UC Berkeley
Department of Electrical Engineering and Computer Science
EECS 227A
Nonlinear and Convex Optimization
Solutions 7
Fall 2009
Problem 7.1
(a) The derivatives are
f
0
(
x
) =

1
/x
2
,f
00
(
x
) = 2
/x
3
,f
000
(
x
) =

6
/x
4
so the selfconcordance condition is
6
x
4
≤
2
±
2
x
3
²
3
/
2
=
4
√
2
x
4
√
x
which holds if
√
x
≤
4
√
2
/
6 =
p
8
/
9
(b) If we make an aﬃne change of variables
y
i
= 8(
b
i

a
T
i
x
)
/
(9
α
), then
y
i
<
8
/
9 for all
x
∈
dom
f
. The function
f
reduces to
∑
m
i
=1
1
/y
i
, which is self concordant by the result
in (a).
Problem 7.2
(a) We have
d
dx
f
00
(
x
)

1
/
2
= (

1
/
2)
f
000
(
x
)
f
00
(
x
)
3
/
2
Integrating
d
dx
f
00
(
x
)

1
/
2
= 1
gives
f
(
x
) =

log(
x
+
c
0
) +
c
1
x
+
c
2
. Integrating
d
dx
f
00
(
x
)

1
/
2
=

1
gives
f
(
x
) =

log(

x
+
c
0
) +
c
1
x
+
c
2
(b) Suppose
f
00
(0)
>
0
,f
00
(¯
x
) = 0 for ¯
x >
0, and
f
00
(
x
)
>
0 on the interval between 0 and
¯
x
. The inequality

1
≤
d
dx
f
00
(
x
)

1
/
2
≤
1
holds for
x
between 0 and ¯
x
. Integrating gives
f
00
(¯
x
)

1
/
2

f
00
(0)

1
/
2
≤
¯
x
which contradicts
f
00
(¯
x
) = 0
1
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View Full DocumentProblem 7.3
(a) The composition rules show that
tf
0
(
x
)+
φ
h
(
x
) is convex in
x
, since
h
is increasing and
convex, and
f
i
are convex.
(b) The minimizer of
tf
0
(
x
) +
φ
h
(
x
)
,z
=
x
*
(
t
), satisﬁes
t
∇
f
0
(
z
) +
∇
φ
(
z
) = 0. Expanding
this we get
t
∇
f
0
(
z
) +
m
X
i
=1
h
0
(
f
i
(
z
))
∇
f
i
(
z
) = 0
This shows that
z
minimizes the Lagrangian
f
0
(
z
) +
∑
m
i
=1
λ
i
f
i
(
z
), for
λ
i
=
h
0
(
f
i
(
z
))
/t,i
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 Spring '11
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 Electrical Engineering, Derivative, Convex function, Diag, log det

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