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+--- Thread: Inverse super-composition (/showthread.php?tid=1104)
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RE: Inverse super-composition - Xorter - 01/12/2017
Okey, I have found something looking like the solution. I feel I am closer than ever before.
For the following functional equation
\( 2x ^{oN} = exp(x) \)
my method gives: \( N = 0.5250548915-0.2859572213x+0.3455886342x^2-0.0759083804x^3+0.0201418181x^4-0.0045151851x^5+0.0006517298x^6+... \)
Let us check it:
\( 2x ^{o0.5250548915-0.2859572213x+0.3455886342x^2-0.0759083804x^3+0.0201418181x^4-0.0045151851x^5+0.0006517298x^6+...} \) 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?
RE: Inverse super-composition - Xorter - 05/26/2018
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!