07/19/2012, 10:16 PM
(07/19/2012, 04:44 PM)JmsNxn Wrote: We start by defining a sequence of analytic functions that obey the following rules:
\( \vartheta_n : \mathbb{C} \to \mathbb{C} \)
\( \vartheta_n(n) = 1 \) and \( k \in \mathbb{N}_0\,\,\,\vartheta_n(k) = 0 \,\,\Leftrightarrow\,\, k \neq n \)
Necessarily; these functions can be arbitrary and they pose the true uniqueness criterion. Nonetheless we suggest these functions and give a plausible solution.
\( {\bf I}:\,\,\,\,\vartheta_n(s) = \frac{s}{n} e^{e^s-e^n} \prod_{k=1}^{\infty}\frac{1-\frac{s}{k}}{1-\frac{n}{k}}e^{ \frac{n - s}{k}}\,\,\,;\,\,\,n \ge 1 \)
\( \vartheta_0(s) = e^{e^s-1} \prod_{k=1}^{\infty}(1-\frac{s}{k})e^{-\frac{s}{k}} \)
Reason for this is it has fast convergence and satisfies the required conditions.
Now we perform a trick. Considering hyper operators; which are written by the following:
\( x\,\,\bigtriangleup_{s}\,\,(x \,\,\bigtriangleup_{s+1}\,\,y) = x \,\,\bigtriangleup_{s+1}\,\,(y+1) \)
\( x \,\,\bigtriangleup_0\,\,y = x + y \)
and the identity of \( \bigtriangleup_n \) for any natural \( n \) is \( 1 \).
We get the usual sequence where zero is addition, one is multiplication, two is exponentiation, etc... We then write that complex operators are products of natural operators with complex exponents. Here; \( \vartheta_n \) is as before.
\( {\bf II}:\,\,\,\,x \,\,\bigtriangleup_s\,\, y = \prod_{n=0}^{\infty}(x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)} = (x + y)^{\vartheta_0(s)} \cdot (x\cdot y)^{\vartheta_1(s)} \cdot (x^y) ^{\vartheta_2(s)} \cdot (^y x)^{\vartheta_3(s)} \cdot ... \)
Dont you need to prove the identity of the infinite product ?
e.g. does the infinite product formula hold for x + y and x*y ??
regards
tommy1729
