óptica-física y psico-neurología-extraterrestre y arte-matemático y análisis-matemático y termodinámica

Ley:

d_{rw}[f(w,x)]+d_{x}[f(w,x)] = a·( sin(2arw)+(-1)·(1/(ax))^{n} )

f(w,x) = ( sin(arw) )^{2}+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

d_{rw}[g(w,x)]+d_{x}[g(w,x)] = a·( (-1)·sin(2arw)+(1/(ax))^{n} )

g(w,x) = ( cos(arw) )^{2}+( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

( f(w,x)+( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) )·...

... ( g(w,x)+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ) = (1/4) <==> ...

... w = (1/(ar))·(pi/4)

( f(w,x)+( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) )·...

... ( g(w,x)+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ) = (3/16) <==> ...

... ( w = (1/(ar))·(pi/3) || w = (1/(ar))·(pi/6) )

Deducción:

( sin(arw) )^{2}·( cos(arw) )^{2} = (1/4)

( cos(arw) )^{2} = 1+(-1)·( sin(arw) )^{2}

( sin(arw) )^{4}+(-1)·( sin(arw) )^{2}+(1/4) = 0

( sin(arw) )^{2} = (1/2)·( 1+( 1+(-1) )^{(1/2)}) = (1/2)

arw = arc-sin( (1/2)^{(1/2)} ) = (pi/4)

Ley:

d_{rw}[f(w,x)]+d_{x}[f(w,x)] = a·( ( sin(arw) )^{2}+(-1)·(1/(ax))^{n} )

f(w,x) = (1/2)·arw+(-1)·(1/4)·sin(2arw) )+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

d_{rw}[g(w,x)]+d_{x}[g(w,x)] = a·( ( cos(arw) )^{2}+(1/(ax))^{n} )

g(w,x) = (1/2)·arw+(1/4)·sin(2arw) )+( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

int[ f(w,x)+g(w,x) ]d[arw] = 2pi^{2} <==> w = (1/(ar))·2pi

Deducción:

int[ arw ]d[arw] = (1/2)·(arw)^{2} = 2pi^{2}

(arw)^{2} = 4pi^{2} = (2pi)^{2}

Principio: [ de la primera directriz ]

Hay contacto extraterrestre,

con motor de curvatura,

siendo el próximo,

pudiendo ir a su planeta.

No hay contacto extraterrestre,

sin motor de curvatura,

siendo el prójimo,

no pudiendo ir a su planeta.

Ley:

No puede haber contacto extraterrestre,

saltando-te la primera directriz,

porque te crees un dios del universo.

Puede haber contacto extraterrestre,

no saltando-te la primera directriz,

porque no te crees un dios del universo.

Ley:

Pensamiento peligroso:

Te crees un dios del universo 

y entonces también te crees que caminas solo sin estar allí.

Pensamiento seguro:

Quizás te cree un dios del universo

pero no te crees que caminas solo sin estar allí.

Principio: [ de la segunda directriz ]

No puede haber contacto extraterrestre des-ascendido,

con un mundo ascendido,

porque no se puede estar en un mundo des-ascendido,

con testimonio del evangelio,

siendo el prójimo de ti el mundo des-ascendido.

Puede haber contacto extraterrestre des-ascendido,

con un mundo des-ascendido, 

porque se puede estar en un mundo des-ascendido,

sin testimonio del evangelio,

siendo el próximo de ti el mundo des-ascendido.

Ley:

Todos los hombres que se creen dioses del universo,

son de la Tierra,

y no de Cygnus-Kepler,

porque se han saltado la segunda directriz.

Todos los hombres que no se creen dioses del universo,

son de Cygnus-Kepler,

y no de la Tierra,

porque no se han saltado la segunda directriz.

Esto tiene cura,

como vas a ser un dios del universo,

sin saber su Ley de la primera y la segunda directriz.

Análisis matemático 2:

Arte:

Sea Z(s) = sum[n = 1]-[oo][ (1/n)^{s} ] ==>

[Es][ sum[n = 1]-[oo][ (s+(-1))·(1/n)^{s} ] = ( Z(s)/Z(s+(-1)) ) ]

Exposición:

s = 1 

( Z(s)/Z(s+(-1)) ) = ln(2)

f(s+(-1)) = ( 1/(s+(-1)) )

Id(s+(-1)) = ( 1/(s+(-1)) ) <==> s = 2

g( h(s+(-1)) ) = Z(s+(-1))

Id( h(s+(-1)) ) = Z(s+(-1)) <==> h = Z

sum[n = 1]-[oo][ (s+(-1))·(1/n)^{s} ] = (s+(-1))·sum[n = 1]-[oo][ (1/n)^{s} ] = ...

.... (s+(-1))·Z(s) = f(s+(-1))·Z(s) = ( 1/(s+(-1)) )·Z(s) = ...

... ( 1/(g o h)(s+(-1)) ) )·Z(s) = ( 1/g( h(s+(-1)) ) )·Z(s) = ( Z(s)/Z(s+(-1)) )

Arte:

Sea H(s) = sum[n = 1]-[oo][ (1+(1/n))^{s} ] ==>

[Es][ sum[n = 1]-[oo][ 0s·(1+(1/n))^{s} ] = ( H(s)/H(s+(-1)) ) ]

Exposición:

s = 1

( H(s)/H(s+(-1)) ) = 1+ln(2)

Arte:

Sea Z(s) = sum[n = 1]-[oo][ (1/n)^{s} ] ==>

[Es][ sum[n = 1]-[oo][ 0·[ n // s ]·(1/n)^{s} ] = (s+(-1))·Z(s+(-1)) ]

Exposición:

s = 1

u(s) = 1

v(1) = s

sum[n = 1]-[oo][ 0·[ n // s ]·(1/n)^{s} ] = sum[n = 1]-[oo][ 0·[ n // u(s) ]·(1/n)^{u(s)} ] = ...

... sum[n = 1]-[oo][ 0·[ n // 1 ]·(1/n) ] = sum[n = 1]-[oo][ 0n·(1/n) ] = sum[n = 1]-[oo][ 0 ] = 1 = ...

... 0·oo = 0·Z(0) = (1+(-1))·Z(1+(-1)) = (v(1)+(-1))·Z(v(1)+(-1)) = (s+(-1))·Z(s+(-1))

Arte:

Sea Z(s) = sum[n = 1]-[oo][ (1/n)^{s} ] ==>

[Es][ sum[n = 1]-[oo][ (1/2)^{n+(-1)}·[ n // s ]·(1/n)^{s} ] = 2s·(s+(-1))·Z(s+(-1)) ]

Exposición:

s = 1

u(s) = 1

v(1) = s

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

... sum[n = 1]-[oo][ (1/2)^{n+(-1)}·[ n // u(s) ]·(1/n)^{u(s)} ] = ...

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

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

... (v(1)+v(1))·(v(1)+(-1))·Z(v(1)+(-1)) = (s+s)·(s+(-1))·Z(s+(-1)) = 2s·(s+(-1))·Z(s+(-1))

Análisis matemático 1:

[%] Derivación

Continuidad

Cuerpos ordenados

Sucesiones

Análisis matemático 2:

[%] Integración y producto integral

Integral definida

Euler Falsus Infinitorum

Teoremas y Artes de series

Análisis matemático 3:

[%] Derivadas parciales

[%] Optimización

Continuidad

Análisis matemático 4:

[%] Integrales múltiples

[%] Integrales de línea

Integrales impropias

Análisis matemático 5:

Sucesiones de funciones

Integral de Lebesgue

Funciones Error y Series de potencies

Análisis matemático 6:

Transformada integral exponencial

Arte método de Euler

Arte series de Laurent

Teorema:

Sea d_{x}[F(x)] = f(x) ==>

F(x) es continua <==> f(x) es continua

Demostración:

Sea s > 0 ==>

Sea d < (s/2) ==>

| F(x+h)+(-1)·F(x) | < d

| F(x+h)+(-1)·F(x) | = 0

| f(x+h)+(-1)·f(x) | = 0^{2} = 2·0 < 2d < s

Sea d > 0 ==>

Sea s > 0 ==>

| f(x+h)+(-1)·f(x) | < s < 2s

| f(x+h)+(-1)·f(x) | = 2·0 = 0^{2}

| F(x+h)+(-1)·F(x) | = 0

| F(x+h)+(-1)·F(x) | < d

Teorema:

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

Demostración:

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

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

Teorema:

lim[n = oo][ sum[k = 1]-[n][ ( n/(n^{2}+k^{2}) ) ] ] = (pi/4)

Teorema:

lim[n = oo][ sum[k = 1]-[n][ ( 1/(n^{p}+k^{p}) )·pk^{p+(-1)} ] ] = ln(2)

Teorema:

lim[n = oo][ sum[k = 1]-[n][ (npk^{p+(-1)}+k^{p})·e^{(k/n)}·(1/n)^{p+1} ] ] = e

Teorema:

lim[n = oo][ sum[k = 1]-[n][ f(k/n)·(1/n) ] ] = F(1)+(-1)·F(0)

Demostración:

lim[n = oo][ sum[k = 1]-[n][ f(k/n)·(1/n) ] ] = int[x = 0]-[1][ f(x) ]d[x] = F(1)+(-1)·F(0)

Teorema:

lim[n = oo][ sum[k = 1]-[n][ (p+1)·k^{p}·f( (k/n)^{p+1} )·(1/n)^{p+1} ] ] = F(1)+(-1)·F(0)

Teorema:

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

Demostración:

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

lim[n = oo][ sum[k = 1]-[n][ (k/n)^{p}·e^{(k/n)} ]·(1/n) ] = int[x = 0]-[1][ x^{p}·e^{x} ]d[x]

k = j+(-1)

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

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

... x^{k+1 [o] 0·(k+1)^{1}+1+p} ] = ...

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

... x^{k+1 [o] 0·(k+1)^{1}+1+p} ] = ...

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

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

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

... x^{k+1 [o] 1·(k+1)^{0}+p} ] = ...

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

... x^{k+1 [o] 1·(k+1)^{0}+p} ] = ...

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

Teorema:

int[ x^{p}·e^{x} ]d[x] = x^{p}·er-h[p](x)

er-h[p](0) = (1/(1+p))

Demostración:

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

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

... x^{k [o] ( (1+(-x))·(k+1)^{x}+x+(-1) )+p} ] ]d[x] =

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

... (1/( (1+(-x))·(k+1)^{x}+p+x ))...

... x^{k+1 [o] (1+(-x))·(k+1)^{x}+p+x} ] = ...

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

... x^{k+1 [o] (1+(-x))·(k+1)^{x}+p+x} ] = ...

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

... x^{j [o] (1+(-x))·j^{x}+p+x} ] = x^{p}·er-h[p](x)

Teorema:

d_{x}[ er-h[p](x) ] = (1/x)·er-h[p](x)+er-h[p+1](x)

Teorema:

e^{x} = (p+1)·(1/x)·er-h[p](x)+er-h[p+1](x)

p = (-1)

Demostración:

x^{p}·e^{x} = px^{p+(-1)}·er-h[p](x)+x^{p}·d_{x}[er-h[p](x)] = ...

... px^{p+(-1)}·er-h[p](x)+x^{p+(-1)}·er-h[p](x)+x^{p}·er-h[p+1](x)

Teorema:

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

... (1/(1+p))·0^{p}+(-1)·(1/e)·(1/(p+1))

int[ x^{p}·e^{(-x)} ]d[x] = (-1)·x^{p}·er-h[p](-x)

Teorema:

lim[n = oo][ sum[k = 1]-[n][ k^{p}·cos(k/n)·(1/n)^{p+1} ] ] = ...

... (1/(1+p))·sin(1)+(-1)·(1/(1+p))·0^{p}·sin(0)

Demostración:

lim[n = oo][ sum[k = 1]-[n][ k^{p}·cos(k/n)·(1/n)^{p+1} ] ] = ...

lim[n = oo][ sum[k = 1]-[n][ (k/n)^{p}·cos(k/n) ]·(1/n) ] = int[x = 0]-[1][ x^{p}·cos(x) ]d[x]

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

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

... x^{2k+1 [o] 0·(2k+1)^{1}+1+p} ] = 

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

... x^{2k+1 [o] 0·(2k+1)^{1}+1+p} ] = (1/(1+p))·sin(1)

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

... x^{2k^{0} [o] ( 1·(2k)^{0}+(-1) )+p} ] ]d[x] = ...

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

... x^{2k+1 [o] 1·(2k+1)^{0}+p} ] = ...

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

... x^{2k+1 [o] 1·(2k+1)^{0}+p} ] = (1/(1+p))·0^{p}·sin(0)

Teorema:

lim[n = oo][ sum[k = 1]-[n][ k^{p}·sin(k/n)·(1/n)^{p+1} ] ] = ...

... (1/(1+p))·0^{p}·cos(0)+(-1)·(1/(1+p))·cos(1)

Demostración:

lim[n = oo][ sum[k = 1]-[n][ k^{p}·sin(k/n)·(1/n)^{p+1} ] ] = ...

lim[n = oo][ sum[k = 1]-[n][ (k/n)^{p}·sin(k/n) ]·(1/n) ] = int[x = 0]-[1][ x^{p}·sin(x) ]d[x]

k = j+(-1)

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

... x^{2k+1 [o] 0·(2k+1)^{1}+p} ] ]d[x] = ...

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

... x^{2k+2 [o] 0·(2k+2)^{1}+1+p} ] = ...

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

... x^{2j [o] 0·(2j)^{1}+1+p} ] = (-1)·(1/(1+p))·cos(1)

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

... x^{2k+1 [o] 1·(2k+1)^{0}+(-1)+p} ] ]d[x] = ...

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

... ( 1/(1·(2k+2)^{0}+p) )·x^{2k+2 [o] 1·(2k+2)^{0}+p} ] = ...

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

... x^{2k+2 [o] 1·(2k+2)^{0}+p} ] = ...

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

... x^{2j [o] 1j^{0}+p} ] = (-1)·(1/(1+p))·0^{p}·cos(0)

Troyan Sedor pleyadense es mejor que se vaya,

porque va estallar su nave por la primera directriz,

como se acerque mucho a Cygnus-Kepler.

Ley: [ de ejemplo de teoría ]

Si d_{V}[P_{0}]·V^{2}+PV+(-1)·d_{P}[k]·TP = 0 ==>

V_{min} = (-1)·(1/2)·( 1/d_{V}[P_{0}] )·P

(1/4)·( 1/d_{V}[P_{0}] )·P^{2}+d_{P}[k]·TP = 0

P_{min} = (-1)·2·d_{P}[k]·T·d_{V}[P_{0}]

(PV)_{min} = d_{P}[k]·TP

d_{P}[T(P)]·p = qR <==> p = qR·( 1/(PV)_{min} )·d_{P}[k]·(-1)·P^{2}

Deducción:

d_{V}[ d_{V}[P_{0}]·V^{2}+PV+(-1)·d_{P}[k]·TP ] = ...

... d_{V}[d_{V}[P_{0}]·V^{2}]+d_{V}[PV]+d_{V}[ (-1)·d_{P}[k]·TP ] = ...

... d_{V}[d_{V}[P_{0}]·V^{2}]+d_{V}[PV]+0 = d_{V}[d_{V}[P_{0}]·V^{2}]+d_{V}[PV] = ...

... d_{V}[P_{0}]·d_{V}[V^{2}]+P·d_{V}[V] = d_{V}[P_{0}]·2V+P

d_{V}[P_{0}]·2V+P = 0

d_{V}[P_{0}]·2V = d_{V}[P_{0}]·2V+0 = d_{V}[P_{0}]·2V+(P+(-P)) = ...

... ( d_{V}[P_{0}]·2V+P )+(-P) = 0+(-P) = (-P)·

V = ( (1/2)·(1/d_{V}[P_{0}])·(d_{V}[P_{0}]·2) )·V = ...

... (1/2)·(1/d_{V}[P_{0}])·( (d_{V}[P_{0}]·2)·V ) = (-1)·(1/2)·(1/d_{V}[P_{0}])·P

d_{P}[T(P)] = (PV)_{min}·(1/d_{P}[k])·(-1)·(1/P)^{2}

Ley:

Si ( P+d_{xyz}^{3}[q(x,y,z)]·gh )·V = kT ==>

q(x,y,z) = kT·(1/(gh))·(1/V)·xyz+(-1)·P·(1/(gh))·xyz

Ley:

Si ( P+d_{xy}^{2}[q(x,y)]·g )·V = kT ==>

q(x,y) = kT·(1/g)·(1/V)·xy+(-1)·P·(1/g)·xy

Ley:

Si ( P+d_{xy}^{2}[q(x,y)]·g )·V = kT·xya^{2} ==>

q(x,y) = kT·(1/g)·(1/V)·(1/4)·(axy)^{2}+(-1)·P·(1/g)·xy

Ley:

Si ( P+d_{x}[m(x)]·u^{2} )·V = kT ==>

m(x) = kT·(1/u)^{2}·(1/V)·x+(-1)·P·(1/u)^{2}·x

El libro de problemas seleccionados de la física elemental es todo algebraico,

y hay pocos que se puedan pasar a Irodov problems

La gente habla de extraterrestres creyendo-se superior,

y son inferiores saltando-se la primera directriz.

Yo no me proyecto en los elfos para estar con mi mujer,

porque no tengo motor de curvatura,

y no la veré nunca en esta vida,

como solo tenga electricidad y no gravedad.

