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[March for reflection|Maynard]mindset and method (continued)

已有 374 次阅读 2022-7-11 15:30 |个人分类:牛津大学|系统分类:科研笔记

[注:下文是群邮件的内容,标题是原有的。内容是学习一篇数学文章的笔记。]

["Terms of awareness /use" folded below] On going is to read a paper of primes to increase generic understanding on mathematics.

Knowledge is not knowledge until needed in circumstance.

 ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ Λ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁻⁰ 1

To give a brief indication as to why such a principle might hold, we see that by inclusion-exclution on the smallest prime factor P⁻(n) of n we have

#{ ∈ Ap > x^1/2} = #{ n ∈ A: P⁻(n) > x^1/4} - Σ(p) #{ n ∈ A: P⁻(n) = }.

Note: Σ(p) temporarily denotes Σ(x^1/4<p<=x^1/2).

---- The first term on the right is to count the members n in A, such that P⁻(n) > x^1/4.

----  P⁻(n) is the smallest prime factor of n.

---- Say, in the case of 15 = 3·5, one has P⁻(15) = 3.

---- P⁻(2^k) = 2, P⁻(7·11·23) = 7, etc..

---- That is, for a composite number n, one can always sort its prime factors to pick up the smallest one.

---- While P⁻(n) is not an obscure concept, but what is the idea behind? How did it occur at first?

---- The key point is that, n may or may not be a composite number!

---- In the case of n being a prime, one has P⁻(n) = n.

---- To proceed, I denote the set ∈ Ap > x^1/2 } as A(ξ)p, meaning the collection of primes greater than ξ in A.

---- Similarly, { n ∈ A: P⁻(n) > x^1/4 } is denoted as A(η)n.

---- For n being primes, A(η)n is the same as A(η)p.

---- So, one apparently has A(ξ)p ⊆ A(η)p ⊆ A(η)n, assuming η < ξ.

---- As p can be written as P⁻(p), one can view P⁻(p) as a general form of p.

---- Under this view, A(ξ)p  and A(η)n are of the same form.

---- By the way, primes are the invariant points of the function P⁻(·).

---- On the level of counting, the total counts of A(ξ)p are included in the total counts of A(η)n.

---- So, to setup an equality between #A(ξ)p and #A(η)n, one needs to remove the addtional counts in the later.

---- To this end, it appears useful to examine a specific case of A.

(Also, the second term on the right appears puzzling)

---- Take A = [1, x], with x = 100. List the factorated members of A —— 

1, 23, 2·2, 5, 2·3, 7, 2·2·2, 3·3, 2·5,

11, 2·2·3, 13, 2·7, 3·5, 2^4, 17, 2·3·3, 19, 2·2·5,

3·7, 2·11, 23, 2^3·3, 5·5, 2·13, 3^3, 2^2·7, 29, 2·3·5

31, 2^5, 3·11, 2·17, 5·7, 2^2·3^3, 37, 2·19, 3·13, 2^3·5,

41, 2·3·7, 43, 2^2·11, 3·3·5, 2·23, 47, 2^4·3, 7·7, 2·5·5,

3·17, 2^2·13, 53, 2·3^3, 5·11, 2^3·7, 3·19, 2·29, 59, 2^2·3·5,

61, 2·31, 3^3·7, 2^6, 5·13, 2·3·11, 67, 2^2·17, 3·23, 2·5·7,

71, 2^3·3^2, 73, 2·37, 3·5^2, 2^2·19, 7·11, 2·3·13, 79, 2^4·5,

3^4, 2·41, 83, 2^2·3·7, 5·17, 2·43, 3·29, 2^3·11, 89, 2·3^2·5,

7·13, 2^2·23, 3·31, 2·47, 5·19, 2^5·3, 97, 2·7^2, 3^2·11, 2^2·5^2.

Note: I resorted to a computer command for the numbers greater than 30.

---- There are totally 25 primes in A.

---- Let ξ = x^1/2 (=10), η = x^1/4 (= 3.162..), one has

---- A(ξ)p = { p ∈ Ap > ξ } = {11, 13, 17, 19; 23, 29; 31, 37; 41, 43, 47; 53, 59; 61, 67; 71, 73, 79; 83, 89; 97}.

---- #A(ξ)p  = 21

---- A(η)n = { n ∈ A: P⁻(n) > η } = A(ξ)p ∪ {5, 7 } ∪ {5·5, 5·7, 7·7, 5·11, 5·13, 7·11, 5·17, 7·13, 5·19}.

---- #A(η)n = 21 + (2 + 9) = 32.

Observation on A(η)n: in the third set, the members fall into two groups whose leading factors are in the second set {5, 7}.

---- Now, I'm in the position to observe the second term on the right of the counting formula ——

Σ(p) #{ n ∈ A: P⁻(n) = }

where (p) is the short hand for η < p ≤ ξ, denoting the range of p.

---- There are only two primes 5 and 7 for the present case.

---- As 5 or 7 are used as the leading factors of n, the other factors of n are not smaller than 5 or 7.

---- So, Σ(p) #{ n ∈ A: P⁻(n) = } #{ n ∈ A: P⁻(n) = 5 } + #{ n ∈ A: P⁻(n) = 7 }.

---- For a better notation, I define A(p)n:= { n ∈ A: P⁻(n) = p }.

---- Note: as p and n meet in the notation, one gains the signal to use "=" (instead of ">") in the condition.

---- At last, I give P⁻(n) the name of leading prime factor, or simply leading factor.

---- Now, the inclusion-exclusion formula takes the form of ——

#A(ξ)p = #A(η)n - Σ(p)#A(p)n.

Note: the trick is to count n (in A) with a conditional leading factor, such as P⁻(n) > η or P⁻(n) = p.

.

Comments: the specific case of A is helpful in understanding the inclusion-exclusion formula.

Exercise: try other cases of A with other cases of ξ and η.

 ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ Λ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁻⁰ 1

.

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