Math 462: HW5 Solutions
Due on August 15, 2014
Jacky Chong
1

Jacky Chong
Remark : We are working in the context of Riemann Integrals.
Problem 1
5.1.4
Find the Fourier cosine series of the functions | sin x | in the interval ( − π, π ). Use it to find the sums
∞
X
1 n =1
4 n 2 − 1 and
∞
X n =1
4
( − 1) n n 2 − 1
.
Solution : Since | sin x | is an even function on ( − π, π ), then it has a Fourier cosine series given by
| sin x | ∼
1
2
A
0
+
∞
X n =1
A n cos nx where the A n s can be readily computed by the integral formula
A n
=
=
=
1 Z
π
π
− π
| sin x | cos nx dx
1 Z
0
π
π
2
0
Z
π sin x cos nx dx − sin x cos nx dx
π
1
π
Z
0
− π sin x cos nx dx
Consider the trigonometric identity sin[( n + 1) x ] − sin[( n − 1) x ]
= sin x cos nx,
2 then it follows
A n
=
(
− 4
π
0 n 2
1
− 1 if n is even if n is odd for all n ∈
N
. Moreover, it is trivial to check that A
0
=
4
π
. Hence the Fourier cosine series is given by
| sin x | ∼
2
π
−
4
π
X
1 even n n 2 − 1
2 cos nx =
π
4
−
π
∞
X
1 n =1
4 n 2 − 1 cos 2 nx.
Assume the Fourier cosine series converges pointwise to | sin x | on ( − π, π ), then we have that
| sin x | =
2
π
−
4
∞
X
1
π n =1
4 n 2 − 1 cos 2 nx.
Set x = 0, we get that
∞
X
1 n =1
4 n 2 − 1
=
1
2
.
Set x =
π
2
, we get that
1 =
2
π
−
4
∞
X
( − 1) n
π n =1
4 n 2 − 1 or
∞
X
( − 1) n n =1
4 n 2 − 1
=
1
2
−
π
.
4
Page 2 of 9

Jacky Chong Problem 1
Problem 2
5.2.11
Find the full Fourier series of e x on ( − l, l ) in its real and complex forms.
Solution : The complex form of the full Fourier series of e x on ( − l, l ) is given by e x ∼
∞
X n = −∞ c n e inπx/l where c n can be readily compute via the following formula which yields c n
=
1
2 l
Z l
− l exp 1 − inπ l x dx = e l − inπ − e
− l + inπ
.
2( l − inπ )
Hence it follows e x ∼
∞
X n = −∞ e l − inπ − e
− l + inπ e inπx/l
2( l − inπ )
=
∞
X n = −∞
( − 1) n sinh l l 2 l + inπ
+ n 2 π 2 e inπx/l
.
The real form is given by e x ∼
∞
X n = −∞
( − 1) n sinh l l 2 l + inπ
+ n 2 π 2 e inπx/l
= sinh l l
=
=
= sinh l sinh l sinh l l l l
+
− 1
X n = −∞
( − 1) n sinh l l 2 l + inπ
+ n 2 π 2 e inπx/l
+
∞
X
( − 1) n sinh l l 2 l + inπ
+ n 2 π 2 e inπx/l n =1
+
∞
X
( − 1) n sinh l l 2 l − inπ
+ n 2 π 2 e
− inπx/l n =1
+
∞
X
( − 1) n l 2 n =1 sinh l
+ n 2 π 2 l ( e inπx/l
+
∞
X
( − 1) n n =1 sinh l l 2 l + inπ
+ n 2 π 2 e inπx/l
+ e
− inπx/l
) − nπ e inπx/l − e
− inπx/l i
+ 2
∞
X
( − 1) n l 2 n =1 sinh l
+ n 2 π 2 h l cos nπx l
− nπ sin nπx l i
.
Problem 3
5.3.10
( The Gram-Schmidt orthogonalization procedure ) If X
1
, X
2
, . . .
is any sequence (finite or infinite) of linearly independent vectors in any vector space with an inner product, it can be replaced by a sequence of linear combinations that are mutually orthogonal. The idea is that at each step one subtracts off the
= X
1
/ k X
1 k .
components parallel to the previous vectors. The procedure is as follows. First, we let Z
1
Second, we define
Y
2
= X
2
− ( X
2
, Z
1
) Z
1 and Z
2
=
Y
2 k Y
2 k
.
Third, we define
Y
3
= X
3
− ( X
3
, Z
2
) Z
2
− ( X
3
, Z
1
) Z
1 and Z
3
=
Y
3 k Y
3 k
, and so on.
(a) Show that all the vectors Z
1
, Z
2
, Z
3
, . . .
are orthogonal to each other.
(b) Apply the procedure to the pair of functions cos x + cos 2 x and 3 cos x − 4 cos 2 x in the interval (0 , π ) to
Problem 3 continued on next page. . .
Page 3 of 9

