Metamath Proof Explorer


Theorem aks4d1p1p1

Description: Exponential law for finite products, special case. (Contributed by metakunt, 22-Jul-2024)

Ref Expression
Hypotheses aks4d1p1p1.1 ⊢ φ → A ∈ ℝ +
aks4d1p1p1.2 ⊢ φ → N ∈ ℕ
Assertion aks4d1p1p1 ⊢ φ → ∏ k = 1 N A k = A ∑ k = 1 N k

Proof

Step Hyp Ref Expression
1 aks4d1p1p1.1 ⊢ φ → A ∈ ℝ +
2 aks4d1p1p1.2 ⊢ φ → N ∈ ℕ
3 1 rpcnd ⊢ φ → A ∈ ℂ
4 3 adantr ⊢ φ ∧ k ∈ 1 … N → A ∈ ℂ
5 1 rpne0d ⊢ φ → A ≠ 0
6 5 adantr ⊢ φ ∧ k ∈ 1 … N → A ≠ 0
7 elfzelz ⊢ k ∈ 1 … N → k ∈ ℤ
8 7 zcnd ⊢ k ∈ 1 … N → k ∈ ℂ
9 8 adantl ⊢ φ ∧ k ∈ 1 … N → k ∈ ℂ
10 4 6 9 3jca ⊢ φ ∧ k ∈ 1 … N → A ∈ ℂ ∧ A ≠ 0 ∧ k ∈ ℂ
11 cxpef ⊢ A ∈ ℂ ∧ A ≠ 0 ∧ k ∈ ℂ → A k = e k ⁢ log ⁡ A
12 10 11 syl ⊢ φ ∧ k ∈ 1 … N → A k = e k ⁢ log ⁡ A
13 12 prodeq2dv ⊢ φ → ∏ k = 1 N A k = ∏ k = 1 N e k ⁢ log ⁡ A
14 eqid ⊢ ℤ ≥ 1 = ℤ ≥ 1
15 nnuz ⊢ ℕ = ℤ ≥ 1
16 2 15 eleqtrdi ⊢ φ → N ∈ ℤ ≥ 1
17 eluzelcn ⊢ k ∈ ℤ ≥ 1 → k ∈ ℂ
18 17 adantl ⊢ φ ∧ k ∈ ℤ ≥ 1 → k ∈ ℂ
19 3 adantr ⊢ φ ∧ k ∈ ℤ ≥ 1 → A ∈ ℂ
20 5 adantr ⊢ φ ∧ k ∈ ℤ ≥ 1 → A ≠ 0
21 19 20 logcld ⊢ φ ∧ k ∈ ℤ ≥ 1 → log ⁡ A ∈ ℂ
22 18 21 mulcld ⊢ φ ∧ k ∈ ℤ ≥ 1 → k ⁢ log ⁡ A ∈ ℂ
23 14 16 22 fprodefsum ⊢ φ → ∏ k = 1 N e k ⁢ log ⁡ A = e ∑ k = 1 N k ⁢ log ⁡ A
24 fzfid ⊢ φ → 1 … N ∈ Fin
25 3 5 logcld ⊢ φ → log ⁡ A ∈ ℂ
26 24 25 9 fsummulc1 ⊢ φ → ∑ k = 1 N k ⁢ log ⁡ A = ∑ k = 1 N k ⁢ log ⁡ A
27 26 eqcomd ⊢ φ → ∑ k = 1 N k ⁢ log ⁡ A = ∑ k = 1 N k ⁢ log ⁡ A
28 27 fveq2d ⊢ φ → e ∑ k = 1 N k ⁢ log ⁡ A = e ∑ k = 1 N k ⁢ log ⁡ A
29 24 9 fsumcl ⊢ φ → ∑ k = 1 N k ∈ ℂ
30 3 5 29 cxpefd ⊢ φ → A ∑ k = 1 N k = e ∑ k = 1 N k ⁢ log ⁡ A
31 30 eqcomd ⊢ φ → e ∑ k = 1 N k ⁢ log ⁡ A = A ∑ k = 1 N k
32 28 31 eqtrd ⊢ φ → e ∑ k = 1 N k ⁢ log ⁡ A = A ∑ k = 1 N k
33 23 32 eqtrd ⊢ φ → ∏ k = 1 N e k ⁢ log ⁡ A = A ∑ k = 1 N k
34 13 33 eqtrd ⊢ φ → ∏ k = 1 N A k = A ∑ k = 1 N k