This is homework 5, due Friday November 18 at the beginning of lecture.
First of all, some of you did not realize that the first question in the previous homework set was indeed a question, and skipped it. If that was the case, please solve it now and turn it in. The sooner you do so, the sooner I’ll be able to grade it and return it with the rest of your homework 4. The question was as follows:
- Suppose that is monotonically increasing. Then and exist for all . Moreover,
for all .
To be explicit: You need to prove all the assertions in the paragraph above.
The new set follows.
As we are seeing in lecture, with series and in general with limits the order of operations matters. Perhaps you have heard of Fubini’s theorem, that we will cover next term. This result tells us that, under appropriate conditions,
i.e., if is “sufficiently nice”, we can exchange the order of integration in a double integral. The corresponding result for series is that if a doubly-indexed sequence of real numbers
is “sufficiently nice”, then
One of the customary notions of “sufficiently nice” is that all the be non-negative. In this case, this is (a particular case of) Tonelli’s theorem. First, let’s see that some assumptions are needed:
if , if , and if
(It may help if you display the numbers so that is the entry on the -th row and -th column of an infinite square array.) Prove that . Find (with proof) , and conclude that
- (In this problem .) Prove Tonelli’s theorem for series: If all the are non-negative, then
The sum in the middle is defined as in the previous homework. Note that it is possible that one of these series diverges to . In that case, you must prove that the other two series diverge as well. On the other hand, you must show that if all the series converge, then they all converge to the same number.
Here is one possible approach. I’ll concentrate on the equality of the first two series. To show that they are equal, it suffices to show that each series is less than or equal to the other one. To show that , recall that the left hand side is defined as a supremum of sums over finite sets, and any finite is contained in for a sufficiently large .
To show that , explain first why it suffices to show the inequality where on the left hand side is restricted to range over for some integer . And to show this inequality, explain why it suffices to show it when is restricted to range over for some integer .
- Suppose now that , but the do not need to be non-negative. Show that converges, and
- We proved that any power series converges inside some open interval centered at and diverges outside the closure of the interval, . Of course, if we instead consider a series , the same result holds but now the interval of convergence is centered at rather than . Show that if , then the function indeed admits an expansion of the form , and that the corresponding interval (call it ) has radius at least the closest distance from to an end-point of . Also, find the series when and . Explain how one can in fact get a series expansion of this form for any .
Here is one possible approach: Write
If we could exchange the two summation symbols, explain why this series would become
This is a series of the required form, provided that the coefficients are actually finite. Show that this is indeed the case, and that the exchange of order is valid in this case, resulting in a series that converges as long as , where is the closest distance of to an end-point of . (Of course, it is possible that the series actually converges in a larger interval.)
- A function defined on some non-empty open subset of is said to be real analytic iff for any there is an open interval centered at and contained in such that can be written as a power series for all . Show that is real analytic on . Suppose that and are real analytic on some interval , say and . Let be the set of points where . Suppose that contains at least one limit point. Show that for all and therefore for all .
Here is one way of showing this: Let be the set of points in that are limit points of elements of . By assumption, . Show that is closed as a subset of , that is, any limit point of that belongs to also belongs to . If we show that is open, we are done, because we showed in a previous homework set that the only subsets of that are simultaneously open and closed are and , and the same argument shows that must either be or , but . (If you are not sure about this step, please provide a proof for it as well.)
Explain why we may as well assume that , and our goal is to show that for all . First, show that if for all , then for all . For this, suppose otherwise, so for at least one . Pick the smallest such index, say . Show that we can write for some function such that , and therefore for but sufficiently small.
So we are left with the task of showing that is open. For this, pick a point . Consider the series expansion of about , which exists because of the previous problem. Show (by a similar argument to the one in the previous paragraph) that all the coefficients in this expansion are 0, or else is not a limit point of . Since power series converge on intervals, conclude that is indeed open.
- In lecture we gave an example that depended on the power series expansion of for real and . Prove the validity of this expansion, namely, that for we have
where for and for .
Here is one possible approach: First show that the series converges for . Call the function defined by the series. We want to show that . Show first that , for all , and . This last equation means that . Show that this means that , and conclude from here.