The " helping function " from the OP was x^2.
Notice it is hard or even impossible to get a new helping function from An old one :
Let g(x) be analytic with fixpoints 0,1.
Then
g^[-1]( g(x)^2 )
Has the same fixpoints with the same derivatives at them.
So that does not work.
Formula's (addition , product ) for trig and hyperbolic trig function also do not give the Desired results.
Also the selfreference is not in our favour :
This type of lim form for a super requires a helper , but that helper needs a super ... With in the case of using this type of lim , again a helper.
---
So , putting this aside for a while , Maybe relaxing " closed form " helps.
Like Sums , integrals , products.
Not many product forms are discovered or written for functional equations and it seems rather hard.
Gottried solved
F(x+1) = e F(x) with the pxp(x).
Since this OP is more focused on x^2 I Will solve
T(x^2) = 2 T(x).
For x>0.
The aim is to generalize these products to get more helpers.
Or show it can not be done, which would be hard ; a very general statement is always hard to disprove without An example.
For x>0 and the product over n > 0 :
T(x) = (x-1) \( \Pi \) 2 (x^{2^(-n)} + 1)^{-1}.
I call pxp the superproduct and T(x) the abelproduct.
You can easily prove that T(x^2) = 2 T(x) by simply plugging in x^2.
In fact T(xy) = T(x) + T(y) for Re(x,y) > 0.
[ Re(x,y) = shorthand for Re(x),Re(y). ]
Actually T(x) is just ln(x) for x>0.
[yes , because we have a product for ln for positive x , this relates to fake function theory ... Notice we also have x^{1/2}, x^{1/4} etc in the Abelproduct , for which we have fakes as well. But that is a bit off topic here. In a way trivial because of the connection between products , zero's and fake function theory. However this is yet another way to arrive at a fake ln ( 3 or 2 ways , depending on how you count equivalent ) , so it is worth a comment imho. ]
I already expressed the aim with this , but without Trying hard , a logical guess seems to be finding Abelproduct to solve
2 k(x) = k(a x^2 + b x + c).
Notice a product that contains
(a x^2 + b x + c)^[m] for integer m is acceptable.
I consider a new thread , but this connects so strongly.
Otherwise it might seem a curious coincidence of little importance or potential.
Im also intrested in C^oo solutions.
I was not able to find nowhere analytic C^oo Abelproducts , but then again i lack time.
Have to go.
Regards
Tommy1729
Notice it is hard or even impossible to get a new helping function from An old one :
Let g(x) be analytic with fixpoints 0,1.
Then
g^[-1]( g(x)^2 )
Has the same fixpoints with the same derivatives at them.
So that does not work.
Formula's (addition , product ) for trig and hyperbolic trig function also do not give the Desired results.
Also the selfreference is not in our favour :
This type of lim form for a super requires a helper , but that helper needs a super ... With in the case of using this type of lim , again a helper.
---
So , putting this aside for a while , Maybe relaxing " closed form " helps.
Like Sums , integrals , products.
Not many product forms are discovered or written for functional equations and it seems rather hard.
Gottried solved
F(x+1) = e F(x) with the pxp(x).
Since this OP is more focused on x^2 I Will solve
T(x^2) = 2 T(x).
For x>0.
The aim is to generalize these products to get more helpers.
Or show it can not be done, which would be hard ; a very general statement is always hard to disprove without An example.
For x>0 and the product over n > 0 :
T(x) = (x-1) \( \Pi \) 2 (x^{2^(-n)} + 1)^{-1}.
I call pxp the superproduct and T(x) the abelproduct.
You can easily prove that T(x^2) = 2 T(x) by simply plugging in x^2.
In fact T(xy) = T(x) + T(y) for Re(x,y) > 0.
[ Re(x,y) = shorthand for Re(x),Re(y). ]
Actually T(x) is just ln(x) for x>0.
[yes , because we have a product for ln for positive x , this relates to fake function theory ... Notice we also have x^{1/2}, x^{1/4} etc in the Abelproduct , for which we have fakes as well. But that is a bit off topic here. In a way trivial because of the connection between products , zero's and fake function theory. However this is yet another way to arrive at a fake ln ( 3 or 2 ways , depending on how you count equivalent ) , so it is worth a comment imho. ]
I already expressed the aim with this , but without Trying hard , a logical guess seems to be finding Abelproduct to solve
2 k(x) = k(a x^2 + b x + c).
Notice a product that contains
(a x^2 + b x + c)^[m] for integer m is acceptable.
I consider a new thread , but this connects so strongly.
Otherwise it might seem a curious coincidence of little importance or potential.
Im also intrested in C^oo solutions.
I was not able to find nowhere analytic C^oo Abelproducts , but then again i lack time.
Have to go.
Regards
Tommy1729