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10.21
Warning.
By the definition of differentiablity
given in Math 111, the domain of a function was required to contain
some interval
![$(a-\varepsilon, a+\varepsilon)$](img1099.gif)
in order for the function be differentiable at
![$a$](img590.gif)
.
In definition
10.1 this condition has been replaced
by requiring
![$a$](img590.gif)
to be a limit point of the domain of the function.
Now a function whose domain is a closed interval
![$[a,b]$](img935.gif)
may be differentiable at
![$a$](img590.gif)
and/or
![$b$](img954.gif)
.
10.22
Definition (Critical point.)
Let
![$f$](img13.gif)
be a complex function, and let
![$a\in\mbox{{\bf C}}$](img86.gif)
. If
![$f$](img13.gif)
is differentiable at
![$a$](img590.gif)
and
![$f'(a)=0$](img1100.gif)
, we call
![$a$](img590.gif)
a
critical point for
![$f$](img13.gif)
.
10.23
Theorem (Critical Point Theorem.)
Let
be a function.
Suppose
has a maximum at some point
, and that
contains an interval
where
.
If
is differentiable at
, then
. The theorem also holds if we
replace `` maximum" by `` minimum."
Proof: Suppose
has a maximum at
,
Define two sequences
,
in
by
Clearly
and
, and
and
for all
.
We have
By the inequality theorem,
Also,
so
Since
, we conclude that
. The proof for minimum points
is left to you.
10.24
Theorem (Rolle's Theorem.)
Let
with
and let
be a function that is
continuous on
and differentiable on
. Suppose that
.
Then there is a number
such that
.
Proof: We know from the extreme value theorem that
has a maximum at some point
. If
, then the critical point theorem says
, and we
are finished. Suppose
. We know there is a point
such
that
has a minimum at
. If
we get
by the critical
point theorem, so suppose
. Then since
and
, we have
, and it follows that
is a
constant function on
, and in this case
for all
.
This theorem says that the tangent to the graph of
at some point
is parallel to the chord joining
to
.
Proof: Let
so the equation of the line joining
to
is
, and
Let
Then
and
is continuous on
and differentiable on
. By Rolle's
theorem, there is some
such that
; i.e.,
; i.e.,
10.26
Remark.
The mean value theorem does not hold for complex valued functions. Let
Then
so
But
so
![$F'(t) = 0 \hspace{1ex}\Longleftrightarrow\hspace{1ex}t = -i$](img1144.gif)
, and there is no point in
![$t \in (-1,1)$](img1145.gif)
with
![$F'(t) = 0.$](img1146.gif)
10.27
Definition (Interior point.)
Let
![$J$](img1147.gif)
be an interval in
![$\mbox{{\bf R}}$](img4.gif)
. A number
![$a \in J$](img1148.gif)
is an
interior point
of
![$J$](img1147.gif)
if and only if
![$a$](img590.gif)
is not an end point
of
![$J$](img1147.gif)
. The set of all interior points of
![$J$](img1147.gif)
is called the
interior of
![$J$](img1147.gif)
and is denoted by
![$\mbox{\rm int}(J)$](img1149.gif)
.
10.28
Examples.
If
![$a<b$](img933.gif)
, then
If
is an interval, and
are points in
with
, then every point in
is in the interior of
.
10.29
Theorem.
Let
be an interval in
, and let
be a continuous
function on
. Then:
- a)
- If
for all
, then
is increasing on
.
- b)
- If
for all
, then
is strictly increasing on
.
- c)
- If
for all
, then
is decreasing on
.
- d)
- If
for all
, then
is strictly decreasing on
.
- e)
- If
for all
, then
is constant on
.
Proof: All five statements have similar proofs. I'll prove only part a).
Suppose
for all
. Then for all
with
we have
is continuous on
and differentiable
on
, so by the mean value theorem
Hence,
is increasing on
.
10.30
Exercise.
Prove part e) of the previous theorem; i.e., show that if
![$J$](img1147.gif)
is an interval in
![$\mbox{{\bf R}}$](img4.gif)
and
![$f\colon J\to\mbox{{\bf R}}$](img1154.gif)
is continuous and satisfies
![$f'(t)=0$](img1164.gif)
for all
![$t\in \mbox{\rm int}(J)$](img1165.gif)
, then
![$f$](img13.gif)
is
constant on
![$J$](img1147.gif)
. [It is sufficient to show that
![$f(s)=f(t)$](img1166.gif)
for all
![$s,t\in J$](img1161.gif)
.]
Proof:
Let
be any sequence in
such that
. Then
is a sequence in
,
and hence
It follows that
I've shown that
10.33
Definition (Path, line segment.)
If
![$a,b\in\mbox{{\bf C}}$](img1184.gif)
, then the
path joining
to ![$b$](img954.gif)
is the function
and the set
is called the
line segment joining
![$a$](img590.gif)
to
![$b$](img954.gif)
.
10.34
Example.
We showed in example
10.9 that the function
![$\mbox{\rm conj}: z \mapsto z^*$](img1190.gif)
is a nowhere differentiable function on
![$\mbox{{\bf C}}$](img2.gif)
.
