[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。]

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Seemingly plain usage might create high power at the right place.

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Story - Armies were trained. 

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Recall that Sect.5 is of Fourier anaysis on digit functions. The task is to work out the detailed derivation ——

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^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. 

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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.

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---- 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).

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---- 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)).

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Now, let's see the grouped terms one by one ——

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   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)).

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   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)).

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   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)).

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   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)).

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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)))

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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)) ]

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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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