sábado, 19 de junio de 2021

cinemàtica y mecànica

Ley: [ Encuentro de dos barcas con velocidad constante ]

Una baja por la corriente desde x_{0} = 0.

Una sube contra-corriente desde x_{0} = x.

[Et][ (u+v)·t = (u+(-v))·t+x ]

t = (x/(2v))


Ley: [ encuentro de dos barcas con aceleración constante ]

Una barca baja desde x_{0} = 0.

Una barca sube desde x_{0} = x.

[Et][ (1/2)·(a+s)·t^{2} = (1/2)·(a+(-s))·t^{2}+x ]

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


Ley:

Avance de un móvil por un móvil frenando.

Avance de un móvil por un móvil en marcha.

[Et][ (1/2)·at^{2}+ut = vt & a < 0 ] & [Et][ at+u = 0 & a < 0 ]

( t = ( (2·(v+(-u)))/a ) > 0 <==> v < u ) & t = (-1)·(u/a)


(-1)·(u/a) < ( (2·(v+(-u)))/a )

0 < ( (2v+(-u))/a )

Si 2v < u ==> Hay avance frenado.

(-1)·(u/a) > ( (2·(v+(-u)))/a )

0 > ( (2v+(-u))/a )

Si 2v > u ==> Hay avance en marcha.


Ley:

Caída de una piedra desde un puente de altura h,

encima de un móvil en marcha con velocidad constante.

[Et][ (1/2)·(q/m)·gt^{2} = h & g > 0 ] & [Et][ vt = x ]

t = ( ((2h)/g)·(m/q) )^{(1/2)} & x = v·( ((2h)/g)·(m/q) )^{(1/2)}

Ley:

Caída de una piedra desde un puente de altura h,

encima de un móvil en marcha con aceleración constante.

[Et][ (1/2)·(q/m)·gt^{2} = h & g > 0 ] & [Et][ (1/2)·at^{2} = x ]

t = ( ((2h)/g)·(m/q) )^{(1/2)} & x = a·( (h/g)·(m/q) )


Ley:

Estirar dos cajas hasta que se rompa la cuerda que las une [ con s·e^{at} ].

[Et][ ...

... m_{1}·d_{tt}^{2}[x] = se^{at}+(-T) & ...

... m_{2}·d_{tt}^{2}[y] = (-s)·e^{at}+T & ...

... d_{tt}^{2}[x] = (-1)·d_{tt}^{2}[y] ...

... ]

(m_{1}+m_{2})·T = (m_{1}+m_{2})·se^{at}

t = (1/a)·ln(T/s)

Velocidades iniciales de separación:

u_{k} = (1/m_{1})·(T/a)·( ln(T/s)+(-1) )

v_{k} = (1/m_{2})·(T/a)·( ln(T/s)+(-1) )

Se mueve lo sistema antes de romperse la cuerda:

Posiciones iniciales de separación:

x_{k} = (1/m_{1})·(T/a^{2})·( (1/2)·( ln(T/s) )^{2}+(-1) )

y_{k} = (1/m_{2})·(T/a^{2})·( (1/2)·( ln(T/s) )^{2}+(-1) )

Posiciones iniciales:

x_{0} = (1/m_{1})·(s/a^{2})

y_{0} = (1/m_{2})·(s/a^{2})


Ley:

Estirar dos cajas hasta que se rompa la cuerda que las une [ con at ].

[Et][ ...

... m_{1}·d_{tt}^{2}[x] = at+(-T) & ...

... m_{2}·d_{tt}^{2}[y] = (-1)·at+T & ...

... d_{tt}^{2}[x] = (-1)·d_{tt}^{2}[y] ...

... ]

(m_{1}+m_{2})·T = (m_{1}+m_{2})·at

t = (T/a)

Velocidades iniciales de separación:

u_{k} = (1/m_{1})·( (a/2)·(T/a)^{2}+(-T)·(T/a) )

v_{k} = (1/m_{2})·( (a/2)·(T/a)^{2}+(-T)·(T/a) )

Posiciones iniciales de separación:

x_{k} = (1/m_{1})·( (a/6)·(T/a)^{3}+(-1)·(T/2)·(T/a)^{2} )

y_{k} = (1/m_{2})·( (a/6)·(T/a)^{3}+(-1)·(T/2)·(T/a)^{2} )

ecuacions diferencials

d_{x}[ pluf[n]-arc-sin(x) ] = ...

