12/01/2010, 06:14 PM
f(x) = f(a(x)) = f(b(x))
when a(x) and b(x) are non-linear it "seems" that a(x) and b(x) need to commute. (*)
in fact P(x) = 2-periodic function ( period a and b )
S(x) => S(a(x)) = S(x) + a , S(b(x)) = S(x) + b
hence S(a(b(x))) = S(b(a(x))) = S(x) + a + b (*)
and
f(x) = P(S(x))
"seems" the only solution.
note that we must have poles in this case because of the 2-periodic function.
however , i mentioned "seems" , because i dont know many counterexamples.
but if
f(x) = f(exp(x))
then f(x) = f(x + 2pi i)
and i excluded that by writing ' nonlinear ' above.
but thats just because i know the following :
f(x) = 1-periodic [(slog(x))]
slog(x) = slog(exp(x)) - 1
slog(x + 2pi i) = slog(exp(x + 2pi i)) -1 = slog(x)
hence f(x) = 1-periodic [(slog(x))] = 1-periodic [(slog(x + 2pi i))]
thus f(x) = f(x + 2pi i)
in the OP i wrote :
**
f(x) = f(g(x))
its a simple equation.
and i talked about it before. as did e.g. gottfried.
in particular for a certain given g(x) the equation is solvable for f(x) by using the super of g(x) and a 1-periodic function...
but that is not the end of the story.
what if f(x) is given and we want to find g(x) ?
well for starters g(x) might not be unique.
if f(x) = f(exp(x)) then we must also have f(x) = f(x + 2pi i)
the pattern is clear
f(x) = f(g(x))
=> g(x) = g(g1(x)) => f(x) = f(g1(x))
and we can now replace g(x) with g1(x) , g1(x) with g2(x) and repeat ...
until we arrive at a moebius function.... if we ever ...
i guess we could call g_n(x) the invariants of f(x).
**
so it seems the g_n are a family of invariants.
now back to the modified (!) equation
f(x) = f(a_n(x)) = f(b_n(x))
where a_n and b_n are not iterations or invariants of eachother.
now it seems that if a_n and b_n are not cyclic iterations , then indeed
P(x) = 2-periodic function ( period a and b )
S(x) => S(a(x)) = S(x) + a , S(b(x)) = S(x) + b
hence S(a(b(x))) = S(b(a(x))) = S(x) + a + b (*)
and
f(x) = P(S(x))
"seems" the only solution....
or not ?
does this rule out ( with the similar above conditions )
f(x) = f(a_n(x)) = f(b_n(x)) = f(c_n(x)) then ?
i think so.
but what about algebraic functions then ?
consider for instance
f(x) = f(alg(x)) and the related alg^[n]alg(x).
more specifly if they have closed forms , such as the following ( with thanks to "achille" )
For p, r, s with p > 0,
Let Q(x) be (p^2-1)*x^2 + 2*(p+1)*r*x + s,
then
f(x) = p*x + sqrt(Q(x)) + r;
=> f^[n](x) = T_n(p)*x + U_{n-1}(p)*sqrt(Q(x)) + r*(T_n(p)-1)/(p-1);
where T_n(p) and U_n(p) are the n-th Chebyshev's
polynomial of first and second kind.
In particular, for (p,r,s) = (3,1,20)
f(x) = 3*x+ 2*sqrt(2*x^2+2*x+5) + 1
=> f^[2](x) = 17*x + 12*sqrt(2*x^2+2*x+5) + 8
f^[3](x) = 99*x + 70*sqrt(2*x^2+2*x+5) + 49
...
f^[n](x) = T_n(3)*x + 2*U_{n-1}(3)*sqrt(2*x^2+2*x+5) + (T_n(3)-1)/2
makes one wonder about equations like
f(x) = f(3*x+ 2*sqrt(2*x^2+2*x+5) + 1)
or more complicated ones !
