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datadataeverywhere comments on Approaching Infinity - Less Wrong Discussion

-1 Post author: Psychohistorian 01 February 2011 08:11AM

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Comment author: datadataeverywhere 01 February 2011 04:28:25PM *  1 point [-]

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.

Comment author: Vladimir_Nesov 01 February 2011 07:45:52PM 1 point [-]

(2) is the limit of F(2, x) as x approaches infinity

It's not. As I understand from the post, in your notation, (2)=F(2,[2])=F(2,F(2,4)).

Comment author: datadataeverywhere 01 February 2011 08:03:19PM 0 points [-]

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.

Comment author: calef 01 February 2011 06:07:38PM *  1 point [-]

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.

Comment author: datadataeverywhere 01 February 2011 08:05:24PM *  0 points [-]

Count me wrong. You understood correctly the first time. See Vladimir's comment; the notation is confusing, but it is a finite process.

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