sábado, 10 de abril de 2021

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)

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