Todos estos son maricones,

tienen motor de curvatura,

y se proyectan en hombres,

en vez de estar proyectados con su mujer,

y son maricones aunque no miren pichas.

Rezo al Mal:

Los hombres no tienen motor de curvatura,

y no pueden ir a ver a su mujer,

pero no son maricones.

Los extraterrestres tienen motor de curvatura,

y pueden ir a ver a su mujer,

pero son maricones.

electro-magnetismo y mecánica-ingeniería y ecuaciones-en-derivadas-parciales y mecánica-física y análisis-matemático-6 y medicina

Examen de electro-magnetismo:

Principio:

E(x,y,z) = qk·(1/r)^{3}·a·< x^{2},y^{2},z^{2} >

E(yz,zx,xy) = qk·(1/r)^{4}·a^{2}·< (yz)^{2},(zx)^{2},(xy)^{2} >

Ley:

div[ E(x,y,z) ] = ?

Anti-div[ E(yz,zx,xy) ] = ?

Ley:

Anti-Potencial[ E(x,y,z) ] = ?

Potencial[ E(yz,zx,xy) ] = ?

Ley: [ de corrección del examen ]

div[ E(x,y,z) ] = d_{x(yz)}^{2}[ Anti-Potencial[ E(x,y,z) ] ]

Anti-div[ E(yz,zx,xy) ] = d_{x(yz)}^{2}[ Potencial[ E(yz,zx,xy) ] ]

Ley:

R·d_{t}[q(t)]+(-C)·p(t) = W·f(ut)·e^{ut}

p(t) = W·( 1/(uR·d_{ut}[f(ut)]+(-C)·f(ut)) )·f(ut)·e^{ut}

q(t) = W·( ut /o(ut)o/ (uR·f(ut)+(-C)·int[ f(ut) ]d[ut]) ) [o(ut)o] f(ut) [o(ut)o] e^{ut}

Ley:

R·d_{t}[q(t)]+C·p(t) = W·f(ut)·e^{(-1)·ut}

p(t) = W·( 1/((-u)·R·d_{ut}[f(ut)]+C·f(ut)) )·f(ut)·e^{(-1)·ut}

q(t) = W·( ut /o(ut)o/ ((-u)·R·f(ut)+C·int[ f(ut) ]d[ut]) ) [o(ut)o] f(ut) [o(ut)o] e^{(-1)·ut}

Ley:

Sea ( d_{t}[ I_{cx} ] = 0 & d_{t}[ I_{cy} ] = 0 ) ==>

Si d[M_{1}(t)] = (1/2)·mgx·(1/s)^{2}·cos(nw)·d[w] ==>

M_{1}(t) = (1/2)·mg·(x/n)·(1/s)^{2}·sin(nw)

Si d[ d[M_{2}(t)] ] = mg·(1/s)^{2}·sin(nw)·cos(nw)·d[y]d[w] ==>

M_{2}(t) = mg·(y/n)·(1/s)^{2}·(1/2)·( sin(nw) )^{2}

M_{1}(t) = M_{2}(t) <==> ( w(t) = (1/n)·arc-sin( I_{cx}/I_{cy} ) & I_{cx} [< I_{cy} )

Ley:

Sea d_{t}[ I_{c} ] = 0 ==>

Si d[M_{1}(t)] = (1/2)·I_{c}·u^{2}·cos(nw)·d[w] ==>

M_{1}(t) = (1/2)·I_{c}·u^{2}·(1/n)·sin(nw)

Si d[ d[M_{2}(t)] ] = I_{c}·u^{2}·(1/x)·sin(nw)·cos(nw)·d[x]d[w] ==>

M_{2}(t) = I_{c}·u^{2}·ln(ax)·(1/n)·(1/2)·( sin(nw) )^{2}

M_{1}(t) = M_{2}(t) <==> ( w(t) = (1/n)·arc-sin( ( 1/ln( aI_{c}·(1/(md)) ) ) ) & aI_{c} >] md·e )

Ley: [ del calor electro-magnético ]

div[ E_{e}(x,y,t) ] = (-2)·(1/c)·B_{e}(x,y,t)

Deducción:

E_{e}(x,y,t)+int[ B_{e}(x,y,t) ]d[t] = 0 = m·d_{tt}^{2}[ < x,y > ]

x(t) = ct·( cos(w) )^{2}

y(t) = ct·( sin(w) )^{2}

div[ E_{e}(x,y,t) ]+div[ inr[ B_{e}(x,y,t) ]d[t] ] = 0^{2}

div[ int[ B_{e}(x,y,t) ]d[t] ] = ( 1/(d[x]+d[y]) )·(d[x]+d[y]) [o] div[ int[ B_{e}(x,y,t) ]d[t] ]

div[ E_{e}(x,y,t) ]+2·(1/c)·B_{e}(x,y,t) = 0^{2}

div[ E_{e}(x,y,t) ] = (-2)·(1/c)·B_{e}(x,y,t)

Ley: [ del calor gravito-magnético ]

div[ E_{g}(x,y,t) ] = (-2)·(1/c)·B_{g}(x,y,t)

Teorema:

d_{x}[u(x,y,t)]+d_{y}[u(x,y,t)] = (-2)·(1/c)·d_{t}[u(x,y,t)]

u(x,y,0) = H(ax,ay)

u(x,y,(1/u)) = K(ax,ay)

u(x,y,t) = ( (1+(-1)·ut)·H(ax,ay)+ut·K(ax,ay) || 1 )·e^{ax+ay+(-1)·act || 0}

Teorema:

d_{x}[u(x,y,t)]+d_{y}[u(x,y,t)] = 2·(1/c)·d_{t}[u(x,y,t)]

u(x,y,0) = H(ax,ay)

u(x,y,(1/u)) = K(ax,ay)

u(x,y,t) = ( (1+(-1)·ut)·H(ax,ay)+ut·K(ax,ay) || 1 )·e^{ax+ay+act || 0}

Teorema:

d_{x}[u(x,y,t)]+d_{y}[u(x,y,t)] = (-2)·(1/c)·d_{t}[u(x,y,t)]

u(0,0,t) = f(ut)

u(p,q,t) = g(ut)

u(x,y,t) = ...

... ( (1/2)·( (1+(-1)·(x/p))·f(ut)+(1+(-1)·(y/q))·f(ut) )+(1/2)·( (x/p)·g(ut)+(y/q)·g(ut) ) || 1 )·...

... e^{ax+ay+(-1)·act || 0}

Teorema:

d_{x}[u(x,y,t)]+d_{y}[u(x,y,t)] = 2·(1/c)·d_{t}[u(x,y,t)]

u(0,0,t) = f(ut)

u(p,q,t) = g(ut)

u(x,y,t) = ...

... ( (1/2)·( (1+(-1)·(x/p))·f(ut)+(1+(-1)·(y/q))·f(ut) )+(1/2)·( (x/p)·g(ut)+(y/q)·g(ut) ) || 1 )·...

... e^{ax+ay+act || 0}

Teorema:

d_{x}[u(x,y,t)]+d_{y}[u(x,y,t)] = (-2)·(1/c)·d_{t}[u(x,y,t)]

u(0,q,t) = f(ut)

u(p,0,t) = g(ut)

u(x,y,t) = ?

Teorema:

d_{x}[u(x,y,t)]+d_{y}[u(x,y,t)] = 2·(1/c)·d_{t}[u(x,y,t)]

u(0,q,t) = f(ut)

u(p,0,t) = g(ut)

u(x,y,t) = ?

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = (-2)·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,0) = H(ax,ay)

d_{t}[u(x,y,0)] = 0

u(x,y,t) = ...

... (1/2)·( e^{ax+ay+ac·it || ln( H(ax,ay) )+act}+e^{ax+ay+ac·it || ln( H(ax,ay) )+(-1)·act} )

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = 2·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,0) = H(ax,ay)

d_{t}[u(x,y,0)] = 0

u(x,y,t) = ...

... (1/2)·( e^{ax+ay+act || ln( H(ax,ay) )+act}+e^{ax+ay+act || ln( H(ax,ay) )+(-1)·act} )

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = (-2)·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(0,y,0) = F(ay)

u(r,y,0) = G(ay)

d_{t}[u(x,y,0)] = 0

u(x,y,t) = ?

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = 2·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(0,y,0) = F(ay)

u(r,y,0) = G(ay)

d_{t}[u(x,y,0)] = 0

u(x,y,t) = ?

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = (-2)·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,0) = 0

d_{t}[u(x,y,0)] = h(ax,ay)

u(x,y,t) = (1/2)·sum[k = 1]-[oo][ ...

... int[h(ax,ay)+(-1)·act·0 || (4t)^{(1/2)}]-[h(ax,ay)+act·0 || (4t)^{(1/2)}][ w ]d[w] ]·e^{ax+ay+ac·it || 0}

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = 2·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,0) = 0

d_{t}[u(x,y,0)] = h(ax,ay)

u(x,y,t) = (1/2)·sum[k = 1]-[oo][ ...

... int[h(ax,ay)+(-1)·act·0 || (4t)^{(1/2)}]-[h(ax,ay)+act·0 || (4t)^{(1/2)}][ w ]d[w] ]·e^{ax+ay+act || 0}

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = (-2)·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,0) = 0

d_{t}[u(0,y,0)] = ac·f(ay)

d_{t}[u(r,y,0)] = ac·g(ay)

u(x,y,t) = ?

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = 2·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,0) = 0

d_{t}[u(0,y,0)] = ac·f(ay)

d_{t}[u(r,y,0)] = ac·g(ay)

u(x,y,t) = ?

Motores a combustión de explosión acotada:

Ley:

Sea d[I_{c}] = (1/s)^{2}·Mrv·d[t] ==>

Si (I_{c}/2)·d_{t}[w]^{2} = qgh·cos(ut) ==>

x(t) = (M/(md))·(1/s)^{2}·rvt

w(t) = (1/u)·( 2qgh·(1/(Mrv))·us^{2}·( ln(ut) [o(ut)o] sin(ut) ) )^{[o(ut)o] (1/2)}

(1/u) [< t [< (pi/u)

Ley:

Sea d[I_{c}] = (1/s)^{2}·Mrgt·d[t] ==>

Si (I_{c}/2)·d_{t}[w]^{2} = qgh·sin(ut) ==>

x(t) = (M/(md))·(1/s)^{2}·rg·(1/2)·t^{2}

w(t) = (1/u)·( 4qgh·(1/(Mrg))·(us)^{2}·( (1/(ut)) [o(ut)o] cos(ut) ) )^{[o(ut)o] (1/2)}

(1/u) [< t [< (pi/(2u))

Teorema:

( cos(w) )^{4}+(-1)·( sin(w) )^{4}+i·sin(2w) = e^{2iw}

Teorema:

( cos(w) )^{4}+( sin(w) )^{4}+(1/2)·( sin(2w) )^{2} = 1

Ley:

Sea ( d_{t}[ I_{cx} ] = 0 & d_{t}[ I_{cy} ] = 0 ) ==>

Si d[ d[M(t)] ] = qg·(1/s)^{2}·sin(nw)·cos(nw)·d[x]d[w] ==>

M(t) = qg·(x/n)·(1/s)^{2}·(1/2)·( sin(nw) )^{2}

(I_{c}/2)·d_{t}[w]^{2} = qgx·(1/(ns))^{2}·(1/4)·( nw+(-1)·(1/2)·sin(2nw) )

x(t) = I_{c}·(1/(md))

w(t) = (1/n)·Anti-[ ( s /o(s)o/ ( (1/4)·s^{2}+(1/8)·cos(2s) ) )^{[o(s)o](1/2)}]-( ...

... ( (1/(md))·qg )^{(1/2)}·(1/s)·t )

Ley:

Sea d_{t}[ I_{c} ] = 0 ==>

Si d[ d[M(t)] ] = I_{c}·u^{2}·(1/x)·sin(nw)·cos(nw)·d[x]d[w] ==>

M(t) = I_{c}·u^{2}·ln(ax)·(1/n)·(1/2)·( sin(nw) )^{2}

(I_{c}/2)·d_{t}[w]^{2} = I_{c}·u^{2}·ln(ax)·(1/n)^{2}·(1/4)·( nw+(-1)·(1/2)·sin(2nw) )

x(t) = I_{c}·(1/(md))

w(t) = (1/n)·Anti-[ ( s /o(s)o/ ( (1/4)·s^{2}+(1/8)·cos(2s) ) )^{[o(s)o](1/2)}]-( ...

... ( ln( aI_{c}·(1/md) ) )^{(1/2)}·ut )

Ecuaciones de densidades:

Leyes de agua y aceite:

Ley: 

d_{x}[u(x,y)]+d_{y}[u(x,y)] = (m/V)·xy

u(0,y) = m·F(ay)

u(r,y) = m·G(ay)

u(x,y) = ( (1+(-1)·(x/r))·F(ay)+(x/r)·G(ay) || 1 )·( (m/(4V))·yx^{2} || (m/(4V))·xy^{2} || m )

Ley:

d_{x}[u(x,y)]+d_{y}[u(x,y)] = (-V)·m·( 1/(xy) )^{2}

u(x,0) = m·F(ax)

u(x,r) = m·G(ax)

u(x,y) = ( (1+(-1)·(y/r))·F(ax)+(y/r)·G(ax) || 1 )·( (V/2)·( m/(xy^{2}) ) || (V/2)·( m/(yx^{2}) ) || m )

Ley: [ de ola de mar ]

d_{x}[u(x,y)]+d_{y}[u(x,y)] = m·(1/a)·(1/(xy))

u(0,y) = m·F(ay)

u(r,y) = m·G(ay)

u(x,y) = ( (1+(-1)·(x/r))·F(ay)+(x/r)·G(ay) || 1 )·( (1/2)·(m/(ay))·ln(ax) || (1/2)·(m/(ax))·ln(ay) || m )

Ley: 

d_{x}[u(x,y)]+d_{y}[u(x,y)]+a·u(x,y) = (m/V)·xy

u(0,y) = m·F(ay)

u(r,y) = m·G(ay)

u(x,y) = ( (1+(-1)·(x/r))·F(ay)+(x/r)·G(ay) || 1 )·....

... ( (m/(6V))·yx^{2} || (m/(6V))·xy^{2} || (1/(3V))·(m/a)·xy || m )

Ley:

d_{x}[u(x,y)]+d_{y}[u(x,y)]+a·u(x,y) = (-V)·m·( 1/(xy) )^{2}

u(x,0) = m·F(ax)

u(x,r) = m·G(ax)

u(x,y) = ( (1+(-1)·(y/r))·F(ax)+(y/r)·G(ax) || 1 )·...

... ( (V/3)·( m/(xy^{2}) ) || (V/3)·( m/(yx^{2}) ) || (-1)·(V/3)·(m/a)·( 1/(xy) )^{2} || m )

Arte:

Sea u(x) = e^{(-x)} ==>

[Ax][ f(a)·(1/u)^{0} = f(a) ]

[Ex][ (-1)^{k}·(k+(-1))!·d_{a...a}^{k}[f(a)]·(1/u)^{k} = d_{a...a}^{k}[f(a)] ]

Exposición:

x = (-1)·(1/k)·ln( (-1)^{k}·(k+(-1))! )

Sea z(x) = e^{(-x)}+a ==>

Sea u(x) = e^{(-x)} ==>

d[u] = d[z]

s(u) = 1

d[u] = d[s(u)] = d[1] ==>

Caso 1:

int[x = 0]-[1][ f(a)/(a+(-z)) ]d[z] = int-int[ (-1)·d_{a}[f(a)]·(1/u) ]d[u]d[a] = f(a)

int[ (-1)·d_{a}[f(a)]·(1/u) ]d[u] = d_{a}[f(a)]

(-1)·d_{a}[f(a)]·(1/z) = d_{a}[f(a)]

Caso 2:

int-int[x = 0]-[1][ f(a)/(a+(-z))^{2} ]d[z]d[z] = ...

... int-int-int-int[ d_{aa}^{2}[f(a)]·(1/u)^{2} ]d[u]d[u]d[a]d[a] = f(a)

int-int[ d_{aa}^{2}[f(a)]·(1/u)^{2} ]d[u]d[u] = d_{aa}^{2}[f(a)]

d_{aa}^{2}[f(a)]·(1/z)^{2} = d_{aa}^{2}[f(a)]

Caso 3:

int-int-int[x = 0]-[1][ 2·f(a)/(a+(-z))^{3} ]d[z]d[z]d[z] = ...

