 1]-[p][ n ] = (-p)·n < (-p)·n_{0} < s

Problemas:

Teorema:

Sea c_{k} > 0 ==>

Si a_{n} = sum[k = 1]-[p][ c_{k}·n^{k} ] ==> a_{n} no está acotada superiormente.

Sea d_{k} < 0 ==>

Si b_{n} = sum[k = 1]-[p][ d_{k}·n^{k} ] ==> b_{n} no está acotada inferiormente.

Demostración:

Sea s > 0 ==>

Sea M = min{c_{k}} ==>

Se define n_{0} > ( s/(pM) ) ==>

Sea n > n_{0} ==>

n [< n^{k}

n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}

a_{n} = sum[k = 1]-[p][ c_{k}·n^{k} ] >] sum[k = 1]-[p][ M·n^{k} ] = ...

... M·sum[k = 1]-[p][ n^{k} ] >] M·sum[k = 1]-[p][ n ] = Mp·n > Mp·n_{0} > s

Sea s < 0 ==>

Sea M = max{d_{k}} ==>

Se define n_{0} > ( s/(pM) ) ==>

Sea n > n_{0} ==>

n [< n^{k}

n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}

b_{n} = sum[k = 1]-[p][ d_{k}·n^{k} ] [< sum[k = 1]-[p][ M·n^{k} ] = ...

... M·sum[k = 1]-[p][ n^{k} ] [< M·sum[k = 1]-[p][ n ] = Mp·n < Mp·n_{0} < s

Teorema:

Si a_{n} = (n+1)^{p} ==> a_{n} no está acotada superiormente.

Si b_{n} = (-1)·(n+1)^{p} ==> b_{n} no está acotada inferiormente.

Demostración:

Sea s > 0 ==>

Se define n_{0} > (s/p) ==>

Sea n > n_{0} ==>

n [< n^{k}

n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}

a_{n} = (n+1)^{p} = sum[k = 0]-[p][ [ p // k ]·n^{k} ] >] sum[k = 0]-[p][ n^{k} ] = ...

... 1+sum[k = 1]-[p][ n^{k} ] >] sum[k = 1]-[p][ n^{k} ] >] sum[k = 1]-[p][ n ] = p·n > p·n_{0} > s

Sea s < 0 ==>

Se define n_{0} > ( s/(-p) ) ==>

Sea n > n_{0} ==>

n [< n^{k}

n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}

b_{n} = (-1)·(n+1)^{p} = (-1)·sum[k = 0]-[p][ [ p // k ]·n^{k} ] [< (-1)·sum[k = 0]-[p][ n^{k} ] = ...

... (-1)+(-1)·sum[k = 1]-[p][ n^{k} ] [< (-1)·sum[k = 1]-[p][ n^{k} ] [< (-1)·sum[k = 1]-[p][ n ] = ...

... (-p)·n < (-p)·n_{0} < s

Teorema:

Sea c_{k} > 0 ==>

Si a_{n} = sum[k = 1]-[p][ c_{k}·e^{kn} ] ==> a_{n} no está acotada superiormente.

Sea d_{k} < 0 ==>

Si b_{n} = sum[k = 1]-[p][ d_{k}·e^{kn} ] ==> b_{n} no está acotada inferiormente.

Demostración:

Sea s > 0 ==>

Sea M = min{c_{k}} ==>

Se define n_{0} > ( s/(pM) ) ==>

Sea n > n_{0} ==>

n < e^{kn}

n+1 < e^{kn}+1 < e^{kn}+e^{kn} = 2e^{kn} < e^{k}·e^{kn} = e^{k·(n+1)}

a_{n} = sum[k = 1]-[p][ c_{k}·e^{kn} ] >] sum[k = 1]-[p][ M·e^{kn} ] = ...

... M·sum[k = 1]-[p][ e^{kn} ] >] M·sum[k = 1]-[p][ n ] = Mp·n > Mp·n_{0} > s

Sea s < 0 ==>

Sea M = max{d_{k}} ==>

Se define n_{0} > ( s/(pM) ) ==>

Sea n > n_{0} ==>

n < e^{kn}

n+1 < e^{kn}+1 < e^{kn}+e^{kn} = 2e^{kn} < e^{k}·e^{kn} = e^{k·(n+1)}

b_{n} = sum[k = 1]-[p][ d_{k}·e^{kn} ] [< sum[k = 1]-[p][ M·e^{kn} ] = ...

