Consider the problem
min
a
g
(
a
)
=
∥
A
(
ω
;
a
)

D
(
ω
)
∥
p
min
a
g
(
a
)
=
∥
A
(
ω
;
a
)

D
(
ω
)
∥
p
(8)for a∈RM+1a∈RM+1. Problem Equation 8 is equivalent to the better posed problem
min
a
f
(
a
)
=
g
(
a
)
p
=
∥
A
(
ω
;
a
)

D
(
ω
)
∥
p
p
=
∑
i
=
0
L
∣
C
i
a

D
i
∣
p
min
a
f
(
a
)
=
g
(
a
)
p
=
∥
A
(
ω
;
a
)

D
(
ω
)
∥
p
p
=
∑
i
=
0
L
∣
C
i
a

D
i
∣
p
(9)where Di=D(ωi)Di=D(ωi), ωi∈[0,π]ωi∈[0,π], Ci=[Ci,0,...,Ci,M]Ci=[Ci,0,...,Ci,M], and
C
=
C
0
⋮
C
L
C
=
C
0
⋮
C
L
(10)The ijijth element of CC is given by Ci,j=cos ωi(Mj)Ci,j=cos ωi(Mj), where 0≤i≤L0≤i≤L and 0≤j≤M0≤j≤M. From Equation 9 we have that
∇
f
(
a
)
=
∂
∂
a
0
f
(
a
)
⋮
∂
∂
a
M
f
(
a
)
∇
f
(
a
)
=
∂
∂
a
0
f
(
a
)
⋮
∂
∂
a
M
f
(
a
)
(11)where ajaj is the jjth element of a∈RM+1a∈RM+1 and
∂
∂
a
j
f
(
a
)
=
∂
∂
a
j
∑
i
=
0
L
∣
C
i
a

D
i
∣
p
=
∑
i
=
0
L
∂
∂
a
j
∣
C
i
a

D
i
∣
p
=
p
∑
i
=
0
L
∣
C
i
a

D
i
∣
p

1
·
∂
∂
a
j
∣
C
i
a

D
i
∣
∂
∂
a
j
f
(
a
)
=
∂
∂
a
j
∑
i
=
0
L
∣
C
i
a

D
i
∣
p
=
∑
i
=
0
L
∂
∂
a
j
∣
C
i
a

D
i
∣
p
=
p
∑
i
=
0
L
∣
C
i
a

D
i
∣
p

1
·
∂
∂
a
j
∣
C
i
a

D
i
∣
(12)Now,
∂
∂
a
j
∣
C
i
a

D
i
∣
=
sign
(
C
i
a

D
i
)
·
∂
∂
a
j
(
C
i
a

D
i
)
=
C
i
,
j
sign
(
C
i
a

D
i
)
∂
∂
a
j
∣
C
i
a

D
i
∣
=
sign
(
C
i
a

D
i
)
·
∂
∂
a
j
(
C
i
a

D
i
)
=
C
i
,
j
sign
(
C
i
a

D
i
)
(13)where
sign
(
x
)
=
1
x
>
0
0
x
=
0

1
x
<
0
sign
(
x
)
=
1
x
>
0
0
x
=
0

1
x
<
0
(15)Therefore the Jacobian of f(a)f(a) is given by
∇
f
(
a
)
=
p
∑
i
=
0
L
C
i
,
0
∣
C
i
a

D
i
∣
p

1
sign
(
C
i
a

D
i
)
⋮
p
∑
i
=
0
L
C
i
,
M

1
∣
C
i
a

D
i
∣
p

1
sign
(
C
i
a

D
i
)
∇
f
(
a
)
=
p
∑
i
=
0
L
C
i
,
0
∣
C
i
a

D
i
∣
p

1
sign
(
C
i
a

D
i
)
⋮
p
∑
i
=
0
L
C
i
,
M

1
∣
C
i
a

D
i
∣
p

1
sign
(
C
i
a

D
i
)
(16)The Hessian of f(a)f(a) is the matrix ∇2f(a)∇2f(a) whose jmjmth element (0≤j,m≤M0≤j,m≤M) is given by
∇
j
,
m
2
f
(
a
)
=
∂
a
2
∂
a
j
∂
a
m
f
(
a
)
=
∂
∂
a
m
∂
∂
a
j
f
(
a
)
=
∑
i
=
0
L
p
C
i
,
j
∂
∂
a
m
∣
D
i

