funcions expansives

Sigui A totalment ordenat

Si < f: A ---> A & a_{k} --> f(a_{k}) = max{a_{1},...,a_{n}} > ==> a_{k} [< f(a_{k})

Si < f: A ---> A & a_{k} --> f(a_{k}) = min{a_{1},...,a_{n}} > ==> a_{k} >] f(a_{k})

Si < f: A ---> A & a_{k} --> f(a_{k}) = sup{a_{1},...,a_{n}} > ==> a_{k} < f(a_{k})

Si < f: A ---> A & a_{k} --> f(a_{k}) = inf{a_{1},...,a_{n}} > ==> a_{k} > f(a_{k})

Si < f: A ---> A & x --> f_{n}(x) = x+n > ==> x [< f_{n}(x)

x [< x+n [< x+(n+1)

Si < f: A ---> A & x --> f_{n}(x) = x+(-n) > ==> x >] f_{n}(x)

Si < f: A ---> A & x --> f_{n}(x) = x+np > ==> x [< f_{n}(x)

x [< x+np [< x+np+p = x+(n+1)·p

Si < f: A ---> A & x --> f_{n}(x) = x+(-n)·p > ==> x >] f_{n}(x)

Si < f: A ---> A & x --> f_{n}(x) = x+p^{n} > ==> x < f_{n}(x)

x [< x+p^{n} [< x+p^{n}+...(p)...+p^{n} = x+p^{(n+1)}

Si < f: A ---> A & x --> f_{n}(x) = x+(-1)·p^{n} > ==> x > f_{n}(x)

funcions transitives

{x} ---> {f( {x} )} ---> {g( {f( {x} )} )}

}x{ ---> }f( }x{ ){ ---> }g( }f( }x{ ){ ){

A[ & ]{x} ---> A[ & ]{f( A[ & ]{x} )} ---> A[ & ]{g( A[ & ]{f( A[ & ]{x} )} )}

A[ & ]}x{ ---> A[ & ]}f( A[ & ]}x{ ){ ---> A[ & ]}g( A[ & ]}f( A[ & ]}x{ ){ ){

A[ || ]{x} ---> A[ || ]{f( A[ || ]{x} )} ---> A[ || ]{g( A[ || ]{f( A[ || ]{x} )} )}

A[ || ]}x{ ---> A[ || ]}f( A[ || ]}x{ ){ ---> A[ || ]}g( A[ || ]}f( A[ || ]}x{ ){ ){

{x_{1},...,x_{n}} ---> {f_{1}( {x_{1},...,x_{n}} ),...,f_{n}( {x_{1},...,x_{n}} )} ---> ...

... {

... g_{1}( {f_{1}( {x_{1},...,x_{n}} ),...,f_{n}( {x_{1},...,x_{n}} )} ),...,

... g_{n}( {f_{1}( {x_{1},...,x_{n}} ),...,f_{n}( {x_{1},...,x_{n}} )} )

... }

}x_{1},...,x_{n}{ ---> }f_{1}( }x_{1},...,x_{n}{ ),...,f_{n}( }x_{1},...,x_{n}{ ){ ---> ...

... }

... g_{1}( }f_{1}( }x_{1},...,x_{n}{ ),...,f_{n}( }x_{1},...,x_{n}{ ){ ),...,

... g_{n}( }f_{1}( }x_{1},...,x_{n}{ ),...,f_{n}( }x_{1},...,x_{n}{ ){ )

... {

laboratori de problemes:

A[ & ]{x_{1},...,x_{n}}

A[ & ]}x_{1},...,x_{n}{

A[ || ]{x_{1},...,x_{n}}

A[ || ]}x_{1},...,x_{n}{

desigualtat

constructor:

Si ( k€N & p€N ) ==> 1 [< p^{k}

1 [< p^{k} [< p^{k}+...(p)...+p^{k} = p^{(k+1)}

destructor:

Si ( k€N & p€N ) ==> p^{k} < 1

constructor:

Si ( k€N & p€N ) ==> 1 [< p^{k}+kn

1 [< p^{k}+kn [< p^{(k+1)}+kn [< p^{(k+1)}+kn+n = p^{(k+1)}+(k+1)·n

destructor:

Si ( k€N & p€N ) ==> p^{k}+kn < 1

negació de desigualtat

Sigui A totalment ordenat.

