Metamath Proof Explorer


Theorem ef4p

Description: Separate out the first four terms of the infinite series expansion of the exponential function. (Contributed by Paul Chapman, 19-Jan-2008) (Revised by Mario Carneiro, 29-Apr-2014)

Ref Expression
Hypothesis ef4p.1 ⊢ F = n ∈ ℕ 0 ⟼ A n n !
Assertion ef4p ⊢ A ∈ ℂ → e A = 1 + A + A 2 2 + A 3 6 + ∑ k ∈ ℤ ≥ 4 F ⁡ k

Proof

Step Hyp Ref Expression
1 ef4p.1 ⊢ F = n ∈ ℕ 0 ⟼ A n n !
2 df-4 ⊢ 4 = 3 + 1
3 3nn0 ⊢ 3 ∈ ℕ 0
4 id ⊢ A ∈ ℂ → A ∈ ℂ
5 ax-1cn ⊢ 1 ∈ ℂ
6 addcl ⊢ 1 ∈ ℂ ∧ A ∈ ℂ → 1 + A ∈ ℂ
7 5 6 mpan ⊢ A ∈ ℂ → 1 + A ∈ ℂ
8 sqcl ⊢ A ∈ ℂ → A 2 ∈ ℂ
9 8 halfcld ⊢ A ∈ ℂ → A 2 2 ∈ ℂ
10 7 9 addcld ⊢ A ∈ ℂ → 1 + A + A 2 2 ∈ ℂ
11 df-3 ⊢ 3 = 2 + 1
12 2nn0 ⊢ 2 ∈ ℕ 0
13 df-2 ⊢ 2 = 1 + 1
14 1nn0 ⊢ 1 ∈ ℕ 0
15 5 a1i ⊢ A ∈ ℂ → 1 ∈ ℂ
16 1e0p1 ⊢ 1 = 0 + 1
17 0nn0 ⊢ 0 ∈ ℕ 0
18 0cnd ⊢ A ∈ ℂ → 0 ∈ ℂ
19 1 efval2 ⊢ A ∈ ℂ → e A = ∑ k ∈ ℕ 0 F ⁡ k
20 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
21 20 sumeq1i ⊢ ∑ k ∈ ℕ 0 F ⁡ k = ∑ k ∈ ℤ ≥ 0 F ⁡ k
22 19 21 eqtr2di ⊢ A ∈ ℂ → ∑ k ∈ ℤ ≥ 0 F ⁡ k = e A
23 22 oveq2d ⊢ A ∈ ℂ → 0 + ∑ k ∈ ℤ ≥ 0 F ⁡ k = 0 + e A
24 efcl ⊢ A ∈ ℂ → e A ∈ ℂ
25 24 addlidd ⊢ A ∈ ℂ → 0 + e A = e A
26 23 25 eqtr2d ⊢ A ∈ ℂ → e A = 0 + ∑ k ∈ ℤ ≥ 0 F ⁡ k
27 eft0val ⊢ A ∈ ℂ → A 0 0 ! = 1
28 27 oveq2d ⊢ A ∈ ℂ → 0 + A 0 0 ! = 0 + 1
29 0p1e1 ⊢ 0 + 1 = 1
30 28 29 eqtrdi ⊢ A ∈ ℂ → 0 + A 0 0 ! = 1
31 1 16 17 4 18 26 30 efsep ⊢ A ∈ ℂ → e A = 1 + ∑ k ∈ ℤ ≥ 1 F ⁡ k
32 exp1 ⊢ A ∈ ℂ → A 1 = A
33 fac1 ⊢ 1 ! = 1
34 33 a1i ⊢ A ∈ ℂ → 1 ! = 1
35 32 34 oveq12d ⊢ A ∈ ℂ → A 1 1 ! = A 1
36 div1 ⊢ A ∈ ℂ → A 1 = A
37 35 36 eqtrd ⊢ A ∈ ℂ → A 1 1 ! = A
38 37 oveq2d ⊢ A ∈ ℂ → 1 + A 1 1 ! = 1 + A
39 1 13 14 4 15 31 38 efsep ⊢ A ∈ ℂ → e A = 1 + A + ∑ k ∈ ℤ ≥ 2 F ⁡ k
40 fac2 ⊢ 2 ! = 2
41 40 oveq2i ⊢ A 2 2 ! = A 2 2
42 41 oveq2i ⊢ 1 + A + A 2 2 ! = 1 + A + A 2 2
43 42 a1i ⊢ A ∈ ℂ → 1 + A + A 2 2 ! = 1 + A + A 2 2
44 1 11 12 4 7 39 43 efsep ⊢ A ∈ ℂ → e A = 1 + A + A 2 2 + ∑ k ∈ ℤ ≥ 3 F ⁡ k
45 fac3 ⊢ 3 ! = 6
46 45 oveq2i ⊢ A 3 3 ! = A 3 6
47 46 oveq2i ⊢ 1 + A + A 2 2 + A 3 3 ! = 1 + A + A 2 2 + A 3 6
48 47 a1i ⊢ A ∈ ℂ → 1 + A + A 2 2 + A 3 3 ! = 1 + A + A 2 2 + A 3 6
49 1 2 3 4 10 44 48 efsep ⊢ A ∈ ℂ → e A = 1 + A + A 2 2 + A 3 6 + ∑ k ∈ ℤ ≥ 4 F ⁡ k