C
i
a
∣
p

1
sign
(
D
i

C
i
a
)
=
∑
i
=
0
L
α
∂
∂
a
m
b
(
a
)
d
(
a
)
∇
j
,
m
2
f
(
a
)
=
∂
a
2
∂
a
j
∂
a
m
f
(
a
)
=
∂
∂
a
m
∂
∂
a
j
f
(
a
)
=
∑
i
=
0
L
p
C
i
,
j
∂
∂
a
m
∣
D
i

C
i
a
∣
p

1
sign
(
D
i

C
i
a
)
=
∑
i
=
0
L
α
∂
∂
a
m
b
(
a
)
d
(
a
)
(17)where adequate substitutions have been made for the sake of simplicity. We have
∂
∂
a
m
b
(
a
)
=
∂
∂
a
m
∣
C
i
a

D
i
∣
p

1
=
(
p

1
)
C
i
,
m
∣
C
i
a

D
i
∣
p

2
sign
(
C
i
a

D
i
)
∂
∂
a
m
d
(
a
)
=
∂
∂
a
m
sign
(
D
i

C
i
a
)
=
0
∂
∂
a
m
b
(
a
)
=
∂
∂
a
m
∣
C
i
a

D
i
∣
p

1
=
(
p

1
)
C
i
,
m
∣
C
i
a

D
i
∣
p

2
sign
(
C
i
a

D
i
)
∂
∂
a
m
d
(
a
)
=
∂
∂
a
m
sign
(
D
i

C
i
a
)
=
0
(18)Note that the partial derivative of d(a)d(a) at DiCia=0DiCia=0 is not defined. Therefore
∂
∂
a
m
b
(
a
)
d
(
a
)
=
b
(
a
)
∂
∂
a
m
d
(
a
)
+
d
(
a
)
∂
∂
a
m
b
(
a
)
=
(
p

1
)
C
i
,
m
∣
C
i
a

D
i
∣
p

2
sign
2
(
C
i
a

D
i
)
∂
∂
a
m
b
(
a
)
d
(
a
)
=
b
(
a
)
∂
∂
a
m
d
(
a
)
+
d
(
a
)
∂
∂
a
m
b
(
a
)
=
(
p

1
)
C
i
,
m
∣
C
i
a

D
i
∣
p

2
sign
2
(
C
i
a

D
i
)
(19)Note that sign2(CiaDi)=1sign2(CiaDi)=1 for all DiCia≠0DiCia≠0 where it is not defined. Then
∇
j
,
m
2
f
(
a
)
=
p
(
p

1
)
∑
i
=
0
L
C
i
,
j
C
i
,
m
∣
C
i
a

D
i
∣
p

2
∇
j
,
m
2
f
(
a
)
=
p
(
p

1
)
∑
i
=
0
L
C
i
,
j
C
i
,
m
∣
C
i
a

D
i
∣
p

2
(20)except at DiCia=0DiCia=0 where it is not defined.
Based on Equation 16 and Equation 20, one can apply Newton's method to problem Equation 8 as follows,
 Given a0∈RM+1a0∈RM+1, D∈RL+1D∈RL+1, C∈RL+1×M+1C∈RL+1×M+1
 For i=0,1,...i=0,1,...
 Find ∇f(ai)∇f(ai).
 Find ∇2f(ai)∇2f(ai).
 Solve ∇2f(ai)s=∇f(ai)∇2f(ai)s=∇f(ai) for ss.
 Let a+=ai+sa+=ai+s.
 Check for convergence and iterate if necessary.
Note that for problem Equation 8 the Jacobian of f(a)f(a) can be written as
∇
f
(
a
)
=
p
C
T
y
∇
f
(
a
)
=
p
C
T
y
(21)where
y
=
∣
C
a
i

