EDIT: I misunderstood the op, as can be seen from this post and the child.
I don’t understand why no one else is objecting to treating (2) as a number.
If F(x, s) = {s-sided function of x}, e.g. F(2, 3) = /2\, F(2,5) = [2>, then clearly F(2,x) > 2^x for x > 3.
(2) is the limit of F(2, x) as x approaches infinity; just as 2^x is infinite in the limit, so is (2). I’m not even sure whether ((2)) is well-defined, because we haven’t been told how it approaches the limits, and it’s not clear to me that all methods yield the same function.
Oh actually you’re right. I didn’t interpret the op correctly. I thought it was just some weird extension of Knuth’s up arrow notation but now I see what’s going on.
In that sense, (2) isn’t a real number, as infinity isn’t a real number, it’s an extended real number.
And I think you’re right again, ((2)) isn’t well defined I don’t think.
Hmm, now I think you might be right, and that I misunderstood the poster’s original intention. The paragraph currently reads
…
I’ll spare the next [X] operators, and go right to (X) (“circle-X”). (X) follows the process that took us from triangle to square to pentagon, iterated an additional [X] times.
Is that an edit? I do not remember the phrase following the last comma. The notation is at least confusing, in that triangle->square->pentagon->...->circle ought to represent a limiting process, rather than a finite one.
Thank you. I would ask the op to use a less confusing notation, but I will go ahead and edit my other objections.
EDIT: I misunderstood the op, as can be seen from this post and the child.
I don’t understand why no one else is objecting to treating (2) as a number.
If F(x, s) = {s-sided function of x}, e.g. F(2, 3) = /2\, F(2,5) = [2>, then clearly F(2,x) > 2^x for x > 3.
(2) is the limit of F(2, x) as x approaches infinity; just as 2^x is infinite in the limit, so is (2). I’m not even sure whether ((2)) is well-defined, because we haven’t been told how it approaches the limits, and it’s not clear to me that all methods yield the same function.
Oh actually you’re right. I didn’t interpret the op correctly. I thought it was just some weird extension of Knuth’s up arrow notation but now I see what’s going on.
In that sense, (2) isn’t a real number, as infinity isn’t a real number, it’s an extended real number.
And I think you’re right again, ((2)) isn’t well defined I don’t think.
Count me wrong. You understood correctly the first time. See Vladimir’s comment; the notation is confusing, but it is a finite process.
It’s not. As I understand from the post, in your notation, (2)=F(2,[2])=F(2,F(2,4)).
Hmm, now I think you might be right, and that I misunderstood the poster’s original intention. The paragraph currently reads
Is that an edit? I do not remember the phrase following the last comma. The notation is at least confusing, in that triangle->square->pentagon->...->circle ought to represent a limiting process, rather than a finite one.
Thank you. I would ask the op to use a less confusing notation, but I will go ahead and edit my other objections.