maybe i just need a good book on invariants or alike :p
regards
tommy1729
when a(x) and b(x) are non-linear it "seems" that a(x) and b(x) need to commute. (*)
in fact P(x) = 2-periodic function ( period a and b )
S(x) => S(a(x)) = S(x) + a , S(b(x)) = S(x) + b
hence S(a(b(x))) = S(b(a(x))) = S(x) + a + b (*)
and
f(x) = P(S(x))
"seems" the only solution.
note that we must have poles in this case because of the 2-periodic function.
however , i mentioned "seems" , because i dont know many counterexamples.
but if
f(x) = f(exp(x))
then f(x) = f(x + 2pi i)
and i excluded that by writing ' nonlinear ' above.
but thats just because i know the following :
f(x) = 1-periodic [(slog(x))]
slog(x) = slog(exp(x)) - 1
slog(x + 2pi i) = slog(exp(x + 2pi i)) -1 = slog(x)
hence f(x) = 1-periodic [(slog(x))] = 1-periodic [(slog(x + 2pi i))]
thus f(x) = f(x + 2pi i)
in the OP i wrote :
**
f(x) = f(g(x))
its a simple equation.
and i talked about it before. as did e.g. gottfried.
in particular for a certain given g(x) the equation is solvable for f(x) by using the super of g(x) and a 1-periodic function...
but that is not the end of the story.
what if f(x) is given and we want to find g(x) ?
well for starters g(x) might not be unique.
if f(x) = f(exp(x)) then we must also have f(x) = f(x + 2pi i)
the pattern is clear
f(x) = f(g(x))
=> g(x) = g(g1(x)) => f(x) = f(g1(x))
and we can now replace g(x) with g1(x) , g1(x) with g2(x) and repeat ...
until we arrive at a moebius function.... if we ever ...
i guess we could call g_n(x) the invariants of f(x).
**
so it seems the g_n are a family of invariants.
now back to the modified (!) equation
f(x) = f(a_n(x)) = f(b_n(x))
where a_n and b_n are not iterations or invariants of eachother.
now it seems that if a_n and b_n are not cyclic iterations , then indeed
P(x) = 2-periodic function ( period a and b )
S(x) => S(a(x)) = S(x) + a , S(b(x)) = S(x) + b
hence S(a(b(x))) = S(b(a(x))) = S(x) + a + b (*)
and
f(x) = P(S(x))
"seems" the only solution....
or not ?
does this rule out ( with the similar above conditions )
f(x) = f(a_n(x)) = f(b_n(x)) = f(c_n(x)) then ?
i think so.
but what about algebraic functions then ?
consider for instance
f(x) = f(alg(x)) and the related alg^[n]alg(x).
more specifly if they have closed forms , such as the following ( with thanks to "achille" )
For p, r, s with p > 0,
Let Q(x) be (p^2-1)*x^2 + 2*(p+1)*r*x + s,
then
f(x) = p*x + sqrt(Q(x)) + r;
=> f^[n](x) = T_n(p)*x + U_{n-1}(p)*sqrt(Q(x)) + r*(T_n(p)-1)/(p-1);
where T_n(p) and U_n(p) are the n-th Chebyshev's
polynomial of first and second kind.
In particular, for (p,r,s) = (3,1,20)
f(x) = 3*x+ 2*sqrt(2*x^2+2*x+5) + 1
=> f^[2](x) = 17*x + 12*sqrt(2*x^2+2*x+5) + 8
f^[3](x) = 99*x + 70*sqrt(2*x^2+2*x+5) + 49
...
f^[n](x) = T_n(3)*x + 2*U_{n-1}(3)*sqrt(2*x^2+2*x+5) + (T_n(3)-1)/2
makes one wonder about equations like
f(x) = f(3*x+ 2*sqrt(2*x^2+2*x+5) + 1)
or more complicated ones !
maybe i just need a good book on invariants or alike :p
regards
tommy1729