... int-int-int-int-int-int[ (-1)·2·d_{aaa}^{2}[f(a)]·(1/u)^{3} ]d[u]d[u]d[u]d[a]d[a]d[a] = f(a)

int-int-int[ (-1)·2·d_{aaa}^{3}[f(a)]·(1/u)^{3} ]d[u]d[u]d[u] = d_{aaa}^{3}[f(a)]

(-1)·2·d_{aaa}^{3}[f(a)]·(1/u)^{3} = d_{aaa}^{3}[f(a)]

Artes: [ de series de Laurent ]

Sea z(x) = e^{(-x)} ==>

Exposición:

x = 0

Arte:

[Ex][ e^{x} = 1+sum[k = 1]-[oo][ (-1)^{k}·(1/k)·( xe^{x} )^{k} ] ]

[Ex][ e^{(-x)} = 1+sum[k = 1]-[oo][ (1/k)·( xe^{(-x)} )^{k} ] ]

Arte:

[Ex][ ( 1/(1+(-x)) ) = 1+sum[k = 1]-[oo][ k!·(1/k)·( xe^{(-x)} )^{k} ] ]

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

Arte:

[Ex][ e-pos[m](x) = m+sum[k = 1]-[oo][ (-1)^{k}·( 1+m·(1/k) )·( xe^{x} )^{k} ] ]

[Ex][ e-neg[m](x) = (-m)+sum[k = 1]-[oo][ (-1)^{k}·( 1+(-m)·(1/k) )·( xe^{x} )^{k} ] ]

Arte:

[Ex][ octopus(x) = 1+sum[k = 1]-[oo][ (-1)^{k}·(k+1)!·(1/k)·( xe^{x} )^{k} ] ]

[Ex][ d_{x}[ octopus(x) ] = 2+sum[k = 1]-[oo][ (-1)^{k}·(k+2)!·(1/k)·( xe^{x} )^{k} ] ]

Arte:

[Ex][ ln(1+x) = (-x)·e^{x}+sum[k = 2]-[oo][ (-1)·k!·(1/k)^{2}·( xe^{x} )^{k} ] ]

[Ex][ ln(1+(-x)) = xe^{(-x)}+sum[k = 2]-[oo][ (-1)^{k+1}·k!·(1/k)^{2}·( xe^{(-x)} )^{k} ] ]

(-0) = 0 = ln(1+0) = ln(1)

Enfermedad de centro de dos mandamientos duales a densidad de carga constante:

Ley:

d_{x}[f(x)] = qaie^{axi}

d_{x}[g(x)] = (-1)·qaie^{(-1)·ayi}

s(y) = x

Robar la intimidad,

sin conexión de luz eléctrica:

No puede duchar-se con cortina opaca.

Ley:

d_{x}[f(x)] = iqa·cos(ax)

d_{x}[g(x)] = (-1)·qa·sin(ax)

f(x)+g(x) = qe^{axi}

Robar la libertad,

sin conexión de luz eléctrica:

No puede salir lloviendo o nublado.

Ley:

d_{x}[f(x)] = (-i)·qa·cos(ax)

d_{x}[g(x)] = (-1)·qa·sin(ax)

f(x)+g(x) = qe^{(-1)·axi}

Terapia con constructor:

Ley:

d_{x}[f(x)] = qae^{ax}

d_{x}[g(x)] = (-1)·qae^{(-1)·ay}

s(y) = x

No robar la intimidad,

con visita de algoritmo interno:

Ley:

d_{x}[f(x)] = qa·cosh(ax)

d_{x}[g(x)] = qa·sinh(ax)

f(x)+g(x) = qe^{ax}

No robar la libertad,

con visita de algoritmo externo:

Ley:

d_{x}[f(x)] = (-1)·qa·cosh(ax)

d_{x}[g(x)] = qa·sinh(ax)

f(x)+g(x) = qe^{(-1)·ax}

Enfermedad de centro de dos mandamientos duales a densidad de carga variable:

Ley:

d_{x}[f(x)] = d_{x}[q(x)]·ie^{axi}

d_{x}[g(x)] = (-1)·d_{x}[q(x)]·ie^{(-1)·ayi}

s(y) = x

Deducción:

int[ d_{x}[q(x)] ]d[x] [o(x)o] int[ ie^{axi} ]d[x] = int[ d_{x}[q(x)] ]d[x] [o(ax)o] int[ ie^{axi} ]d[ax]

Ley:

d_{x}[f(x)] = i·d_{x}[q(x)]·cos(ax)

d_{x}[g(x)] = (-1)·d_{x}[q(x)]·sin(ax)

f(x)+g(x) = q(x) [o(ax)o] e^{axi}

Ley:

d_{x}[f(x)] = (-i)·d_{x}[q(x)]·cos(ax)

d_{x}[g(x)] = (-1)·d_{x}[q(x)]·sin(ax)

f(x)+g(x) = q(x) [o(ax)o] e^{(-1)·axi}

Terapia con constructor:

Ley:

d_{x}[f(x)] = d_{x}[q(x)]·e^{ax}

d_{x}[g(x)] = (-1)·d_{x}[q(x)]·e^{(-1)·ay}

s(y) = x

Ley:

d_{x}[f(x)] = d_{x}[q(x)]·cosh(ax)

d_{x}[g(x)] = d_{x}[q(x)]·sinh(ax)

f(x)+g(x) = q(x) [o(ax)o] e^{ax}

Ley:

d_{x}[f(x)] = (-1)·d_{x}[q(x)]·cosh(ax)

d_{x}[g(x)] = d_{x}[q(x)]·sinh(ax)

f(x)+g(x) = q(x) [o(ax)o] e^{(-1)·ax}

Principio: [ de oftalmología de imagen y sonido ]

Vista sana:

d_{x}[q( (pi/(2a)) )]·d_{y}[p( (-1)·(pi/(2a)) )]+d_{x}[p( (pi/a) )]·d_{y}[q( (0/a) )] = pqa^{2}

Oída sana:

d_{x}[q( (pi/(2a))·i )]·d_{y}[p( (-1)·(pi/(2a))·i )]+d_{x}[p( (pi/a)·i )]·d_{y}[q( (0/a)·i )] = pqa^{2}

Principio: [ de definición de lentes ]

Lentes de Miopía:

f(ax) = (-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

Lentes de Hipermetropía:

g(ay) = ( ay /o(ay)o/ (1/(n+1))·(ay)^{n+1} )

Ley: [ de gafas de miopía ]

q(x) = qe^{(-1)·(1/(n+1))·(ax)^{n+1} [o(ax)o] sin(ax) [o(ax)o] f(ax)} = qe^{sin(ax)}

p(x) = pe^{(-1)·(1/(n+1))·(ax)^{n+1} [o(ax)o] cos(ax) [o(ax)o] f(ax)} = pe^{cos(ax)}

Ley: [ de gafas de hipermetropía ]

p(y) = pe^{(1/(n+1))·(ay)^{n+1} [o(ay)o] sin(ay) [o(ay)o] g(ay)} = pe^{sin(ay)}

q(y) = qe^{(1/(n+1))·(ay)^{n+1} [o(ay)o] cos(ay) [o(ay)o] g(ay)} = qe^{cos(ay)}

Ley: [ de sonotone de miopía ]

q(x) = qe^{(-1)·(1/(n+1))·(ax)^{n+1} [o(ax)o] sinh(ax) [o(ax)o] f(ax)} = qe^{sinh(ax)}

p(x) = pe^{(-1)·(1/(n+1))·(ax)^{n+1} [o(ax)o] i·cosh(ax) [o(ax)o] f(ax)} = pe^{i·cosh(ax)}

Ley: [ de sonotone de hipermetropía ]

p(y) = pe^{(1/(n+1))·(ay)^{n+1} [o(ay)o] sinh(ay) [o(ay)o] g(ay)} = pe^{sinh(ay)}

q(y) = qe^{(1/(n+1))·(ay)^{n+1} [o(ay)o] i·cosh(ay) [o(ay)o] g(ay)} = qe^{i·cosh(ay)}

Principio: [ de ecuación de la lente ]

Miopía:

d_{z}[f(z,x)]+d_{x}[f(z,x)] = d_{z}[p(z)]+a·(-1)·(1/(ax))^{n}

f(z,x) = p(z)+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

Hipermetropía:

d_{z}[g(z,y)]+d_{y}[g(z,y)] = d_{z}[q(z)]+a·(1/(ay))^{n}

g(z,y) = q(z)+( ay /o(ay)o/ (1/(n+1))·(ay)^{n+1} )

Ley:

d_{z}[f(z,x)]+d_{x}[f(z,x)] = a·( 1+(-1)·(1/(ax))^{n} )

f(z,x) = az+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

d_{z}[g(z,x)]+d_{x}[g(z,x)] = a·( 1+(1/(ax))^{n} )

g(z,x) = az+( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

f(z,x)+g(z,x) = n·(n+1) <==> z = (1/(2a))·n·(n+1)

Si n = 2k ==> (1/2)·n·(n+1) € N

Si n = 2k+1 ==> (1/2)·n·(n+1) € N

Deducción:

d_{z}[f(z,x)] = d_{z}[ az+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ] = ...

... d_{z}[ az ]+d_{z}[ (-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ] = ...

... d_{z}[az]+0 = d_{z}[az] = a·d_{z}[z] = a

d_{x}[f(z,x)] = d_{x}[ az+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ] = ...

... d_{x}[ az ]+d_{x}[ (-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ] = ...

... 0+d_{x}[ (-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ] = ...

... d_{x}[ (-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ] = ...

... a·d_{ax}[ (-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ] = ...

... a·(-1)·d_{ax}[ ( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} ) ] = a·(-1)·(1/(ax))^{n}

Ley:

d_{z}[f(z,y)]+d_{y}[f(z,y)] = a·( 2+(-1)·(1/(ay))^{n} )

f(z,y) = 2az+(-1)·( ay /o(ay)o/ (1/(n+1))·(ay)^{n+1} )

d_{z}[g(z,y)]+d_{y}[g(z,y)] = a·( 2+(1/(ay))^{n} )

g(z,y) = 2az+( ay /o(ay)o/ (1/(n+1))·(ay)^{n+1} )

( f(z,y)+g(z,y) )^{(1/2)} = n·(n+1) <==> z = (1/(4a))·n^{2}·(n^{2}+2n+1)

Si n = 2k ==> (1/4)·n^{2}·(n^{2}+2n+1) € N

Si n = 2k+1 ==> (1/4)·n^{2}·(n^{2}+2n+1) € N

Deducción:

Si n = 2k+1 ==>

(n^{2}+2n+1) = (n+1)^{2} = (2k+2)^{2} = 4k^{2}+4k+4 = 4·(k^{2}+k+1)

Ley:

d_{z}[f(z,x)]+d_{x}[f(z,x)] = a·( (1/(az))+(-1)·(1/(ax))^{n} )

f(z,x) = ln(az)+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

d_{z}[g(z,x)]+d_{x}[g(z,x)] = a·( (-1)·( 1/(1+(-1)·(az)) )+(-1)·(1/(ax))^{n} )

g(z,x) = ln(1+(-1)·(az))+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

f(z,x) = g(z,x) <==> z = (1/(2a))

Deducción:

ln(az) = ln(1+(-1)·(az))

az = 1+(-1)·(az)

2az = 1

z = (1/(2a))

Ley:

d_{z}[f(z,y)]+d_{y}[f(z,y)] = a·( (1/2)·(1/(az))+(1/(ay))^{n} )

f(z,y) = (1/2)·ln(az)+( ay /o(ay)o/ (1/(n+1))·(ay)^{n+1} )

d_{z}[g(z,y)]+d_{y}[g(z,y)] = a·( (-1)·( 1/((3/4)+(-1)·(az)) )+(1/(ay))^{n} )

g(z,y) = ln((3/4)+(-1)·(az))+( ay /o(ay)o/ (1/(n+1))·(ay)^{n+1} )

f(z,y) = g(z,y) <==> ( z = (1/(4a)) con raíz positiva  || z = (9/(4a)) con raíz negativa )

Deducción:

(1/2)·ln(az) = ln((3/4)+(-1)·(az))

(az)^{(1/2)} = (3/4)+(-1)·(az)

az = (9/16)+(-1)·(3/2)·az+(az)^{2}

0 = (9/16)+(-1)·(5/2)·az+(az)^{2}

az = (1/2)·( (5/2)+(-1)·( (25/4)+(-1)·(9/4) )^{(1/2)} ) = (1/2)·( (5/2)+(-2) ) = (1/4)

z = (1/(4a))

az = (1/2)·( (5/2)+( (25/4)+(-1)·(9/4) )^{(1/2)} ) = (1/2)·( (5/2)+2 ) = (9/4)

z = (9/(4a))

Principio: [ de refracción de la lente ]

sin(arw)+(-1)·cos(ars) = sw·(k+(-j))

(-1)·sin(arw)+cos(ars) = sw·(j+(-k))

w = (pi/2) <==> s = 0

w = 0 <==> s = (pi/2)

Ley:

Si k = j ==> sin(arw) = cos(ars)

Si j = k ==> cos(ars) = sin(arw)

Deducción:

sin(arw)+(-1)·cos(ars) = sw·(k+(-j)) = sw·(k+(-k)) = sw·0

sin(arw) = sin(arw)+0 = sin(arw)+( (-1)·cos(ars)+cos(ars) ) = ( sin(arw)+(-1)·cos(ars) )+cos(ars) = ...

... sw·0+cos(ars) = cos(ars)

Ley:

d_{rw}[f(w,x)]+d_{x}[f(w,x)] = a·( cos(arw)+(-1)·(1/(ax))^{n} )

f(w,x) = sin(arw)+(-1)·( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

d_{rs}[g(s,x)]+d_{x}[g(s,x)] = a·( sin(ars)+(1/(ax))^{n} )

g(s,x) = (-1)·cos(ars)+( ax /o(ax)o/ (1/(n+1))·(ax)^{n+1} )

f(w,x)+g(s,x) = sw·(k+(-j))

Ley:

d_{rw}[f(w,y)]+d_{y}[f(w,y)] = a·( (-1)·cos(arw)+(1/(ay))^{n} )

f(w,y) = (-1)·sin(arw)+( ay /o(ay)o/ (1/(n+1))·(ay)^{n+1} )

d_{rs}[g(s,y)]+d_{y}[g(s,y)] = a·( (-1)·sin(ars)+(-1)·(1/(ay))^{n} )

g(s,y) = cos(ars)+(-1)·( ay /o(ay)o/ (1/(n+1))·(ay)^{n+1} )

f(w,y)+g(s,y) = sw·(j+(-k))

Óptica de miopía de imagen:

Sea n la dioptría ==>

d_{x}[q(x)] = (-1)·q(x)·cos(ax)·a·(ax)^{n}

d_{x}[p(x)] = p(x)·sin(ax)·a·(ax)^{n}

Operación Láser de longitud de onda x = rojo 

f(x) = e^{( ( 1/(n+1) )·(ax)^{n+1}+ax ) [o(ax)o] sin(ax) }

g(x) = e^{( ( 1/(n+1) )·(ax)^{n+1}+ax ) [o(ax)o] cos(ax) }

Óptica de hipermetropía de imagen:

Sea n la dioptría ==>

d_{y}[p(y)] = p(y)·cos(ay)·a·(ay)^{n}

d_{y}[q(y)] = (-1)·q(y)·sin(ay)·a·(ay)^{n}

Operación Láser de longitud de onda y = verde 

f(y) = e^{( (-1)·( 1/(n+1) )·(ay)^{n+1}+ay ) [o(ay)o] sin(ay) }

g(y) = e^{( (-1)·( 1/(n+1) )·(ay)^{n+1}+ay ) [o(ay)o] cos(ay) }

Óptica de miopía de sonido:

Sea n la dioptría ==>

d_{x}[q(x)] = (-1)·q(x)·cosh(ax)·a·(ax)^{n}

d_{x}[p(x)] = (-i)·p(x)·sinh(ax)·a·(ax)^{n}

Operación Láser de longitud de onda x = rojo 

f(x) = e^{( ( 1/(n+1) )·(ax)^{n+1}+ax ) [o(ax)o] sinh(ax) }

g(x) = e^{( ( 1/(n+1) )·(ax)^{n+1}+ax ) [o(ax)o] i·cosh(ax) }

Óptica de hipermetropía de sonido:

Sea n la dioptría ==>

d_{y}[p(y)] = p(y)·cosh(ay)·a·(ay)^{n}

d_{y}[q(y)] = i·q(y)·sinh(ay)·a·(ay)^{n}

Operación Láser de longitud de onda y = verde 

f(y) = e^{( (-1)·( 1/(n+1) )·(ay)^{n+1}+ay ) [o(ay)o] sinh(ay) }

g(y) = e^{( (-1)·( 1/(n+1) )·(ay)^{n+1+ay ) [o(ay)o] i·cosh(ay) }

Catarata de miopía de imagen:

d_{x}[q(x)] = (-1)·q(x)·cos(ax)·a·(ax)^{10}

d_{x}[p(x)] = p(x)·sin(ax)·a·(ax)^{10}

Operación Láser de longitud de onda x = rojo

f(x) = e^{( (1/11)·(ax)^{11}+ax ) [o(ax)o] sin(ax) }

g(x) = e^{( (1/11)·(ax)^{11}+ax ) [o(ax)o] cos(ax) }

Catarata de hipermetropía de imagen ( ceguera ):

d_{y}[p(y)] = p(y)·cos(ay)·a·(ay)^{10}

d_{y}[q(y)] = (-1)·q(y)·sin(ay)·a·(ay)^{10}

Operación Láser de longitud de onda y = verde 

f(y) = e^{( (-1)·(1/11)·(ay)^{11}+ay ) [o(ay)o] sin(ay) }

g(y) = e^{( (-1)·(1/11)·(ay)^{11}+ay ) [o(ay)o] cos(ay) }

Ley: [ de Grado en Medicina Teoría Homologada ]

Matemáticas 1: Cálculo diferencial.

Química.

Matemáticas 2: Cálculo integral.

Física: Termodinámica y Cabal sanguíneo.

Espectroscopia de fluido corporal.

