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Lecture 7
We
continue
our
proof
of
the
Inverse Function
Theorem.
As before,
we
let
U
be
an open
set in
R
n
, and
we assume that 0
∈
U
.
We let
f
:
U
→
R
n
be
a
C
1
map,
and we assume
f
(0) =
0
and
that
Df
(0) =
I
.
We
summarize
what
we
have
proved
so
far in
the following
theorem.
Theorem 2.18.
There
exists
a
neighborhood
U
0
of
0
in
U
and a neighborhood
V
of
0
in
R
n
such
that
1.
f
maps
U
0
bijectively
onto
V
2.
f
−
1
:
V
U
0
is
continuous,
→
3.
f
−
1
is
differentiable
at
0
.
Now,
we
let
U
be
an open set
in
R
n
, and
we let
f
:
U
→
R
n
be a
C
2
map, as
before,
but
we
return to our
original
assumptions
that
a
∈
U
,
b
=
f
(
a
), and
Df
(
a
) :
R
n
R
n
→
is bijective.
We
prove
the
following
theorem.
Theorem 2.19.
There
exists
a
neighborhood
U
0
of
a
in
U
and a neighborhood
V
of
b
in
R
n
such
that
1.
f
maps
U
0
bijectively
onto
V
2.
f
−
1
:
V
U
0
is
continuous,
→
3.
f
−
1
is
differentiable
at
b
.
Proof.
The
map
f
:
U
→
R
n
maps
a
to
b
. Define
U
�
=
U
−
a
=
{
x
−
a
:
x
∈
U
}
. Also
define
f
1
:
U
�
→
R
n
by
f
1
(
x
) =
f
(
x
+
a
)
−
b
, so
that
f
1
(0) = 0 and
Df
1
(0) =
Df
(
a
)
(using
the
Chain Rule).
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 Fall '04
 unknown
 Riemann integral, Riemann sum, Inverse Function Theorem, det Jf

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