Honors Introduction to Analysis I
Homework IX
Solution
April 21, 2009
Problem 1
Suppose
f
is
C
1
on an interval and that
f
0
satisfies the H¨older condition of order
α
,

f
0
(
x
)

f
0
(
y
)
 ≤
M

x

y

α
for all
x, y
in the interval, where
α
∈
(0
,
1]
is fixed. Show that

Δ
2
h
f
(
x
)
 ≤
c

h

1+
α
. How does the constant
c
relate to
M
?
Solution.
We apply the mean value theorem to
g
(
t
) =
f
(
t
+
h
)

f
(
t
) on the interval [
x, x
+
h
]. Then
Δ
2
h
f
(
x
) =
f
(
x
+ 2
h
)

2
f
(
x
+
h
) +
f
(
x
) =
f
(
x
+ 2
h
)

f
(
x
+
h
)

(
f
(
x
+
h
)

f
(
x
)) =
g
(
x
+
h
)

g
(
x
)
and
g
0
(
t
) =
f
0
(
t
+
h
)

f
0
(
t
).
By the MVT we get that
g
(
x
+
h
)

g
(
x
) =
hg
0
(
x
0
) for some
x
0
∈
(
x, x
+
h
), which is equivalent to saying
Δ
2
h
f
(
x
) =
h
(
f
0
(
x
0
+
h
)

f
0
(
x
0
))
so

Δ
2
h
f
(
x
)

=

h

f
0
(
x
0
+
h
)

f
0
(
x
0
)
 ≤ 
h

M

x
0
+
h

x
0

α
=
M

h

1+
α
≤
c

h

1+
α
for
c
≥
M
.
Hence

Δ
2
h
f
(
x
)
 ≤
c

h

1+
α
for
c
≥
M
(the minimal value for
c
, for which this is true, is
M
)
Problem 2
Suppose that
f
has a zero of order
j
at
x
0
and that
g
has a zero of order
k
at
x
0
. What can you say
about the order of zero of the function
f
+
g
,
f
·
g
and
f/g
at
x
0
?
Solution.
By definition
f
(
x
) =
a
j
(
x

x
0
)
j
+
o
(

x

x
0

j
) and
g
(
x
) =
a
k
(
x

x
0
)
j
+
o
(

x

x
0

k
), or
f
(
n
)
(
x
0
) = 0
for
n
= 0
, ..., j

1 and
f
(
j
)
(
x
0
)
6
= 0 and
g
(
n
)
(
x
0
) = 0 for
n
= 0
, ..., k

1 and
g
(
k
)
(
x
0
)
6
= 0.
•
Let
m
= min
{
j, k
}
, then (
f
+
g
)
(
n
)
(
x
0
) = 0 for
n
= 0
, ..., m

1 and (
f
+
g
)
(
m
)
(
x
0
)
6
= 0. Or (
f
+
g
)(
x
) =
a
m
(
x

x
0
)
m
+
o
(

x

x
0

m
), so
f
+
g
has a zero of order min
{
j, k
}
at
x
0
.
•
For any
n
, (
fg
)
(
n
)
(
x
) =
n
∑
i
=0
a
i
f
(
i
)
(
x
)
g
(
n

i
)
(
x
) for
a
i
=
n
i
. From this we notice that (
fg
)
(
n
)
(
x
0
) = 0 for
n
= 0
, ..., k
+
j

1 and (
fg
)
(
k
+
j
)
(
x
0
) =
k
+
j
j
f
(
j
)
(
x
0
)
g
(
k
)
(
x
0
)
6
= 0, so
fg
has a zero of order
j
+
k
.
Equivalently one can see that (
fg
)(
x
) =
a
k
a
j
(
x

x
0
)
j
+
k
+
o
(

x

x
0

k
+
j
).
•
Let’s first analyze the case when
j < k
.
We write:
f
(
x
)
g
(
x
)
=
f
(
j
)
(
x
0
)(
x

x
0
)
j
+
...
+
a
j
f
(
k
)
(
x
0
)(
x

x
0
)
k
+
o
(

x

x
0

k
)
g
(
k
)
(
x
0
)(
x

x
0
)
k
+
o
(

x

x
0

k
)
or
f
(
x
)
g
(
x
)
=
f
(
j
)
(
x
0
) +
...
+
a
j
f
(
k
)
(
x
0
)(
x

x
0
)
k

j
+
o
(

x

x
0

k

j
)
g
(
k
)
(
x
0
)(
x

x
0
)
k

j
+
o
(

x

x
0

k

j
)
where
a
j
comes from the Taylor expansion, after dividing by 1
/k
!.
1
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The first term will be
f
(
j
)
(
x
0
)
g
(
k
)
(
x
0
)(
x

x
0
)
k

j
+
o
(

x

x
0

k

j
)
, which gives us a singularity at
x
0
, so the function is not
defined there. Hence in order to be possible to have zeroes at
x
0
we need that
j
≥
k
. In that case we have
that
f
(
x
)
g
(
x
)
=
f
(
j
)
(
x
0
)(
x

x
0
)
j
+
o
(

x

x
0

j
)
g
(
k
)
(
x
0
)(
x

x
0
)
k
+
o
(

x

x
0

k
)
or
f
(
x
)
g
(
x
)
=
f
(
j
)
(
x
0
)(
x

x
0
)
j

k
+
o
(

x

x
0

j

k
)
g
(
k
)
(
x
0
) +
o
(1)
o
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 '08
 PROTSAK
 Taylor Series, zeroes, k=0, real zeroes, Euler–Maclaurin formula, Bernoulli polynomials

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