[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.
What is Erdos's motivation to consider the missing digit problem?
♖ ♘ 7 5
♔ ♗ 2 3
Story - Let there be a king, noble said. And there was a king. "A manager". Noble sent a bishop. Army was orgnized. So established the kingdom...
♙ ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ Λ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁺⁻⁰ 1
In particular, this shows that there are infinitely many primes which have no digit equal to 7 in their decimal expansion.
---- It is wondering who will fund the game of without the noble...
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We aim to give a unified treatment to simplified forms of these results, emphasising the properties we are using to count primes.
---- Unification is of understanding. (TOM)
---- I would expect a map.
---- "simplified forms"? It appears the forms are already very simple...
---- "properties" is of counting. (Not "methods"?)
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In particular, we aim to give fairly complete proofs of Theorem 2.1 and Theorem 2.3 in the case when the base b is a sufficiently large constant, and give a sketch of Bourgain's result when working with a base b which is sufficiently large.
---- Sufficiently large base is of special concern.
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Specifically, we prove the following:
Theorem 2.4 (Weak Theorem 2.1). Let b sufficiently large. Let sb(n) denote the sum of digits of n in base b. Then for any fixed choice of (m, b-1) = 1, we have
#{ p < b^k : ≡ a (mod m) } = (1 + (1))b^k / mklogb
as k --> ∞ through the integers.
---- From the appearence, the weak form just replaces x with b^k compared to the original form.
---- The infinitely large k is considered.
---- Apparently, there are infinitely many such primes.
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Theorem 2.5 (Weak Theorem 2.3). Let b be sufficiently large. Then for any choice of a0 ∈ {0, ..., b - 1} and for any x > b^k (?), we have
#{ p < b^k : p has no base b digit equal to a0} = (κb(a0) + o(1)) (b - 1)^k / klogb,
as k -->∞ through the integers, where
/ b / (b-1), if (a0, b) ≠ 1,
κb(a0) =
\ b(φ(b) - 1) / (b-1)φ(b), if (a0, b) = 1.
---- klogb on the denominator is a replacement of logx.
---- The original numerator is x^log(b-1)/logb.
---- By replacing x with b^k, one has
---- (b^k)^log(b-1)/logb.
---- Recall that, (2^2)^3 = 2^(2·3) = 2^6.
---- So, (b^k)^log(b-1)/logb = b^(k·log(b-1)/logb).
---- This is not the case in Theorem 2.5.
---- For a very large b, one agrees that b^k is close to (b-1)^k.
---- If one views b^k as x, one may view (b-1)^k as x closely.
---- That is to say, x^log(b-1)/logb is replaced by ω·x or ω·(b - 1)^k.
---- Actually, log(b-1)/logb is very close to 1 (from the left) for a big b.
---- Say, log(b - 1)/ logb ≈ 0.954 for b = 10;
---- ........log(b - 1)/ logb ≈ 0.998 for b = 100;
----.........log(b - 1)/ logb ≈ 0.99986 for b = 1000.
---- So, one is able to linearize x^log(b-1)/logb by ω·x with a properly chosen ω.
---- Maynard chose ω = (κb(a0) + o(1)).
---- This is what has happened on the numerator.
---- Verification: set b = 10, k =2; a0=2.
---- The estimation is of (10/9 + 0)·9^2/log100 = 19.5.
---- The primes with digit 2 are counted as 22.
---- The error is of - 9.09%.
---- The accuracy is improved by a factor of 2.
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Comments: Maynard's estimation appears less satisfactory, but the problem is of missing digits (very different from the problems in Th2.1 and Th2.2). According to Nath (2021, preprint), the missing digit problem can be dated back to Erdos (1998). In a more recent paper by Maynard (2021), a general estimation for multiple missing digits has been setup.
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In both cases, extra ideas are required to show the full theorem which go beyond a simple refinement of the method presented here.
---- The method shown in this paper is a simple refinement.
---- It appears the "full theorem" is elsewhere.
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Nevertheless, the proofs of Theorems 2.4 and 2.5 still contain most of the qualitatively important aspects, which also appear in Theorem 2.2.
---- The proofs show some important aspects that cover a seemingly different problem (Th2.2).
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The overall approach has much in common with work of Drmota, Mauduit and Rivat on the sum of digits function for polynomials.[4]
---- To mention the source of inspiration.
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Homework: watch VR movies at home.
♙ 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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