Math 293 Solutions to Problem Set 11.
§
10.3 #1
. Both terms of
x
3
+sin 2
x
are odd, so the whole function is odd. In other words,
(

x
)
3
+ sin(

2
x
) =

x
3

sin 2
x
=

(
x
3
+ sin 2
x
).
#2
. sin
2
(

x
) = (

sin
x
)
2
= (sin
x
)
2
= sin
2
x
, so this function is even.
#3
. Replacing
x
by

x
in (1

x
2
)

1
/
2
does not change the value of this function, so it is
even.
#10
.
The function
f
(
x
) =

x

is even, so its Fourier series will have only cosine terms
(and the constant term
a
0
2
). We have
a
n
=
1
π
R
π

π
f
(
x
) cos
nx dx
=
2
π
R
π
0
x
cos
nx dx
since
the function is even.
Using integration by parts, we get
2
π
£
x
sin
nx
n
fl
fl
π
0

R
π
0
sin
nx dx
/
.
Evaluating the first term at 0 and
π
gives 0, so we are left with
a
n
=

2
π
R
π
0
sin
nx dx
=
2
π
(
cos
nx
n
2
fl
fl
π
0
)
=
2
πn
2
(
cos
nπ

1
)
.
The factor cos
nπ

1 is 0 if
n
is even and

2 if
n
is odd.
Thus the nonzero
a
n
’s are
a
2
n

1
=

4
π
(2
n

1)
2
.
When
n
= 0 these formulas
do not work since they would involve dividing by 0, so we do a separate calculation:
a
0
=
2
π
R
π
0
x dx
=
· · ·
=
π
. Now we can write the final answer, the Fourier cosine series
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 Spring '07
 TERRELL,R
 Math, Differential Equations, Equations, Fourier Series, Periodic function, Leonhard Euler, Joseph Fourier

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