05/07/2011, 10:45 PM

(05/07/2011, 08:41 PM)JmsNxn Wrote: This proof involves the use of a new operator:

and it's inverse:

and the little differential operator:

(The notation is more unambiguous than in your previous thread )

But your operator can be expressed with the classical differentiation, see:

Or purely functional with the composition operation :

PS: when you write ln with backslash in front:

Code:

`[tex]\ln(x)[/tex]`