... ( pluf[n]-arc-sin(x) )^{n}·( 1+(-1)·( pluf[n]-arc-sin(x) )^{2} )^{(1/2)}

d_{x}[ pluf[n]-arc-cos(x) ] = ...

... (-1)·( pluf[n]-arc-cos(x) )^{n}·( 1+(-1)·( pluf[n]-arc-cos(x) )^{2} )^{(1/2)}


d_{x}[y(x)] = ( sin(y(x)) )^{n}

y(x) = arc-sin( pluf[n]-arc-sin(x) )

d_{x}[y(x)] = ( cos(y(x)) )^{n}

y(x) = arc-cos( pluf[n]-arc-cos(x) )


d_{x}[ pluf[n]-arc-tan(x) ] = ...

... ( pluf[n]-arc-tan(x) )^{n}·( 1+( pluf[n]-arc-tan(x) )^{2} )

d_{x}[ pluf[n]-arc-cot(x) ] = ...

... (-1)·( pluf[n]-arc-cot(x) )^{n}·( 1+( pluf[n]-arc-cot(x) )^{2} )


d_{x}[y(x)] = ( tan(y(x)) )^{n}

y(x) = arc-tan( pluf[n]-arc-tan(x) )

d_{x}[y(x)] = ( cot(y(x)) )^{n}

y(x) = arc-cot( pluf[n]-arc-cot(x) )


arc-cos(x)+arc-sin(-x) = (3/2)·pi

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

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

pi+(pi/2) = 0+(3/2)·pi = 2pi+(-1)·(pi/2) = ...

... (pi/4)+(5/4)·pi = (3/4)·pi+(3/4)·pi = (7/4)·pi+(-1)·(pi/4) = ...

... (1/3)·pi+(7/6)·pi = (1/6)·pi+(4/3)·pi


arc-cot(x)+arc-tan(x) = k

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

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

arc-coth(x)+(-1)·arc-tanh(x) = k

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


int[ arc-tanh(x) ] d[x] = x·arc-tanh(x)+(1/2)·( ln(1+(-1)·x^{2}) )

int[ arc-coth(x) ] d[x] = x·arc-coth(x)+(1/2)·( ln(1+(-1)·x^{2}) )

jueves, 17 de junio de 2021

càlcul diferencial y integral

[ES(x)][ f(x)+(-1)·g(x) = d_{x}[ S(x) ] ]

f(x)+(-1)·f(x) = d_{x}[ 1 ]

f(x)+(-1)·g(x) = d_{x}[ P(x) ] <==> g(x)+(-1)·f(x) = d_{x}[ (-1)·P(x) ]

Si f(x)+(-1)·g(x) = d_{x}[ P(x) ] & g(x)+(-1)·h(x) = d_{x}[ Q(x) ] ==>

f(x)+(-1)·h(x) = d_{x}[ P(x)+Q(x) ]


[ES(x)][ f(x)+(-1)·g(x) = int[ S(x) ] d[x] ]

f(x)+(-1)·f(x) = int[ 0^{2} ] d[x]

f(x)+(-1)·g(x) = int[ P(x) ] d[x] <==> g(x)+(-1)·f(x) = int[ (-1)·P(x) ] d[x]

Si f(x)+(-1)·g(x) = int[ P(x) ] d[x] & g(x)+(-1)·h(x) = int[ Q(x) ] d[x] ==>

f(x)+(-1)·h(x) = int[ P(x)+Q(x) ] d[x]


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


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

e^{ln(x)}·d_{x}[ln(x)] = 1

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


f(x) = x^{n}

ln(f(x)) = ln(x^{n}) = n·ln(x)

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


Teoremas:

d_{x}[1] = (1/h)·( (x+h)^{0}+(-1)·x^{0} ) = 0·(h/h) = 0

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

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

d_{x}[x^{n}] = ...

