# April 19, 2005 Lecturer Dmitri Zaitsev Hilary Term 2005 ```April 19, 2005
Lecturer Dmitri Zaitsev
Hilary Term 2005
Course 2E1 2004-05 (SF Engineers &amp; MSISS &amp; MEMS)
S h e e t 20
Due: in the tutorial sessions first Wednesday/Thursday in the next term
Exercise 1
Find the Fourier series of the function f (x):
1 if − π &lt; x &lt; 0
(i) f (x) =
0 if 0 &lt; x &lt; π;
Solution.
1
a0 =
2π
1
an =
π
Z
1
bn =
π
Z
π
−π
1
f (x) cosnx dx =
π
−π
π
Z
1
,
2
f (x) dx =
Z
1
f (x) sinnx dx =
π
−π
π
0
cosnx dx = 0,
−π
Z
0
sinnx dx =
−π
0
−1 + cosnπ
1 −cosnx −1 + (−1)n
=
=
,
π
n −π
nπ
nπ
so we have the Fourier Series expansion:
∞
1 X −1 + (−1)n
f (x) = +
sinnx.
2 n=1
nπ
(ii) f (x) = x, −π &lt; x &lt; π;
Solution. The function is odd (f (−x) = −f (x)), so we only need to calculate bn :
1
bn =
π
Z
π
1
x sinnx dx =
π
−π
Z
π
u dv,
−π
where u = x and dv = sinnx dx, hence v = − cosnnx . Thus
1
1
π
bn = (uv)|−π −
π
π
Z
π
Z
1 −xcosnx 1 π cosnx
vdu =
−
dx =
−π
π
n
n
−π
−π
−π
π
−
2(−1)n+1
2 cosnπ
+0=
,
n
n
so we have the Fourier Series expansion:
∞
X
2(−1)n+1
sinnx.
f (x) =
n
n=1
(iii) f (x) =
n
x if − π &lt; x &lt; 0
−x if 0 &lt; x &lt; π.
Solution. Now the function is even (f (−x) = f (x)), so we only need to calculate an . Furthermore, if g(x) is even, we can simplify our calculation by the formula
Rπ
Rπ
g(x) dx = 2 0 g(x) dx. We have
−π
1
a0 =
2π
Z
1
f (x) dx = 2
2π
−π
1
an =
π
π
Z
π
Z
π
0
1
f (x) dx =
π
1
f (x) cosnx dx = 2
π
−π
Z
Z
π
x dx =
0
π
.
2
π
x cosnx dx = ...
0
The end of calculation is similar to the previous case using integration by parts and is
omitted.
```