This
level-5 vital article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
There is no need to "agree that Ω(1) = 0" since that follows from the defintion: 1 has no prime factors. Also, there is no need to present two definitions for ω(n), with a verbose summation formula, since it can be expressed in a short sentence in words, just like with Ω(n). AxelBoldt
Well, for any additive function f, we have f(1)=0. This is because 1 is coprime to 1, therefore f(1) = f(1*1) = f(1) + f(1), hence f(1) = 0. In particular for any additive f, the function g = 2^f is multiplicative ... there is no need to consider 2^(f(n)-f(1)) ... in fact I am going to change this now. —Preceding unsigned comment added by 96.21.17.171 ( talk) 19:12, 27 June 2009 (UTC)
The straight definition of an additive function is that it preserves the addition operation. That does not imply a numeric function and certainly not a polynomial. There is no need for a multiplication to be defined. For example the string concatenation (often denoted by "+") supports additive functions. For example UPCASE. Your definitions are unnecessarily limiting. − Woodstone 21:39, 27 January 2006 (UTC)
how is it additive? or rather, how exactly do you limit the function to make it additive, if even possible?-- Ghazer ( talk) 23:47, 9 February 2008 (UTC)
As of 23Apr2011, the article lists
BigOmega(54,032,858,972,279)
as 3. However, my CAS lists 54,032,858,972,279 as factoring as
11^1 * 1993^2 * 1236661^1
So BigOmega(54,032,858,972,279) = 1 + 2 + 1 = 4.
72.196.124.204 ( talk) 04:18, 23 April 2011 (UTC) Jonathan King, Math Dept, Univ. of Florida
This
level-5 vital article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
There is no need to "agree that Ω(1) = 0" since that follows from the defintion: 1 has no prime factors. Also, there is no need to present two definitions for ω(n), with a verbose summation formula, since it can be expressed in a short sentence in words, just like with Ω(n). AxelBoldt
Well, for any additive function f, we have f(1)=0. This is because 1 is coprime to 1, therefore f(1) = f(1*1) = f(1) + f(1), hence f(1) = 0. In particular for any additive f, the function g = 2^f is multiplicative ... there is no need to consider 2^(f(n)-f(1)) ... in fact I am going to change this now. —Preceding unsigned comment added by 96.21.17.171 ( talk) 19:12, 27 June 2009 (UTC)
The straight definition of an additive function is that it preserves the addition operation. That does not imply a numeric function and certainly not a polynomial. There is no need for a multiplication to be defined. For example the string concatenation (often denoted by "+") supports additive functions. For example UPCASE. Your definitions are unnecessarily limiting. − Woodstone 21:39, 27 January 2006 (UTC)
how is it additive? or rather, how exactly do you limit the function to make it additive, if even possible?-- Ghazer ( talk) 23:47, 9 February 2008 (UTC)
As of 23Apr2011, the article lists
BigOmega(54,032,858,972,279)
as 3. However, my CAS lists 54,032,858,972,279 as factoring as
11^1 * 1993^2 * 1236661^1
So BigOmega(54,032,858,972,279) = 1 + 2 + 1 = 4.
72.196.124.204 ( talk) 04:18, 23 April 2011 (UTC) Jonathan King, Math Dept, Univ. of Florida