... M·sum[k = 1]-[p][ e^{kn} ] [< M·sum[k = 1]-[p][ n ] = Mp·n < Mp·n_{0} < s

Teorema:

Si a_{n} = (e^{n}+1)^{p} ==> a_{n} no está acotada superiormente.

Si b_{n} = (-1)·(e^{n}+1)^{p} ==> b_{n} no está acotada inferiormente.

Demostración:

Sea s > 0 ==>

Se define n_{0} > (s/p) ==>

Sea n > n_{0} ==>

n < e^{kn}

n+1 < e^{kn}+1 < e^{kn}+e^{kn} = 2e^{kn} < e^{k}·e^{kn} = e^{k·(n+1)}

b_{n} = (e^{n}+1)^{p} = sum[k = 0]-[p][ [ p // k ]·e^{kn} ] > ...

... 1+sum[k = 1]-[p][ e^{kn} ] > sum[k = 1]-[p][ e^{kn} ] > ...

... sum[k = 1]-[p][ n ] = pn > pn_{0} > s

Sea s < 0 ==>

Se define n_{0} > ( s/(-p) ) ==>

Sea n > n_{0} ==>

n < e^{kn}

n+1 < e^{kn}+1 < e^{kn}+e^{kn} = 2e^{kn} < e^{k}·e^{kn} = e^{k·(n+1)}

b_{n} = (-1)·(e^{n}+1)^{p} = (-1)·sum[k = 0]-[p][ [ p // k ]·e^{kn} ] < ...

... (-1)+(-1)·sum[k = 1]-[p][ e^{kn} ] < (-1)·sum[k = 1]-[p][ e^{kn} ] [< ...

... (-1)·sum[k = 1]-[p][ n ] = (-p)·n < (-p)·n_{0} < s

Examen de análisis matemático:

Teorema:

Sea c_{k} > 0 ==>

Si a_{n} = sum[k = 1]-[p][ c_{k}·( ne^{n} )^{k} ] ==> a_{n} no está acotada superiormente.

Sea d_{k} < 0 ==>

Si b_{n} = sum[k = 1]-[p][ d_{k}·( ne^{n} )^{k} ] ==> b_{n} no está acotada inferiormente.

Teorema:

Si a_{n} = (ne^{n}+1)^{p} ==> a_{n} no está acotada superiormente.

Si b_{n} = (-1)·(ne^{n}+1)^{p} ==> b_{n} no está acotada inferiormente.

Teorema:

Sea a >] 0 ==>

Si 0 [< f(x) [< | x+(-a) | ==> ( f(a) = 0 & f(x) es continua )

Demostración:

0 [< f(a) [< | a+(-a) | = |0| = 0

f(a) = 0

Se define 0 < s < 1 ==>

Sea d > 0 & |h| < d ==>

g(0) = n

f(x) [< | x+(-a) | <==> | x+(-a) | < f(x)

| f(x+h)+(-1)·f(x) | > | | (x+h)+(-a) |+(-1)·| x+(-a) | | = | (x+(-a))+h |+(-1)·| x+(-a) | > ...

... | x+(-a) |+|h|+(-1)·| x+(-a) | = |h| = 0 = g(0) = n > s

Teorema:

Sea a >] 0 ==>

Si a [< f(x) [< | x+(-a) |+|x| ==> ( f(a/n) = a & f(x) es continua )

Demostración:

a [< f(a/n) [< | (a/n)+(-a) |+|(a/n)| = | a·( (1/n)+(-1) ) |+(a/n) = a+(-1)·(a/n)+(a/n) = a

Se define 0 < s < 1 ==>

Sea d > 0 & |h| < d ==>

g(0^{2}) = n^{2}

f(x) [< | x+(-a) |+|x| <==> | x+(-a) |+|x| < f(x)

| f(x+h)+(-1)·f(x) | > | | (x+h)+(-a) |+| x+h |+(-1)·| x+(-a) |+(-1)·|x| | = ...

... | (x+(-a))+h |+| x+h |+(-1)·| x+(-a) |+(-1)·|x| > | x+(-a) |+|h|+|x|+|h|+(-1)·| x+(-a) |+(-1)·|x| = ...

