Nota on: note ℝ as real numbers, ℤ as integers, and ℕ as natural numbers, i.e. all non-nega ve integers. In this
ar cle, 𝑖, 𝑗, 𝑘, 𝑚, 𝑛 will be natural numbers, while 𝑟, 𝑠 are real numbers, 𝑥, 𝑦 are variables.
Given sequences 𝑎 ≔ {𝑎 }
, 𝑏 ≔ {𝑏 }
, define convolu on 𝑎 ∘ 𝑏 as sequence 𝑐 where 𝑐 = ∑
Define: Given sequences 𝑎 ≔ {𝑎 } , then its genera ng func on is 𝑓(𝑥, 𝑀) ≔ ∑
𝑀 = +∞, we get 𝑓(𝑥) ≔ ∑
𝑎 𝑥 , i.e. 𝑎 → 𝑓(𝑥).
Lemma of convolu on: Given two sequences 𝑎 ≔ {𝑎 }
, 𝑏 ≔ {𝑏 }
𝑎𝑏
.
𝑎 𝑥 , i.e. 𝑎 → 𝑓(𝑥, 𝑀); when
, the convolu on 𝑎 ∘ 𝑏, and their genera ng
func ons 𝑎 → 𝑓(𝑥, 𝑀), and 𝑏 → 𝑔(𝑥, 𝑀), and 𝑐 ⎯⎯ ℎ(𝑥, 𝑀 + 𝑁) , then 𝑓(𝑥, 𝑀)𝑔(𝑥, 𝑁) = ℎ(𝑥, 𝑀 + 𝑁).
Lemma 1: if lim 𝑎 = 𝛼, i.e. 𝑎 → 𝛼, then lim
→
→
∑
𝑎 = 𝛼, i.e. 𝑎 → 𝛼.
Lemma 2: if lim 𝑎 = 𝛼, i.e. 𝑎 → 𝛼, lim 𝑏 = 𝛽, i.e. 𝑏 → 𝛽, then lim
→
→
→
(𝑎 ∘ 𝑏) = 𝛼𝛽, i.e. 𝑎 ∘ 𝑏 → 𝛼𝛽.
Abel’s theorem: Given sequences 𝑎 ≔ {𝑎 } , 𝑏 ≔ {𝑏 } ; define 𝐴 ≔ ∑ 𝑎 , i.e. 𝐴 ≔ 𝑎 ∘ 1; 𝐵 ≔ ∑ 𝑏 , i.e.
𝐵 ≔ 𝑏 ∘ 1; 𝑐 ≔ (𝑎 ∘ 𝑏) , i.e. 𝑐 ≔ 𝑎 ∘ 𝑏; 𝐶 ≔ ∑ 𝑐 , i.e. 𝐶 ≔ 𝑐 ∘ 1; and 𝐴 → 𝛼, 𝐵 → 𝛽, 𝐶 → 𝛾 as 𝑛 → ∞, then
𝛼𝛽 = 𝛾, i.e. (𝑎 ∘ 1)(𝑏 ∘ 1) → 𝑎 ∘ 𝑏 ∘ 1.
Proof: 𝐶 = 𝑐 ∘ 1 = (𝑎 ∘ 𝑏) ∘ 1 = 𝑎 ∘ (𝑏 ∘ 1) = 𝑎 ∘ 𝐵 = 𝐴 ∘ 𝑏. So ∑ 𝐶 = 𝐶 ∘ 1 = (𝐴 ∘ 𝑏) ∘ 1 = 𝐴 ∘ (𝑏 ∘ 1) = 𝐴 ∘
𝐵, Lemma 2 says 𝐶̅ → 𝛼𝛽. However, Lemma 1 says 𝐶̅ → 𝛾. So 𝛼𝛽 = 𝛾.
𝑟
𝑟
(
)⋯(
) 𝑟
Define: Given 𝑘 ∈ ℕ, 𝑟 ∈ ℝ, de ine 𝑟 choose 𝑘 as
≔
,
≔ 1. Notice
= 0 iff ℕ ∋ 𝑟 < 𝑘.
!
𝑘
0
𝑘
𝑛
Binomial Theorem: Given 𝑛 ∈ ℕ, we have (1 + 𝑥) = ∑
𝑥 .
𝑘
𝑚
𝑛
𝑚
𝑛
𝑚+𝑛
𝑚+𝑛
Lemma 3: Given 𝑘, 𝑚, 𝑛 ∈ ℕ, we have
=
∘
, i.e.
=∑
.
⋅
⋅
𝑖 𝑘−𝑖
⋅
𝑘
𝑚
𝑛
𝑚
𝑚
𝑛
Now, view 𝑚 and 𝑛 as variables, then
and
are polynomials, i.e.
∈ ℕ[𝑚], deg
= 𝑖,
∈
𝑖
𝑘−𝑖
𝑖
𝑖
𝑘−𝑖
𝑛
𝑚
𝑛
𝑚+𝑛
ℕ[𝑛], deg
= 𝑘 − 𝑖. Since
=∑
holds for all posi ve integers of 𝑚 and 𝑛, it is actually
𝑘−𝑖
𝑖 𝑘−𝑖
𝑘
𝑟
𝑠
𝑟+𝑠
a polynomial iden ty in ℕ[𝑚, 𝑛]. Thus, renaming variables to 𝑟 and 𝑠,
≡∑
is also a
𝑖 𝑘−𝑖
𝑘
polynomial iden ty in ℝ[𝑟, 𝑠], which holds for all real values of 𝑟 and 𝑠, for any given integer 𝑘. Hence we have:
𝑟
𝑠
𝑟
𝑠
𝑟+𝑠
𝑟+𝑠
Lemma 4: Given 𝑘 ∈ ℕ, real numbers 𝑟, 𝑠 ∈ ℝ , we s ll have
=
∘
, i.e.
=∑
⋅
⋅
𝑖 𝑘−𝑖
𝑘
𝑘
𝑟
𝑟
Lemma 5: Exponen al property of
series. Given 𝑟, 𝑠 ∈ ℝ ,
→ 𝑓(𝑥, 𝑟), we have 𝑓(𝑥, 𝑟)𝑓(𝑥, 𝑠) = 𝑓(𝑥, 𝑟 + 𝑠).
⋅
⋅
𝑟
𝑟
𝑟
General Binomial Theorem. Given real number 𝑟, (1 + 𝑥) = ∑
𝑥 =1+
𝑥+
𝑥 + ⋯.
𝑘
1
2
Proof: 𝑓(𝑥, 1) = 1 + 𝑥, from Corollary 5, 𝑓 (𝑥, 𝑟) = 𝑓(𝑥, 𝑛𝑟), so 𝑓(𝑥, 𝑚/𝑛) = (1 + 𝑥) / ,𝑓(𝑥, 𝑟) = (1 + 𝑥) .