Nth root when N is colossal

[chasrob]

I was informed about this equation when I asked about distribution.
Distribute N points throughout a "p-dimensional" ball uniformly and select a point; what is the distance between that point and the nearest point,on average?
[distance] = R * (1 - 1/2^1/N)^1/3

As I understand it, 1/N is taking the Nth root of, in this case, 2. Correct?If N is enormous, say 4→7→3→3 in chained arrow notation, wouldn’t the expression in parentheses reduce to 0 and therefore make no sense?

I'm a sci-fi writer and this question comes up in a story I'm working on. BTW, R equals 2→3→2→2→2→2→2!

 

[Deleted member 4993]

I was informed about this equation when I asked aboutdistribution.
Distribute N points throughout a "p-dimensional" ball uniformly andselect a point; what is the distance between that point and the nearest point,on average?
[distance] = R * (1 - 1/2^1/N)^1/3

As I understand it, 1/N is taking the Nth root of, in this case, 2. Correct?If N is enormous, say 4→7→3→3 in chained arrow notation, wouldn’t the expression in brackets reduce to 0 and therefore make no sense?
I'm a sci-fi writer and this question comes up in a story I'm working on. BTW, R equals 2→3→2→2→2→2→2!

Why?

If you have very large number points inside a finite circle - the average "nearest" distance would become very small.
\(\displaystyle \displaystyle{\lim_{N\to \infty}R * \left(1 - \dfrac{1}{2^{\frac{1}{N}}}\right)^{\frac{1}{3}}}\)

=\(\displaystyle \displaystyle{R * \left(1 - \dfrac{1}{2^{\frac{1}{\infty}}}\right)^{\frac{1}{3}}}\)

=\(\displaystyle \displaystyle{R * \left(1 - \dfrac{1}{2^{0}}\right)^{\frac{1}{3}}}\)

 

[chasrob]

Why?

If you have very large number points inside a finite circle - the average "nearest" distance would become very small.
\(\displaystyle \displaystyle{\lim_{n\to \infty}R * \left(1 - \dfrac{1}{2^{\frac{1}{N}}}\right)^{\frac{1}{3}}}\)

=\(\displaystyle \displaystyle{R * \left(1 - \dfrac{1}{2^{\frac{1}{\infty}}}\right)^{\frac{1}{3}}}\)

=\(\displaystyle \displaystyle{R * \left(1 - \dfrac{1}{2^{0}}\right)^{\frac{1}{3}}}\)

Yes the circle/ball is finite, but as I mentioned, R = 2→3→2→2→2→2→2 in that linked notation. A very, very large number, large enough to make N (which dwarfs Graham's Number) unnoticeable, IE in-computably larger than N.

What if N → 0? Result be R = R * 1?

Thanks for your help.

 

[mmm4444bot]

… infinity times zero = infinity …

I would say that infinity times anything is an indeterminant form (aka: undefined).

Are you really thinking the distance between two points will become infinitely large, when the number of points distributed in the sphere becomes infinity large?

 

[chasrob]

I would say that infinity times anything is an indeterminant form (aka: undefined).

Are you really thinking the distance between two points will become infinitely large, when the number of points distributed in the sphere becomes infinity large?

I got that from a page on the web, where infinity times any number is still infinity... including zero. My algebra teacher, many years ago, agreed with you--its undefined. Matter of opinion I guess.

What I'm surmising is that when the sphere is in-computably larger than the number of points, the average distance is R (which is the sphere's rough size in the short-hand notation that must be used).

 

[mmm4444bot]

I got that from a page on the web, where infinity times any number is still infinity... including zero … Matter of opinion I guess.

Sure. I'd need to see what they were discussing. Some "gimmicks" in math turn out to be useful conventions (like electrical engineers viewing zero as a negative number; helps them out, somehow).

What I'm surmising is that when the sphere is in-computably larger than the number of points, the average distance is R …

Oops, I misunderstood. I had been thinking that the sphere was fixed and then N headed toward infinity. I missed the comparison above. Please excuse me.

 

[chasrob]

My bad, the title itself is misleading. The R in the OP equation is what's colossal, compared to the N. So I surmise that N would actually go zero, and the distance would be equal to R * 1; R?
Thanks for your help.