... (1/h)·( x^{n}+nx^{n+(-1)}h+...+h^{n}+(-1)·x^{n} ) = ...

... nx^{n+(-1)}+...+h^{n+(-1)} = nx^{n+(-1)}


d_{x}[sin(x)] = (1/h)·( sin(x+h)+(-1)·sin(x) ) = ...

... (1/h)·( sin(x)cos(h)+cos(x)sin(h)+(-1)·sin(x) ) = cos(x)·( sin(h)/h ) = cos(x)

d_{x}[cos(x)] = (1/h)·( cos(x+h)+(-1)·cos(x) ) = ...

... (1/h)·( cos(x)cos(h)+(-1)·sin(x)sin(h)+(-1)·cos(x) ) = (-1)·sin(x)·( sin(h)/h ) = (-1)·sin(x)


int[ arc-tan(x) ] d[x] = x·arc-tan(x)+(-1)·(1/2)·ln( 1+x^{2} )

int[ arc-cot(x) ] d[x] = x·arc-cot(x)+(1/2)·ln( 1+x^{2} )


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

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


int[ x^{n}·arc-sin(x) ] d[x] = ...

... (1/(n+1))·x^{n+1}·arc-sin(x)+...

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

int[ x^{n}·arc-cos(x) ] d[x] = ... 

... (1/(n+1))·x^{n+1}·arc-cos(x)+...

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

miércoles, 16 de junio de 2021

morfosintaxis

Ayer:

-un hombre de nombre Jûan comió una pizza en su casa.-

[ [t] es ayer ]-

[E$1$ [x] ][ [x] es hombre de nombre [n] ]-[ [n] es Jûan ]-

[ [t] : - [x] comió [y] en [z] - ]-

[E$1$ [y] ][ [y] es pizza ]-[ [z] es casa de [x] ]

Ayer:

-un hombre de nombre Jûan no comió una pizza en su casa.-

[ [t] es ayer ]-

[E$1$ [x] ][ [x] es hombre de nombre [n] ]-[ [n] es Jûan ]-

[ [t] : - [x] no comió [y] en [z] - ]-

[E$1$ [y] ][ [y] es pizza ]-[ [z] es casa de [x] ]


Jûan dijo:

-esas tías son feas-.

[ [x] es Jûan ]-

[ [x] dijo : - P([y]) - ]

P([y]) <==> [E?...? [y] ][ [y] es tías ]-[ [y] son [z] ]-[ [z] es feas ]

Jûan dijo:

-esas tías son guapas-.

[ [x] es Jûan ]-

[ [x] dijo : - Q([y]) - ]

Q([y]) <==> [E?...? [y] ][ [y] es tías ]-[ [y] son [z] ]-[ [z] es guapas ]

martes, 15 de junio de 2021

probabilitats

P[ 0 [< y+(-x) [< nx ] = (1/(n+1)) [< 1

x [< y [< (n+1)·x


P[ 0 [< y+(-1) [< f(x) ] = ( 1/(f(x)+1) ) [< 1

1 [< y [< f(x)+1

0 [< f(x)


P[ 0 [< y+(-x) [< f(x) & 0 [< x ] = ( x/(f(x)+x) ) [< 1

x [< y [< f(x)+x

0 [< f(x)


P[ 0 [< y+(-1)·x^{n} [< f(x) & 0 [< x^{n} ] = ( x^{n}/(f(x)+x^{n}) ) [< 1

x^{n} [< y [< f(x)+x^{n}

0 [< f(x)


P[ (-x)+c = x+(-a) & (-y)+c = y & 0 [< a ] = ( c/(c+a) ) [< 1

(-x)+c = x+(-a) <==> x = ( (c+a)/2 )

(-y)+c = y <==> y = ( c/2 )


P[ (-x)+c = x+(-a) & (-y)+c = y+(-b) & 0 [< b [< a ] = ( (c+b)/(c+a) ) [< 1

(-x)+c = x+(-a) <==> x = ( (c+a)/2 )

