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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.[blog]
Seemingly plain usage might create high power at the right place.
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♙ ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ Λ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁺⁻⁰ 1
Recall that Sect.5 is of Fourier anaysis on digit functions. The task is to work out the detailed derivation ——
.
^gb^k (θ) = ∑(n) e(nθ)e(αsb(n)) ~> П(i)(∑(ni)e(ni(α+b^i·θ)))
Shorthand notation: (n) refers to n < b^k;
(ni) refers to 0≤n0, ..., nk-1< b;
(i) refers to i = 0, ..., k - 1.
.
To be specific, set k = 2, b = 10. As such, "∑(n)" refers to sum of 99 +1 terms ——
∑(n) e(nθ)e(αsb(n)) =
e(0·θ)e(α·sb(0)) +
e(1·θ)e(α·sb(1)) + e(2·θ)e(α·sb(2)) + e(3·θ)e(α·sb(3)) + ...
+ e(28·θ)e(α·sb(28)) + e(29·θ)e(α·sb(29)) + e(30·θ)e(α·sb(30)) + ...
+ e(97·θ)e(α·sb(97)) + e(98·θ)e(α·sb(98)) + e(99·θ)e(α·sb(99)).
Note: n = 0 has to be included for a technical reason; see below.
.
---- Now, each instance of n is listed with better explicity.
---- Remember that the base b expansion of n = ∑(i)ni·b^i is applied to each instance of n.
---- By definition, the digit function sb(n) = sb(∑(i)ni·b^i ) = ∑(i)ni.
---- There are two kinds* of instances of n in the above 99 terms.
---- # Take a look for the one* digit instance at first.
---- It is representative to take n = 3 for the instance of one digit* ——
---- 3 = n = ∑(i)ni·b^i = n0·b^0 + n1·b^1 (with n0 = 3, n1 = 0 ).
---- sb(3) = sb(n) = sb(∑(i)ni·b^i) = ∑(i)ni = n0 + n1 = 3.
Note for stars: for the upper bound b^k, each number should be viewed as of k digits as one kind, crutial for a unified treatment.
---- Now consider the term ——
---- e(3·θ)e(α·sb(3)) = e((n0·b^0 + n1·b^1)·θ)e(α·(n0 + n1)) = e(n0(α + b^0·θ))·e(n1(α + b^1·θ)).
---- Apparently, for any instance of n of one digit, one has ——
---- e(nθ)e(αsb(n)) = e(n0(α + b^0·θ))·e(n1(α + b^1·θ)).
---- # Take a look for the two digit instance as follows.
---- It is representative to take n = 29 for the instance of two digits ——
---- 29 = n = ∑(i)ni·b^i = n0·b^0 + n1·b^1.
---- sb(29) = sb(n) = sb(∑(i)ni·b^i) = ∑(i)ni = n0 + n1 = 9 + 2 = 11.
---- Now consider the term ——
---- e(29·θ)e(α·sb(29))
= e((n0·b^0 + n1·b^1)·θ)e(α·(n0 + n1))
= e(α·(n0 + n1)+(n0·b^0 + n1·b^1)·θ)
= e(n0·(α + b^0·θ) + n1·(α + b^1·θ))
= e(n0·(α + b^0·θ))·e(n1·(α + b^1·θ))
---- Apparently, for any instance of n of two digits, one has ——
---- e(nθ)e(αsb(n)) = e(n0·(α + b^0·θ))·e(n1·(α + b^1·θ)).
Note: all instances of n are viewed as one kind, with the same form (cf. fruit and fruit).
.
---- Now, return to the Fourier expansion ——
∑(n) e(nθ)e(αsb(n)) =
e(0·θ)e(α·sb(0)) +
e(1·θ)e(α·sb(1)) + e(2·θ)e(α·sb(2)) + e(3·θ)e(α·sb(3)) + ...
+ e(28·θ)e(α·sb(28)) + e(29·θ)e(α·sb(29)) + e(30·θ)e(α·sb(30)) + ...
+ e(97·θ)e(α·sb(97)) + e(98·θ)e(α·sb(98)) + e(99·θ)e(α·sb(99)).
---- Replace the terms using the fruit and fruit with specific ni ——
∑(n) e(nθ)e(αsb(n))
= e(0·(α + b^0·θ))·e(0·(α + b^1·θ)) ⬅ term for n = 0
+ e(1·(α + b^0·θ))·e(0·(α + b^1·θ))
+ e(2·(α + b^0·θ))·e(0·(α + b^1·θ))
...