Teoría genética de infecciones víricas.

Teoría genética de infecciones bacteria-lógicas.

Quimioterapia de desintegración genética.

Óptica de imagen y sonido.

Psico-neurología de negación de voces esquizofrénicas.

Psico-neurología de doble mandamiento dual.

Neurología de resonancia eléctrica.

Neurología de anti-resonancia eléctrica.

Ley:

Un familiar de un matemático o físico tiene convalidada la teoría de medicina,

porque tiene ya la energía para esas o aquellas medicaciones que se derivan de la teoría,

y solo le faltan las asignaturas de practica de atención y cirugía.

Termodinámica de Medicina:

Fiebre y Termómetro:

Ley:

PV = kT

d_{P}[T(P,V)]·p = qR <==> p = ?

d_{V}[T(P,V)]·v = qR <==> v = ?

Ley:

d_{V}[P_{0}]·V^{2}+d_{P}[V_{0}]·P^{2} = kT

d_{P}[T(P,V)]·p = qR <==> p = ?

d_{V}[T(P,V)]·v = qR <==> v = ?

Ley:

d_{V}[P_{0}]·V^{2}+d_{P}[V_{0}]·P^{2} = kT

d_{PP}^{2}[T(P,V)]·p^{2} = qR <==> p = ?

d_{VV}^{2}[T(P,V)]·v^{2} = qR <==> v = ?

Ley:

PV = d_{T}[k]·T^{2}

d_{P}[T(P,V)]·p = qR <==> p = ?

d_{V}[T(P,V)]·v = qR <==> v = ?

Deducción:

d_{P}[T(P,V)] = d_{P}[ ( ( 1/d_{T}[k] )·PV )^{(1/2)} ] = ...

... (1/2)·( ( 1/d_{T}[k] )·PV )^{(-1)·(1/2)}·( V/d_{T}[k] )

física-en-ingeniería y mecánica-de-fluidos y cálculo-integral-geometría y mecánica-en-física

Preliminares:

Principio:

[Ev][ d_{t}[x] = v ]

Ley:

x(t) = vt+h

Deducción:

x(t) = int[ d_{t}[x] ]d[t] = int[ v ]d[t] = v·int[ d[t] ] = vt+h

Principio:

[Eg][ d_{tt}^{2}[x] = g ]

Ley:

Si g = 0 ==> [Ev][ d_{t}[x] = v ]

Ley:

d_{t}[x] = gt+v

x(t) = g·(1/2)·t^{2}+vt+h

Deducción:

d_{t}[x] = int[ d_{tt}^{2}[x] ]d[t] = int[ g ]d[t] = g·int[ d[t] ] = gt+v

x(t) = int[ d_{t}[x] ]d[t] = int[ gt+v ]d[t]  = int[ gt ]d[t]+int[ v ]d[t] = g·int[ t ]d[t]+v·int[ d[t] ] = ...

... g·(1/2)·t^{2}+vt+h

Ley:

Sea x(t) = ( r^{n}+(vt)^{n} )^{(1/n)} ==>

d_{t}[x] = ( r^{n}+(vt)^{n} )^{(1/n)+(-1)}·(vt)^{n+(-1)}·v

Deducción:

d_{t}[ ( r^{n}+(vt)^{n} )^{(1/n)} ] = ...

... d_{r^{n}+(vt)^{n}}[ ( r^{n}+(vt)^{n} )^{(1/n)} ]·...

... d_{vt}[ r^{n}+(vt)^{n} ]·d_{t}[vt] = ...

... d_{r^{n}+(vt)^{n}}[ ( r^{n}+(vt)^{n} )^{(1/n)} ]·...

... ( d_{vt}[ r^{n} ]+d_{vt}[ (vt)^{n} ] )·d_{t}[vt] = ...

... d_{r^{n}+(vt)^{n}}[ ( r^{n}+(vt)^{n} )^{(1/n)} ]·...

... ( 0+d_{vt}[ (vt)^{n} ] )·d_{t}[vt] = ...

... d_{r^{n}+(vt)^{n}}[ ( r^{n}+(vt)^{n} )^{(1/n)} ]·d_{vt}[ (vt)^{n} ]·d_{t}[vt]

Ley:

Sea d_{t}[x] = ( c^{(1/n)}+(gt)^{(1/n)} )^{n} ==>

x(t) = ( n/(n+1) )·( c^{(1/n)}+(gt)^{(1/n)} )^{n+1} [o(gt)o] ...

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

Deducción:

g·d_{gt}[ ( ( n/(n+1) )·( c^{(1/n)}+(gt)^{(1/n)} )^{n+1} [o(gt)o] ...

... ( 1/(2+(-1)·(1/n)) )·(gt)^{2+(-1)·(1/n)} ) [o(gt)o] t ] = ...

g·d_{gt}[ ( ( n/(n+1) )·( c^{(1/n)}+(gt)^{(1/n)} )^{n+1} ]·...

... d_{gt}[ ( 1/(2+(-1)·(1/n)) )·(gt)^{2+(-1)·(1/n)} ]·d_{gt}[t] ] = ...

... g·( c^{(1/n)}+(gt)^{(1/n)} )^{n}·d_{gt}[t] = ( c^{(1/n)}+(gt)^{(1/n)} )^{n}

Ley:

Sea y(t,x) = ax^{2} ==>

Si d_{t}[x] = v ==>

d_{tt}^{2}[y(t,x)] = 2av^{2}

d_{t}[y(t,x)] = 2av^{2}·t

y(t,x) = a·(vt)^{2}

Deducción:

d[y(t,x)] = 2ax·d[x]

d_{t}[y(t,x)] = 2ax·d_{t}[x]

d_{tt}^{2}[y(t,x)] = 2a·( d_{t}[x]^{2}+x·d_{tt}^{2}[x] )

x(t) = int[ d_{t}[x] ]d[t] = int[ v ]d[t] = v·int[ d[t] ] = vt 

Ley:

Sea y(t,x) = ax^{2} ==>

Si d_{tt}^{2}[x] = g ==>

d_{tt}^{2}[y(t,x)] = 6ag^{2}·(1/2)·t^{2}

d_{t}[y(t,x)] = 2ag·(1/2)·t^{2}·gt

y(t,x) = a·( g^{2}·(1/4)·t^{4} )

Deducción:

d[y(t,x)] = 2ax·d[x]

d_{t}[y(t,x)] = 2ax·d_{t}[x]

d_{tt}^{2}[y(t,x)] = 2a·( d_{t}[x]^{2}+x·d_{tt}^{2}[x] )

d_{t}[x] = int[ d_{tt}^{2}[x] ]d[t] = int[ g ]d[t] = g·int[ d[t] ] = gt

x(t) = int[ d_{t}[x] ]d[t] = int[ gt ]d[t] = g·int[ t ]d[t] = g·(1/2)·t^{2}

t = (2x/g)^{(1/2)}

Ley:

Sea 2·( sin(w) )^{2} = ( 1+(-1)·cos(2w) ) ==>

Sea ( d_{t}[x] = v·sin(ut) & d_{t}[y] = v·( 1+(-1)·cos(ut) ) ==>

d_{t}[r] = 2v·sin((ut)/2)

r(0,2pi) = 8v·(1/u)

Ley:

Sea 2·( cos(w) )^{2} = ( 1+cos(2w) ) ==>

Sea ( d_{t}[x] = v·sin(ut) & d_{t}[y] = v·( 1+cos(ut) ) ==>

d_{t}[r] = 2v·cos((ut)/2)

r((-pi),pi) = 8v·(1/u)

Globos y Drones:

Ley:

Sea d_{t}[x] = uy ==>

Si d_{t}[y] = v ==>

x(y) = u·(1/v)·(1/2)·y^{2}

d_{tt}^{2}[x] = uv

Ley:

Sea d_{t}[x] = uy ==>

Si d_{t}[y] = gt ==>

x(y) = ug·(1/6)·( (2/g)·y )^{(3/2)}

d_{tt}^{2}[x] = ugt

---------------------------------------------------

Procedimiento en coordenadas cartesianas:

---------------------------------------------------

Principio: [ de Fuerza ]

[EF_{k}][ m·d_{tt}^{2}[x] = sum[k = 1]-[n][ F_{k} ] ]

Ley: [ de Impulsión ]

[Ep_{k}][ m·d_{t}[x] = sum[k = 1]-[n][ p_{k} ] ]

Deducción:

int[ m·d_{tt}^{2}[x] ]d[t] = m·int[ d_{tt}^{2}[x] ]d[t] = m·d_{t}[x]

int[ sum[k = 1]-[n][ F_{k} ] ]d[t] = sum[k = 1]-[n][ int[ F_{k} ]d[t] ] = sum[k = 1]-[n][ p_{k} ]

Ley: [ de Energía ]

[EU_{k}][ (m/2)·d_{t}[x]^{2} = sum[k = 1]-[n][ U_{k} ] ]

Deducción:

int[ m·d_{tt}^{2}[x] ]d[x] = int[ m·d_{tt}^{2}[x]·d_{t}[x] ]d[t] = (m/2)·d_{t}[x]^{2}

int[ sum[k = 1]-[n][ F_{k} ] ]d[x] = sum[k = 1]-[n][ int[ F_{k} ]d[x] ] = sum[k = 1]-[n][ U_{k} ]

Ley: [ de Potencia ]

[EN_{k}][ (m/u)·d_{t}[x]^{[o(ut)o] 2} = sum[k = 1]-[n][ N_{k} ] ]

Deducción:

int[ m·d_{tt}^{2}[x] ]d[ d_{t}[x] ] = int[ (m/u)·d_{tt}^{2}[x]^{2} ]d[ut] = (m/u)·d_{t}[x]^{[o(ut)o] 2}

int[ sum[k = 1]-[n][ F_{k} ] ]d[ d_{t}[x] ] = sum[k = 1]-[n][ int[ F_{k} ]d[ d_{t}[x] ] ] = ...

... sum[k = 1]-[n][ N_{k} ]

Fuerza constante:

Ley:

Sea F(t) = F ==>

d_{t}[x] = (F/m)·t

x(t) = (F/m)·(1/2)·t^{2}

Deducción:

d_{t}[x] = int[ d_{tt}^{2}[x] ]d[t] = int[ (F/m) ]d[t] = (F/m)·int[ d[t] ] = (F/m)·t

x(t) = int[ d_{t}[x] ]d[t] = int[ (F/m)·t ]d[t] = (F/m)·int[ t ]d[t] = (F/m)·(1/2)·t^{2}

Ley:

Sea F(t) = F ==>

U(x) = Fx

(m/2)·d_{t}[x]^{2} = U(x)

Deducción:

U(x) = int[ F ]d[x] = F·int[ d[x] ] = Fx

(m/2)·d_{t}[x]^{2} = (m/2)·( (F/m)·t )^{2} = F·( (F/m)·(1/2)·t^{2} ) = Fx = U(x)

Ley:

Sea F(t) = F ==>

N(d_{t}[x]) = F·d_{t}[x]

(m/u)·d_{t}[x]^{[o(ut)o] 2} = N(d_{t}[x])

Deducción:

N(d_{t}[x]) = int[ F ]d[ d_{t}[x] ] = F·int[ d[ d_{t}[x] ] ] = F·d_{t}[x]

(m/u)·d_{t}[x]^{[o(ut)o] 2} = (m/u)·( (F/m)·t )^{[o(ut)o] 2} = int[ (m/u)·(F/m)^{2} ]d[ut] = ...

... F·( (F/m)·t ) = F·d_{t}[x] = N(d_{t}[x])

Fuerza lineal de carga variable:

Ley:

Sea F(t) = Itg ==>

d_{t}[y] = (1/m)·Ig·(1/2)·t^{2}

y(t) = (1/m)·Ig·(1/6)·t^{3}

Deducción:

d_{t}[y] = int[ d_{tt}^{2}[y] ]d[t] = int[ (1/m)·Itg ]d[t] = (1/m)·Ig·int[ t ]dt = (1/m)·Ig·(1/2)·t^{2}

y(t) = int[ d_{t}[y] ]d[t] = int[ (1/m)·Ig·(1/2)·t^{2} ]d[t] = (1/m)·Ig·(1/2)·int[ t^{2} ]d[t] = ...

... (1/m)·Ig·(1/6)·t^{3}

Ley:

Sea F(t) = Itg ==>

U(y) = (1/m)·(Ig)^{2}·(1/8)·t^{4}

(m/2)·d_{t}[y]^{2} = U(y)

Deducción:

U(y) = int[ Itg ]d[y] = int[ Itg d_{t}[y] ]d[t] = int[ (1/m)·( Ig )^{2}·(1/2)·t^{3}· ]d[t] = ...

... (1/m)·( Ig )^{2}·(1/8)·t^{4}

(m/2)·d_{t}[y]^{2} = (m/2)·( ( (1/m)·Ig )·(1/2)·t^{2} )^{2} = (1/m)·( Ig )^{2}·(1/8)·t^{4} = U(y)

Ley:

Sea F(t) = Itg ==>

N(d_{t}[y]) = (1/m)·( Ig )^{2}·(1/3)·t^{3}

(m/u)·d_{t}[y]^{[o(ut)o] 2} = N(d_{t}[y])

Deducción:

N(d_{t}[y]) = int[ Itg ]d[ d_{t}[y] ] = int[ (1/m)·Itg d_{tt}^{2}[y] ]d[t] = ...

... int[ (1/m)·( Itg )^{2} ]d[t] = (1/m)·( Ig )^{2}·(1/3)·t^{3}

(m/u)·d_{t}[y]^{[o(ut)o] 2} = (m/u)·( ( (1/m)·Ig )·(1/2)·t^{2} )^{[o(ut)o] 2} = ...

... int[ (1/m)·( Ig )^{2}·t^{2} ]d[t] = (1/m)·( Ig )^{2}·(1/3)·t^{3} = N(d_{t}[y])

Fuerzas de amortiguación y de resistencia de fluido:

Horizontal:

Ley:

m·d_{tt}^{2}[x] = (-k)·x

x(t) = re^{(k/m)^{(1/2)}·it}

Deducción:

m·d_{tt}^{2}[x] = m·d_{t}[ d_{t}[x] ] = m·d_{t}[ d_{t}[ re^{(k/m)^{(1/2)}·it} ] ] = ...

... m·d_{t}[ r·d_{t}[ e^{(k/m)^{(1/2)}·it} ] ] = m·d_{t}[ re^{(k/m)^{(1/2)}·it}·(k/m)^{(1/2)}·i ] = ...

... mr·(k/m)^{(1/2)}·i·d_{t}[ e^{(k/m)^{(1/2)}·it} ] = ...

... mr·( (k/m)^{(1/2)}·i )^{2} e^{(k/m)^{(1/2)}·it} = ...

... (-k)·re^{(k/m)^{(1/2)}·it} = (-k)·x

Ley

m·d_{tt}^{2}[x] = (-b)·d_{t}[x]

d_{t}[x] = ve^{(-1)·(b/m)·t}

Deducción:

m·d_{tt}^{2}[x] = m·d_{t}[ d_{t}[x] ] = m·d_{t}[ ve^{(-1)·(b/m)·t} ] = ...

... mv·d_{t}[ e^{(-1)·(b/m)·t} ] = mv·e^{(-1)·(b/m)·t}·(-1)·(b/m) = ...

... (-b)·ve^{(-1)·(b/m)·t} = (-b)·d_{t}[x]

Ley:

U(x) = (-k)·(1/2)·x^{2}

(m/2)·d_{t}[x]^{2} = U(x)

Deducción:

U(x) = int[ (-k)·x ]d[x] = (-k)·int[ x ]d[x] = (-k)·(1/2)·x^{2}

(m/2)·d_{t}[x]^{2} = (m/2)·( i·(k/m)^{(1/2)}·re^{(k/m)^{(1/2)}·it} )^{2} = ...

... (-k)·(1/2)·( r^{2}·e^{(k/m)^{(1/2)}·2it} ) = (-k)·(1/2)·x^{2} = U(x)

Ley:

N(d_{t}[x]) = (-b)·(1/2)·d_{t}[x]^{2}

(m/u)·d_{t}[x]^{[o(ut)o] 2} = N(d_{t}[x])

Deducción:

N(d_{t}[x]) = int[ (-b)·d_{t}[x] ]d[ d_{t}[x] ] = ...

... (-b)·int[ d_{t}[x] ]d[ d_{t}[x] ] = (-b)·(1/2)·d_{t}[x]^{2}

(m/u)·d_{t}[x]^{[o(ut)o] 2} = (m/u)·( ve^{(-1)·(b/m)·t} )^{[o(ut)o] 2} = ...

... (m/u)·int[ ( (-1)·(b/m)·v )^{2}·e^{(-2)·(b/m)·t} ]d[ut] = ...

... (-b)·(1/2)·( v^{2}·e^{(-2)·(b/m)·t} ) = (-b)·(1/2)·d_{t}[x]^{2} = N(d_{t}[x])

Fuerzas de amortiguación y de resistencia de fluido:

Vertical:

Ley:

m·d_{tt}^{2}[y] = (-k)·y+qg

y(t) = re^{(k/m)^{(1/2)}·it}+(1/k)·qg

Deducción:

m·d_{tt}^{2}[y] = m·d_{t}[ d_{t}[y] ] = m·d_{t}[ d_{t}[ re^{(k/m)^{(1/2)}·it}+(1/k)·qg ] ] = ...

