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?
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