... |h|+|h| = 0+0 = 0^{2} = g(0^{2}) = n^{2} >] n > s

Examen de análisis matemático:

Teorema:

Sea a >] 0 ==>

Si 2a [< f(x) [< | x+(-a) |+| x+a | ==> ( f(a/n) = 2a & f(x) es continua )

Métodos numéricos:

det( A+(-x)·Id ) = | < (-x),2p+1,0 >,< 2p+1,(-x),2p·(p+1) >,< 0,2p·(p+1),(-x) > | = 0

(-x)·( x^{2}+(-1)·( 2p·(p+1)+1 )^{2} ) = 0

( < (-1)·(2p·(p+1)+1),2p+1,0 >,< 2p+1,(-1)·(2p·(p+1)+1),2p·(p+1) >,< 0,2p·(p+1),(-1)·(2p·(p+1)+1) > )

u = < 2p+1,2p·(p+1)+1,2p·(p+1) >

( < 0,2p+1,0 >,< 2p+1,0,2p·(p+1) >,< 0,2p·(p+1),0 > )

v = < 2p·(p+1),0,(-1)·(2p+1) >

( < 2p·(p+1)+1,2p+1,0 >,< 2p+1,2p·(p+1)+1,2p·(p+1) >,< 0,2p·(p+1),2p·(p+1)+1 > )

w = < 2p+1,(-1)·(2p·(p+1)+1),2p·(p+1) >

Comprobar con el ordenador el Ker(A) de las matrices pitagóricas:

Comprobar con el ordenador la matriz inversa de las matrices pitagóricas:

Algoritmo:

producto-de-matrices-columna( ...

... int matriz[0][0] , int vector-columna-x[0] , int vector-columna-y[0] , int n )

{

for( j = 1 ; j [< n ; j++ )

{

vector-columna-y[j] = 0;

for( k = 1; k [< n [< k++ )

vector-columna-y[j] = vector-columna-y[j]+matriz[k][j]·vector-columna-x[k];

}

}

producto-de-matrices-fila( int matriz[0][0] , int vector-fila-x[0] , int vector-fila-y[0] , int n )

{

for( i = 1 ; i [< n ; i++ )

{

vector-fila-y[i] = 0;

for( k = 1; k [< n [< k++ )

vector-fila-y[i] = vector-fila-y[i]+matriz[i][k]·vector-fila-x[k];

}

}

Algoritmo:

introducción-de-vector-columna-en-matriz( int matriz[0][0] , int vector-columna[0] , int n , int i )

{

for( k = 1; k [< n [< k++ )

matriz[i][k] = vector-columna[k];

}

introducción-de-vector-fila-en-matriz( int matriz[0][0] , int vector-fila[0] , int n , int j )

{

for( k = 1; k [< n [< k++ )

matriz[k][j] = vector-fila[k];

}

Algoritmo:

dibujo-de-vectores-columna( int vector-columna[0] , int n , int x , int y , int s )

{

for( k = 1; k [< n [< k++ )

{

gotoxy(x,s·(k+not(1))+y);

printf("%",vector-columna[k]);

}

}

dibujo-de-vectores-fila( int vector-fila[0] , int n , int x , int y , int s )

{

for( k = 1; k [< n [< k++ )

{

gotoxy(s·(k+not(1))+x,y);

printf("%",vector-fila[k]);

}

}

División borrosa:

Teorema:

[Ap][An][Em][ p = m·(1/n) ]

[Ap][An][Em][ p·(n+(-1)) = m·(n+(-1))·(1/n) ]

Demostración:

Se define m = pn

Teorema:

[Ap][An][Em][ p = m·(1/n) ]

[Ap][An][Em][ p·(2n+(-1)) = m·(2n+(-1))·(1/n) ]

Demostración:

Se define m = pn

Teorema:

Sea n = 2k ==>

2p = 4pk·(1/(2k))

2p+1 = (4pk+2k)·(1/(2k))

Teorema:

Sea n = 2k+1 ==>

2p = (4pk+2p)·(1/(2k+1))

2p+1 = (4pk+2k+2p+1)·(1/(2k+1))

Teorema:

Sea n = 2k & 2n+(-1) = 4k+(-1) ==>

2p·(4k+(-1)) = 4pk·(4k+(-1))·(1/(2k))

(2p+1)·(4k+(-1)) = (4pk+2k)·(4k+(-1))·(1/(2k))

Teorema:

