1. Introduction
Let
. Iff is continuous on
and differentiable on
, then the mean value theorem for derivatives states that [1] [2], there is
such that
If f is continuous on
, the mean value theorem for definite integrals states that there is
such that
Generally speaking, the c in the mean value theorem for the derivative need not equal the
in the mean value theorem for the definite integral. It is possible that
, for example,
. In this case,
. In this note, we determine all functions for which the two mean value theorems give the same c.
Our results rely heavily on the following identity theorem for real analytic functions, its proof can be found in [3].
Theorem If f and g are real analytic functions on an open interval U and there is an open set
such that
then
2. Main Result
Lemma 1 Suppose f is analytic on an open interval U, f and
have inverses and for any
with
, there is
such that
then
, where
,
and
are arbitrary real numbers.
Proof. By the hypothesis,
Fix
. If we denote
, then
and the above relation becomes
Taking the derivative of both sides, we obtain
To compute
, we use the derivative of an inverse function and the chain rule,
Replacing
by
gives,
We obtain
(1)
We need to find f satisfying the above relation. Let W be an open neighborhood of a such that
Then (1) becomes
Combining like terms and cross multiplying, we obtain
(2)
If
, then it is easy to see that
. In this case,
, which cannot happen because we assume that f is invertible. Thus,
. Equating the respective coefficients of
and
on both sides of (2), we obtain
In general, for
,
, for some
. To find
, we observe that
satisfies (1). Hence the power series of
should satisfy (2). But in the power series expansion of
at
, the coefficient
.
That is,
. In other words, for f to satisfy (1), the power series expansion of
at
must be
where
are real numbers. If
, then
, violating the fact that
has an inverse. Thus,
By the identity theorem,
on U.
Finally, we can relax the requirement that both f and
have inverse.
Theorem 1 If f is analytic on an open interval U, and
with
, then the two mean value theorems give the same c if and only if
or
, where
,
and
are any real numbers.
Proof. The necessary condition is obvious. We only prove the sufficient condition. If f is constant on U, then we are done. Suppose f is not constant on U, then there is some
such that
. Since
is continuous at
, there is an open interval I of
with
such that
on I. Hence,
exists on I.
Case 1.
is constant on I, then on I,
for some
with
. By the identity theorem,
on U.
Case 2
is not constant on I, i.e.
for some
. But
is continuous at
, so there is an open interval J containing
with
such that
on J. Then, bothf and
have inverses on J, and by the Lemma above,
on J, for some
, and
. But the identity theorem implies that,
on U.
3. Conclusion
In this work, we find that if
and f is an analytic function, there is
satisfies both
then f must be a linear function or an exponential function. The proof relies heavily on the fact that f is analytic, we don’t know if this condition can be weakened.
Acknowledgements
The author is indebted to Alexander Kasiukov for many helpful discussions, without which this project would not be possible