... m·d_{t}[ d_{t}[ re^{(k/m)^{(1/2)}·it} ]+d_{t}[ (1/k)·qg ] ] = ...

... m·d_{t}[ d_{t}[ re^{(k/m)^{(1/2)}·it} ]+0 ] = m·d_{t}[ d_{t}[ re^{(k/m)^{(1/2)}·it} ] = ...

... m·d_{tt}^{2}[ re^{(k/m)^{(1/2)}·it} ] = (-k)·re^{(k/m)^{(1/2)}·it} = ...

... (-k)·re^{(k/m)^{(1/2)}·it}+(-1)·qg+qg = (-k)·( re^{(k/m)^{(1/2)}·it}+(1/k)·qg )+qg = (-k)·y+qg

Ley:

m·d_{tt}^{2}[y] = (-b)·d_{t}[y]+qg

d_{t}[y] = ve^{(-1)·(b/m)·t}+(1/b)·qg

Deducción:

m·d_{tt}^{2}[y] = m·d_{t}[ d_{t}[y] ] = m·d_{t}[ ve^{(-1)·(b/m)·t}+(1/b)·qg ] = ...

... m·( d_{t}[ ve^{(-1)·(b/m)·t} ]+d_{t}[ (1/b)·qg ] ) = ...

... m·( d_{t}[ ve^{(-1)·(b/m)·t} ]+0 ) = m·d_{t}[ ve^{(-1)·(b/m)·t} ] = (-b)·ve^{(-1)·(b/m)·t} = ...

... (-b)·ve^{(-1)·(b/m)·t}+(-1)·qg+qg = (-b)·( ve^{(-1)·(b/m)·t}+(1/b)·qg )+qg = (-b)·d_{t}[y]+qg

Obertura de hombros y caderas robótica:

Por amortiguador de retorno con empuje de obertura de fluido.

Ley:

m·d_{tt}^{2}[y] = (-k)·y+( sin(w)·qg+sin(s)·pg )

y(t) = re^{(k/m)^{(1/2)}·it}+(1/k)·( sin(w)·qg+sin(s)·pg )

Ley:

m·d_{tt}^{2}[y] = (-b)·d_{t}[y]+( sin(w)·qg+sin(s)·pg )

d_{t}[y] = ve^{(-1)·(b/m)·t}+(1/b)·( sin(w)·qg+sin(s)·pg )

Estiramiento de rodillas y codos robótica:

Por amortiguador de retorno con empuje de obertura de fluido.

Ley:

m·d_{tt}^{2}[y] = (-k)·y+( F+(-1)·qg )

y(t) = re^{(k/m)^{(1/2)}·it}+(1/k)·( F+(-1)·qg )

Ley:

m·d_{tt}^{2}[y] = (-b)·d_{t}[y]+( F+(-1)·qg )

d_{t}[y] = ve^{(-1)·(b/m)·t}+(1/b)·( F+(-1)·qg )

Motores robóticos que aumentan la fuerza según la carga:

Ley:

m·d_{tt}^{2}[y] = (-k)·y+( Itg+qg )

y(t) = re^{(k/m)^{(1/2)}·it}+(1/k)·( Itg+qg )

Ley:

m·d_{tt}^{2}[y] = (-b)·d_{t}[y]+( Itg+qg )

d_{t}[y] = ve^{(-1)·(b/m)·t}+(1/b)·( Itg+qg )+(-1)·(1/b)^{2}·Ig

----------------------------------------------

Procedimiento en coordenadas polares:

----------------------------------------------

Principio: [ de Inercia angular ]

[EI_{ck}][ mdx = sum[k = 1]-[n][ I_{ck} ] ]

[ d ] = ( metro / ( radian )^{2} )

Ley: [ de Impulsión angular ]

[EL_{k}][ md·d_{t}[x] = sum[k = 1]-[n][ L_{k} ] ]

Deducción:

d_{t}[ mdx ] = md·d_{t}[x] = md·d_{t}[x]

d_{t}[ sum[k = 1]-[n][ I_{ck} ] ] = sum[k = 1]-[n][ d_{t}[ I_{ck} ] ] = ...

... sum[k = 1]-[n][ L_{k} ]

Ley: [ de Fuerza angular ]

[EH_{k}][ md·d_{tt}^{2}[x] = sum[k = 1]-[n][ H_{k} ] ]

Deducción:

d_{t}[ md·d_{t}[x] ] = md·d_{t}[ d_{t}[x] ] = md·d_{tt}^{2}[x]

d_{t}[ sum[k = 1]-[n][ L_{k} ] ] = sum[k = 1]-[n][ d_{t}[ L_{k} ] ] = ...

... sum[k = 1]-[n][ H_{k} ]

Principio: [ de Energía angular ]

[EU_{k}][ I_{c}·(1/2)·d_{t}[w]^{2} = sum[k = 1]-[n][ U_{k} ] ]

Ley:

Sea U(w) = U ==>

Si I_{c} = M·(r/s)^{2} ==> 

x(t) = (M/m)·(1/d)·(r/s)^{2}

w(t) = ( (2/M)·U )^{(1/2)}·(s/r)·t

Deducción:

mdx = M·(r/s)^{2}

I_{c}·(1/2)·d_{t}[w]^{2} = M·(r/s)^{2}·(1/2)·d_{t}[ ( (2/M)·U )^{(1/2)}·(s/r)·t ]^{2} = ...

... M·(r/s)^{2}·(1/2)·( ( (2/M)·U )^{(1/2)}·(s/r)·d_{t}[ t ] )^{2} = ...

... M·(r/s)^{2}·(1/2)·( ( (2/M)·U )^{(1/2)}·(s/r) )^{2} = U

Principio: [ Fundamental de la dinámica angular ]

[EM_{k}][ d_{w}[ U(w) ] = sum[k = 1]-[n][ M_{k} ] ]

[ M_{k} ] = ( Joule / Radian )

Ley:

L(t)·(1/2)·d_{t}[w]+I_{c}·d_{tt}^{2}[w] = sum[k = 1]-[n][ M_{k} ]

Deducción:

d_{w}[ I_{c}·(1/2)·d_{t}[w]^{2} ] = (1/d_{t}[w])·d_{t}[ I_{c}·(1/2)·d_{t}[w]^{2} ] ...

... (1/d_{t}[w])·( d_{t}[ I_{c} ]·(1/2)·d_{t}[w]^{2}+I_{c}·d_{t}[ (1/2)·d_{t}[w]^{2} ] ) = ...

... L(t)·(1/2)·d_{t}[w]+I_{c}·d_{tt}^{2}[w]

Ley: [ de Momento de Fuerza ]

Si d_{t}[ I_{c} ] = 0 ==> 

[EM_{k}][ I_{c}·d_{tt}^{2}[w] = sum[k = 1]-[n][ M_{k} ] ]

Ley: [ de Momento de Impulsión ]

Si d_{t}[ I_{c} ] = 0 ==>

[EK_{k}][ I_{c}·d_{t}[w] = sum[k = 1]-[n][ K_{k} ] ]

Deducción:

int[ I_{c}·d_{tt}^{2}[w] ]d[t] = I_{c}·int[ d_{tt}^{2}[x] ]d[t] = I_{c}·d_{t}[w]

int[ sum[k = 1]-[n][ M_{k} ] ]d[t] = sum[k = 1]-[n][ int[ M_{k} ]d[t] ] = ...

... sum[k = 1]-[n][ K_{k} ]

Ley: [ de Potencia angular ]

Si d_{t}[ I_{c} ] = 0 ==>

[EN_{k}][ (I_{c}/u)·d_{t}[w]^{[o(ut)o] 2} = sum[k = 1]-[n][ N_{k} ] ]

Deducción:

int[ I_{c}·d_{tt}^{2}[w] ]d[ d_{t}[w] ] = int[ (I_{c}/u)·d_{tt}^{2}[w]^{2} ]d[ut] = ...

.. (I_{c}/u)·d_{t}[w]^{[o(ut)o] 2}

int[ sum[k = 1]-[n][ M_{k} ] ]d[ d_{t}[w] ] = sum[k = 1]-[n][ int[ M_{k} ]d[ d_{t}[w] ] ] = ...

... sum[k = 1]-[n][ N_{k} ]

Ley:

Si ( d_{t}[ I_{c} ] = 0 & M(w) = F·(x/s) ) ==> 

d_{t}[w] = (1/I_{c})·F·(x/s)·t

w(t) = (1/I_{c})·F·(x/s)·(1/2)·t^{2}

Deducción:

Problema.

Ley:

Si ( d_{t}[ I_{c} ] = 0 & M(w) = F·(x/s) ) ==> 

U(w) = F·(x/s)·w

I_{c}·d_{t}[w]^{2} = U(w)

Deducción:

U(w) = int[ M(w) ]d[w] = int[ F·(x/s) ]d[w] = F·(x/s)·int[ d[w] ] = F·(x/s)·w

(I_{c}/2)·d_{t}[w]^{2} = (I_{c}/2)·( (1/I_{c})·F·(x/s)·t )^{2} = ...

... F·(x/s)·( (1/I_{c})·F·(x/s)·(1/2)·t^{2} ) = F·(x/s)·w = U(w)

Ley:

Si ( d_{t}[ I_{c} ] = 0 & M(w) = F·(x/s) ) ==> 

N(d_{t}[w]) = F·(x/s)·d_{t}[w]

(I_{c}/u)·d_{t}[w]^{[o(ut)o] 2} = N(d_{t}[w])

Deducción:

N(d_{t}[w]) = int[ M(w) ]d[ d_{t}[w] ] = int[ F·(x/s)·d_{tt}^{2}[w] ]d[t] = ...

... int[ (1/I_{c})·( F·(x/s) )^{2} ]d[t] = (1/I_{c})·( F·(x/s) )^{2}·int[ d[t] ] = ...

... (1/I_{c})·( F·(x/s) )^{2}·t = F·(x/s)·d_{t}[w]

(I_{c}/u)·d_{t}[w]^{[o(u)o] 2} = (I_{c}/u)·( (1/I_{c})·F·(x/s)·t )^{[o(ut)o] 2} = ...

... (I_{c}/u)·int[ ( (1/I_{c})·F·(x/s) )^{2} ]d[ut] = ...

...(1/I_{c})·( F·(x/s) )^{2}·int[ d[t] ] = (1/I_{c})·( F·(x/s) )^{2}·t = ...

... F·(x/s)·d_{t}[w] = N(d_{t}[w])

Inercias angulares constantes:

Principio:

[Ef][ I_{c} = int[ ( (r·f(n))/s )^{2}·d_{n}[m(n)] ]d[n] ]

Ley:

Si f(n) = n^{0} ==> I_{c} = int[ (r/s)^{2}·d_{n}[m(n)] ]d[n]

Si f(n) = (n/r) ==> I_{c} = int[ (n/s)^{2}·d_{n}[m(n)] ]d[n]

Principio:

[Ef][Eg][ I_{c} = int-int[ (1/2)·( ( (r·f(p))/s )^{2}+( (r·g(q))/s )^{2} )·d_{pq}^{2}[m(p,q)] ]d[p]d[q] ]

Ley:

Si ( f(p) = p^{0} & g(q) = q^{0} ) ==> I_{c} = int-int[ (r/s)^{2}·d_{pq}^{2}[m(p,q)] ]d[p]d[q]

Si ( f(p) = (p/r) & g(q) = (q/r) ) ==> ...

... I_{c} = int-int[ (1/2)·( (p/s)^{2}+(q/s)^{2} )·d_{pq}^{2}[m(p,q)] ]d[p]d[q]

Ley:

Sea ( U(w) = U & F(x) = int[ f(x) ]d[x] ) ==>

Si I_{c} = int[ (r/s)^{2}·Ma·f(an) ]d[n] ==> 

I_{c} = (r/s)^{2}·M·F(an)

x(t) = (1/(md))·(r/s)^{2}·M·F(an)

w(t) = ( 2U )^{(1/2)}·(s/r)·( 1/(M·F(an)) )^{(1/2)}·t

K(t) = ( 2U )^{(1/2)}·(r/s)·( M·F(an) )^{(1/2)}

Ley:

Sea U(w) = U ==>

Si I_{c} = int[ (n/s)^{2}·Ma ]d[n] ==> 

I_{c} = (1/3)·(n/s)^{2}·Man

x(t) = (1/(md))·(1/3)·(n/s)^{2}·Man

w(t) = ( 6U )^{(1/2)}·(s/n)·( 1/(Man) )^{(1/2)}·t

K(t) = ( 6U )^{(1/2)}·(1/3)·(n/s)·( Man )^{(1/2)}

Ley:

Sea ( U(w) = U & F(x) = int[ f(x) ]d[x] & G(x) = int[ g(x) ]d[x] ) ==>

Si I_{c} = int-int[ (r/s)^{2}·Ma^{2}·f(ap)·g(aq) ]d[p]d[q] ==> 

I_{c} = (r/s)^{2}·M·F(ap)·G(aq)

x(t) = (1/(md))·(r/s)^{2}·M·F(ap)·G(aq)

w(t) = ( 2U )^{(1/2)}·(s/r)·( 1/(M·F(ap)·G(aq)) )^{(1/2)}·t

K(t) = ( 2U )^{(1/2)}·(r/s)·( M·F(an)·G(aq) )^{(1/2)}

Deducción:

I_{c} = int-int[ (r/s)^{2}·M·f(ap)·g(aq) ]d[ap]d[aq] = (r/s)^{2}·M·int-int[ f(ap)·g(aq) ]d[ap]d[aq] = ...

... (r/s)^{2}·M·int[ g(aq)·int[ f(ap) ]d[ap] ]d[aq] = (r/s)^{2}·M·int[ g(aq)·F(ap) ]d[aq] = ...

... (r/s)^{2}·M·F(ap)·int[ g(aq) ]d[aq] = (r/s)^{2}·M·F(ap)·G(aq)

Ley:

Sea ( U(w) = U & F(x) = int[ f(x) ]d[x] & G(x) = int[ g(x) ]d[x] ) ==>

Si I_{c} = int-int[ (r/s)^{2}·Ma^{2}·( f(ap)+g(aq) ) ]d[p]d[q] ==> 

I_{c} = (r/s)^{2}·M·( F(ap)·aq+ap·G(aq) )

x(t) = (1/(md))·(r/s)^{2}·M·( F(ap)·aq+ap·G(aq) )

w(t) = ( 2U )^{(1/2)}·(s/r)·( 1/(M·( F(ap)·aq+ap·G(aq) )) )^{(1/2)}·t

K(t) = ( 2U )^{(1/2)}·(r/s)·( M·( F(ap)·aq+ap·G(aq) ) )^{(1/2)}

Ley:

Sea ( U(w) = U & p = r·sin(nw) & q = r·cos(nw) ) ==>

Si I_{c} = int-int[ (1/2)·( (p/s)^{2}+(q/s)^{2} )·Ma^{2} ]d[p]d[q] ==> 

I_{c} = (r/s)^{2}·M·(1/4)·npi·(ar)^{2}

x(t) = (1/(md))·(r/s)^{2}·M·(1/4)·npi·(ar)^{2}

w(t) = ( (8/(npi))·U )^{(1/2)}·(s/r)·(1/(ar))·(1/M)^{(1/2)}·t

K(t) = ( (8/(npi))·U )^{(1/2)}·(r/s)·(ar)·M^{(1/2)}

Deducción:

d[p] = nr·cos(nw)·d[w] & d[q] = nr·(-1)·sin(nw)·d[w]

d[p]d[q] = (nr)^{2}·cos(nw)·(-1)·sin(nw)·d[w]d[w]

Ley:

Sea ( U(w) = U & p = (r/i)·sinh(nw) & q = r·cosh(nw) ) ==>

Si I_{c} = int-int[ (1/2)·( (p/s)^{2}+(q/s)^{2} )·Ma^{2} ]d[p]d[q] ==> 

I_{c} = (r/s)^{2}·M·(1/4)·npi·(ar)^{2}

Motores de rotación.

Ley:

Sea U(w) = U ==>

Si d[I_{c}] = (1/s)^{2}·Mrv·d[t] ==>

x(t) = (1/(md))·(1/s)^{2}·Mrvt

w(t) = ( (8/M)·(1/(rv))·U )^{(1/2)}·st^{(1/2)}

Ley:

Sea U(w) = U ==>

Si d[I_{c}] = (1/s)^{2}·Mrgt·d[t] ==>

x(t) = (1/(md))·(1/s)^{2}·Mrg·(1/2)·t^{2}

w(t) = ( (4/M)·(1/(rg))·U )^{(1/2)}·s·ln(ut)

Articulaciones robóticas y de vehículo.

Ley:

Sea U(w) = U ==>

Si d[L(t)] = (1/s)^{2}·Mv·d[x] ==>

x(t) = re^{(M/m)·(v/d)·(1/s)^{2}·t}

w(t) = ( (8/(mdr))·U )^{(1/2)}·(-1)·(m/M)·(d/v)·s^{2}·e^{(-1)·(1/2)·(M/m)·(v/d)·(1/s)^{2}·t}

Ley:

Sea U(w) = U ==>

Si d[L(t)] = (1/s)^{2}·Mg·d[tx] ==>

x(t) = re^{(M/m)·(g/d)·(1/s)^{2}·(1/2)·t^{2}}

w(t) = ( (8/(mdr))·U )^{(1/2)}·...

