where is a nonnegative, increasing function defined for all .
Use this to deduce the inequalities
by taking .
Proof. Since is increasing we know (where denotes the greatest integer less than or equal to ) for all . Define step functions and by
Then and are constant on the open subintervals of the partition
Since is integrable we must have
for every pair of step functions . Hence,
Next, if we take (which is nonnegative and increasing on ) we have
From this we then get two inequalities
Consider a bounded, monotonic, real-valued function on the interval . The define sequences
- Prove that
- Prove that the two sequences and converge to .
- State and prove a generalization of the above to interval .
- Proof.First, we define two step functions,
where denotes the greatest integer less than or equal to . Then we define a partition of ,
For any we have
So, and are constant on the open subintervals of the partition .
Since is monotonically increasing and for all (by the definition of and ) we have
- Proof. From part (a) we have
since . Since does not depend on we have
- Claim: If is a real-valued function that is monotonic increasing and bounded on the interval , then
for and defined as follows:
be a partition of the interval . Then, define step functions and with and for . By these definitions we have for all (since is monotonic increasing). Since is bounded and monotonic increasing it is integrable, and
And, since , and we have
(Note: I cannot get the bounds Apostol asks for. I prove a different set below. I cannot figure out if it is a mistake in the book or not.)
Proof. Using the algebraic identity, valid for ,
Therefore, integrating term by term,
Furthermore, we have
Taking , we then have
From the inequality for this integral we then have
Prove that there exists a number with such that
Proof. For we have
(Since for we know .) Therefore, integrating the terms in the inequality from 0 to 1,
Prove that the error term in the Taylor expansion of satisfies the following inequality.
Proof. To prove this we will work directly from the definition of the error as an integral,
We know for we have, (we need for the expansion of to be valid),
Therefore we have
So, we can bound the error term by bounding the integral,