EC3070 FINANCIAL DERIVATIVES
The Mean Value Theorem
Rolle’s Theorem.
If
f
(
x
) is continuous in the closed interval [
a, b
] and
differentiable in the open interval (
a, b
), and if
f
(
a
) =
f
(
b
) = 0, then there
exists a number
c
∈
(
a, b
) such that
f
(
c
) = 0.
When it is represented geometrically, this theorem should strike one as obvious;
and to prove it formally may seem a waste of time. Nevertheless, in proving
it, we can show how visual logic may be converted into verbal or algebraic
logic. Some of the more punctilious issues of Mathematical Analysis cannot be
represented adequately in diagrams, and it is for this reason that we prefer to
rely upon algebraic methods when seeking firm proofs of analytic propositions.
To prove Rolle’s theorem algebraically, we should invoke the result that
the function
f
(
x
) must achieve an upper bound or a lower bound in the interval
(
a, b
). Or course, the function might have both an upper and a lower bound in
(
a, b
). However, imagine that the function rises above the line in the interval,
and that it cuts the line only at the points
a, b
which are not included in the
open interval. Then it has an upper bound but not a lower bound. If the
function is horizontal over the interval, then every point in (
a, b
) is both an
upper bound and a lower bound.
Since, in that, case
f
(
x
) = 0 for every
x
∈
(
a, b
) there is nothing to prove.
We shall prove the theorem for the case where there is an upper bound in
(
a, b
) which corresponds to the point
c
. Then, by assumption,
f
(
c
)
≥
f
(
c
+
h
)
for small values of
h >
0 which do not carry us outside the interval. It follows
that
f
(
c
+) = lim
h
→
0+
f
(
c
+
h
)
−
f
(
c
)
h
≤
0
.
Here the symbolism associated with the limit indicates that
h
tends to 0 from
above.
Now let
h
be a small
negative
number such that
c
+
h < c
remains in the
interval (
a, b
). Then
f
(
c
−
) = lim
h
→
0
−
f
(
c
+
h
)
−
f
(
c
)
h
≥
0
,
where the symbolism associated with the limit indicates that
h
tends to 0
from below. Now the assumption that
f
(
x
) is continuous in (
a, b
) implies that
f
(
c
+) =
f
(
c
−
); and the only way in which this can be reconciled with the
inequalities above is if
f
(
c
) = 0. This proves the theorem in part. The rest of
the proof, which concerns the case where there is a lower bound, follows along
the same lines.
D.S.G. Pollock:
stephen
[email protected]
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THE MEAN VALUE THEOREM
The following theorem, which is of prime importance in Mathematical
Analysis, represents a generalisation of Rolle’s theorem and it has a similar
visual or geometric interpretation:
The Mean Value Theorem.
If
f
(
x
) is continuous in the interval [
a, b
] and
differentiable in the open interval (
a, b
), then there exists a point
c
∈
(
a, b
)
such that
f
(
b
)
−
f
(
a
)
b
−
a
=
f
(
c
)
.
Proof.
The line passing through the coordinates
{
a, f
(
a
)
}
and
{
b, f
(
b
)
}
of
the function
f
(
x
) has the equation
(
x
) =
f
(
a
) +
f
(
b
)
−
f
(
a
)
b
−
a
(
x
−
a
)
.
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 Spring '12
 D.S.G.Pollock
 Calculus, Intermediate Value Theorem, Rolle's theorem

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