I've been experimenting looking at these operators and I've become much more familiar with its structure. I can prove convergence now.
Let's suppose by contradiction that; for sufficiently large n:
\( \frac{x\,\,\bigtriangleup_n\,\,y}{(n\,\,\bigtriangleup_n\,\,n)^{n-s}}\,\, > \,\frac{1}{n^2} \)
However. It is clear that for some \( n\,\, >\,\, x,y \)
1: \( n\,\,\bigtriangleup_n\,\,n\,\, >\,\, x\,\,\bigtriangleup_n\,\,y \)
Therefore we write:
\( (n\,\,\bigtriangleup_n\,\,n)^{n-s}\,\, <\,\, n^2(x\,\,\bigtriangleup_n\,\,y) \)
Now we know that \( n^2\,\, <\,\, x\,\,\bigtriangleup_n\,\,y \)
Therefore:
\( (n\,\,\bigtriangleup_n\,\,n)^{n-s}\,\, <\,\, (x\,\,\bigtriangleup_n\,\,y)^2 \)
However this is a contradiction because for sufficiently large \( n \) the left equation becomes much larger than the right. This is easy to deduce by the relation 1 above. Therefore to prove convergence we just need add the claim:
\( \ln(x\,\,\bigtriangleup_n\,\,y)\,\, < \,\,x\,\,\bigtriangleup_n\,\,y \)
\( |\vartheta_n(s)| \,\,\le \,\,|(n\,\,\bigtriangleup_n\,\,n)^{s-n}| \)
YES! We have convergence for all s.
\( x\,\,\bigtriangleup_s\,\,y = \prod_{n=0}^{\infty} (x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)} \)
\( \vartheta_n(s) = \frac{\sin(\pi(s-n))}{\pi(s-n)}(n\,\,\bigtriangleup_n\,\,n)^{s-n} \psi_n(s) \)
An important theorem I have to prove is the following, I consider it a stern requirement of hyperoperators. For all \( \Re(s) \ge 0 \) and \( \epsilon \,\,>\,\, 0 \) and \( x,y\,\,> 1 \)
\( |x\,\,\bigtriangleup_{s+\epsilon}\,\,y| > |x\,\,\bigtriangleup_{s}\,\,y| \)
I'll mull over that for awhile.
also; hopefully:
\( \frac{d^n}{ds^n} x \,\,\bigtriangleup_s\,\,y\,\,\,\,> 0 \) for at least \( \Re(s) > 0 \)
Let's suppose by contradiction that; for sufficiently large n:
\( \frac{x\,\,\bigtriangleup_n\,\,y}{(n\,\,\bigtriangleup_n\,\,n)^{n-s}}\,\, > \,\frac{1}{n^2} \)
However. It is clear that for some \( n\,\, >\,\, x,y \)
1: \( n\,\,\bigtriangleup_n\,\,n\,\, >\,\, x\,\,\bigtriangleup_n\,\,y \)
Therefore we write:
\( (n\,\,\bigtriangleup_n\,\,n)^{n-s}\,\, <\,\, n^2(x\,\,\bigtriangleup_n\,\,y) \)
Now we know that \( n^2\,\, <\,\, x\,\,\bigtriangleup_n\,\,y \)
Therefore:
\( (n\,\,\bigtriangleup_n\,\,n)^{n-s}\,\, <\,\, (x\,\,\bigtriangleup_n\,\,y)^2 \)
However this is a contradiction because for sufficiently large \( n \) the left equation becomes much larger than the right. This is easy to deduce by the relation 1 above. Therefore to prove convergence we just need add the claim:
\( \ln(x\,\,\bigtriangleup_n\,\,y)\,\, < \,\,x\,\,\bigtriangleup_n\,\,y \)
\( |\vartheta_n(s)| \,\,\le \,\,|(n\,\,\bigtriangleup_n\,\,n)^{s-n}| \)
YES! We have convergence for all s.
\( x\,\,\bigtriangleup_s\,\,y = \prod_{n=0}^{\infty} (x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)} \)
\( \vartheta_n(s) = \frac{\sin(\pi(s-n))}{\pi(s-n)}(n\,\,\bigtriangleup_n\,\,n)^{s-n} \psi_n(s) \)
An important theorem I have to prove is the following, I consider it a stern requirement of hyperoperators. For all \( \Re(s) \ge 0 \) and \( \epsilon \,\,>\,\, 0 \) and \( x,y\,\,> 1 \)
\( |x\,\,\bigtriangleup_{s+\epsilon}\,\,y| > |x\,\,\bigtriangleup_{s}\,\,y| \)
I'll mull over that for awhile.
also; hopefully:
\( \frac{d^n}{ds^n} x \,\,\bigtriangleup_s\,\,y\,\,\,\,> 0 \) for at least \( \Re(s) > 0 \)
