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 Inverse super-composition Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 01/12/2017, 04:19 PM 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? Xorter Unizo Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 05/26/2018, 12:00 AM 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 « Next Oldest | Next Newest »

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