Talk:Natural density

The set of all square free integers has density (pi^2)/6.

i added this line under examples. i haven't figured out how to type math —Preceding unsigned comment added by Jesusonfire (talk • contribs) 10:40, 16 December 2007 (UTC)

i am sry i a too drunk. i meant 6/pi squared —Preceding unsigned comment added by Jesusonfire (talk • contribs) 10:44, 16 December 2007 (UTC)

Asymptotic Density

I have seen the term asymptotic density be used to refer to more than just the density of subsets of the natural numbers. For example, see the text Number Theoretic Density and Logical Limit Laws by Stanley N. Burris. Adammanifold (talk) 18:04, 28 January 2010 (UTC)

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Set theory notation

Would it be possible to remove the "too technical" banner if a natural language explanation of the "definition" section were given, along side an overview of some of the set theory notation signs?Edaham (talk) 05:23, 10 January 2017 (UTC)

Define?

Could the article define "upper density" and "lower density", instead of just give an example? Bubba73 ^{You talkin' to me?} 19:51, 20 August 2019 (UTC)

Never mind - they are defined as "upper asymptotic density", etc. Bubba73 ^{You talkin' to me?} 19:55, 20 August 2019 (UTC)

Example "set of numbers whose binary expansion contains an odd number of digits"

Can someone please explain how we got the 2 different equations for upper and lower densities in that example? How (or why) does it use 2^(2m+1) in the denominator for upper, but 2^(2m+2) in the denominator for lower? Ateista (talk) 13:39, 28 February 2024 (UTC)

I guess, binary expansion of any number from $[2^{2m+1};2^{2m+2}-1]$ contains an even number of digits, so the numerators in this two fractions are the same.

Denominator in the first fraction corresponds to the beginning of the interval, and to the end of the interval in the second case.
Pepka-prygni (talk) 10:39, 17 May 2024 (UTC)

Lower asymptotic density

Hello!

I think these two definitions $${\underline {d_{1}}}(A)=\liminf _{n\rightarrow \infty }{\frac {a(n)}{n}}$$ $${\underline {d_{2}}}(A)=\liminf _{n\rightarrow \infty }{\frac {n}{a_{n}}}$$ are not equivalent. Consider A as a union of two arithmetical progressions $A=\{3\cdot n\}\cup \{5\cdot n\},n\in \mathbb {N}$.

Then

${\underline {d_{1}}}(A)={\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{15}}={\frac {7}{15}}$

${\underline {d_{2}}}(A)\leq {\frac {1}{5}}$
Pepka-prygni (talk) 08:09, 17 May 2024 (UTC)