+ e(9·(α + b^0·θ))·e(0·(α + b^1·θ))
+ e(0·(α + b^0·θ))·e(1·(α + b^1·θ))
+ e(1·(α + b^0·θ))·e(1·(α + b^1·θ))
+ e(2·(α + b^0·θ))·e(1·(α + b^1·θ))
...
+ e(9·(α + b^0·θ))·e(1·(α + b^1·θ))
+ e(0·(α + b^0·θ))·e(2·(α + b^1·θ))
+ e(1·(α + b^0·θ))·e(2·(α + b^1·θ))
+ e(2·(α + b^0·θ))·e(2·(α + b^1·θ))
...
+ e(9·(α + b^0·θ))·e(2·(α + b^1·θ))
.
.
.
+ e(0·(α + b^0·θ))·e(9·(α + b^1·θ))
+ e(1·(α + b^0·θ))·e(9·(α + b^1·θ))
+ e(2·(α + b^0·θ))·e(9·(α + b^1·θ))
...
+ e(9·(α + b^0·θ))·e(9·(α + b^1·θ)).
.
Now, let's see the grouped terms one by one ——
.
e(0·(α + b^0·θ))·e(0·(α + b^1·θ)) ⬅ term for n = 0
+ e(1·(α + b^0·θ))·e(0·(α + b^1·θ))
+ e(2·(α + b^0·θ))·e(0·(α + b^1·θ))
...
+ e(9·(α + b^0·θ))·e(0·(α + b^1·θ))
= [ e(0·(α + b^0·θ)) + e(1·(α + b^0·θ)) + e(2·(α + b^0·θ)) +...+ e(9·(α + b^0·θ)) ]·e(0·(α + b^1·θ))
= [ ∑(ni) e(ni·(α + b^0·θ)) ]·e(0·(α + b^1·θ)).
.
e(0·(α + b^0·θ))·e(1·(α + b^1·θ))
+ e(1·(α + b^0·θ))·e(1·(α + b^1·θ))
+ e(2·(α + b^0·θ))·e(1·(α + b^1·θ))
...
+ e(9·(α + b^0·θ))·e(1·(α + b^1·θ))
= [ ∑(ni) e(ni·(α + b^0·θ)) ]·e(1·(α + b^1·θ)).
.
e(0·(α + b^0·θ))·e(2·(α + b^1·θ))
+ e(1·(α + b^0·θ))·e(2·(α + b^1·θ))
+ e(2·(α + b^0·θ))·e(2·(α + b^1·θ))
...
+ e(9·(α + b^0·θ))·e(2·(α + b^1·θ))
= [ ∑(ni) e(ni·(α + b^0·θ)) ]·e(2·(α + b^1·θ)).
.
e(0·(α + b^0·θ))·e(9·(α + b^1·θ))
+ e(1·(α + b^0·θ))·e(9·(α + b^1·θ))
+ e(2·(α + b^0·θ))·e(9·(α + b^1·θ))
...
+ e(9·(α + b^0·θ))·e(9·(α + b^1·θ)).
= [ ∑(ni) e(ni·(α + b^0·θ)) ]·e(9·(α + b^1·θ)).
.
Now, we are arriving the target ——
∑(n) e(nθ)e(αsb(n)) =
[ ∑(ni) e(ni·(α + b^0·θ)) ]·e(0·(α + b^1·θ)) +
[ ∑(ni) e(ni·(α + b^0·θ)) ]·e(1·(α + b^1·θ)) +
[ ∑(ni) e(ni·(α + b^0·θ)) ]·e(2·(α + b^1·θ)) +
... ... ...
[ ∑(ni) e(ni·(α + b^0·θ)) ]·e(9·(α + b^1·θ))
= [ ∑(ni) e(ni·(α + b^0·θ)) ]·[ e(0·(α + b^1·θ)) + e(1·(α + b^1·θ)) + e(2·(α + b^1·θ)) +...+ e(9·(α + b^1·θ) ]
= [ ∑(ni) e(ni·(α + b^0·θ)) ]·[ ∑(ni) e(ni·(α + b^1·θ)) ]
= П(i)(∑(ni)e(ni(α+b^i·θ)))
.
Note: looking at the final form, one realizes that the index i of b^i is fixed inside, not running with "∑(ni)". For the general case of k digits, the final form is viewed as ——
[ ∑(ni)e(ni(α+b^0·θ)) ]·[ ∑(ni)e(ni(α+b^1·θ)) ]· ... ·[ ∑(ni)e(ni(α+b^k-1·θ)) ]
.
Comment: the final form is elegant, justifying the digital representation of n with leading zeros like 0...01 (=1). As the digit function sb(n) is involved in the Fourier transform, the final form is of a characteristic mode. It has to be a key finding.
♙ 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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