The infinite operator is simply defined as:

\( a\,\,\bigtriangleup_{0}\,\,b = a + b\\

\\a\,\,\bigtriangleup_{1}\,\,b = a\cdot b\\

\\a\,\,\bigtriangleup_{2}\,\,b = a^b\\

...\\

\lim_{n\to\infty} a\,\,\bigtriangleup_{n}\,\,b = a\,\,\bigtriangleup_{\infty}\,\,b

\)

I know that for negative integers values of b (and zero) it returns to successorship, i.e

\( a\,\, \bigtriangleup_{\infty}\,\, 0 = 1\\

\\a\,\, \bigtriangleup_{\infty}\,\, -1 = 0\\

\\a\,\, \bigtriangleup_{\infty}\,\, -2 = -1\\

... \)

but we also have an interesting formula, since the euler constants and the eta constants converge to 2 and 4, and since infinite iteration of eta constant x with hyper operator x results in euler constant x we have this very odd formula:

\( 2\,\, \bigtriangleup_{\infty} \,\,(2\,\, \bigtriangleup_{\infty} \,\,(2\,\, \bigtriangleup_{\infty} \,\,(2... \,\, = 4 \)

but we also know that:

\( 2\,\,\bigtriangleup_{\infty}\,\,2 = 4 \)

and that

\( 2 = 4\,\,\bigtriangledown_{\infty}\,\,4 = 2 \)

if this is \( \bigtriangledown_{\infty} \) is the root inverse of the infinite operator.

we find that

\( f(x) = 2 \bigtriangleup_{\infty} x \) equals 4 at both 2 and 4. This means, if we want the infinite operator to be analytic and continuous, we have at least one zero in the second derivative. But we'd probably want two so that f(x) is increasing towards infinity.

Also, I was wondering, since it's completely possible to take the superfunction of the infinite operator, how would we denote that? Since it can't be \( \infty + 1 \) should we make it something like \( \infty + k \) where k is just another unit but denotes a seperate branch of superfunctions on top of the infinite operator?

\( g(x) =2\,\,\bigtriangleup_{\infty+k}\,\,x \) \( g(x) = 4 \) for all \( x\ge2 \)

This means we have a flatline for the infinite operators superfunction. Interesting.

\( a\,\,\bigtriangleup_{0}\,\,b = a + b\\

\\a\,\,\bigtriangleup_{1}\,\,b = a\cdot b\\

\\a\,\,\bigtriangleup_{2}\,\,b = a^b\\

...\\

\lim_{n\to\infty} a\,\,\bigtriangleup_{n}\,\,b = a\,\,\bigtriangleup_{\infty}\,\,b

\)

I know that for negative integers values of b (and zero) it returns to successorship, i.e

\( a\,\, \bigtriangleup_{\infty}\,\, 0 = 1\\

\\a\,\, \bigtriangleup_{\infty}\,\, -1 = 0\\

\\a\,\, \bigtriangleup_{\infty}\,\, -2 = -1\\

... \)

but we also have an interesting formula, since the euler constants and the eta constants converge to 2 and 4, and since infinite iteration of eta constant x with hyper operator x results in euler constant x we have this very odd formula:

\( 2\,\, \bigtriangleup_{\infty} \,\,(2\,\, \bigtriangleup_{\infty} \,\,(2\,\, \bigtriangleup_{\infty} \,\,(2... \,\, = 4 \)

but we also know that:

\( 2\,\,\bigtriangleup_{\infty}\,\,2 = 4 \)

and that

\( 2 = 4\,\,\bigtriangledown_{\infty}\,\,4 = 2 \)

if this is \( \bigtriangledown_{\infty} \) is the root inverse of the infinite operator.

we find that

\( f(x) = 2 \bigtriangleup_{\infty} x \) equals 4 at both 2 and 4. This means, if we want the infinite operator to be analytic and continuous, we have at least one zero in the second derivative. But we'd probably want two so that f(x) is increasing towards infinity.

Also, I was wondering, since it's completely possible to take the superfunction of the infinite operator, how would we denote that? Since it can't be \( \infty + 1 \) should we make it something like \( \infty + k \) where k is just another unit but denotes a seperate branch of superfunctions on top of the infinite operator?

\( g(x) =2\,\,\bigtriangleup_{\infty+k}\,\,x \) \( g(x) = 4 \) for all \( x\ge2 \)

This means we have a flatline for the infinite operators superfunction. Interesting.