D
∣
p

1
sign
(
C
a
i

D
)
=
∣
C
a
i

D
∣
p

2
(
C
a
i

D
)
y
=
∣
C
a
i

D
∣
p

1
sign
(
C
a
i

D
)
=
∣
C
a
i

D
∣
p

2
(
C
a
i

D
)
(22)Also,
∇
j
,
m
2
f
(
a
)
=
p
(
p

1
)
C
j
T
Z
C
m
∇
j
,
m
2
f
(
a
)
=
p
(
p

1
)
C
j
T
Z
C
m
(23)where
Z
=
diag
∣
C
a
i

D
∣
p

2
Z
=
diag
∣
C
a
i

D
∣
p

2
(24)and
C
j
=
C
0
,
j
⋮
C
L
,
j
C
j
=
C
0
,
j
⋮
C
L
,
j
(25)Therefore
∇
2
f
(
a
)
=
(
p
2

p
)
C
T
Z
C
∇
2
f
(
a
)
=
(
p
2

p
)
C
T
Z
C
(26)From Equation 26, the Hessian
∇2f(a)∇2f(a) can be expressed as
∇
2
f
(
a
)
=
(
p
2

p
)
C
T
W
T
W
C
∇
2
f
(
a
)
=
(
p
2

p
)
C
T
W
T
W
C
(27)where
W
=
diag
∣
C
a
i

D
∣
p

2
2
W
=
diag
∣
C
a
i

D
∣
p

2
2
(28)The matrix C∈R(L+1)×(M+1)C∈R(L+1)×(M+1) is given by
C
=
cos
M
ω
0
cos
(
M

1
)
ω
0
⋯
cos
(
M

j
)
ω
0
⋯
cos
ω
0
1
cos
M
ω
1
cos
(
M

1
)
ω
1
⋯
cos
(
M

j
)
ω
1
⋯
cos
ω
1
1
⋮
⋮
⋱
⋮
⋮
⋮
cos
M
ω
i
cos
(
M

1
)
ω
i
⋯
cos
(
M

j
)
ω
i
⋯
cos
ω
i
1
⋮
⋮
⋮
⋱
⋮
⋮
cos
M
ω
L

1
cos
(
M

1
)
ω
L

1
⋯
cos
(
M

j
)
ω
L

1
⋯
cos
ω
L

1
1
cos
M
ω
L
cos
(
M

1
)
ω
L
⋯
cos
(
M

j
)
ω
L
⋯
cos
ω
L
1
C
=
cos
M
ω
0
cos
(
M

1
)
ω
0
⋯
cos
(
M

j
)
ω
0
⋯
cos
ω
0
1
cos
M
ω
1
cos
(
M

1
)
ω
1
⋯
cos
(
M

j
)
ω
1
⋯
cos
ω
1
1
⋮
⋮
⋱
⋮
⋮
⋮
cos
M
ω
i
cos
(
M

1
)
ω
i
⋯
cos
(
M

j
)
ω
i
⋯
cos
ω
i
1
⋮
⋮
⋮
⋱
⋮
⋮
cos
M
ω
L

1
cos
(
M

1
)
ω
L

1
⋯
cos
(
M

j
)
ω
L

1
⋯
cos
ω
L

1
1
cos
M
ω
L
cos
(
M

1
)
ω
L
⋯
cos
(
M

j
)
ω
L
⋯
cos
ω
L
1
(29)The matrix H=∇2f(a)H=∇2f(a) is positive definite (for p>1p>1). To see this, consider H=KTKH=KTK where K=WCK=WC. Let z∈RM+1z∈RM+1, z≠0z≠0. Then
z
T
H
z
=
z
T
K
T
K
z
=
∥
K
z
∥
2
2
>
0
z
T
H
z
=
z
T
K
T
K
z
=
∥
K
z
∥
2
2
>
0
(30)unless z∈N(K)z∈N(K). But since WW is diagonal and CC is full column rank, N(K)=0N(K)=0 . Thus zTHz≥0zTHz≥0 (identity only if z=0z=0) and so HH is positive definite.