Metamath Proof Explorer


Theorem coeval

Description: Value of the coefficient function. (Contributed by Mario Carneiro, 22-Jul-2014)

Ref Expression
Assertion coeval ( 𝐹 ∈ ( Poly ‘ 𝑆 ) → ( coeff ‘ 𝐹 ) = ( 𝑎 ∈ ( ℂ ↑m0 ) ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝐹 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) )

Proof

Step Hyp Ref Expression
1 plyssc ( Poly ‘ 𝑆 ) ⊆ ( Poly ‘ ℂ )
2 1 sseli ( 𝐹 ∈ ( Poly ‘ 𝑆 ) → 𝐹 ∈ ( Poly ‘ ℂ ) )
3 eqeq1 ( 𝑓 = 𝐹 → ( 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ↔ 𝐹 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) )
4 3 anbi2d ( 𝑓 = 𝐹 → ( ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ↔ ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝐹 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) )
5 4 rexbidv ( 𝑓 = 𝐹 → ( ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ↔ ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝐹 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) )
6 5 riotabidv ( 𝑓 = 𝐹 → ( 𝑎 ∈ ( ℂ ↑m0 ) ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) = ( 𝑎 ∈ ( ℂ ↑m0 ) ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝐹 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) )
7 df-coe coeff = ( 𝑓 ∈ ( Poly ‘ ℂ ) ↦ ( 𝑎 ∈ ( ℂ ↑m0 ) ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝑓 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) )
8 riotaex ( 𝑎 ∈ ( ℂ ↑m0 ) ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝐹 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) ∈ V
9 6 7 8 fvmpt ( 𝐹 ∈ ( Poly ‘ ℂ ) → ( coeff ‘ 𝐹 ) = ( 𝑎 ∈ ( ℂ ↑m0 ) ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝐹 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) )
10 2 9 syl ( 𝐹 ∈ ( Poly ‘ 𝑆 ) → ( coeff ‘ 𝐹 ) = ( 𝑎 ∈ ( ℂ ↑m0 ) ∃ 𝑛 ∈ ℕ0 ( ( 𝑎 “ ( ℤ ‘ ( 𝑛 + 1 ) ) ) = { 0 } ∧ 𝐹 = ( 𝑧 ∈ ℂ ↦ Σ 𝑘 ∈ ( 0 ... 𝑛 ) ( ( 𝑎𝑘 ) · ( 𝑧𝑘 ) ) ) ) ) )