Metamath Proof Explorer


Theorem taylfvallem

Description: Lemma for taylfval . (Contributed by Mario Carneiro, 30-Dec-2016)

Ref Expression
Hypotheses taylfval.s ( 𝜑𝑆 ∈ { ℝ , ℂ } )
taylfval.f ( 𝜑𝐹 : 𝐴 ⟶ ℂ )
taylfval.a ( 𝜑𝐴𝑆 )
taylfval.n ( 𝜑 → ( 𝑁 ∈ ℕ0𝑁 = +∞ ) )
taylfval.b ( ( 𝜑𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ) → 𝐵 ∈ dom ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) )
Assertion taylfvallem ( ( 𝜑𝑋 ∈ ℂ ) → ( ℂfld tsums ( 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ↦ ( ( ( ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) ‘ 𝐵 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝑋𝐵 ) ↑ 𝑘 ) ) ) ) ⊆ ℂ )

Proof

Step Hyp Ref Expression
1 taylfval.s ( 𝜑𝑆 ∈ { ℝ , ℂ } )
2 taylfval.f ( 𝜑𝐹 : 𝐴 ⟶ ℂ )
3 taylfval.a ( 𝜑𝐴𝑆 )
4 taylfval.n ( 𝜑 → ( 𝑁 ∈ ℕ0𝑁 = +∞ ) )
5 taylfval.b ( ( 𝜑𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ) → 𝐵 ∈ dom ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) )
6 cnfldbas ℂ = ( Base ‘ ℂfld )
7 cnring fld ∈ Ring
8 ringcmn ( ℂfld ∈ Ring → ℂfld ∈ CMnd )
9 7 8 mp1i ( ( 𝜑𝑋 ∈ ℂ ) → ℂfld ∈ CMnd )
10 cnfldtps fld ∈ TopSp
11 10 a1i ( ( 𝜑𝑋 ∈ ℂ ) → ℂfld ∈ TopSp )
12 ovex ( 0 [,] 𝑁 ) ∈ V
13 12 inex1 ( ( 0 [,] 𝑁 ) ∩ ℤ ) ∈ V
14 13 a1i ( ( 𝜑𝑋 ∈ ℂ ) → ( ( 0 [,] 𝑁 ) ∩ ℤ ) ∈ V )
15 1 2 3 4 5 taylfvallem1 ( ( ( 𝜑𝑋 ∈ ℂ ) ∧ 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ) → ( ( ( ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) ‘ 𝐵 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝑋𝐵 ) ↑ 𝑘 ) ) ∈ ℂ )
16 15 fmpttd ( ( 𝜑𝑋 ∈ ℂ ) → ( 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ↦ ( ( ( ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) ‘ 𝐵 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝑋𝐵 ) ↑ 𝑘 ) ) ) : ( ( 0 [,] 𝑁 ) ∩ ℤ ) ⟶ ℂ )
17 6 9 11 14 16 tsmscl ( ( 𝜑𝑋 ∈ ℂ ) → ( ℂfld tsums ( 𝑘 ∈ ( ( 0 [,] 𝑁 ) ∩ ℤ ) ↦ ( ( ( ( ( 𝑆 D𝑛 𝐹 ) ‘ 𝑘 ) ‘ 𝐵 ) / ( ! ‘ 𝑘 ) ) · ( ( 𝑋𝐵 ) ↑ 𝑘 ) ) ) ) ⊆ ℂ )