... ( (-1)·(m/M)·(d/g)·s^{2}·ln(ut) ) [o(t)o] e^{(-1)·(1/2)·(M/m)·(g/d)·(1/s)^{2}·(1/2)·t^{2}}

Ley:

Sea U(w) = U ==>

Si d[L(t)] = (1/s)^{2}·Mav·2x·d[x] ==>

x(t) = ( (-1)·(M/m)·(v/d)·(1/s)^{2}·at )^{(-1)}

w(t) = ( (2/(md))·U )^{(1/2)}·( (-1)·(M/m)·(v/d)·(1/s)^{2}·a )^{(1/2)}·(2/3)·t^{(3/2)}

Deducción:

d_{x}[ x^{2} ] = 2x

d_{x}[ x^{2} ]·d[x] = 2x·d[x]

d[ x^{2} ] = 2x·d[x]

L(t) = int[ d[L(t)] ] = int[ (1/s)^{2}·Mav·2x·d[x] ] = ...

... int[ (1/s)^{2}·Mav ]d[x^{2}] =  (1/s)^{2}·Mav·int[ d[x^{2}] ] =  (1/s)^{2}·Mavx^{2}

Ley:

Sea U(w) = U ==>

Si d[L(t)] = (1/s)^{2}·Mag·( d[t]·x^{2}+t·2x·d[x] ) ==>

x(t) = ( (-1)·(M/m)·(g/d)·(1/s)^{2}·a·(1/2)·t^{2} )^{(-1)}

w(t) = ( (2/(md))·U )^{(1/2)}·( (-1)·(M/m)·(g/d)·(1/s)^{2}·a·(1/2) )^{(1/2)}·(1/2)·t^{2}

Deducción:

d_{t}[ tx^{2} ] = d_{t}[t]·x^{2}+t·2x·d_{t}[x]

d_{t}[ tx^{2} ]·d[t] = ( d_{t}[t]·x^{2}+t·2x·d_{t}[x] )·d[t]

d[ tx^{2} ] = d[t]·x^{2}+t·2x·d[x]

L(t) = int[ d[L(t)] ] = int[ (1/s)^{2}·Mag·( d[t]·x^{2}+t·2x·d[x] ) ] = ...

... int[ (1/s)^{2}·Mag ]d[tx^{2}] =  (1/s)^{2}·Mag·int[ d[tx^{2}] ] =  (1/s)^{2}·Magtx^{2}

Ley: [ de rezo al Mal ]

Los hombres no están atacando,

a los xtraterrestres.

Los xtraterrestres no están atacando,

a los hombres.

Ley:

Se matan entre ellos en su mundo.

Cometen adulterio entre ellos en su mundo.

Principio:

U(x,y,z) = Potencial[ Q(x,y,z) ]

U(yz,zx,xy) = Anti-Potencial[ Q(yz,zx,xy) ]

Principio:

div-exp[ U(x,y,z) ] = sum[k = 1]-[3][ d_{xyz}^{3}[ e^{U_{k}(x,y,z)} ]

Si div-exp[ U(x,y,z) ] = 0 ==>

div-exp[ U(x,y,z) ] = d_{xyz}^{3}[ e^{sum[k = 1]-[3][ U_{k}(x,y,z) ]} ]

Principio:

Anti-div-exp[ U(yz,zx,xy) ] = sum[k = 1]-[3][ d_{kij}^{2}[ e^{U_{k}(yz,zx,xy)} ] ]

Si Anti-div-exp[ U(yz,zx,xy) ] = 0 ==>

Anti-div-exp[ U(yz,zx,xy) ] = d_{kij}^{2}[ e^{sum[k = 1]-[3][ U_{k}(yz,zx,xy) ]} ]

Ley:

Si Q(x,y,z) = U·< (1/x),(1/y),(1/z) > ==> 

U(x,y,z) = U·( ln(ax)+ln(ay)+ln(az) )

div-exp[ U(x,y,z) ] = Ua^{3}

F(z) = int-int[ div-exp[ U(x,y,z) ] ]d[x]d[x]+int-int[ div-exp[ U(x,y,z) ] ]d[y]d[y] = ...

... Ua^{3}·(1/2)·( x^{2}+y^{2} )

Ley:

Si Q(yz,zx,xy) = U·< (1/(yz)),(1/(zx)),(1/(xy)) > ==> 

U(yz,zx,xy) = U·( ln(byz)+ln(bzx)+ln(bxy) )

Anti-div-exp[ U(yz,zx,xy) ] = Ub^{3}·4xyz

F(z) = int-int[ Anti-div-exp[ U(yz,zx,xy) ] ]d[x]d[y]+int-int[ Anti-div-exp[ U(yz,zx,xy) ] ]d[y]d[x] = ...

... Ub^{3}·2z·(xy)^{2}

Ley:

Si Q(x,y,z) = aU·< ((y+z)/x),((z+x)/y),((x+y)/z) > ==> 

U(x,y,z) = U·( (ay+az)·ln(ax)+(az+ax)·ln(ay)+(ax+ay)·ln(az) )

div-exp[ U(x,y,z) ] = Ua^{3}·( ...

... (ax)^{ay+az+(-1)}·ln(ax)·( 2+(ay+az)·ln(ax) )+...

... (ay)^{az+ax+(-1)}·ln(ay)·( 2+(az+ax)·ln(ay) )+...

... (az)^{ax+ay+(-1)}·ln(az)·( 2+(ax+ay)·ln(az) ) )

Ley:

E(x_{k}) = int-int[ div-exp[ U_{k}(x,y,z) ] ]d[(1/a)^{2}·(i+j)] = ...

... U·(ak)^{ai+aj+(-1)}·[o(ai+aj)o] ( 2+(1/2)·(ai+aj)·ln(ak) )·(ai+aj)

x_{k}(t) = ...

... (1/a)·Anti-[ ( s /o(s)o/ ...

... int[ (as)^{ai+aj+(-1)}·[o(ai+aj)o] ( 2+(1/2)·(ai+aj)·ln(as) )·(ai+aj) ]d[s] )^{[o(s)o] (1/2)} ]-( ...

... ( (2/m)·U )^{(1/2)}·at )

Deducción:

d_{ax}[ (ax)^{ay+az} ] = (ay+az)·(ax)^{ay+az+(-1)} 

d_{ay}[ (ay+az)·(ax)^{ay+az+(-1)} ] = (ax)^{ay+az+(-1)}·( 1+ay·ln(ax)+az·ln(ax) )

d_{az}[ (ax)^{ay+az+(-1)}·( 1+(ay+az)·ln(ax) ) ] = (ax)^{ay+az+(-1)}·ln(ax)·( 2+(ay+az)·ln(ax) )

Ley:

Si Q(yz,zx,xy) = bU·< ((zx+xy)/(yz)),((xy+yz)/(zx)),((yz+zx)/(xy)) > ==> 

U(yz,zx,xy) = U·( (bzx+bxy)·ln(byz)+(bxy+byz)·ln(bzx)+(byz+bzx)·ln(bxy) )

Anti-div-exp[ U(yz,zx,xy) ] = Ub·( ...

... (byz)^{bzx+bxy+(-1)}·(bz+by)·( 1+ln(byz) )+...

... (bzx)^{bxy+byz+(-1)}·(bx+bz)·( 1+ln(bzx) )+...

... (bxy)^{byz+bzx+(-1)}·(by+bx)·( 1+ln(bxy) ) )

Ley:

E(x_{k}) = int[ Anti-div-exp[ U_{k}(yz,zx,xy) ] ]d[(1/b)·k] = U·(bij)^{bik+bkj+(-1)}·( ( 1/ln(bij) )+1 )

x_{k}(t) = (a/b)·Anti-[ ( s /o(s)o/ int[ (bij)^{ais+asj+(-1)}·( ( 1/ln(bij) )+1 ) ]d[s] )^{[o(s)o] (1/2)} ]-( ...

... ( (2/m)·U )^{(1/2)}·(b/a)·t )

Deducción:

d_{byz}[ (byz)^{bzx+bxy} ] = (bzx+bxy)·(byz)^{bzx+bxy+(-1)}

d_{x}[ (bzx+bxy)·(byz)^{bzx+bxy+(-1)} ] = (byz)^{bzx+bxy+(-1)}·(bz+by)·( 1+ln(byz) )

Áreas y Volumenes:

Teorema:

x = r·cos(w)

y = r·sin(w)

d[x]d[y] = (1/2)·( d[x]d[y]+d[x]d[y] ) = ...

... (1/2)·( d_{r}[x]·d_{w}[y]+(-1)·d_{w}[x]·d_{r}d[y] )·d[r]d[w]

Teorema:

Área de un círculo:

A(r) = int[w = 0]-[2pi][ inr[r = 0]-[r][ r ]d[r] ]d[w] = ...

... int[w = 0]-[2pi][ (1/2)·r^{2} ]d[w] = (1/2)·r^{2}·int[w = 0]-[2pi][ d[w] ] = pi·r^{2}

Perímetro de un círculo:

B(r) = d_{r}[ A(r) ] = d_{r}[ pi·r^{2} ] = 2pi·r

Teorema:

Área de un sector circular:

A(r,w) = int[w = 0]-[w][ inr[r = 0]-[r][ r ]d[r] ]d[w] = ...

... int[w = 0]-[w][ (1/2)·r^{2} ]d[w] = (1/2)·r^{2}·int[w = 0]-[w][ d[w] ] = (1/2)·wr^{2}

Perímetro de un sector circular:

B(r,w) = d_{r}[ A(r,w) ] = d_{r}[ (1/2)·wr^{2} ] = wr

Teorema:

d[z] = r·sin(s)·d[s]

x = r·cos(2w)

y = r·sin(2w)

d[x]d[y]d[z] = (1/2)·( d[x]d[y]d[z]+d[x]d[y]d[z] ) = ...

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

Teorema:

Volumen de una esfera:

A(r) = int[s = 0]-[pi][ int[w = 0]-[2pi][ int[r = 0]-[r][ r^{2}·sin(s) ]d[r] ]d[w] ]d[s] = ...

... int[s = 0]-[pi][ int[w = 0]-[2pi][ (1/3)·r^{3}·sin(s) ]d[w] ]d[s] = ...

... (1/3)·r^{3}·int[s = 0]-[pi][ sin(s)·int[w = 0]-[2pi][ d[w] ] ]d[s] = ...

... (1/3)·r^{3}·int[s = 0]-[pi][ sin(s)·2pi ]d[s] = (2/3)·pi·r^{3}·int[s = 0]-[pi][ sin(s) ]d[s] = ...

... (2/3)·pi·r^{3}·(-1)·( cos(pi)+(-1)·cos(0) ) = (4/3)·pi·r^{3}

Superficie de una esfera:

B(r) = d_{r}[ A(r) ] = d_{r}[ (4/3)·pi·r^{3} ] = 4pi·r^{2}

Teorema:

Volumen de un hemisferio:

A(r,w) = int[s = 0]-[pi][ int[w = 0]-[w][ int[r = 0]-[r][ r^{2}·sin(s) ]d[r] ]d[w] ]d[s] = ...

... int[s = 0]-[pi][ int[w = 0]-[w][ (1/3)·r^{3}·sin(s) ]d[w] ]d[s] = ...

... (1/3)·r^{3}·int[s = 0]-[pi][ sin(s)·int[w = 0]-[w][ d[w] ] ]d[s] = ...

... (1/3)·r^{3}·int[s = 0]-[pi][ sin(s)·w ]d[s] = (2/3)·wr^{3}·int[s = 0]-[pi][ sin(s) ]d[s] = ...

... (2/3)·wr^{3}·(-1)·( cos(pi)+(-1)·cos(0) ) = (2/3)·wr^{3}

Superficie de un hemisferio:

B(r,w) = d_{r}[ A(r,w) ] = d_{r}[ (2/3)·wr^{3} ] = 2wr^{2}

--------------------------------

Mecánica Teórica y Ondas:

--------------------------------

Teorema:

Sea d_{x}[F(x)] = f(x) ==>

d_{x}[ Anti-[F(s)]-(x) ] = ( 1/f( Anti-[F(s)]-(x) ) )

Demostración:

d_{y}[ Anti-[F(s)]-( F(y) ) ] = d_{y}[y] = 1

d[ Anti-[F(s)]-( F(y) ) ] = d[y]

d_{F(y)}[ Anti-[F(s)]-( F(y) ) ] = d_{F(y)}[y] = ( 1/d_{y}[F(y)]) = ( 1/f(y) )

Sea y = Anti-[F(s)]-(x) ==>

d_{F( Anti-[F(s)]-(x) )}[ Anti-[F(s)]-( F( Anti-[F(s)]-(x) ) ) ] = ( 1/f( Anti-[F(s)]-(x) ) )

Teorema:

Sea d_{x}[F(x)] = f(x) ==>

d_{x}[ Anti-[( s /o(s)o/ F(s) )]-(x) ] = f( Anti-[( s /o(s)o/ F(s) )]-(x) )

Demostración:

d_{x}[ Anti-[( s /o(s)o/ F(s) )]-(x) ] = ( 1/( 1/f( Anti-[( s /o(s)o/ F(s) )]-(x) ) ) ) = ...

... f( Anti-[( s /o(s)o/ F(s) )]-(x) )

Teorema:

d_{x}[ arc-sin(x) ] = ( 1/(1+(-1)·x^{2})^{(1/2)} )

Demostración:

d_{x}[ arc-sin(x) ] = ( 1/cos( arc-sin(x) ) ) = ( 1/(1+(-1)·(sin( arc-sin(x) ))^{2})^{(1/2)} )

Teorema:

d_{x}[ arc-cos(x) ] = (-1)·( 1/(1+(-1)·x^{2})^{(1/2)} )

Demostración:

d_{x}[ arc-cos(x) ] = (-1)·( 1/sin( arc-cos(x) ) ) = ( 1/(1+(-1)·(cos( arc-cos(x) ))^{2})^{(1/2)} )

Racionamiento antiguo:

Teorema:

d_{x}[ e^{x} ] = e^{x}

Demostración:

d_{x}[ e^{x} ] = ( 1/(1/y) ) = y = e^{x}

Teorema:

d_{x}[ ln(x) ] = (1/x)

Demostración:

d_{x}[ ln(x) ] = ( 1/e^{y} ) = ( 1/e^{ln(x)} ) = (1/x)

Teorema:

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

(1/2)·d_{t}[x(t)]^{2} = F(s)·u^{2}

d_{tt}^{2}[x(t)] = (1/2)·( F(s) )^{(-1)·(1/2)}·f(s)·( F(s) )^{(1/2)}·2u^{2} = f(s)·u^{2}

Demostración:

d_{t}[ F(s) ] = d_{s}[F(s)]·d_{t}[s]

Teorema:

Sea x(t) = Anti-[ ( s /o(s)o/ int[ H(ut) [o(ut)o] F(s) ]d[s] )^{[o(s)o] (1/2)} ]-( 2^{(1/2)}·ut ) ] ==>

(1/2)·d_{t}[x(t)]^{2} = ( H(ut) [o(ut)o] F(s) )·u^{2}

d_{tt}^{2}[x(t)] = ...

... (1/2)·( H(ut) [o(ut)o] F(s) )^{(-1)·(1/2)}·h(ut)·f(s)·( H(ut) [o(ut)o] F(s) )^{(1/2)}·2u^{2} = ...  

... h(ut)·f(s)·u^{2}

Demostración:

u·d_{ut}[ H(ut) [o(ut)o] F(s) ] = u·d_{ut}[H(ut)]·d_{ut}[F(s)] = ...

... u·h(ut)·d_{s}[F(s)]·d_{ut}[s] = h(ut)·d_{s}[F(s)]·d_{t}[s]

Teorema:

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

(1/2)·d_{t}[x(t)]^{2} = ( (ut) [o(ut)o] F(s) )·u^{2}

d_{tt}^{2}[x(t)] = ...

... (1/2)·( (ut) [o(ut)o] F(s) )^{(-1)·(1/2)}·f(s)·( (ut) [o(ut)o] F(s) )^{(1/2)}·2u^{2} = f(s)·u^{2}

Ley:

d[H(t)] = (1/pi)^{2}·MgI·d[tx]

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

... ( 2·(M/m)·gI·(1/u)·(1/d) )^{(1/2)}·(1/pi)·t )

Deducción:

H(t) = int[ d[H(t)] ] = int[ (1/pi)^{2}·MgI ]d[tx] = (1/pi)^{2}·MgI·int[ d[tx] ] = (1/pi)^{2}·MgItx

md·d_{tt}^{2}[x] = (1/pi)^{2}·MgItx

md·d_{tt}^{2}[x]·d[x] = (1/pi)^{2}·MgI·(1/u)·(ut)·x·d_{ut}[x]·d[ut]

md·(1/2)·d_{t}[ax]^{2} = (1/pi)^{2}·MgI·(1/u)·(1/2)·(ut)^{2} [o(ut)o] (1/2)·(ax)^{2}

Ley:

d[H(t)] = (1/pi)^{2}·Mv·d[(1/t)·x]

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

... ( 2·(M/m)·(vu)·(1/d) )^{(1/2)}·(1/pi)·t )

Deducción:

H(t) = int[ d[H(t)] ] = int[ (1/pi)^{2}·Mv ]d[(1/t)·x] = (1/pi)^{2}·Mv·int[ d[(1/t)·x] ] = ...

