Is it just me, or does it happen to everyone? When I try to correct someone else’s arithmetic, I can count on making an error of my own. For instance: I was reviewing William Goldbloom Bloch’s clever and provocative book The Unimaginable Mathematics of Borges’ Library of Babel. I noted in passing a numerical error, trivial in significance even though it’s enormous in magnitude. Bloch had occasion to introduce the number:
To give a sense of this number’s size, he mentioned that it has 33 million digits. I pointed out that the true digit count is exponentially larger: 1033,013,740. Now I’ve received a correction to my correction, in a note from Michael Rosenthal:
The author [of the book] stated 33 million digits as the length, but Hayes getting closer states that the length is (10 to the 33,013,730). Isn’t the actual number of digits ((10 to the 33,013,730) + 1)?
Of course Rosenthal is right. There are 10n n-digit decimal numbers, but 10n is not one of them. On the other hand, Rosenthal also gives a double-barrelled demonstration of my point about the perils of correcting other people’s arithmetic: In the course of setting me straight, he makes a typographical slip of his own, twice mistranscribing 33,013,740.
My correspondence with Rosenthal provoked me to take a closer look at where that enormous number came from in the first place. This has led me into some diverting adventures with factorials and logarithms. But it has also put me in the position of proposing another correction to a published number, which means I am in grave danger of making still more mistakes.
Bloch’s book is a mathematical companion to “The Library of Babel,” a brief ficción by the Argentinian fabulist Jorge Luis Borges. In the Library of Babel all books are written in an alphabet of just 25 symbols. Each book has 410 pages, and each page has 40 lines of 80 characters, for a total of 1,312,000 characters per volume. We are told that the library possesses one copy of every possible volume composed in this way. Thus the number of volumes on the library’s shelves is 251,312,000. For convenience of reference let us call this number B (for Book, or Babel, or Borges, or Bloch).
The much larger number mentioned in my review enters Bloch’s story when he asks how many ways we might arrange the books on the library shelves. The answer, of course, is the factorial of B. If we imagine that the shelves have discrete slots that hold one book each, then we have B choices for the first slot, B–1 choices for the second slot, B–2 for the third, and so on, until we come to the final slot where there’s just one book left to choose. The product of all these numbers is B!.
So what sort of a number is B!? Can we have a look at it? In this age of computational wonders, surely we can just fire up Sage or Mathematica, or write a four-liner in Lisp, and have the digits of that number served to us on a silver platter, no?
No. We can’t.
After three decades of playing with computers, I am still childishly agog at the ease with which I can tap a few keys and calculate a gargantuan quantity such as 52!, the number of permutations of a standard deck of playing cards:
8065817517094387857166063685640376
6975289505440883277824000000000000
I can even calculate 10,000!, unleashing a Niagara of digits that spill out onto my screen in a long gray torrent—some 35,660 of them by the time the flood abates. (Digression: How many of those digits are trailing zeros? This is a question Fred Gruenberger would always ask when he talked about factorials in his old Popular Computing newsletter, back in the days when computing that many digits took a bit of doing.)
Computing has come a long way, but even with so much numerical horsepower at our fingertips, the direct calculation of B! is still totally unthinkable. If Bloch had been right about the 33 million digits, the computation might be feasible, if tedious, but computing a number with 1033,013,740 digits is just ridiculous, with or without the extra digit I forgot.
If we can’t compute B! outright, maybe we can at least get its logarithm. For a rough estimate of the magnitude—in other words, for counting the digits—the realm of logarithms is the right place to be. Since the factorial of n is a product,
the logarithm of the factorial has to be a sum:
I don’t know a quick and easy way to perform this summation, but I do know how to evaluate a slightly different sum, one where every term on the right hand side of the equation is log n. Since there are n of these terms, the sum is obviously just n log n. What we have calculated in this way is the logarithm of nn. Thus nn can be taken as a crude approximation—a very sloppy upper bound—to the value of n!. For our particular case we have log BB = B log B. When I plug in the numerical value of B, and use base-10 logarithms, I come up with
Exponentiating both sides gives this answer:
Hmm. There’s a problem here. This is supposed to be a loose upper bound, but it’s smaller than the value we were expecting for B!. Clearly, BB can’t be less than B!.
Bloch reports that he came to his estimate of B! via Stirling’s approximation. If you look up this bit of mathematical technology, you’ll find a formula something like this:
It would be easy to translate the formula directly into computer code, but it’s pointless to do so if the aim is to calculate B!. Again the only practical way to proceed is to take logarithms of both sides. Then the product in the formula becomes a sum of three terms. The first term, log nn, we already know: It’s n log n. The second term, log e–n, is simply –n if we choose to work with natural logarithms; if we want to stick with base-10 logs, it’s –n/log 10. And the third term we don’t need to worry about; at the level of precision we can hope to achieve in these calculations, ½ log 2πn is too small to notice. So, keeping just the first two terms, we have a rough-and-ready approximation to Stirling’s approximation, which ought to work well enough for very large n. B certainly seems to qualify as a large n; when I plug in its value, here’s what I get:
I have plodded my way slowly and methodically through the steps of this computation in an effort to persuade myself that I might finally have it right. At this point I’m fairly sure that log B! is closer to 101,834,103 than it is to 1033,013,740, but I’m chary of making a stronger claim. With all my crossing back and forth between the safe world of logarithms and the uninhabitable wasteland of exponentials, I worry that I’ve misplaced a factor of e or, worse, mistaken the logarithm of a number for the number itself. If I’ve goofed again somewhere in this calculation, I trust that readers will let me know.
I want to emphasize that my aim in telling this story is certainly not to embarrass Bloch, whose work I admire. His book does a splendid job of explaining ideas in combinatorics, number theory and even topology, in a literary context where such themes don’t get much exposure. His little numerical error—if it is an error—has no impact on the meaning of the passage where it appears.
Indeed, it’s hard to imagine a circumstance where the difference between the two proposed values of B! could have any earthly consequence. Both numbers are always going to be ghostly presences in our universe—actors whose voice we hear from offstage but who never show their face under the lights. A certain breed of radical constructivist would argue that the numbers we’re talking about don’t exist at all. I don’t find that doctrine helpful or persuasive, but I do think there’s a distinction worth making between numbers we can actually hold in our hands and these strange, outsized objects that come equipped with their own exclamation points. Computing with numbers this large is like working with toxic or radioactive materials. You never get to touch anything directly; all manipulations are performed by remote control from behind a blast shield. The experience can be exciting, but afterwards you don’t come away with a sense that you really know what you’re doing.
Fred Gruenberger would have wanted to know how many trailing zeros there are in B!. Can we answer?