Metamath Proof Explorer


Theorem etransclem19

Description: The N -th derivative of H is 0 if N is large enough. (Contributed by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses etransclem19.s ⊢ φ → S ∈ ℝ ℂ
etransclem19.x ⊢ φ → X ∈ TopOpen ⁡ ℂ fld ↾ 𝑡 S
etransclem19.p ⊢ φ → P ∈ ℕ
etransclem19.1 ⊢ H = j ∈ 0 … M ⟼ x ∈ X ⟼ x − j if j = 0 P − 1 P
etransclem19.J ⊢ φ → J ∈ 0 … M
etransclem19.n ⊢ φ → N ∈ ℤ
etransclem19.7 ⊢ φ → if J = 0 P − 1 P < N
Assertion etransclem19 ⊢ φ → S D n H ⁡ J ⁡ N = x ∈ X ⟼ 0

Proof

Step Hyp Ref Expression
1 etransclem19.s ⊢ φ → S ∈ ℝ ℂ
2 etransclem19.x ⊢ φ → X ∈ TopOpen ⁡ ℂ fld ↾ 𝑡 S
3 etransclem19.p ⊢ φ → P ∈ ℕ
4 etransclem19.1 ⊢ H = j ∈ 0 … M ⟼ x ∈ X ⟼ x − j if j = 0 P − 1 P
5 etransclem19.J ⊢ φ → J ∈ 0 … M
6 etransclem19.n ⊢ φ → N ∈ ℤ
7 etransclem19.7 ⊢ φ → if J = 0 P − 1 P < N
8 0red ⊢ φ → 0 ∈ ℝ
9 6 zred ⊢ φ → N ∈ ℝ
10 nnm1nn0 ⊢ P ∈ ℕ → P − 1 ∈ ℕ 0
11 3 10 syl ⊢ φ → P − 1 ∈ ℕ 0
12 11 nn0red ⊢ φ → P − 1 ∈ ℝ
13 3 nnred ⊢ φ → P ∈ ℝ
14 12 13 ifcld ⊢ φ → if J = 0 P − 1 P ∈ ℝ
15 11 nn0ge0d ⊢ φ → 0 ≤ P − 1
16 15 adantr ⊢ φ ∧ J = 0 → 0 ≤ P − 1
17 iftrue ⊢ J = 0 → if J = 0 P − 1 P = P − 1
18 17 eqcomd ⊢ J = 0 → P − 1 = if J = 0 P − 1 P
19 18 adantl ⊢ φ ∧ J = 0 → P − 1 = if J = 0 P − 1 P
20 16 19 breqtrd ⊢ φ ∧ J = 0 → 0 ≤ if J = 0 P − 1 P
21 3 nnnn0d ⊢ φ → P ∈ ℕ 0
22 21 nn0ge0d ⊢ φ → 0 ≤ P
23 22 adantr ⊢ φ ∧ ¬ J = 0 → 0 ≤ P
24 iffalse ⊢ ¬ J = 0 → if J = 0 P − 1 P = P
25 24 eqcomd ⊢ ¬ J = 0 → P = if J = 0 P − 1 P
26 25 adantl ⊢ φ ∧ ¬ J = 0 → P = if J = 0 P − 1 P
27 23 26 breqtrd ⊢ φ ∧ ¬ J = 0 → 0 ≤ if J = 0 P − 1 P
28 20 27 pm2.61dan ⊢ φ → 0 ≤ if J = 0 P − 1 P
29 8 14 9 28 7 lelttrd ⊢ φ → 0 < N
30 8 9 29 ltled ⊢ φ → 0 ≤ N
31 elnn0z ⊢ N ∈ ℕ 0 ↔ N ∈ ℤ ∧ 0 ≤ N
32 6 30 31 sylanbrc ⊢ φ → N ∈ ℕ 0
33 1 2 3 4 5 32 etransclem17 ⊢ φ → S D n H ⁡ J ⁡ N = x ∈ X ⟼ if if J = 0 P − 1 P < N 0 if J = 0 P − 1 P ! if J = 0 P − 1 P − N ! ⁢ x − J if J = 0 P − 1 P − N
34 7 iftrued ⊢ φ → if if J = 0 P − 1 P < N 0 if J = 0 P − 1 P ! if J = 0 P − 1 P − N ! ⁢ x − J if J = 0 P − 1 P − N = 0
35 34 mpteq2dv ⊢ φ → x ∈ X ⟼ if if J = 0 P − 1 P < N 0 if J = 0 P − 1 P ! if J = 0 P − 1 P − N ! ⁢ x − J if J = 0 P − 1 P − N = x ∈ X ⟼ 0
36 33 35 eqtrd ⊢ φ → S D n H ⁡ J ⁡ N = x ∈ X ⟼ 0