Jacky Chong get an orthogonal pair.
Problem 3 (continued)
Solution :
(a) We shall prove the statement by strong induction.
The base case is trivially true.
Now, suppose
Z
1
, . . . , Z k − 1 are mutually orthogonal, i.e.
( Z i
, Z j
) =
(
1 if i = j
0 otherwise for i = 1 , . . . , k − 1 and j = 1 , . . . , k − 1. Consider
Z k
=
X k k X k
−
−
P k − 1 i =1
( X k
, Z i
) Z i
P k − 1 i =1
( X k
, Z i
) Z i k then for a fix 1 ≤ l ≤ k − 1 we have that
( Z k
, Z l
) =
=
=
( X k
, Z l
) − k X k
−
P k − 1 i =1
( X k
P k − 1 i =1
( X k
, Z i
)( Z i
, Z l
)
, Z i
) Z i k
( X k k X
, Z l k
−
) − P k − 1 i =1
( X k
, Z i
) δ il
P k − 1 i =1
( X k
, Z i
) Z i k k X
( X k
, Z l
) − ( X k
, Z l
) k
− P k − 1 i =1
( X k
, Z i
) Z i k
= 0 .
Hence Z k is orthogonal to all Z l where l = 1 , . . . , k − 1. Moreover, we also have that
( Z k
, Z k
) =
( X k
− P k − 1 i =1
( X k
, Z i
) Z i
, X k k X k
−
− P k − 1 i =1
( X k
P k − 1 i =1
( X k
, Z i
) Z i k 2
, Z i
) Z i
)
= 1 .
(b) Observe then it follows k cos x + cos 2 x k 2
2
=
=
=
=
Z
π
| cos x + cos 2 x | 2
0
Z
π dx cos
2 x + 2 cos x cos 2 x + cos
2
2 x dx
0
Z
π
0
Z
π
1 + cos 2
2 x
1 + cos x +
+ cos 3 x + cos x +
1
1 + cos 4 cos 2 x + cos 3 x +
1
2 x dx cos 4 x dx = π,
0
2 2 cos x + cos 2 x
Z
1
= k cos x + cos 2 x k
2
=
1
√
π
(cos x + cos 2 x ) .
Next, observe
( X
2
, Z
1
) =
=
=
1
√
π
1
√
π
2
√
π
Z
π
0
Z
π
0
Z
π
0
(cos x + cos 2 x )(3 cos x − 4 cos 2 x ) dx
3 cos
2 x − cos x cos 2 x − 4 cos
2
2 x dx
− 1 − cos x + 3 cos 2 x − cos 3 x − 4 cos 4 x dx = −
√
π
2
Page 4 of 9

Jacky Chong which means
Y
2
= X
2
− ( X
2
, Z
1
) Z
1
=
7
2
(cos x − cos 2 x ) .
Computing the norm of Y
2 yields k Y
2 k 2
2
=
49
4
Z
π
(cos x − cos 2 x )
2
0 dx =
49
π
4 then it follows
Z
2
=
Y
2 k Y
2 k
2
=
1
√
π
(cos x − cos 2 x ) .
Problem 4
5.3.12
Prove Green’s first identity : For every pair of functions f ( x ) , g ( x ) on ( a, b ),
Z b f
00
( x ) g ( x ) dx = −
Z b f
0
( x ) g
0
( x ) dx + f
0 g a a b a
.
Problem 3
Solution : Assume f ∈ C
2
[ a, b ] and g ∈ C
1
[ a, b ], then f
0 g ∈ C
1
[ a, b ]. In particular, we may apply the
Fundamental Theorem of Calculus for Riemann Integral to get f
0
( x ) g ( x ) b a
=
Z b a d ds
( f
0
( s ) g ( s ))
0 ds =
Z b f
00
( s ) g ( s ) + f
0
( s ) g
0
( s ) ds.
a
Hence it follows
Z b f
00
( x ) g ( x ) dx = f
0
( x ) g ( x ) a b a
−
Z b f
0
( x ) g
0
( x ) dx.
a
Problem 5
5.4.5
Let φ ( x ) = 0 for 0 < x < 1 and φ ( x ) = 1 for 1 < x < 3.
(a) Find the first four nonzero terms of its Fourier cosine series explicitly.
(b) For each x (0 ≤ x ≤ 3), what is the sum of this series?
(c) Does it converge to φ ( x ) in the L 2 sense? Why?
(d) Put x = 0 to find the sum
1 +
1
2
−
1
4
−
1
5
+
1
7
+
1
8
1
−
10
1
−
11
+ · · · .
Solution :
(a) It’s clear that A
0
=
4
3
. Next, observe
A n
=
2
3
Z
3
1 cos nπx
3
2 dx = − nπ sin nπ
3
Page 5 of 9

