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=n0Ann!
Assertion ef4p AeA=1+A+A22+A36+k4Fk

Proof

Step Hyp Ref Expression
1 ef4p.1 F=n0Ann!
2 df-4 4=3+1
3 3nn0 30
4 id AA
5 ax-1cn 1
6 addcl 1A1+A
7 5 6 mpan A1+A
8 sqcl AA2
9 8 halfcld AA22
10 7 9 addcld A1+A+A22
11 df-3 3=2+1
12 2nn0 20
13 df-2 2=1+1
14 1nn0 10
15 5 a1i A1
16 1e0p1 1=0+1
17 0nn0 00
18 0cnd A0
19 1 efval2 AeA=k0Fk
20 nn0uz 0=0
21 20 sumeq1i k0Fk=k0Fk
22 19 21 eqtr2di Ak0Fk=eA
23 22 oveq2d A0+k0Fk=0+eA
24 efcl AeA
25 24 addlidd A0+eA=eA
26 23 25 eqtr2d AeA=0+k0Fk
27 eft0val AA00!=1
28 27 oveq2d A0+A00!=0+1
29 0p1e1 0+1=1
30 28 29 eqtrdi A0+A00!=1
31 1 16 17 4 18 26 30 efsep AeA=1+k1Fk
32 exp1 AA1=A
33 fac1 1!=1
34 33 a1i A1!=1
35 32 34 oveq12d AA11!=A1
36 div1 AA1=A
37 35 36 eqtrd AA11!=A
38 37 oveq2d A1+A11!=1+A
39 1 13 14 4 15 31 38 efsep AeA=1+A+k2Fk
40 fac2 2!=2
41 40 oveq2i A22!=A22
42 41 oveq2i 1+A+A22!=1+A+A22
43 42 a1i A1+A+A22!=1+A+A22
44 1 11 12 4 7 39 43 efsep AeA=1+A+A22+k3Fk
45 fac3 3!=6
46 45 oveq2i A33!=A36
47 46 oveq2i 1+A+A22+A33!=1+A+A22+A36
48 47 a1i A1+A+A22+A33!=1+A+A22+A36
49 1 2 3 4 10 44 48 efsep AeA=1+A+A22+A36+k4Fk