2008 March « The Unapologetic Mathematician

And it’s a bunt to right field!

From Strange Maps I come across renderings of the Apollo 11 landing site laid over a baseball diamond and a soccer pitch.

March 24, 2008 Posted by John Armstrong | Uncategorized | | No Comments

Sufficient Conditions for Integrability

Let’s consider some conditions under which we’ll know that a given Riemann-Stieltjes integral will exist. First off, we have a straightforward adaptation of our old result that continuous functions are Riemann integrable. Now I assert that any continuous function $f$ on an interval $\left[a,b\right]$ is Riemann-Stieltjes integrable over that interval with respect to any function $\alpha$ of bounded variation on the same interval. In particular, the function $\alpha(x)=x$ is clearly of bounded variation, and so we will recover our old result.

In fact, we can even adapt the old proof. The Heine-Cantor theorem says that the function $f$ , being continuous on the compact interval $\left[a,b\right]$ is uniformly compact. As usual, we can assume that $\alpha$ is increasing on $\left[a,b\right]$ . And now Riemann’s condition tells us to consider the difference

$\displaystyle U_{\alpha,x}(f)-L_{\alpha,x}(f)=\sum\limits_{i=1}^n\left(M_i(f)-m_i(f)\right)\left(\alpha(x_i)-\alpha(x_{i-1})\right)$

We want this difference to go to zero as we choose finer and finer partitions $x$

By uniform continuity we can pick a small enough $\delta$ (depending only on $\epsilon$ ) so that when $|x-y|<\delta$ we’ll have $|f(x)-f(y)|<\frac{\epsilon}{V}$ , where $V$ is the total variation $\alpha(b)-\alpha(a)$ . Then picking a partition whose subintervals are thinner than $\delta$ makes it so that $M_i(f)-m_i(f)<\frac{\epsilon}{V}$ , which we can then pull out of the sum. What remains sums to exactly the total variation $V$ , and so the difference $U_{\alpha,x}(f)-L_{\alpha,x}(f)$ is below $\epsilon$ , and our theorem holds.

Immediately from this result and integration by parts we come up with another set of sufficient conditions. If $\alpha$ is continuous and $f$ is of bounded variation on an interval $\left[a,b\right]$ then $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ over $\left[a,b\right]$ . Since the integrator $\alpha(x)=x$ is also continuous, this tells us that any function of bounded variation is Riemann integrable!

Of course, these conditions are just sufficient. That is, if they hold then we know that the integral exists. However, if an integral exists, we can’t use these to conclude anything about either the integrand or the integrator. For that we need necessary conditions.

March 24, 2008 Posted by John Armstrong | Analysis, Calculus | | 2 Comments

If only

The webcomic Saturday Morning Breakfast Cereal presents a proof of the infinitude of primes. A note: while that one comic is innocuous, there’s a lot of them that aren’t really that safe for work.

March 24, 2008 Posted by John Armstrong | Uncategorized | | No Comments

Sunday Samples 61

Today is Easter Sunday. I know a lot of you in the math/science regions of the internet are actively anti-religion, but I just can’t muster up that sort of disdain. Sure, Bible literalism is stupid (like I’ve said before), but there are perfectly orthodox (and Orthodox) modes of belief that don’t entail any form of Creationism or literalistic interpretation of scripture. I can’t condemn that sort of thing, particularly because it would include some of my closest friends and family. And I can’t tell whether anti-religionists don’t realize that there are such systems or if they want me to denounce and reject those people I know who do adhere to them.

And when it comes down to it, I have to recognize that there are some intensely powerful archetypes in there that seem designed to resonate with human minds. From a purely psychological standpoint, there’s a lot to be said for theology. Maybe there is and maybe there isn’t a deity, but if a given mind behaves as if there is, then it’s a useful construct to speak in terms of. Remember, there’s no directory tree in your computer either, but we all talk as if there is because that’s how your operating system behaves. Operationally, theism works for a lot of systems.