Jacky Chong Problem 5 where sin nπ
3
=




0
√
3



−
2
√
3
2 if if n n
≡
≡
0
4
,
,
3
5 mod 6 if n ≡ 1 , 2 mod 6 mod 6
.
Hence, it follows
A n
=







0
−
√
3
√
3 nπ nπ if if n n
≡
≡
0
4
,
,
3
5 mod 6 if n ≡ 1 , 2 mod 6 mod 6
.
This gives us the following Fourier cosine series
φ ( x ) ∼
2
3
−
√
3
π cos
πx
3
−
√
3
2 π cos
2 πx
3
+
√
3
4 π cos
4 πx
3
+
√
3 cos
5 π
5 πx
− . . .
3
(b) By Theorem 4(ii), the Fourier cosine series converges pointwise everywhere on
R
. Moreover, let us extend to an even periodic function then for each fixed x ∈ [0 , 3] we have that
1
2
A
0
+
∞
X
A n n =1 cos nπx
3
=
1
2
[ φ ext
( x +) + φ ext
( x − )] where
1
2
[ φ ext
( x +) + φ ext
( x − )] =




0 if 0 ≤ x < 1
1
2 if x = 1 , 3



1 if 1 < x < 3
(c) Since φ ( x ) is L
2 in the L 2 integrable, then by Theorem 3 we know the Fourier cosine series does indeed converge sense to φ .
(d) Set x = 0 yields or
0 =
2
3
−
√
3
π
−
√
3
2 π
+
√
3
4 π
+
√
3
5 π
−
√
3
7 π
−
√
3
8 π
+ . . .
3
2 π
√
3
= 1 +
1
2
−
1
4
−
1
5
+
1
7
+
1
8
1
−
10
1
−
11
+ · · · .
Problem 6
5.4.15
Let φ ( x ) ≡ 1 for 0 < x < π . Expand
1 =
∞
X
B n cos n +
1
2 n =0 x .
(a) Find B n
.
(b) Let − 2 π < x < 2 π . For which such x does this series converges? For each such x , what is the sum of the series?
(c) Apply Parseval’s equality to this series. Use it to calculate the sum
1 +
1
3 2
1
+
5 2
+ · · · .
Problem 6 continued on next page. . .
Page 6 of 9

Jacky Chong Problem 6 (continued)
Solution :
(a) First, consider the following trigonometric identity cos n +
1
2 x cos m +
1
2 x = cos [( n + m + 1) x ] + cos [( n − m ) x ]
2 then we have that
Z
π
0 cos n +
1
2 x cos m +
1
2 x dx =
(
0 if n = m
π
2 if n = m
.
Therefore, the coefficients of the above trigonometric series can be readily computed using the following formula
B n
2
=
π
Z
π
0 cos n +
1
2 x dx =
4
π
( − 1) n
2 n + 1
.
(b) Consider the series
4
π
∞
X
( − 1) n
2 n + 1 cos n + n =0
1
2 x .
By Theorem 4, we know the above trigonometric series converges pointwise on (0 , π ) to 1. If x ∈ ( − π, 0) then we have
4
∞
X
π n =0
( − 1) n
2 n + 1 cos n +
1
2
4 x = −
π
∞
X
( − 1) n n =0
2 n + 1 cos n +
1
2
| x | = − 1 .
Lastly, when x = 0, we have the series
4
∞
X
( − 1) n
π n =0
2 n + 1 which is convergent by Leibniz’s Test (Alternating Series Test). To evaluate the series, we use the fact arctan x =
∞
X n =0
( − 1) n x 2 n +1
.
2 n + 1 for all x ∈ [ − 1 , 1], which means
∞
X
( − 1) n n =0
2 n + 1
=
π
4 or
4
∞
X
( − 1) n
π n =0
2 n + 1
= 1
(c) Apply Parseval’s equality to the series yields
16
π 2
∞
X n =0
1
(2 n + 1) 2
Z
π cos
2
0 or n +
∞
X
1 n =0
(2 n + 1) 2
=
π
2
.
8
1
2 x dx = π
Page 7 of 9