... (1/pi)^{2}·Mv·(1/t)·x

md·d_{tt}^{2}[x] = (1/pi)^{2}·Mv·(1/t)·x

md·d_{tt}^{2}[x]·d[x] = (1/pi)^{2}·Mvu·(1/(ut))·x·d_{ut}[x]·d[ut]

md·(1/2)·d_{t}[ax]^{2} = (1/pi)^{2}·Mvu·ln(ut) [o(ut)o] (1/2)·(ax)^{2}

Ley:

Le tenéis que decir a Esquerra Republicana,

que queréis el título de la universidad de Stroniken como el mío,

enseñando el testimonio de uno mismo con Dios,

escrito con vuestra letra:

Jûan Garriga Peralta-Peraltotzak:

Filósofo de la ciencia matemática y de la ciencia lógica,

por la universidad de Stroniken.

Y que vos lo envíen.

Ley: [ de onda electro-magnética plana de superficie ]

Lap[ E_{e}(x,y,t) ] = (-2)·(1/c)^{2}·d_{tt}^{2}[ int[ B_{e}(x,y,t) ]d[t] ]

Deducción:

E_{e}(x,y,t)+int[ B_{e}(x,y,t) ]d[t] = 0 = m·d_{tt}^{2}[ < x,y > ]

x(t) = ct·cos(w) 

y(t) = ct·sin(w)

Lap[ int[ B_{e}(x,y,t) ]d[t] ] = ...

... ( 1/(d[x]^{2}+d[y]^{2}) )·(d[x]^{2}+d[y]^{2}) [o] Lap[ int[ B_{e}(x,y,t) ]d[t] ]

Lap[ E_{e}(x,y,t) ]+Lap[ int[ B_{e}(x,y,t) ]d[t] ]= 0^{3}

Lap[ E_{e}(x,y,t) ]+2·(1/c)^{2}·d_{tt}^{2}[ int[ B_{e}(x,y,t) ]d[t] ] = 0^{3}

Lap[ E_{e}(x,y,t) ] = (-2)·(1/c)^{2}·d_{tt}^{2}[ int[ B_{e}(x,y,t) ]d[t] ]

Ley: [ de onda gravito-magnética plana de superficie ]

Lap[ int[ B_{g}(x,y,t) ]d[t] ] = (-2)·(1/c)^{2}·d_{tt}^{2}[ E_{g}(x,y,t) ]

Deducción:

int[ B_{g}(x,y,t) ]d[t]+E_{g}(x,y,t) = 0 = m·d_{tt}^{2}[ < x,y > ]

x(t) = ct·cos(w) 

y(t) = ct·sin(w)

Lap[ E_{g}(x,y,t) ] = ( 1/(d[x]^{2}+d[y]^{2}) )·(d[x]^{2}+d[y]^{2}) [o] Lap[ E_{g}(x,y,t) ]

Lap[ int[ B_{g}(x,y,t) ]d[t] ]+Lap[ E_{g}(x,y,t) ] = 0^{3}

Lap[ int[ B_{g}(x,y,t) ]d[t] ]+2·(1/c)^{2}·d_{tt}^{2}[ E_{g}(x,y,t) ] = 0^{3}

Lap[ int[ B_{g}(x,y,t) ]d[t] ] = (-2)·(1/c)^{2}·d_{tt}^{2}[ E_{g}(x,y,t) ]

Ecuaciones de ondas elípticas planas de superficie:

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = (-2)·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,t) = (1/2)·( ...

... e^{ax+ay+acit || ln( H(ax,ay) )+act}+...

... e^{ax+ay+acit || ln( H(ax,ay) )+(-1)·act} )

u(x,y,0) = H(ax,ay)

d_{t}[ u(x,y,0) ] = 0

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = (-2)·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,t) = sum[k = 1]-[oo][ ...

... int[h(ax,ay)+(-1)·(1/2)·act·0 || (2t)^{(1/2)}]-[h(ax,ay)+(1/2)·act·0 || (2t)^{(1/2)}][ w ]d[w] ]·...

... e^{ax+ay+acit || 0}

u(x,y,0) = 0

d_{t}[ u(x,y,0) ] = ac·h(ax,ay)

Ecuaciones de ondas hiperbólicas planas de superficie:

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = 2·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,t) = (1/2)·( ...

... e^{ax+ay+act || ln( H(ax,ay) )+act}+...

... e^{ax+ay+act || ln( H(ax,ay) )+(-1)·act} )

u(x,y,0) = H(ax,ay)

d_{t}[ u(x,y,0) ] = 0

Teorema:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = 2·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,t) = sum[k = 1]-[oo][ ...

... int[h(ax,ay)+(-1)·(1/2)·act·0 || (2t)^{(1/2)}]-[h(ax,ay)+(1/2)·act·0 || (2t)^{(1/2)}][ w ]d[w] ]·...

... e^{ax+ay+act || 0}

u(x,y,0) = 0

d_{t}[ u(x,y,0) ] = ac·h(ax,ay)

Ley:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = (-2)·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,t) = E_{e}·(1/2)·( ...

... e^{ax+ay+acit || ln( (ax)^{2}+(ay)^{2} )+act}+...

... e^{ax+ay+acit || ln( (ax)^{2}+(ay)^{2} )+(-1)·act} )

u(x,y,0) = (ax)^{2}+(ay)^{2}

d_{t}[ u(x,y,0) ] = 0

Ley:

d_{xx}^{2}[u(x,y,t)]+d_{yy}^{2}[u(x,y,t)] = (-2)·(1/c)^{2}·d_{tt}^{2}[u(x,y,t)]

u(x,y,t) = E_{e}·sum[k = 1]-[oo][ ...

... int[(ax+ay)+(-1)·act·0 || (2t)^{(1/2)}]-[(ax+ay)+act·0 || (2t)^{(1/2)}][ w ]d[w] ]·...

... e^{ax+ay+acit || 0}

u(x,y,0) = 0

d_{t}[ u(x,y,0) ] = 2ac·(ax+ay)

producto-tensorial y teoría-de-juegos-en-economía-y-física y análisis-matemático-5 y mecánica-momento-de-inercia

Teorema:

A = k·( < a,0 > % < 0,a >)

B = k·( < a,0 > % < 0,(-a) >)

dim(A) = (1/2)

dim(B) = (1/2)

( < a,0 > % < 0,0 >) = (1/2)·( < a,0 > % < 0,a >)+(1/2)·( < a,0 > % < 0,(-a) >)

Lema: [ de gasto en defensa ]

PIB por cápita 8,400€

inversión del 5%:

(1/20)·32,000+20·1,600 = 33,600€ por cada 4 catalanohablantes.

Compra de armamento de 32,700€ por pack militar:

misil o torpedo o caja de proyectiles de cañón.

Lema:

Estrategia ganadora de venta,

del supremo o igual:

Make America empate again.

h(1) = 32,700€

F(1,1) = 1·1+(1+1) = 3

Estrategia ganadora de no venta,

del ínfimo o igual:

Make America first again.

h(2) = 65,400€

F(2,0) = 2·0+(2+0) = 2

Estrategias perdedoras:

h(1/2) = 16,350€

h(3/2) = 49,050€

F((1/2),(3/2)) = (3/4)+((1/2)+(3/2)) = (2.75)

F((3/2),(1/2)) = (3/4)+((3/2)+(1/2)) = (2.75)

Ley:

Hay gente que no es,

que no hay condenación.

Hay gente que es,

que hay condenación.

Deducción: [ por teoría de juegos ]

Jugadas ganadoras:

Joder a un esclavo infiel:

F(1,(-1)) = (-1)+(1+(-1)) = (-1)

F(2,(-1)) = (-2)+(2+(-1)) = (-1)

Jugadas perdedoras:

Joder a un señor fiel:

F(1,(-2)) = (-2)+(1+(-2)) = (-3)

F(2,(-2)) = (-4)+(2+(-2)) = (-4)

Ley:

Solo se puede conquistar con victoria,

haciendo que los infieles del Gestalt ignoren a los señores.

Solo se puede liberar con victoria,

no haciendo que los fieles del Gestalt ignoren a los señores.

Deducción: [ por teoría de juegos ]

Jugada ganadora:

Ignorar los infieles del Gestalt a los señores:

F(0,(-1)) = 0+(0+(-1)) = (-1)

Los infieles joden a infieles.

Jugada perdedora:

Ignorar los fieles del Gestalt a los señores:

F(0,(-2)) = 0+(0+(-2)) = (-2)

Los infieles joden a fieles.

Dual por dual de sabor:

Basik-kowetch-tate oil.

Acid-kowetch-tate wine.

Teorema:

int[x = 0]-[ln(oo)][ e^{(-1)·x^{n}} ]d[x] = (1/n!)

Teorema:

int[x = 0]-[ln(oo^{m+1})][ x^{m}·e^{(-x)} ]d[x] = m!

int[x = 0]-[ln(oo^{m+1})][ x^{m}·e^{(-1)·x^{n}} ]d[x] = (m!/n!)

Teorema:

int[x = 0]-[ln(oo^{(1/m)+1})][ x^{(1/m)}·e^{(-x)} ]d[x] = m

int[x = 0]-[ln(oo^{(1/m)+1})][ x^{(1/m)}·e^{(-1)·x^{n}} ]d[x] = m·( m!/(mn)! )

Demostración:

m = 0 & n = 1

int[x = 0]-[ln(oo)][ e^{(-1)·x} ]d[x] = [ (-1)·e^{(-1)·x} ]_{x = 0}^{x = ln(oo)} = 1

m = 1 & n = 1

int[x = 0]-[ln(oo^{2})][ xe^{(-1)·x} ]d[x] = ...

... [ (-1)·xe^{(-1)·x} ]_{x = 0}^{x = ln(oo^{2})}+[ (-1)·e^{(-1)·x} ]_{x = 0}^{x = ln(oo)} = 1

Teorema:

ln( (p+1)^{oo} ) = ( oo·log_{p+1}(2^{p}) )·ln(p+1) = ln(2^{p})·oo = ln(oo^{p})

Arte:

[En][ int[x = 0]-[ln(oo)][ e^{(-1)·x^{n}} ]d[x] = (1/n) ]

Exposición:

n = 1

Se define H(t) = int[x = 0]-[ln(oo)][ e^{(-1)·x^{n}·F(t)} ]d[x] ==>

d_{t}[H(t)] = f(t)·int[x = 0]-[ln(oo)][ (-1)·x^{n}·e^{(-1)·x^{n}·F(t)} ]d[x]

F(t) = x

d_{t}[H(t)] = d_{t}[x]·( 1/(n+1) )·(-1)

H(t) = x·( 1/(n+1) )·(-1) = F(t)·( 1/(n+1) )·(-1)

H( F^{o(-1)}(1) ) = F( F^{o(-1)}(1) )·( 1/(n+1) )·(-1) = ( 1/(n+1) )·(-1)

u(1) = (1/n)

v(1/n) = 0

H( F^{o(-1)}(1) ) = ( 1/(n+u(1)) )·(-1)^{u(1)} = ( 1/(n+v(1/n)) )·(-1)^{v(1/n)} = (1/n)

Arte:

[En][Em][ int[x = 0]-[ln(oo^{m+1})][ x^{m}·e^{(-1)·x^{n}} ]d[x] = ( 1/(m+n) ) ]

Exposición:

m = 0 & n = 1

Se define H(t) = int[x = 0]-[ln(oo^{m+1})][ x^{m}·e^{(-1)·x^{n}·F(t)} ]d[x] ==>

d_{t}[H(t)] = f(t)·int[x = 0]-[ln(oo^{m+1})][ (-1)·x^{m+n}·e^{(-1)·x^{n}·F(t)} ]d[x]

F(t) = x^{m+1}

d_{t}[H(t)] = (m+1)·x^{m}·d_{t}[x]·( 1/(m+n+1) )·(-1)

H(t) = x^{m+1}·( 1/(m+n+1) )·(-1) = F(t)·( 1/(m+n+1) )·(-1)

H( F^{o(-1)}(1) ) = F( F^{o(-1)}(1) )·( 1/(m+n+1) )·(-1) = ( 1/(m+n+1) )·(-1)

u(1) = (1/n)

v(1/n) = 0

H( F^{o(-1)}(1) ) = ( 1/(m+n+u(1)) )·(-1)^{u(1)} = ( 1/(m+n+v(1/n)) )·(-1)^{v(1/n)} = ( 1/(m+n) )

Teorema:

int[x = 0]-[oo][ ( sin(nx)/(nx) ) ]d[x] = (pi/n)

Demostración:

Se define H(t) = int[x = 0]-[oo][ e^{(-1)·(nxt)}·( sin(nx)/(nx) ) ]d[x] ==>

y = nx & d[y] = n·d[x]

H(t) = (1/n)·int[y = 0]-[oo][ e^{(-1)·(yt)}·( sin(y)/y ) ]d[y]

d_{t}[H(t)] = (-1)·(1/n)·int[y = 0]-[oo][ e^{(-1)·(yt)}·sin(y) ]d[y]

d_{t}[H(t)] = (-1)·(1/n)·( 1/(t^{2}+1) )

H(t) = (-1)·(1/n)·arc-tan(t)

H(0) = (-1)·(1/n)·(-pi) = (pi/n)

Teorema:

int[x = 0]-[oo][ ( cos(x/n)/x ) ]d[x] = ln(n)

Demostración:

Se define H(t) = int[x = 0]-[oo][ e^{(-1)·(xt)}·( cos(x/n)/x ) ]d[x] ==>

H(t) = int[x = 0]-[oo][ e^{(-1)·(xt)}·( cos(x/n)/x ) ]d[x]

d_{t}[H(t)] = (-1)·int[x = 0]-[oo][ e^{(-1)·(xt)}·cos(x/n) ]d[x]

d_{t}[H(t)] = (-1)·( t/(t^{2}+(1/n)^{2}) )

H(t) = (-1)·(1/2)·ln(t^{2}+(1/n)^{2})

H(0) = (-1)·(1/2)·2·ln(1/n) = ln(n)

Teorema:

int[x = ln(0)]-[0][ int[y = ln(0)]-[0][ e^{(-1)·( x^{n}+y^{n} )} ]d[y] ]d[x] = (1/n!)^{2}

Demostración:

e^{(-1)·( x^{n}+y^{n} )} [o(x || y)o] ( x /o(x || y)o/ x^{n} ) [o(x || y)o] ( y /o(x || y)o/ y^{n} )

Teorema:

int[x = 0]-[ln(oo)][ e^{(-1)·x^{2}} ]d[x] = (1/2!)

Demostración:

int[w = 0]-[2pi][ int[r = 0]-[oo][ e^{(-1)·r^{2}}·(1/4)·2r ]d[r] ]d[w] = (pi/2)

(-1)·e^{(-1)·r^{2}}·(1/4)·w = (-1)·e^{(-1)·( x^{2}+y^{2} )}·(1/4)·arc-tan(x/y)

arc-tan(0/oo)+(-1)·arc-tan(0/(-oo)) = 2pi

arc-tan(oo/(-oo)) = arc-tan(oo·(-0)) = arc-tan(oo·0) = arc-tan(oo/oo)

e^{(-1)·( x^{2}+y^{2} )}·(1/4)·8pi = (pi/2)

4 veces el cuadrante de la exponencial y valores convergentes de arco-tangente.

Teorema:

int[x = 0]-[pi][ e^{(-1)·( sin(x) )^{n}} ]d[x] = (2/n!)

Demostración:

int[x = 0]-[pi][ e^{(-1)·( sin(x) )^{n}}·( 1/d_{x}[sin(x)] ) ]d[sin(x)] = ...

... (-1)·e^{(-1)·( sin(x) )^{n}} [o(sin(x))o] ( sin(x) /o(sin(x))o/ ( sin(x) )^{n} )·( 1/cos(x) )

Teorema:

int[x = (-1)·(pi/2)]-[(pi/2)][ e^{(-1)·( cos(x) )^{n}} ]d[x] = (2/n!)

Demostración:

int[x = (-1)·(pi/2)]-[(pi/2)][ e^{(-1)·( cos(x) )^{n}}·( 1/d_{x}[cos(x)] ) ]d[cos(x)] = ...

... e^{(-1)·( cos(x) )^{n}} [o(cos(x))o] ( cos(x) /o(cos(x))o/ ( cos(x) )^{n} )·( 1/sin(x) )

Teorema:

int[x = (-oo)]-[oo][ ln(x^{n}+1) ]d[x] = n·( ln(2)+pi·i )

Área positiva > 1 + Área negativa < 1 = a la integral impropia

Verificación del método de Euler por Hôpital-Garriga:

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

( ln(x^{n}+1)+(-1) )·(w^{n}+1) [o(x)o] ( w /o(x)o/ (w^{n}+1) ) = ...

