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COMPLEX FOURIER SERIES

The Fourier series of

f (x )

in

(−π , π )

is:

∞

f ( x) = ao + ∑ [a n cos nx + bn sin nx ]
n =1

(1)

We know from Euler’s formula that:

e inx = cos nx + i sin nx
e

− inx

= cos nx − i sin nx

(2)

Equations implies,

e inx + e − inx
cos nx =
2
and

e inx − e − inx − i inx
sin nx =
=
(e − e −inx )
...

2
2

[an = cn + k n ,

bn = i (c n − k n )]
...
cos nxdx

−

π

∫π f ( x)
...

So,

2

(3)

1 ⎡1
1
c n = (a n − ibn ) = ⎢
2 ⎣π
2
1
=
2π

π

π

⎤
∫π f ( x) cos nxdx − π −∫π f ( x) sin nxdx⎥
−
⎦
i

π

1
f ( x)[c n − i sin nx ]dx =
∫π
2π
−

π

∫π

f ( x)e −inx dx

−

(4)
Similarly,

1
kn =
2π

π

∫π f ( x)e

inx

dx
(5)

−

We can combine the formula (4) and (5) as:

k n = c−n
...


−

So, (3) becomes,

3

f ( x) =

∞

∑c e

inx

n

n = −∞

(6)

Where,

1
cn =
2π

π

∫π

f ( x)e −inx dx, n = 0, ± 1, ± 2,
...
For a function of period

2 L , the

complex Fourier series is:

f ( x) =

∞

∑c e

n = −∞

inπx
L

n

and

L

1
cn =
∫L f ( x)e
2L −

− inπx
L

dx, n = 0, ± 1, ± 2,
...
Find Complex Fourier Series of,

f ( x) = e , − π < x < π
...

2
π (1 + n )
n

Thus,

f ( x) =

∞

∑c e

n = −∞

inx

n


...
Find the complex form of the Fourier series expansion
of

1
f (t ) = cos t , (−π , π )
2
and change it into trigonometric form
...
e dt
−

2(−1) n +1
)e −it dt =
2
(4n − 1)π

Therefore,

2(−1) n +1 int 2 4 ∞ (−1) n +1
f (t ) = ∑
cos nt

...

− iπ
⎡ iπ
⎤
inπ
n
−inπ
2
2
= (−1)
...


We might consider applying the general techniques for solution
of differential equations
...


First finding the general solution,

xc = c1 x1 + c 2 x 2

of

the associated homogeneous differential equation
...


Then finding a single particular solution

xp

of the non

homogeneous equation in (1), and

3
...


7

c1

and

c2

so that

The following Fourier series method is more convenient and more
useful
...


Then

f (t ) , if piecewise smooth, has a Fourier series:

nπ
nπ
⎡
f (t ) = Ao + ∑ ⎢ An cos
t + Bn sin
L
L
n =1 ⎣
∞

⎤
t⎥
⎦

(3)

which has coefficients An and Bn that we can and do compute
...
We attempt to
determine the coefficients in equation (1) by first substituting the
series in equation (3) and (4) into the differential equation (1) and then
equating coefficients of like terms
...
Example below
illustrates these processes
...
Find a formal Fourier series solution of the endpoint
value problem
...


The Fourier series of f(t) is:

(−1) n +1
4t = ∑
sin nπt
π n=1 n
8

∞

(3)

We anticipate a sine series solution

∞

x (t ) = ∑ bn sin nπt

(4)

n =1

note that any such series will satisfy (2)
...


(5)

8(−1) n +1
bn =
2 2
nπ (4 − n π )

(6)

Using (6) in (4) we get,

(−1) n +1 sin nπt
x(t ) = ∑
π n =1 n(4 − n 2π 2 )
8

∞

(7)

Example- 4
...

Solution
...

16
nπ
This implies:

8(−1) n +1
4
...

2 2
1 2 2
nπ (64 − n π )
nπ (4 − n π )
16

(6)

Using (6) in (4) we get,

( −1) n +1 sin nπt
x(t ) =
∑ n(64 − n 2π 2 ) , n = 1, 2, 3,

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