Topics in Mathematics 201-BNJ-05
Vincent Carrier
Euler’s Identity
The Maclaurin Series of a function f is given by
f (x) =
∞
X
f (n) (0)
n!
n=0
xn .
Maclaurin Series of sin x
n
0
f (n) (x)
sin x
f (n) (0)
0
sin x =
1
2
3
4
cos x − sin x − cos x sin x
1
−1
0
0
5
6
7
8
cos x − sin x − cos x sin x
1
−1
0
0
···
···
···
0 0 1 1 0 2 1 3 0 4 1 5 0 6 1 7 0 8
x + x + x − x + x + x + x − x + x + ···
0!
1!
2!
3!
4!
5!
6!
7!
8!
sin x = x −
x3
x5
x7
x9
+
−
+
− ···
3!
5!
7!
9!
Maclaurin Series of cos x
n
f (n) (x)
0
1
2
3
cos x − sin x − cos x sin x
f (n) (0)
cos x =
1
0
−1
0
4
5
6
7
cos x − sin x − cos x sin x
1
0
−1
0
8
···
cos x
···
1
···
1 0 0 1 1 2 0 3 1 4 0 5 1 6 0 7 1 8
x + x − x + x + x + x − x + x + x + ···
0!
1!
2!
3!
4!
5!
6!
7!
8!
cos x = 1 −
x2
x4
x6
x8
+
−
+
− ···
2!
4!
6!
8!
Maclaurin Series of ex
f (n) (x) = ex
f (n) (0) = 1
ex = 1 + x +
for n = 0, 1, 2, . . .
x3
x4
x5
x2
+
+
+
+ ···
2!
3!
4!
5!
Maclaurin Series of eix
Let i be a number such that i2 = −1. Then
i1 = i
eix = 1 + ix +
i2 = −1
i3 = −i
i4 = 1
i5 = i
···
(ix)2 (ix)3 (ix)4 (ix)5 (ix)6 (ix)7 (ix)8 (ix)9
+
+
+
+
+
+
+
+ ···
2!
3!
4!
5!
6!
7!
8!
9!
x2
x3 x4
x5 x6
x7 x8
x9
−i
+
+i
−
−i
+
+i
− ···
2!
3!
4!
5!
6!
7!
8!
9!
x2 x4 x 6 x8
x3 x5 x 7 x9
=
1−
+
−
+
− ··· + i x −
+
−
+
− ···
2!
4!
6!
8!
3!
5!
7!
9!
= 1 + ix −
= cos x + i sin x.
Therefore,
eix = cos x + i sin x.
Letting x = π yields
eiπ = cos π + i sin π
= −1 + i(0)
= −1.
Thus,
eiπ + 1 = 0.