Metamath Proof Explorer


Definition df-dvn

Description: Define the n -th derivative operator on functions on the complex numbers. This just iterates the derivative operation according to the last argument. (Contributed by Mario Carneiro, 11-Feb-2015)

Ref Expression
Assertion df-dvn D𝑛 = ( 𝑠 ∈ 𝒫 ℂ , 𝑓 ∈ ( ℂ ↑pm 𝑠 ) ↦ seq 0 ( ( ( 𝑥 ∈ V ↦ ( 𝑠 D 𝑥 ) ) ∘ 1st ) , ( ℕ0 × { 𝑓 } ) ) )

Detailed syntax breakdown

Step Hyp Ref Expression
0 cdvn D𝑛
1 vs 𝑠
2 cc
3 2 cpw 𝒫 ℂ
4 vf 𝑓
5 cpm pm
6 1 cv 𝑠
7 2 6 5 co ( ℂ ↑pm 𝑠 )
8 cc0 0
9 vx 𝑥
10 cvv V
11 cdv D
12 9 cv 𝑥
13 6 12 11 co ( 𝑠 D 𝑥 )
14 9 10 13 cmpt ( 𝑥 ∈ V ↦ ( 𝑠 D 𝑥 ) )
15 c1st 1st
16 14 15 ccom ( ( 𝑥 ∈ V ↦ ( 𝑠 D 𝑥 ) ) ∘ 1st )
17 cn0 0
18 4 cv 𝑓
19 18 csn { 𝑓 }
20 17 19 cxp ( ℕ0 × { 𝑓 } )
21 16 20 8 cseq seq 0 ( ( ( 𝑥 ∈ V ↦ ( 𝑠 D 𝑥 ) ) ∘ 1st ) , ( ℕ0 × { 𝑓 } ) )
22 1 4 3 7 21 cmpo ( 𝑠 ∈ 𝒫 ℂ , 𝑓 ∈ ( ℂ ↑pm 𝑠 ) ↦ seq 0 ( ( ( 𝑥 ∈ V ↦ ( 𝑠 D 𝑥 ) ) ∘ 1st ) , ( ℕ0 × { 𝑓 } ) ) )
23 0 22 wceq D𝑛 = ( 𝑠 ∈ 𝒫 ℂ , 𝑓 ∈ ( ℂ ↑pm 𝑠 ) ↦ seq 0 ( ( ( 𝑥 ∈ V ↦ ( 𝑠 D 𝑥 ) ) ∘ 1st ) , ( ℕ0 × { 𝑓 } ) ) )