Metamath Proof Explorer


Theorem taylplem2

Description: Lemma for taylpfval and similar theorems. (Contributed by Mario Carneiro, 31-Dec-2016)

Ref Expression
Hypotheses taylpfval.s ⊢ ( 𝜑 → 𝑆 ∈ { ℝ , ℂ } )
taylpfval.f ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ ℂ )
taylpfval.a ⊢ ( 𝜑 → 𝐴 ⊆ 𝑆 )
taylpfval.n ⊢ ( 𝜑 → 𝑁 ∈ ℕ0 )
taylpfval.b ⊢ ( 𝜑 → 𝐵 ∈ dom ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑁 ) )
Assertion taylplem2 ( ( ( 𝜑 ∧ 𝑋 ∈ ℂ ) ∧ 𝑘 ∈ ( 0 ... 𝑁 ) ) → ( ( ( ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) ‘ 𝐵 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝑋 − 𝐵 ) ↑ 𝑘 ) ) ∈ ℂ )

Proof

Step Hyp Ref Expression
1 taylpfval.s ⊢ ( 𝜑 → 𝑆 ∈ { ℝ , ℂ } )
2 taylpfval.f ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ ℂ )
3 taylpfval.a ⊢ ( 𝜑 → 𝐴 ⊆ 𝑆 )
4 taylpfval.n ⊢ ( 𝜑 → 𝑁 ∈ ℕ0 )
5 taylpfval.b ⊢ ( 𝜑 → 𝐵 ∈ dom ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑁 ) )
6 0z ⊢ 0 ∈ ℤ
7 4 nn0zd ⊢ ( 𝜑 → 𝑁 ∈ ℤ )
8 fzval2 ⊢ ( ( 0 ∈ ℤ ∧ 𝑁 ∈ ℤ ) → ( 0 ... 𝑁 ) = ( ( 0 [,] 𝑁 ) ∩ ℤ ) )
9 6 7 8 sylancr ⊢ ( 𝜑 → ( 0 ... 𝑁 ) = ( ( 0 [,] 𝑁 ) ∩ ℤ ) )
10 9 eleq2d ⊢ ( 𝜑 → ( 𝑘 ∈ ( 0 ... 𝑁 ) ↔ 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ) )
11 10 adantr ⊢ ( ( 𝜑 ∧ 𝑋 ∈ ℂ ) → ( 𝑘 ∈ ( 0 ... 𝑁 ) ↔ 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ) )
12 11 biimpa ⊢ ( ( ( 𝜑 ∧ 𝑋 ∈ ℂ ) ∧ 𝑘 ∈ ( 0 ... 𝑁 ) ) → 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) )
13 4 orcd ⊢ ( 𝜑 → ( 𝑁 ∈ ℕ0 ∨ 𝑁 = +∞ ) )
14 1 2 3 4 5 taylplem1 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ) → 𝐵 ∈ dom ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) )
15 1 2 3 13 14 taylfvallem1 ⊢ ( ( ( 𝜑 ∧ 𝑋 ∈ ℂ ) ∧ 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ) → ( ( ( ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) ‘ 𝐵 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝑋 − 𝐵 ) ↑ 𝑘 ) ) ∈ ℂ )
16 12 15 syldan ⊢ ( ( ( 𝜑 ∧ 𝑋 ∈ ℂ ) ∧ 𝑘 ∈ ( 0 ... 𝑁 ) ) → ( ( ( ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) ‘ 𝐵 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝑋 − 𝐵 ) ↑ 𝑘 ) ) ∈ ℂ )