¬( x [< y ) <==> y < x

[<==] absurd

x [< y & ( y [< x & y != x )

[==>]

( x [< y || y [< x ) & ¬( x [< y)

y [< x & ¬( x [< y )

y [< x & ( ¬( x [< y ) || ¬( y [< x ) )

y [< x & y != x

destructor algebràic

d_{x}[x^{(n+1)}] = (-1)·(n+1)·x^{n}

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

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

int[x^{n}] d[x] = ((-n)+(-1))·x^{(-n)+(-1)}

cadenes de funcions - topología algebràica

f(x) = y+x

g(x) = (-y)+x

cadena per composició:

   0 ----> f(0)   ---->...(n)...---->   (fo...(n)...of)(0)

(-0) ----> g(-0) ---->...(n)...----> (go...(n)...og)(-0)

f(x) = yx

g(x) = (1/y)·x

   1 ---->    f(1)   ---->...(n)...---->   (fo...(n)...of)(1)

(1/1) ----> g(1/1) ---->...(n)...----> (go...(n)...og)(1/1)

f(x) = ln(x)

g(x) = e^{x} = exp(x)

1 ---->  f(1)   ---->...(n)...---->   (fo...(n)...of)(exp(...(n)...exp(0)...(n)...))

0 ----> g(0)   ---->...(n)...---->    (go...(n)...og)(ln(...(n)...ln(1)...(n)...))

f_{n}(x) = n^{k}+x

g_{n}(x) = (-1)·n^{k}+x

   0 ---->  f_{1}(0)   ---->...(n)...---->   (f_{n}o...(n)...of_{1})(0)

(-0) ----> g_{1}(-0) ---->...(n)...---->    (g_{n}o...(n)...og_{1})(-0)

f_{n}(x) = n^{k}·x

g_{n}(x) = (1/n^{k})·x

1      ---->  f_{1}(1)   ---->...(n)...---->   (f_{n}o...(n)...of_{1})(1)

(1/1) ----> g_{1}(1/1)   ---->...(n)...---->  (g_{n}o...(n)...og_{1})(1/1)

f(x^{n}) = d_{x}[ x^{n} ]

g(x^{(1/n)}) = d_{x}[ x^{(1/n)} ]

1 ---->  f(1)   ---->...(n)...---->   (fo...(n)...of)(1)

1 ----> g(1)   ---->...(n)...---->    (go...(n)...og)(1)

f(x^{(-n)}) = d_{x}[ x^{(-n)} ]

g(x^{(1/(-n))}) = d_{x}[ x^{(1/(-n))} ]

1 ---->  f(1)   ---->...(n)...---->   (fo...(n)...of)(1)

1 ----> g(1)   ---->...(n)...---->    (go...(n)...og)(1)

lavabo-habitación según l'evangelio borroso stronikián

En mi piso tengo tres habitaciones de cuatro compartimentos de casa.

P(x) = (3/4)

|xxx|-|o|

En mi piso tengo un lavabo de cuatro compartimentos de casa.

¬P(x) = (1/4) 

|ooo|-|x|

cons lasers

volum de un con:

int[ 0--> h ][ 2pi·R·(x/h) ] d[x] = pi·R·h

P·pi·R·h = h_{e}f

P·pi·R·h = (-1)·h_{e}f

destructor algebraic

a+...(n)...+a+a+...(m)...+a+a = (n+m)·a+a

(-a)+...(n)...+(-a)+(-a)+...(m)...+(-a)+(-a) = (-1)·(n+m)·a+(-a)

a+...(n)...+a+a+...(m)...+a = (-1)·(n+m)·a

a+...(n)...+a+a+...(m)...+a = (-n)·a+(-m)·a

(-a)+...(n)...+(-a)+(-a)+...(m)...+(-a) = (n+m)·a

(-a)+...(n)...+(-a)+(-a)+...(m)...+(-a) = na+ma

a·...(n)...·a·a·...(m)...·a·a = a^{(n+m)}·a

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

a·...(n)...·a·a·...(m)...·a = a^{(-1)·(n+m)}

a·...(n)...·a·a·...(m)...·a = (1/a^{n})·(1/a^{m})

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

(1/a)·...(n)...·(1/a)·(1/a)·...(m)...·(1/a) = a^{n}·a^{m}

mis nombres

nombre Judío-cristiano:

Sant Jûan l'stronikián.

nombre islámico:

Jûanathád l'stronikián.

dual de Anathana.