The archetype in play today is sacrifice and redemption, and the miracle of forgiveness. We’re only human, and we make mistakes. We hurt (and are hurt by) each other, and we have a need to right the wrongs that stand between us. Our selfish impulse is to carry grudges, or to refuse to admit our errors. But as social creatures we recognize the damage that course does, and we ask, “what can I do to make up for this wrong I’ve done?” And we respond with forgiveness, repairing the social fabric so the community can hold together.

If it makes you uncomfortable, forget the spiritual overtones of the day and just focus on the secular, human realities. The rest is just a story designed to focus our attention here anyway. As a species, H. sapiens has evolved to form these social groupings, and has developed empathy to reinforce them. We feel ourselves the emotional pain we cause to those close to us, and in our anguish we reach out for redemption.

The important thing here is the recognition that we, as individuals, are unable to survive on our own. The closest I’ve seen to a superficially-secular treatment of this is the AA jargon about “hitting bottom”. Sometimes life, the universe, and everything are just too overwhelming for one person, and he has to fall back on the support of the community, or (if you’re not averse to the archtype) on the divine, as personified in the religious community. How marvelous that we’ve developed this network to cope with individual failures, and to stand stronger together than we ever could apart. The capacity for forgiveness, for redemption, and for a resurrection of our social bonds after they’ve been cut down is nothing short of a miracle.

In Hebrew, this urge to celebrate and praise the divine led to the word הַלְלוּיָהּ, which is transliterated (but not translated) as the English “hallelujah”, or “alleluia”. Roughly, it means “praise the Lord”, but it has come to be more of a word of celebration itself than an injunction to celebration. During the season of Lent, Catholics (and possibly other Christians, though I’ve no direct knowledge) abstain from such overt praise. The word isn’t spoken, and the parts of the mass involving it aren’t sung. And then on Easter Sunday, the Church cries out in praise, “Alleluia, alleluia, Christ is risen.”

In 1984, Leonard Cohen wrote a long, rambling song about this sense of powerlessness and the need for redemption by something greater than oneself. It’s been covered and rewritten many times, but easily the best was the one by Jeff Buckley on his 1994 album Grace. It’s a spare, unadorned setting that strips right down to the core of the song, and carries through this sense of loss and longing. And so I present Jeff Buckley’s version of “Hallelujah”.
Read more »

March 23, 2008 Posted by John Armstrong | Sunday Samples | | 1 Comment

Classic math books

Yesterday, someone left a comment, which I’m reinterpreting a bit as a call for help in what to do for self-directed study of mathematics. I could answer this more accurately at the undergraduate-to-graduate transition, but I’m a little weak at the period entering college in the first place. Partly this is because I had a rather eclectic approach, reading all sorts of things throughout my middle and high school career, and most of the math I took formally as an undergraduate was at the graduate level, or was in specialized subjects.

So, what basic, general books get the community’s vote of confidence? What book could a high-schooler get ahold of that would give her a clear introduction to abstract algebra? advanced calculus? topology? combinatorics? number theory? differential geometry? linear algebra? I’m assuming here that having run out of math in high school means having completed the AP calculus sequence.

To put it another way: what books should a professor have at hand on his bookshelf for just this purpose? A bright youngster comes along looking for guidance. You think that independently reading through a clearly-written text at the next level up would be a great idea, and you reach for…

To clarify: I’m not looking for web resources, fora, or communities. Please don’t randomly post links unless they’re to a free copy of a specific book you’re recommending. Along the same lines, if you post a dozen comments one after another, and half of them are links to the same site, I’m going to assume you’re a spammer and not a serious commenter.

March 22, 2008 Posted by John Armstrong | Uncategorized | | 13 Comments

Integrability over subintervals

As I noted when I first motivated bounded variation, we’re often trying to hold down Riemann-Stieltjes sums to help them converge. In a sense, we’re sampling both the integrand $f$ and the variation of the integrator $\alpha$ , and together they’re not big enough to make the Riemann-Stieltjes sums blow up as we take more and more samples. And it seems reasonable that if these sums don’t blow up over the whole interval, then they’ll not blow up over a subinterval.