Jacky Chong Problem 6
Problem 7
5.4.16
Let φ ( x ) = | x | in ( − π, π ). If we approximate it by the function f ( x ) =
1
2 a
0
+ a
1 cos x + b
1 sin x + a
2 cos 2 x + b
2 sin 2 x, what choice of coefficients will minimize the L 2 error?
Solution : Since φ ( x ) = | x | in ( − π, π ) is an even function, then φ ( x ) has a Fourier cosine series representation
φ ( x ) ∼
1
2
A
0
+
∞
X
A n cos nx.
n =1
Moreover, since φ ( x ) is continuous on ( − π, π ) and φ
0
( x ) is piecewise continuous on ( − π, π ), then by Theorem
4 the Fourier series converges pointwise to φ ( x ) on ( − π, π ), i.e.
φ ( x ) =
1
2
A
0
+
∞
X
A n cos nx.
n =1
Let us rewrite f as f ( x ) =
∞
X a n cos nx + b n sin nx n =0 where a n
, b n
= 0 for n ≥ 3. Now, apply Parseval’s equality yields k φ − f k
2
2
=
1
2
( A
0
− a
0
) − b
1 sin x − b
2 sin 2 x +
∞
X n =1
( A n
− a n
) cos nx
= | b
1
| 2 k sin x k 2
2
+ | b
2
| 2 k sin 2 x k 2
2
+
1
2
| A
0
− a
0
| 2
+
∞
X
| A n n =1
2
2
− a n
| 2 k cos nx k 2
2 which is minimized provided b
1
= b
2
= 0 , a
0
= A
0
, a
1
= A
1 and a
2
= A
2
.
Problem 8
5.5.1
Sketch the graph of the Dirichlet kernel
K
N
( θ ) = sin N sin
+
1
2
θ
1
2
θ in the case N = 10. Use a computer graphics program if you wish.
Solution : [left to the reader]
Problem 9
6.3.1
Suppose that u is a harmonic function in the disk D = { r < 2 } and that u = 3 sin 2 θ + 1 for r = 2.
Without finding the solution, answer the following questions
(a) Find the maximum value of u in D .
Problem 9 continued on next page. . .
Page 8 of 9

Jacky Chong
(b) Calculate the value of u at the origin.
Solution :
(a) By the maximum principle of harmonic function, we know that max x ∈ D u ( x ) = max x ∈ ∂D u ( x ) = max
θ ∈ [0 , 2 π )
(3 sin 2 θ + 1) .
However, it’s clear that the maximum occurs when θ =
π
4
, which means max x ∈ D u ( x ) = 4 .
(b) By the mean-value property of harmonic function, we have that u (0) =
1
2 π
Z
2 π
0
3
(3 sin 2 θ + 1) dθ = −
4 π cos 2 θ +
1
2 π
θ
2 π
0
= 1 .
Problem 10
6.3.2
Solve u xx
+ u yy
= 0 in the disk { r < a } with the boundary condition u = 1 + 3 sin θ on r = a.
Problem 9 (continued)
Solution : Using the method of separation of variables, we obtain the following trigonometric series u ( r, θ ) =
1
2
A
0
+
∞
X r n
( A n n =1 cos nθ + B n sin nθ ) .
Applying the boundary condition yields u ( a, θ ) =
1
2
A
0
+
∞
X n =1 a n
( A n cos nθ + B n sin nθ ) = 1 + 3 sin θ.
Then it follows A
0
= 2, A n
= 0 for all n ∈
N and
B n
=
(
3 a
− 1
0 if n = 1 if otherwise
.
Thus, we have that u ( r, θ ) = 1 + 3 a
− 1 r sin θ solves the above Dirichlet problem.
Page 9 of 9