414/514 – Series

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 f is monotonically increasing. Then f(x-) and f(x+) exist for all x\in{\mathbb R}. Moreover,

    f(x-)\le f(x)\le f(x+),

    and

    f(x+)\le f(y-)

    for all x<y.

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,

\displaystyle \int_{I_1}\int_{I_2}f(x,y) dx dy=\int_{I_2}\int_{I_1}f(x,y) dydx,

i.e., if f 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

a_{n,m},\quad n=0,1,\dots; m=0,1,\dots

is “sufficiently nice”, then

\displaystyle \sum_{n=0}^\infty\sum_{m=0}^\infty a_{n,m}=\sum_{m=0}^\infty\sum_{n=0}^\infty a_{n,m}.

One of the customary notions of “sufficiently nice” is that all the a_{n,m} be non-negative. In this case, this is (a particular case of) Tonelli’s theorem. First, let’s see that some assumptions are needed:

  • Let

    a_{i,j}=0 if i<j, a_{i,j}=-1 if i=j, and a_{i,j}=2^{j-i} if j<i.

    (It may help if you display the numbers so that a_{i,j} is the entry on the i-th row and j-th column of an infinite square array.) Prove that \displaystyle \sum_i\sum_j a_{i,j}=-2. Find (with proof) \displaystyle \sum_j\sum_i a_{i,j}, and conclude that \displaystyle \sum_i\sum_j a_{i,j}\ne\sum_j\sum_i a_{i,j}.

  • (In this problem {\mathbb N}=\{0,1,2,\dots\}.) Prove Tonelli’s theorem for series: If all the a_{i,j} are non-negative, then

    \displaystyle \sum_{i=0}^\infty \sum_{j=0}^\infty a_{i,j}=\sum_{(i,j)\in{\mathbb N}^2}a_{i,j}=\sum_{j=0}^\infty\sum_{i=0}^\infty a_{i,j}.

The sum in the middle is defined as in the previous homework. Note that it is possible that one of these series diverges to +\infty. 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 \displaystyle \sum_{(n,m)\in{\mathbb N}^2}a_{n,m}\le \sum_{n\ge0}\sum_{m\ge0} a_{n,m}, recall that the left hand side is defined as a supremum of sums over finite sets, and any finite F\subseteq{\mathbb N}^2 is contained in \{0,1,2,\dots,k\}\times\{0,1,2,\dots,k\} for a sufficiently large k.

To show that \displaystyle \sum_{n\ge0}\sum_{m\ge0}a_{n,m}\le\sum_{(n,m)\in{\mathbb N}^2}a_{n,m}, explain first why it suffices to show the inequality where on the left hand side n is restricted to range over 0,1,2,\dots,N for some integer N. And to show this inequality, explain why it suffices to show it when m is restricted to range over 0,1,2,\dots,M for some integer M.

  • Suppose now that \displaystyle \sum_{n\ge0}\sum_{m\ge0}|a_{n,m}|<+\infty, but the a_{n,m} do not need to be non-negative. Show that \displaystyle \sum_{n\ge0}\sum_{m\ge0}a_{n,m} converges, and

    \displaystyle \sum_{n\ge0}\sum_{m\ge0}a_{n,m}=\sum_{m\ge0}\sum_{n\ge0}a_{n,m}.

  • We proved that any power series f(x)=\sum a_n x^n converges inside some open interval I centered at 0 and diverges outside the closure of the interval, \bar I. Of course, if we instead consider a series \sum b_n (x-\alpha)^n, the same result holds but now the interval J of convergence is centered at \alpha rather than 0. Show that if \alpha\in I, then the function f(x) indeed admits an expansion of the form \sum b_n (x-\alpha)^n, and that the corresponding interval (call it J) has radius at least the closest distance from \alpha to an end-point of I. Also, find the series \sum b_n (x-\alpha)^n when \alpha\in(-1,1) and f(x)=1/(1-x). Explain how one can in fact get a series expansion of this form for any \alpha<1.

Here is one possible approach: Write

\displaystyle \sum_{n\ge0}a_n x^n=\sum_{n\ge0}a_n ((x-\alpha)+\alpha)^n \displaystyle=\sum_{n\ge0}\sum_{m=0}^n a_n \binom{n}{m}(x-\alpha)^m\alpha^{n-m}.

If we could exchange the two summation symbols, explain why this series would become

\displaystyle \sum_{m=0}^\infty\left(\sum_{n=m}^\infty\binom{n}{m}a_n\alpha^{n-m}\right)(x-\alpha)^m.

This is a series of the required form, provided that the coefficients \displaystyle b_m=\sum_{n=m}^\infty\binom{n}{m}a_n\alpha^{n-m} 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 |x-\alpha|<\tau, where \tau is the closest distance of \alpha to an end-point of I. (Of course, it is possible that the series actually converges in a larger interval.)

  • A function f(x) defined on some non-empty open subset U of {\mathbb R} is said to be real analytic iff for any \alpha\in U there is an open interval J centered at \alpha and contained in U such that f(x) can be written as a power series \displaystyle \sum_{n\ge0}b_n(x-\alpha)^n for all x\in J. Show that h(x)=1/(1-x) is real analytic on {\mathbb R}\setminus\{1\}. Suppose that f and g are real analytic on some interval I=(-R,R), say \displaystyle f(x)=\sum_{n\ge0}a_n x^n and \displaystyle g(x)=\sum_{n\ge0}b_n x^n. Let E be the set of points x\in I where f(x)=g(x). Suppose that E contains at least one limit point. Show that a_n=b_n for all n and therefore f(x)=g(x) for all x\in I.

Here is one way of showing this: Let L be the set of points in I that are limit points of elements of E. By assumption, L\ne\emptyset. Show that L is closed as a subset of I, that is, any limit point of L that belongs to I also belongs to L. If we show that L is open, we are done, because we showed in a previous homework set that the only subsets of {\mathbb R} that are simultaneously open and closed are \emptyset and {\mathbb R}, and the same argument shows that L must either be \emptyset or I, but L\ne\emptyset. (If you are not sure about this step, please provide a proof for it as well.)

Explain why we may as well assume that g(x)=0, and our goal is to show that a_n=0 for all n. First, show that if f(x)=0 for all x\in I, then a_n=0 for all n. For this, suppose otherwise, so a_n\ne0 for at least one n. Pick the smallest such index, say n_0. Show that we can write f(x)=x^{n_0}j(x) for some function j(x) such that j(0)\ne0, and therefore f(x)\ne0 for x\ne0 but sufficiently small.

So we are left with the task of showing that L is open. For this, pick a point \alpha\in L. Consider the series expansion of f about \alpha, 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 \alpha is not a limit point of E. Since power series converge on intervals, conclude that L is indeed open.

  • In lecture we gave an example that depended on the power series expansion of (1+x)^\alpha for \alpha real and |x|<1. Prove the validity of this expansion, namely, that for |x|<1 we have

    \displaystyle (1+x)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}x^k,

    where \displaystyle \binom{\alpha}{k}=\frac{\alpha(\alpha-1)\dots(\alpha-k+1)}{k!} for {}0<k and \displaystyle \binom{\alpha}{0}=1 for k=0.

Here is one possible approach: First show that the series converges for |x|<1. Call f(x) the function defined by the series. We want to show that f(x)=(1+x)^\alpha.  Show first that f(0)=1, f(x)>0 for all x, and (1+x)f'(x)=\alpha f(x). This last equation means that f'(x)/f(x)=\alpha/(1+x). Show that this means that (\ln(f(x))'=(\ln((1+x)^\alpha)', and conclude from here.

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