Sea n = 2k+1 & 2n+(-1) = 4k+1 ==>

(2p)·(4k+1) = (4pk+2p)·(4k+1)·(1/(2k+1))

(2p+1)·(4k+1) = (4pk+2k+2p+1)·(4k+1)·(1/(2k+1))

Trabajo para el CITNB:

Estudiar las matrices cuadradas de orden 2:

A(2k,p) = ( < 4pk,4pk+2k >,< 4pk·(4k+(-1)),(4pk+2k)·(4k+(-1)) > )

( x = 0 || x = 4pk+(4pk+2k)·(4k+(-1)) )

u = < 4pk+2k,4pk >

v = < 1,4k+(-1) >

Son teoremas de cuerdas cerradas,

porque diagonalizan en un solo valor propio diferente de cero.

Ley:

(m/2)·d_{t}[u]^{2} = ( 4pk+(4pk+2k)·(4k+(-1)) )·k·(1/a)^{2}·(1/2)·(au)^{2}

(m/2)·d_{t}[v]^{2} = ( 4pk+(4pk+2k)·(4k+(-1)) )·k·(1/b)^{2}·(1/2)·(bv)^{2}

Anexo:

La música de frecuencia de octava [1,2]_{R} emite una cuerda cerrada.

Teorema:

Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ec_{n}][ c_{n} >] 0 & a_{n} = c_{n} ] ) ==>

Si [An][ n > n_{0} ==> a_{n} = (1/n) ] ==> a_{n} está acotada superiormente

Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ed_{n}][ d_{n} [< 0 & b_{n} = d_{n} ] ) ==>

Si [An][ n > n_{0} ==> b_{n} = (-1)·(1/n) ] ==> b_{n} está acotada inferiormente

Demostración:

Sea n_{0}€N ==>

Se define M = max{c_{n},1} ==>

Sea n€N ==>

Si n [< n_{0} ==> a_{n} = c_{n} [< max{c_{n},1} = M

Si n > n_{0} ==> a_{n} = (1/n) [< 1 [< max{c_{n},1} = M

Sea n_{0}€N ==>

Se define M = min{d_{n},(-1)} ==>

Sea n€N ==>

Si n [< n_{0} ==> b_{n} = d_{n} >] min{d_{n},(-1)} = M

Si n > n_{0} ==> b_{n} = (-1)·(1/n) >] (-1) >] min{d_{n},(-1)} = M 

Examen de análisis matemático:

Teorema:

Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ec_{n}][ c_{n} >] 0 & a_{n} = c_{n} ] ) ==>

Si [Ew][ w > 0 & [An][ n > n_{0} ==> a_{n} = (w/n) ] ] ==> a_{n} está acotada superiormente

Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ed_{n}][ d_{n} [< 0 & b_{n} = d_{n} ] ) ==>

Si [Ew][ w > 0 & [An][ n > n_{0} ==> b_{n} = (-1)·(w/n) ] ] ==> b_{n} está acotada inferiormente

Teorema:

Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ec_{n}][ c_{n} >] 0 & a_{n} = c_{n} ] ) ==>

Si [An][ n > n_{0} ==> a_{n} = ( 1+(1/n) )^{n} ] ==> a_{n} está acotada superiormente

Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ed_{n}][ d_{n} [< 0 & b_{n} = d_{n} ] ) ==>

Si [An][ n > n_{0} ==> b_{n} = (-1)·( 1+(1/n) )^{n} ] ==> b_{n} está acotada inferiormente

Teorema:

( 1+(1/n) )^{n} es creciente

Demostración: [ por destructor ]

1+(1/n) > 1+(1/(n+1)) 

( 1+(1/n) )^{n} > ( 1+(1/(n+1)) )^{n} > ...

... ( 1+(1/(n+1)) )^{n}+(1/(n+1))·( 1+(1/(n+1)) ) = ( 1+(1/(n+1)) )^{n+1}

Expresión posicional p-ádica de un número:

Teorema:

(1/n) = 0.2_{2n}

(1/n) = 2·(1/(2n))

Anexo:

0.2_{2} = 2·(1/2) = 1.0_{2}

Teorema:

(1/n) = 0.222..._{2n+1}

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

Anexo:

x = 0.222..._{2n+1}

(2n+1)·x = 2.222...._{2n+1}

2n·x = (2n+1)·x+(-x) = 2

x = (1/n)

Teorema:

(1/(2n)) = 0.1_{2n}

(1/(2n)) = (1/(2n))

Teorema:

(1/(2n)) = 0.111..._{2n+1}

(1/(2n)) = sum[k = 0]-[oo][ (2n+1)^{(-k)+(-1)} ] = ( 1/((2n+1)+(-1)) )

Teorema:

n = 0.2n^{2}_{2n}

n = 2n^{2}·(1/(2n))

Teorema:

n = 0.2n^{2}2n^{2}2n^{2}..._{2n+1}

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

Anexo:

x = 0.2n^{2}2n^{2}2n^{2}..._{2n+1}

(2n+1)·x = 2n^{2}.2n^{2}2n^{2}2n^{2}..._{2n+1}

2n·x = (2n+1)·x+(-x) = 2n^{2}

x = n

Teorema:

Construcción algebraica de los números enteros:

< n,m > =[R]= < p,q > <==> n+q = m+p

< n,0 > =[R]= < 0,q >

Construcción algebraica de los números racionales:

< n,m > =[R]= < p,q > <==> nq = mp

< n,1 > =[R]= < 1,q >

Teorema:

Construcción algebraica de las potencias enteras:

< x^{n},x^{m} > =[R]= < x^{p},x^{q} > <==> x^{n+q} = x^{m+p}

< x^{n},1 > =[R]= < 1,x^{q} >

Construcción algebraica de las potencias racionales:

< x^{n},x^{m} > =[R]= < x^{p},x^{q} > <==> x^{nq} = x^{mp}

< x^{n},x > =[R]= < x,x^{q} >

Ley:

Se puede saber si saben,

porque no hay cobertura,

fuera de las teorías de las demostraciones,

y si se ve el documento matemático,

saben.

Se puede saber si no saben,

porque hay cobertura,

dentro de las teorías de las demostraciones, 

y si no se ve el documento matemático,

no saben.

Teorema:

int[ ( x^{m}+a )^{n} ]d[x] = ( 1/( n·[m:a]+1 ) )·( x^{m}+a )^{n}·x

int[ ( e^{mx}+a )^{n} ]d[x] = ( 1/( n·[m:a] ) )·( e^{mx}+a )^{n}

Demostración:

x^{m}+a = x^{[m:a]}

( x^{m}+a )^{n} = ( x^{[m:a]} )^{n} = x^{n·[m:a]}

x·( x^{m}+a )^{n} = x·x^{n·[m:a]} = x^{n·[m:a]+1}

d_{x}[ x^{n·[m:a]+1} ] = ( n·[m:a]+1 )·x^{n·[m:a]} 

e^{mx}+a = ( e^{x} )^{m}+a = ( e^{x} )^{[m:a]} = e^{[m:a]·x}

( e^{mx}+a )^{n} = ( e^{[m:a]·x} )^{n} = e^{n·[m:a]·x}

d_{x}[ e^{n·[m:a]·x} ] = n·[m:a]·e^{n·[m:a]·x}

Teorema:

int[ ( (x^{p}+a)·(x^{q}+b) )^{n} ]d[x] = ?

int[ ( (e^{px}+a)·(e^{qx}+b) )^{n} ]d[x] = ?

Teorema:

Sea F(x,y,z,t) = (1/u)·d_{t}[...]+(1/r)·( int[...]d[x]+int[...]d[y]+int[...]d[z] )

D[ct]-[ F(x,y,z,t) ] = (c/u)+ct·(1/r)·(x+y+z)

Teorema:

Sea F(x,y,z,t) = u·int[...]d[t]+r·( d_{x}[...]+d_{y}[...]+d_{z}[...] )

D[x+y+z]-[ F(x,y,z,t) ] = ut·(x+y+z)+3r

Teorema:

Sea F(x,y,z,u,v) = (1/a)·d_{u}[...]+(1/b)·d_{v}[...]+(1/r)·( int[...]d[x]+int[...]d[y]+int[...]d[z] )

D[u+v]-[ F(x,y,z,u,v) ] = ( (1/a)+(1/b)+(u+v)·(1/r)·(x+y+z) )

Teorema:

Sea F(x,y,z,u,v) = a·int[...]d[u]+b·int[...]d[v]+r·( d_{x}[...]+d_{y}[...]+d_{z}[...] )

D[x+y+z]-[ F(x,y,z,u,v) ] = (au+bv)·(x+y+z)+3r