( ln(x^{n}+1)+(-1) ) [o(x)o] ( 1 /o(x)o/ 1 ) = [ ln(x^{n}+1)+(-1) ]_{x = (-oo)}^{x = oo}

Demostración:

Se define H(t) = int[x = (-oo)]-[oo][ ln( (xF(t))^{n}+1 ) ]d[x] ==>

d_{t}[H(t)] = f(t)·int[x = (-oo)]-[oo][ ( 1/( (xF(t))^{n}+1 ) )·n·( xF(t) )^{n+(-1)}·x ]d[x]

F(t) = x

d_{t}[H(t)] = d_{t}[x]·int[x = (-oo)]-[oo][ ( 1/(x^{2n}+1) )·nx^{2n+(-1)} ]d[x]

d_{t}[H(t)] = d_{t}[x]·(1/2)·( ln(oo^{2n}+1) )+(-1)·ln((-oo)^{2n}+1) )

d_{t}[H(t)] = d_{t}[x]·(1/2)·2n·( ln(oo)+(-1)·ln(oo)+pi·i ) = d_{t}[x]·n·( ln(2)+pi·i )

H(t) = x·n·( ln(2)+pi·i ) = F(t)·n·( ln(2)+pi·i )

H( F^{o(-1)}(1) ) = F( F^{o(-1)}(1) ) n·( ln(2)+pi·i ) = n·( ln(2)+pi·i )

Arte:

[En][ int[x = (-1)]-[1][ ( 1/(x^{n}+(-1)) ) ]d[x] = (2/n) ]

Exposición:

n = 0

Se define H(t) = int[x = (-1)]-[1][ ( 1/( (xF(t))^{n}+(-1) ) ) ]d[x] ==>

d_{t}[H(t)] = f(t)·int[x = (-1)]-[1][ (-1)·( 1/( (xF(t))^{n}+(-1) ) )^{2}·n·( xF(t) )^{n+(-1)}·x ]d[x]

F(t) = x

d_{t}[H(t)] = d_{t}[x]·int[x = (-1)]-[1][ (-1)·( 1/(x^{2n}+(-1)) )^{2}·nx^{2n+(-1)} ]d[x]

d_{t}[H(t)] = d_{t}[x]·(1/2)·( oo+(-oo) ) = d_{t}[x]·(1/2)

H(t) = x·(1/2) = F(t)·(1/2)

u(1) = m

v(m) = (4/n)

H( F^{o(-1)}(1) ) = F( F^{o(-1)}(1) )·(1/2) = (1/2)·u(1) = (1/2)·v(m) = (2/n)

Momento de inercia Wronskiano:

Ley:

Sea ( x = a·cos(ut) & y = b·sin(ut) ) ==>

M·int[ x·d_{t}[y]+(-y)·d_{t}[x] ]d[t] = M·ab·(ut)

Ley:

M·ab·(ut)·(1/2)·d_{t}[w]^{2} = E·H(w)

w(t) = Anti-[ ( int[ (ut) ]d[s] /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ( (2/M)·(1/(ab) )·E )^{(1/2)}·t )

Ley:

Sea ( x = vt & y = g·(1/2)·t^{2} ) ==>

M·int[ x·d_{t}[y]+(-y)·d_{t}[x] ]d[t] = M·vg·(1/u)^{3}·(1/6)·(ut)^{3}

Ley:

M·vg·(1/u)^{3}·(ut)^{3}·(1/12)·d_{t}[w]^{2} = E·H(w)

w(t) = Anti-[ ( int[ (ut)^{3} ]d[s] /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...

... ( (12/M)·(1/(vg) )·u^{3}·E )^{(1/2)}·t )

Ley:

Sea ( x = vt & y = re^{ut+(-1)} ) ==>

M·int[ x·d_{t}[y]+(-y)·d_{t}[x] ]d[t] = M·vr·(1/u)·(ut+(-1))^{2}·er-h[2]( ut+(-1) )

Ley:

M·vr·(1/u)·(ut+(-1))^{2}·er-h[2]( ut+(-1) )·(1/2)·d_{t}[w]^{2} = E·H(w)

w(t) = Anti-[ ( int[ (ut+(-1))^{2}·er-h[2]( ut+(-1) ) ]d[s] /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...

... ( (2/M)·(1/(vr) )·uE )^{(1/2)}·t )

Ley:

Sea ( x = r·ln(ut) & y = vt ) ==>

M·int[ x·d_{t}[y]+(-y)·d_{t}[x] ]d[t] = M·vr·(1/u)·(ut)·( ln(ut)+(-2) )

Ley:

M·vr·(1/u)·(ut)·( ln(ut)+(-2) )·(1/2)·d_{t}[w]^{2} = E·H(w)

w(t) = Anti-[ ( int[ (ut)·( ln(ut)+(-2) ) ]d[s] /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...

... ( (2/M)·(1/(vr) )·uE )^{(1/2)}·t )

Ley:

Sea d_{t}[x] = ru·(ut+e)^{sin(ut+(pi/2))} ==>

d_{tt}^{2}[x] = ru^{2}·(ut+e)^{sin(ut+(pi/2))}·( cos(ut+(pi/2))·ln(ut+e)+sin(ut+(pi/2))·(1/(ut+e)) )

x(t) = r·(ut+e)^{sin(ut+(pi/2))} [o(ut)o] ( ut /o(ut)o/ sin(ut+(pi/2))·ln(ut+e) )

x(0/u) = re

d_{t}[x(0/u)] = rue

d_{tt}^{2}[x(0/u)] = ru^{2}

Deducción:

ln(f(z)) = ln(g(z)^{h(z)}) = h(z)·ln(g(z))

d_{z}[f(z)] = f(z)·d_{z}[ h(z)·ln(g(z)) ]

Ley:

La verdad implica la felicidad.

La falsedad implica el sufrimiento.

Deducción:

Creer-se una verdad es jugada ganadora:

F(n,1) = n·1+(n+1) = 2n+1

Creer-se una falsedad es jugada perdedora:

F(n,0) = n·0+(n+0) = n

n [< n+n = 2n < 2n+1

Ley:

Un heterosexual que es,

no puede ser homosexual que no es.

Un homosexual que no es,

no puede ser heterosexual que es.

I havere-kate-kute maket-tikaletch-tuted a coment saksahuaketch-kán,

of yu tenotitch-lán.

Yu havere-kate-kute maket-tikaletch-tuted a coment saksahuaketch-kán,

of me tenotitch-lán.

El American-Quetchua tiene 7 dialectos:

Centro americano:

-tikaletch-kal

Sur americano hawsnutch:

-tikaletch-tate

-tikaletch-tute

-tikalet-kazhe

-tikalet-kuzhe

gwzenen plana

-tikalet-huw

yuhened plana

-tikalet-shuw

yushened plana

Ley:

Sea I_{c}·(1/2)·d_{t}[w]^{2} = E·H(w) ==>

Si d[ d[I_{c}] ] = Mrv·d[w]d[t] ==>

w(t) = Anti-[ ( ( (1/2)·s^{2} [o(s)o] int[ (ut) ]d[s] ) /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...

... ( (2/M)·(1/(rv))·uE )^{(1/2)}·t )

Ley:

Sea I_{c}·(1/2)·d_{t}[w]^{2} = E·H(w) ==>

Si d[ d[I_{c}] ] = Mrgt·d[w]d[t] ==>

w(t) = Anti-[ ( ( (1/2)·s^{2} [o(s)o] int[ (ut)^{2} ]d[s] ) /o(s)o/ int[ H(s) ]d[s] )^{[o(s)o] (1/2)} ]-( ...

... ( (4/M)·(1/(rg))·E )^{(1/2)}·ut )

Euskera-Bascotzok-ejército y integrales-de-la-física-matemática y economía-aranceles y productos-conectivos y evangelio-stronikiano

Ley:

Ibai sakona zeharkatzen-sua,

és-de-tek arriskatzi-ten-dut-zatu-dut.

Ibai azalea zeharkatzen-sua,

no és-de-tek arriskatzi-ten-dut-zatu-dut.

Ley:

Sóc-de-tek goiko isilisteko iturri-koak,

en l'Euskera-no-Bascotzok parlatzi-koak.

Sóc-de-tek goiko isilisteko idarra-koak,

en l'Euskera-Bascotzok parlatzi-koak.

Ley:

La ley-tat-koaikek ha-de-tek estatu-dut per mejoria-tat-koaikek,

a berri lendikateko zubi-koak.

La ley-tat-koaikek ha-de-tek estatu-dut per minoria-tat-koaikek,

a berri lendikateko ibai-koak.

Ley:

Ell-lek miratzi-ten-dut-za-tek,

babesten dugu emakum-eskoak.

Ella-lek miratzi-ten-dut-za-tek,

babesten dugu guizon-eskoak.

Ley:

Ye parle ye-de-muá le Françé-de-le-Patuá,

y ele-vut a-vot-má de-le-tom-tambén,

de-le-dans i-çí sa-pé de-le-com.

Tú parle tú-de-tuá le Françé-de-le-Patuá,

y ele-nut a-not-má de-le-tom-tambén,

de-le-dans i-luá sa-pé de-le-com.

Ley:

Nus havoms overuá de-le-dans sa-putch,

tenint-que-pont ritmuá de la lutch,

upuá sape-tutch de-le-dans açutch.

Nus havoms underuá de-le-dans sa-putch,

ne tenint-que-pont ritmuá de la lutch,

dawnuá sape-tutch de-le-dans açutch.

20 años más de soldado del Françé de 1090€ al mes.

Ley:

Haveremitch overesen vihens hofen,

interesen hafens music,

uperesen vihenens of strahen hawsen.

Haveremitch underesen vihens hofen,

awtteresen hafens music,

dawneresen vihenens of strahen hawsen.

Teorema-Ley:

Sea ( y^{2} = d·(r+(-x)) & y = 0 & x = 0 ) ==>

int[x = 0]-[r][ int[y = 0]-[( d·(r+(-x)) )^{(1/2)}][ (k/V)·yx^{2} ]d[y] ]d[x] = ( (kd)/V )·(1/4!)·( r(t) )^{4}

Demostración:

y(r) = 0

int[x = 0]-[r][ int[y = 0]-[( d·(r+(-x)) )^{(1/2)}][ (k/V)·yx^{2} ]d[y] ]d[x] = ...

... (k/V)·int[x = 0]-[r][ (1/2)·d·(r+(-x))·x^{2} ]d[x] = ...

... (k/V)·[ d·(1/2)·( (1/3)·rx^{3}+(-1)·(1/4)·x^{4} ) ]_[x = 0]^{x = r} = ( (kd)/V )·(1/4!)·( r(t) )^{4}

Teorema-Ley:

Sea ( y = (r+(-x)) & y = 0 & x = 0 ) ==>

int[x = 0]-[r][ int[y = 0]-[(r+(-x))][ ka·( (r+(-x))+(-y) ) ]d[y] ]d[x] = ka·(1/3!)·( r(t) )^{3}

Demostración:

y(r) = 0

int[x = 0]-[r][ int[y = 0]-[(r+(-x))][ ka·( (r+(-x))+(-y) ) ]d[y] ]d[x] = ...

... ka·int[x = 0]-[r][ (1/2)·(r+(-x))^{2} ]d[x] = ...

... ka·[ (-1)·(1/2)·(1/3)·(r+(-x))^{3} ]_[x = 0]^{x = r} = ka·(1/3!)·( r(t) )^{3}

Teorema:

Sea ( y = 2·( 1/(1+x^{2}) ) & y = 2+(-x) ) ==>

int[x = 0]-[1][ int[y = 2+(-x)]-[2·( 1/(1+x^{2}) )][ d[y] ] ]d[x] = (pi/2)+(-1)·(3/2)

Demostración:

g(1)+(-1)·f(1) = 0

g(0)+(-1)·f(0) = 0

Teorema:

Sea ( y^{2} = dx & y = x ==>

int[x = 0]-[d][ int[y = x]-[(dx)^{(1/2)}][ d[y] ] ]d[x] = (2/3)·d^{(3/2)}

Demostración:

g(d)+(-1)·f(d) = 0

g(0)+(-1)·f(0) = 0

Teorema:

Sea ( y = sin(x) & y = 1+(-1)·cos(x) ==>

int[x = 0]-[(pi/2)][ int[y = sin(x)]-[1+(-1)·cos(x)][ d[y] ] ]d[x] = (pi/2)

Demostración:

g(pi/2)+(-1)·f(pi/2) = 0

g(0)+(-1)·f(0) = 0

Lema:

p = 2n+( n/(2!+1) ) = (0.30)+(0.30)+(0.10) = (0.70)€

p = 2n+(2!+1)·n = 2·(0.14)+(0.42) = (0.70)€

Pensión de jubilación: 

n = (0.14)€

Arancel:

2!·n = (0.28)€

Al 200% de lo que se gana = n

Lema:

p = 3n+( n/(3!+1) ) = (0.35)+(0.35)+(0.35)+(0.05) = (1.10)€

p = 3n+(3!+1)·n = 3·(0.11)+(0.77) = (1.10)€

Pensión de jubilación: 

n = (0.11)€

Arancel:

3!·n = (0.66)€

Al 600% de lo que se gana = n.

Lema:

p = 4n+( n/(4!+(-8)) ) = (0.64)+(0.64)+(0.64)+(0.64)+(0.04) = (2.60)€

p = 4n+(4!+(-8))·n = 4·(0.13)+(2.08) = (2.60)€

Pensión de jubilación: 

n = (0.13)€

Arancel:

(4!+(-9))·n = (1.95)€

Al 1,500% de lo que se gana = n

Teorema:

¬( {a} [x |-| x] {b} ) = {a} [x <==> x] {b}

Demostración:

¬( { < a,a >,< b,b > } ) = { < a,b >,< b,a > }

Teorema:

¬( {a} [x & x] {b} ) = ¬{a} [x || x] ¬{b}

Demostración:

¬( { < a,b > } ) = { < a,a >,< b,b >,< b,a > }

Teorema:

¬( {a} [x || x] {b} ) = ¬{a} [x & x] ¬{b}

Demostración:

¬( { < a,b >,< a,a >,< b,b > } ) = { < b,a > }

Ley:

Das European Union:

pustesen-hatchteit arancels to:

das United Stateds of das Trump,

one thousand halfen percent,

interesen havens music,

for to spanishen army-zeizen.

Until Trump neguesen-hatchteit,

das five percent of militar gast,

of das spanishen army-zeizen.

Yeik lo harevemitch arreglesen-hatchteited,

cashing das american people-zeizen,

to das spanishen army-zeizen.

Dual:

Canto-pueh-piah-halám,

una canción-heláp de ti.

Cantah-pueh-piah-halám,

una canción-heláp de mi.

Dual:

Yo cantare-pueh-piah-halám,

una cancionen-heláp de ti.

Tú cantare-pueh-piah-halám,

una cancionen-heláp de mi.

Ley: [ islámica stronikiana ]

No se puede-pueh-halám bebere-iba-doh alcohole-iba-hám,

dentro-pueh-halám de misa-iba-hám.

Se puede-pueh-halám bebere-iba-doh alcohole-iba-hám,

fuera-pueh-halám de misa-iba-hám.

Ley: [ cristiana stronikiana ]

Se puede beber alcohol

porque se bebe vino fuera de misa.

No se puede beber alcohol

aunque quizás se bebe vino fuera de misa.

Ley: [ islámica stronikiana ]

Se tiene-pueh-halám que llevare-iba-doh velo-iba-hám

porque se lleva-pueh-halám velo-iba-hám fuera-pueh-halám de misa-iba-hám.

No se tiene-pueh-halám que llevare-iba-doh velo-iba-hám

aunque quizare-iba-doh se lleva-pueh-halám velo-iba-hám fuera-pueh-halám de misa-iba-hám.

Ley: [ cristiana stronikiana ]

Se tiene que llevar velo,

dentro de misa.

No se tiene que llevar velo,

fuera de misa.

Ley:

Los catalanes hemos aceptado las sanciones del presidente americano,

y pagamos el doble en empresas americanas:

Precio mínimo con doble arancel:

2 socios

1.40€

3 socios:

2.20€

4 socios:

5.20€

Si algún país se apunta a esto,

le ponemos aranceles de 4 socios al 1,500%.

Duales de no comercio de España con estados unidos por aranceles:

red-kowetch-tate wine.

green-kowetch-tate wine.

pernatetch-ton of pork.

pernatetch-ton of senglar.

fuetetch-ton of pork.

fuetetch-ton of senglar.

butifarretch-ton of pork.

butifarretch-ton of senglar.

Ley:

Vino verde & levadura ==> Champañe verde y blanco.

Leche blanca & levadura ==> Roquefort blanco y verde.

Principio:

Supongo que estados unidos tiene glorificado el armamento,

para pensiones a soldados, policías y civiles.

Ley:

Le sobran misiles a estados unidos,

porque son pensiones a gente de vuelo.

Le sobran torpedos a estados unidos,

porque son pensiones a marineros.

Le sobran proyectiles de cañón a estados unidos,

porque son pensiones a conductores terrestres.

Ley:

Se puede comprar armamento a estados unidos,

por 32,700€ por pack militar:

Del objeto militar y pensión civil,

en potencia en estados unidos,

al objeto militar y pensión civil,

realizada en el país comprador.

Corolario:

El país comprador no pierde dinero,

teniendo 65,400€ en potencia,

al comprar el pack militar a estados unidos,

y se tiene un beneficio de 32,700€ más 136€ en impuestos.

Jugada ganadora anti-monopolio: 

F(1,1) = 1·1+(1+1) = 3

Es estados unidos el que los pierde,

en vez de hacer los 65,536€ no vendiendo armamento a ninguien.

Jugada perdedora anti-monopolio:

F(2,0) = 2·0+(2+0) = 2