(-y)+c = y+(-b) <==> y = ( (c+b)/2 )


P[ x^{n+1} = ax^{n} & y^{n+1} = by^{n} & 0 [< b [< a  ] = (b/a) [< 1

x = a & y = b


P[ x^{(-n)+(-1)} = ax^{(-n)} & y^{(-n)+(-1)} = by^{(-n)} & 0 [< (1/b) [< (1/a) ] = (a/b) [< 1

x = (1/a) & y = (1/b)


P[ 1 < x ] = ( (x+(-1))/(x+1) ) [< 1

(x+(-1)) [< x [< (x+1)

0 [< (x+(-1))·(x+1) [< x·(x+1) [< (x+1)^{2}


P[ (-1) > x ] = ( (x+1)/(x+(-1)) ) [< 1

(x+(-1)) [< x [< (x+1)

(x+(-1))^{2} >] x·(x+(-1)) >] (x+1)·(x+(-1)) >] 0

domingo, 13 de junio de 2021

polinomis de sumes de factorial

P[n](-1) = (-1)·( 0!+1!+...+n! )

(-1)·P[n](-1) = ( 0!+1!+...+n! )

P[n+1](x) = d_{x}[ P[n](x) ]+(1/x)

P[0](x) = (1/x)

P[1](x) = (-1)·(1/x^{2})+(1/x)

P[2](x) = 2·(1/x^{3})+(-1)·(1/x^{2})+(1/x)

P[3](x) = (-6)·(1/x^{4})+2·(1/x^{3})+(-1)·(1/x^{2})+(1/x)

transitiu borrós

Si ( x [< x_{n} & x_{n} [< x_{n+1} ) ==> x [< x_{n+1}

min{(0.n),(0.1)} [< (0.n+1)

Si ( x [< x_{n} || x_{n} [< x_{n+1} ) ==> x [< x_{n+1}

max{(0.n),(0.1)} [< (0.n+1)


Si ( x >] (-1)·x_{n} || (-1)·x_{n} >] (-1)·x_{n+1} ) ==> x >] (-1)·x_{n+1}

max{(-1)·(0.n),(-1)·(0.1)} >] (-1)·(0.n+1)

(-1)·min{(0.n),(0.1)} >] (-1)·(0.n+1)

min{(0.n),(0.1)} [< (0.n+1)

Si ( x >] (-1)·x_{n} & (-1)·x_{n} >] (-1)·x_{n+1} ) ==> x >] (-1)·x_{n+1}

min{(-1)·(0.n),(-1)·(0.1)} >] (-1)·(0.n+1)

(-1)·max{(0.n),(0.1)} >] (-1)·(0.n+1)

max{(0.n),(0.1)} [< (0.n+1)


Si ( x = x & x [< x_{n} ) ==> x [< x_{n}

min{0,(0.n)} [< (0.n)

Si ( x = x || x [< x_{n} ) ==> x [< x_{n}

max{0,(0.n)} [< (0.n)


Si ( x = x || x >] (-1)·x_{n} ) ==> x >] (-1)·x_{n}

max{(-0),(-1)·(0.n)} >] (-1)·(0.n)

(-1)·min{0,(0.n)} >] (-1)·(0.n)

min{0,(0.n)} [< (0.n)

Si ( x = x & x >] (-1)·x_{n} ) ==> x >] (-1)·x_{n}

min{(-0),(-1)·(0.n)} >] (-1)·(0.n)

(-1)·max{0,(0.n)} >] (-1)·(0.n)

max{0,(0.n)} [< (0.n)

tibidabu

funicular:

(n/40) = (0.15)

socialismo bolivariano:

12€ = n+n = 6+6

social-democracia bolivariana:

6.15€ = n+(n/40) = 6+(0.15)


parque de atracciones:

n = 0.80€

socialismo:

32.80€ = 40n+n = 32+(0.80)

social-democracia:

32.02€ = 40n+(n/40) = 32+(0.02)