I will show that for all
![$a$](img590.gif)
,
![$b$](img954.gif)
in
![$\mbox{{\bf C}}$](img2.gif)
with
![$a \neq b$](img1191.gif)
, the restriction
![$\mbox{\rm conj}\vert _{\Lambda_{ab}}$](img1192.gif)
of conj to the line segment
![$\Lambda_{ab}$](img1187.gif)
is differentiable,
and
Note that all points of
![$\Lambda_{ab}$](img1187.gif)
are limit points
of
![$\Lambda_{ab}$](img1187.gif)
. If
![$z \in \Lambda_{ab}$](img1194.gif)
, then for some real number
![\begin{displaymath}z = a + t(b-a)
\end{displaymath}](img1195.gif) |
(10.35) |
and
![\begin{displaymath}z^* = a^* + t(b^* - a^*).
\end{displaymath}](img1196.gif) |
(10.36) |
If we solve equation (
10.35) for
![$t$](img920.gif)
we get
By using this value for
![$t$](img920.gif)
in equation
(
10.36) we get
Let
![$H_{ab}:\mbox{{\bf C}}\to \mbox{{\bf C}}$](img1199.gif)
be defined by
Then
![$H_{ab}$](img1201.gif)
is differentiable, and
![$\displaystyle {H'(z) = {b^*-a^* \over b-a}}$](img1202.gif)
for all
![$z\in\mbox{{\bf C}}$](img66.gif)
. We have
so by the restriction theorem
10.37
Exercise.
A
Let
![$C(0,1)$](img1205.gif)
denote the unit circle in
![$\mbox{{\bf C}}$](img2.gif)
. Show that
![$\mbox{\rm conj}\vert _{C(0,1)}$](img1206.gif)
is differentiable, and that
In general, the real and imaginary parts of a differentiable function are not
differentiable.
10.38
Example.
If
![$f(z)=z$](img624.gif)
for all
![$z\in\mbox{{\bf C}}$](img66.gif)
, then
![$f$](img13.gif)
is differentiable and
![$f'(z)=1$](img1208.gif)
. However,
![$\mbox{\rm Re}f$](img164.gif)
is nowhere differentiable. In fact, if
![$a\in\mbox{{\bf C}}$](img86.gif)
,
![$\displaystyle {
{{\mbox{\rm Re}(z)-\mbox{\rm Re}(z)}\over {z-a}}}$](img1209.gif)
has no limit at
![$a$](img590.gif)
. To see this, let
![$\displaystyle {a_n=a+{1\over n},\; b_n=a+{i\over n}}$](img1210.gif)
for all
![$n\in\mbox{{\bf Z}}_{\geq 1}$](img117.gif)
. Then
![$\{a_n\}\to a$](img774.gif)
,
![$\{b_n\}\to a$](img1211.gif)
, and
and
Hence, the sequences
![$\displaystyle { \left\{ {{\mbox{\rm Re}(a_n)-\mbox{\rm Re}(a)}\over {a_n-a}}\right\}_{n\geq
1}}$](img1214.gif)
and
![$\displaystyle { \left\{ {{\mbox{\rm Re}( b_n)-\mbox{\rm Re}(a)}\over {b_n-a}}\right\}_{n\geq 1}}$](img1215.gif)
have
different limits, so
![$\displaystyle {\lim_{z\to a}{{\mbox{\rm Re}(z)-\mbox{\rm Re}(a)}
\over {z-a}}}$](img1216.gif)
does not
exist.
However, we do have the following theorem.
10.39
Theorem.
Let
be an interval in
and let
be a function
differentiable at a point
. Write
where
are real
valued. Then
and
are differentiable at
, and
.
Proof: Since
is differentiable at
there is a function
on
such that
is continuous at
and
If
and
, then
and
, so
![\begin{displaymath}
(\mbox{\rm Re}(f))(t)=(\mbox{\rm Re}(f))(a)+(t-a)(\mbox{\rm Re}(D_af))(t)
\mbox{ for all }t\in J
\end{displaymath}](img1226.gif) |
(10.40) |
and
![\begin{displaymath}
(\mbox{\rm Im}(f))(t)=(\mbox{\rm Im}(f))(a)+(t-a)(\mbox{\rm Im}(D_af))(t)
\mbox{ for all }t\in J.
\end{displaymath}](img1227.gif) |
(10.41) |
Since
is continuous at
,
and
are
continuous at
, so equations (10.40) and (10.41) show that
and
are differentiable and
10.42
Example.
Let
![$a\in\mbox{{\bf R}}$](img700.gif)
, and let
![$f(t)=(2t+ia)^3$](img1231.gif)
for all
![$t\in\mbox{{\bf R}}$](img1232.gif)
.
Then
![$f$](img13.gif)
is differentiable and (by
the chain rule),
We have by direct calculation,
so
(This example just illustrates that the theorem is true in a special case.)
10.43
Theorem.
Let
be a complex function and let
, and suppose
contains the
line segment
, and that
for all
. Then
is constant on
; i.e.,
for all
.
Proof: Define a function
by
By the chain rule,
is differentiable on
and
. Since
for all
, we have
for all
. Hence
and hence
If
and
, then
for all
.
Next: 10.3 Trigonometric Functions
Up: 10. The Derivative
Previous: 10.1 Derivatives of Complex
  Index