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Inverse super-composition
#11
Okey, I have found something looking like the solution. I feel I am closer than ever before.
For the following functional equation


my method gives:
Let us check it:
looks like sg like the exoponential function.

Of course, because at x=0 exp x = 1, then x2^n will never go up to 1, so in the reals there is no more beautiful solution for N like mine, I think or I am wrong, ain't I?

What do you think, is this function correct? Or is there better?
Xorter Unizo
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#12
Hi here, again!

I have been thinking about functional logarithm, and I coded it in pari/gp in this way:

Code:
D(f,n)={if(n>0,return(D(deriv(f),n-1)),return(f));};
M(f,n)=matrix(n,n,j,k,1/(k-1)!*subst(D(x*0+f^(j-1),k-1),x,0));
T(A,n)=sum(k=1,n,A[2,k]*x^(k-1));
inv(f,n)=T(M(f,n)^-1,n);
Ln(A,n)=sum(k=1,n,(-1)^(k+1)*(A-1)^k/k);
olog(f,g,n)=T(Ln(M(f,n),n^2)/(0.1^n+Ln(M(g,n),n^2)));

M is the Carleman-matrix, T is a generated taylor-series from the M matrix. Ln is log of a quadratic matrix. And olog is the functional logarithm: olog(f(x),(f^og(x))(x)) = g(x), but somewhy it is not working.
E. g. olog(2x,x*2^(2x),100...) = 2x.
Could help me?
Thank you very much!
Xorter Unizo
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