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[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.
In a genuine progress of mathematics a great cancellation is inevitable.
♖ ♘ 7 5
♔ ♗ 2 3
Story - The sword was ready.
♙ ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ Λ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁺⁻⁰ 1 ⁱ
So comes the last step ——
П(i)(∑(nj)e(nj(α+bⁱ·θ))) ~> П(i)[ e(bα + bⁱ⁺1·θ) - 1 / e(α + bⁱ·θ) - 1 ]
shorthands:
--- ( i ) ~ i = 0, ..., k - 1;
--- ( nj ) ~ 0 ≤ nj < b.
Note: here, the index of ni is replaced by nj for clarity.
The actual target is ——
∑(nj)e(nj(α+bⁱ·θ)) ~> e(bα + bⁱ⁺1·θ) - 1 / e(α + bⁱ·θ) - 1
---- The denominator appears the handle...
---- First guess ——
∑(nj)e(nj(α+bⁱ·θ)) = e(α + bⁱ·θ) - 1·∑(nj)e(nj(α+bⁱ·θ)) / e(α + bⁱ·θ) - 1
---- Now, one focuses on ——
e(α + bⁱ·θ) - 1·∑(nj)e(nj(α+bⁱ·θ))
= ∑(nj)e(α + bⁱ·θ)e(nj(α+bⁱ·θ)) - ∑(nj)e(nj(α+bⁱ·θ)) (#)
---- Now, one focuses on ——
e(α + bⁱ·θ)e(nj(α+bⁱ·θ))
= e(α + bⁱ·θ + nj(α+bⁱ·θ))
= e((1+nj)(α+bⁱ·θ))
---- It appears to have a "right" look.
---- Now, return to (#) for a replacement ——
∑(nj)e((1+nj)(α+bⁱ·θ)) - ∑(nj)e(nj(α+bⁱ·θ))
= e((1+n0)(α+bⁱ·θ)) + e((1+n1)(α+bⁱ·θ)) + ... + e((1+nb-1)(α+bⁱ·θ))
- [ e(n0(α+bⁱ·θ)) + e(n1(α+bⁱ·θ)) + e(nj(α+bⁱ·θ)) + ... + e(nb-1(α+bⁱ·θ)) ] ($)
---- We are ariving at the target ——
---- For the fixed i of (α + bⁱ·θ), the coefficient nj is nothing but the digital number 0, 1, ..., b -1 in order.
Note: One may need to check this point closely in the last note* (near the end).
---- So, 1 + nj is just 1, 2, ..., b, a shift of nj to the right-hand side by one step.
---- In a genuine progress of mathematics, a great cancellation is inevitable.(TOM)
---- So the left terms of ($) are ——
e((1+nb-1)(α+bⁱ·θ)) - e(n0(α+bⁱ·θ))
= e(bα + bⁱ⁺1·θ) - 1.
---- Take it to the top, one has ——
∑(nj)e(nj(α+bⁱ·θ)) = e(bα + bⁱ⁺1·θ) - 1 / e(α + bⁱ·θ) - 1
.
Comment: in retrospect, it was a correct decision to read this paper.
♙ 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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