More specifically, I assert that if $\alpha$ is a function of bounded variation, $f$ is integrable with respect to $\alpha$ on $\left[a,b\right]$ , and $c$ is a point between $a$ and $b$ , then $f$ is integrable with respect to $\alpha$ on $\left[a,c\right]$ .

Then, in the equation expressing “linearity” in the interval

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\int\limits_{\left[a,c\right]}fd\alpha+\int\limits_{\left[c,b\right]}fd\alpha$

we have two of these integrals known to exist. Therefore the third one does as well, and the equation is true$. If we have a subinterval $\left[c,d\right]\subseteq\left[a,b\right]$ , then this theorem states that $f$ is integrable over $\left[c,b\right]$ , and another invocation of the theorem shows that $f$ is integrable over $\left[c,d\right]$ . So being integrable over an interval implies that the function is integrable over any subinterval.

As we said earlier, we can handle all integrators of bounded variation by just considering increasing integrators. And then we can use Riemann’s condition. So, given an $\epsilon>0$ there is a partition $x_\epsilon$ of $\left[a,b\right]$ so that $U_{\alpha,x}(f)-L_{\alpha,x}(f)<\epsilon$ for any partition $x$ finer than $x_\epsilon$ .

We may as well assume that $c$ is a partition point of $x_\epsilon$ , since we can just throw it in if it isn’t already. Then the partition points up to $c$ form a partition $x_\epsilon'$ of $\left[a,c\right]$ . Further, any refinement $x'$ of $x_\epsilon'$ is similarly part of a refinement $x$ of $x_\epsilon$ . So by assumption we know that $U_{\alpha,x}(f)-L_{\alpha,x}(f)<\epsilon$ , and we get down to $x'$ by throwing away terms in the sum for partition points above $c$ . Each of these terms is nonnegative, and so we see that

$U_{\alpha,x'}(f)-L_{\alpha,x'}(f)<U_{\alpha,x}(f)-L_{\alpha,x}(f)<\epsilon$

That is, $f$ satisfies Riemann’s condition with respect to $\alpha$ on $\left[a,c\right]$ , and so it’s integrable.

March 21, 2008 Posted by John Armstrong | Analysis, Calculus | | 1 Comment

Products of integrable functions

From the linearity of the Riemann-Stieltjes integral in the integrand, we know that the collection of functions that are integrable with respect to a given integrator over a given interval form a real vector space. That is, we can add and subtract them and multiply by real number scalars. It turns out that if the integrator is of bounded variation, then they actually form a real algebra — we can multiply them too.

First of all, let’s show that we can square a function. Specifically, if $\alpha$ is a function of bounded variation on $\left[a,b\right]$ , and of $f$ is bounded and integrable with respect to $\alpha$ on this interval, then so is $f^2$ . We know that we can specialize right away to an increasing integrator $\alpha$ . This will work here (unlike for the order properties) because nothing in sight gets broken by subtraction.

Okay, first off we notice that $f(x)^2$ is the same thing as $|f(x)|^2$ , and so they have the same supremum in any subinterval of a partition Then the supremum of $|f(x)|^2$ is the square of the supremum of $|f(x)|$ because squaring is an increasing operation that preserves suprema (and, incidentally, infima). The upshot is that $M_i(f^2)=M_i(|f|)^2$ . Similarly we can show that $m_i(f^2)=m_i(|f|)^2$ . This lets us write

$\displaystyle M_i(f^2)-m_i(f^2)=M_i(|f|)^2-m_i(|f|)^2=\left(M_i(|f|)+m_i(|f|)\right)\left(M_i(|f|)-m_i(|f|)\right)$
$\leq2M\left(M_i(|f|)-m_i(|f|)\right)$

where $M$ is an upper bound for $|f|$ on $\left[a,b\right]$ . Riemann’s condition then tells us that $f^2$ is integrable.

Now let’s take two bounded integrable functions $f$ and $g$ . We’ll write

$f(x)g(x)=\frac{1}{2}\left(\left(f(x)+g(x)\right)^2-f(x)^2-g(x)^2\right)$

and then invoke the previous result and the linearity of integration to show that the product $fg$ is integrable.

