For a differentiable function
let
be the function given by
Iterations of this ``operator''
yield many interesting identities.
For
we have
On the other hand
may be applied to the function
to obtain
Comparing this with the above series expansion, we conclude that
Use Maple to find the sums for
Observe that after
iterations, the resulting function takes the form
for some polynomial
of degree
It is therefore desirous to find the recurrence relation satisfied by the
polynomials
From the identity
(which, again, can be calculated by Maple,) we see that
holds for
For
we have
On the other hand
may be applied to the function
to obtain
Comparing this with the above series expansion, we conclude that
Use Maple to obtain the above expressions.
Observe that after
iterations, the resulting function takes the form
for some polynomial
of degree
The recurrence relation satisfied by the polynomials
may be deduced from the identity
we see that
[Maple
Worksheet]
The series is absolutely convergent if the series converges. (In more formal language, the sequence is absolutely summable if the sequence is summable.)
Theorem. Every absolutely convergent series is convergent. Moreover, a series is absolutely convergent if and only if the series formed from its positive terms and the series formed from its negative terms both converge.
Proof. If
converges, then, by the Cauchy criterion,
Since
it follows that
which shows that
converges.
To prove the second part of the theorem, let
so that
is the series formed from the positive terms of
,
and
is the series formed from the negative terms.
If
and
both converge, then
also converges, so
converges absolutely.
On the other hand, if
converges, then
also converges. Therefore
and
both converge.
Suppose that and that
and that
Then the series
converges.
Proof. We prove that
(1)
(2)
(3) if is even and is odd.
This is so since:
(1)
(2)
(3) If
is even and
is odd, choose
such that
then
Now, the sequence
converges, because it is nondecreasing and is bounded above (by
for any odd
).
Let
Similarly, let
If follows from (3) that
since
it is actually the case that
.
This proves that
Example: The series
is convergent, but not absolutely convergent.
DEFINITION: Let
be a sequence of functions defined on
,
and let
be a function which is also defined on
Then
is called the uniform limit of
on
if for every
there is some
such that for all
in
We also say that converges uniformly to on , or that approaches uniformly on
THEOREM Suppose that
is a sequence of functions which are integrable on
and that
converges uniformly on
to a function
which is integrable on
.
Then
Proof. Let
There is some
such that for all
we have
Since this is true for any
,
it follows that
Theorem. Suppose that is a sequence of functions which are continuous on , and that converges uniformly on to . Then is also continuous on .
Proof. For each in we must prove that is continuous at We will deal only with in ; the cases and require the usual simple modifications.
Let
.
Since
converges uniformly to
on
,
there is some
such that
In particular, for all
such that
is in
,
we have
(1)
(2)
Now is continuous, so there is some such that for we have
(3)
Thus, if
,
then
This proves that is continuous at
Suppose that
is a sequence of functions which are differentiable on
,
with integrable derivatives
,
and that
converges (pointwise) to
Suppose, moreover, that
converges uniformly on
to some continuous function
Then
is differentiable and
Proof. Applying the previous theorem to the interval
,
we see that for each
we have
Since
is continuous, it follows that
for all
in the interval