Metamath Proof Explorer


Theorem iprodefisum

Description: Applying the exponential function to an infinite sum yields an infinite product. (Contributed by Scott Fenton, 11-Feb-2018)

Ref Expression
Hypotheses iprodefisum.1 ⊢ Z = ℤ ≥ M
iprodefisum.2 ⊢ φ → M ∈ ℤ
iprodefisum.3 ⊢ φ ∧ k ∈ Z → F ⁡ k = B
iprodefisum.4 ⊢ φ ∧ k ∈ Z → B ∈ ℂ
iprodefisum.5 ⊢ φ → seq M + F ∈ dom ⁡ ⇝
Assertion iprodefisum ⊢ φ → ∏ k ∈ Z e B = e ∑ k ∈ Z B

Proof

Step Hyp Ref Expression
1 iprodefisum.1 ⊢ Z = ℤ ≥ M
2 iprodefisum.2 ⊢ φ → M ∈ ℤ
3 iprodefisum.3 ⊢ φ ∧ k ∈ Z → F ⁡ k = B
4 iprodefisum.4 ⊢ φ ∧ k ∈ Z → B ∈ ℂ
5 iprodefisum.5 ⊢ φ → seq M + F ∈ dom ⁡ ⇝
6 1 2 3 4 5 isumcl ⊢ φ → ∑ k ∈ Z B ∈ ℂ
7 efne0 ⊢ ∑ k ∈ Z B ∈ ℂ → e ∑ k ∈ Z B ≠ 0
8 6 7 syl ⊢ φ → e ∑ k ∈ Z B ≠ 0
9 efcn ⊢ exp : ℂ ⟶cn ℂ
10 9 a1i ⊢ φ → exp : ℂ ⟶cn ℂ
11 fveq2 ⊢ j = k → F ⁡ j = F ⁡ k
12 eqid ⊢ j ∈ Z ⟼ F ⁡ j = j ∈ Z ⟼ F ⁡ j
13 fvex ⊢ F ⁡ k ∈ V
14 11 12 13 fvmpt ⊢ k ∈ Z → j ∈ Z ⟼ F ⁡ j ⁡ k = F ⁡ k
15 14 adantl ⊢ φ ∧ k ∈ Z → j ∈ Z ⟼ F ⁡ j ⁡ k = F ⁡ k
16 3 4 eqeltrd ⊢ φ ∧ k ∈ Z → F ⁡ k ∈ ℂ
17 15 16 eqeltrd ⊢ φ ∧ k ∈ Z → j ∈ Z ⟼ F ⁡ j ⁡ k ∈ ℂ
18 1 2 17 serf ⊢ φ → seq M + j ∈ Z ⟼ F ⁡ j : Z ⟶ ℂ
19 1 eqcomi ⊢ ℤ ≥ M = Z
20 14 19 eleq2s ⊢ k ∈ ℤ ≥ M → j ∈ Z ⟼ F ⁡ j ⁡ k = F ⁡ k
21 20 adantl ⊢ φ ∧ k ∈ ℤ ≥ M → j ∈ Z ⟼ F ⁡ j ⁡ k = F ⁡ k
22 2 21 seqfeq ⊢ φ → seq M + j ∈ Z ⟼ F ⁡ j = seq M + F
23 climdm ⊢ seq M + F ∈ dom ⁡ ⇝ ↔ seq M + F ⇝ ⇝ ⁡ seq M + F
24 5 23 sylib ⊢ φ → seq M + F ⇝ ⇝ ⁡ seq M + F
25 22 24 eqbrtrd ⊢ φ → seq M + j ∈ Z ⟼ F ⁡ j ⇝ ⇝ ⁡ seq M + F
26 climcl ⊢ seq M + F ⇝ ⇝ ⁡ seq M + F → ⇝ ⁡ seq M + F ∈ ℂ
27 24 26 syl ⊢ φ → ⇝ ⁡ seq M + F ∈ ℂ
28 1 2 10 18 25 27 climcncf ⊢ φ → exp ∘ seq M + j ∈ Z ⟼ F ⁡ j ⇝ e ⇝ ⁡ seq M + F
29 11 cbvmptv ⊢ j ∈ Z ⟼ F ⁡ j = k ∈ Z ⟼ F ⁡ k
30 16 29 fmptd ⊢ φ → j ∈ Z ⟼ F ⁡ j : Z ⟶ ℂ
31 1 2 30 iprodefisumlem ⊢ φ → seq M × exp ∘ j ∈ Z ⟼ F ⁡ j = exp ∘ seq M + j ∈ Z ⟼ F ⁡ j
32 1 2 3 4 isum ⊢ φ → ∑ k ∈ Z B = ⇝ ⁡ seq M + F
33 32 fveq2d ⊢ φ → e ∑ k ∈ Z B = e ⇝ ⁡ seq M + F
34 28 31 33 3brtr4d ⊢ φ → seq M × exp ∘ j ∈ Z ⟼ F ⁡ j ⇝ e ∑ k ∈ Z B
35 fvco3 ⊢ j ∈ Z ⟼ F ⁡ j : Z ⟶ ℂ ∧ k ∈ Z → exp ∘ j ∈ Z ⟼ F ⁡ j ⁡ k = e j ∈ Z ⟼ F ⁡ j ⁡ k
36 30 35 sylan ⊢ φ ∧ k ∈ Z → exp ∘ j ∈ Z ⟼ F ⁡ j ⁡ k = e j ∈ Z ⟼ F ⁡ j ⁡ k
37 15 fveq2d ⊢ φ ∧ k ∈ Z → e j ∈ Z ⟼ F ⁡ j ⁡ k = e F ⁡ k
38 3 fveq2d ⊢ φ ∧ k ∈ Z → e F ⁡ k = e B
39 36 37 38 3eqtrd ⊢ φ ∧ k ∈ Z → exp ∘ j ∈ Z ⟼ F ⁡ j ⁡ k = e B
40 efcl ⊢ B ∈ ℂ → e B ∈ ℂ
41 4 40 syl ⊢ φ ∧ k ∈ Z → e B ∈ ℂ
42 1 2 8 34 39 41 iprodn0 ⊢ φ → ∏ k ∈ Z e B = e ∑ k ∈ Z B