March 20, 2008 Posted by John Armstrong | Analysis, Calculus | | 8 Comments

Increasing Integrators and Order

For what we’re about to do, I’m going to need a couple results about increasing integrators, and how Riemann-Stieltjes integrals with respect to them play nicely with order properties of the real numbers.

When we consider an increasing integrator we have a certain positivity result: if the integrand is nonnegative and the integral exists, then it is nonnegative as well. That is, for $\alpha$ increasing and $f(x)\geq0$ on $\left[a,b\right]$ we have $\int_{\left[a,b\right]}f\,d\alpha\geq0$ as long as it exists. This should be clear, since every Riemann-Stieltjes sum takes the form

$\displaystyle f_{\alpha,x}=\sum\limits_{i=1}^nf(t_i)\left(\alpha(x_i)-\alpha(x_{i-1})\right)\geq0$

where the inequality follows because each value $f(t_i)$ and each difference $\alpha(x_i)-\alpha(x_{i-1})$ is nonnegative. Thus the limit of the sums must be nonnegative as well. From this and the linearity of the integral we see that if $\alpha$ is increasing and $f(x)\geq g(x)$ on $\left[a,b\right]$ , then we have the inequality

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha\geq\int\limits_{\left[a,b\right]}gd\alpha$

as long as both integrals exist.

Now, when we talked about absolute values — the metric for the real numbers — we saw that the absolute value of a sum was always less than the sum of the absolute values. That is, $|x+y|\leq|x|+|y|$ . And since an integral is just a limit of sums, it stands to reason that a similar result would hold here. Specifically, if $\alpha$ is increasing and $f$ is integrable with respect to $\alpha$ on $\left[a,b\right]$ , then so is the function $|f|$ , and further we have the inequality

$\displaystyle\left|\int\limits_{\left[a,b\right]}fd\alpha\right|\leq\int\limits_{\left[a,b\right]}|f|d\alpha$

Indeed, given a partition of $\left[a,b\right]$ the difference $M_i(f)-m_i(f)$ between the supremum and infimum of $f$ on the $i$ th subinterval is the supremum of $f(x)-f(y)$ , where $x$ and $y$ range across $\left[x_{i-1},x_i\right]$ . Then, adapting the above inequality we see that

$||f(x)|-|f(y)||\leq|f(x)-f(y)|$

and so we conclude that

$M_i(|f|)-m_i(|f|)\leq M_i(f)-m_i(f)$

Then we can multiply by $\alpha(x_i)-\alpha(x_{i-1})$ and sum over a partition to find

$U_{\alpha,x}(|f|)-L_{\alpha,x}(|f|)\leq U_{\alpha,x}(f)-L_{\alpha,x}(f)$

Riemann’s condition then tells us that $|f|$ is integrable, and the inequality follows by the previous result.

We might hope to extend these results to integrators of bounded variation, but it won’t work right. This is because we go from increasing functions to functions of bounded variation by subtracting, and this operation will break the order properties.

March 18, 2008 Posted by John Armstrong | Analysis, Calculus | | No Comments

Plans for tomorrow

I’ve been going over notes in preparation for tomorrow’s talk at the University of Pennsylvania (scroll down a bit).

For anyone who happens to be there (Isabel, Charles…) I’ll be heading out from a little south of Baltimore early enough to (hopefully) compensate for the fact that I-95 is closed a little north of the exit for UPenn. I should be there in plenty of time for lunch with Jim Stasheff, and dinner later on. Drop an email (if you remember that I teach at Tulane it’s not too hard to find the address) with any contact information you want to pass along.

March 18, 2008 Posted by John Armstrong | Uncategorized | | 1 Comment

Category Theory Isn’t Useless After All!

Today on the arXiv, we find a posting of an old paper, which uses spans of “reflexive graphs” to give an algebraic framework for describing partita doppia — double-entry bookkeeping.

Now I need to find a follow-on to this paper and start applying to those financial math jobs.

March 18, 2008 Posted by John Armstrong | Algebra, Category theory | | No Comments

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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