Without Machines That Think About Logarithms

I’ve got a few more thoughts about calculating logarithms, based on how the Harvard IBM Automatic Sequence-Controlled Calculator did things, and wanted to share them. I also have some further thoughts coming up shortly courtesy my first guest blogger, which is exciting to me.

The procedure that was used back then to compute common logarithms — logarithms base ten — was built on several legs: that we can work out some logarithms ahead of time, that we can work out the natural (base e) logarithm of a number using an infinite series, that we can convert the natural logarithm to a common logarithm by a single multiplication, and that the logarithm of the product of two (or more) numbers equals the sum of the logarithm of the separate numbers.

From that we got a pretty nice, fairly slick algorithm for producing logarithms. Ahead of time you have to work out the logarithms for 1, 2, 3, 4, 5, 6, 7, 8, and 9; and then, to make things more efficient, you’ll want the logarithms for 1.1, 1.2, 1.3, 1.4, et cetera up to 1.9; for that matter, you’ll also want 1.01, 1.02, 1.03, 1.04, and so on to 1.09. You can get more accurate numbers quickly by working out the logarithms for three digits past the decimal — 1.001, 1.002, 1.003, 1.004, and so on — and for that matter to four digits (1.0001) and more. You’re buying either speed of calculation or precision of result with memory.

The process as described before worked out common logarithms, although there isn’t much reason that it has to be those. It’s a bit convenient, because if you want the logarithm of 47.2286 you’ll want to shift that to the logarithm of 4.72286 plus the logarithm of 10, and the common logarithm of 10 is a nice, easy 1. The same logic works in natural logarithms: the natural logarithm of 47.2286 is the natural logarithm of 4.72286 plus the natural logarithm of 10, but the natural logarithm of 10 is a not-quite-catchy 2.3026 (approximately). You pretty much have to decide whether you want to deal with factors of 10 being an unpleasant number or do deal with calculating natural logarithms and then multiplying them by the common logarithm of e, about 0.43429.

But the point is if you found yourself with no computational tools, but plenty of paper and time, you could reconstruct logarithms for any number you liked pretty well: decide whether you want natural or common logarithms. I’d probably try working out both, since there’s presumably the time, after all, and who knows what kind of problems I’ll want to work out afterwards. And I can get quite nice accuracy after working out maybe 36 logarithms using the formula:

$\log_e\left(1 + h\right) = h - \frac12 h^2 + \frac13 h^3 - \frac14 h^4 + \frac15 h^5 - \cdots$

This will work very well for numbers like 1.1, 1.2, 1.01, 1.02, and so on: for this formula to work, h has to be between -1 and 1, or put another way, we have to be looking for the logarithms of numbers between 0 and 2. And it takes fewer terms to get the result as precise as you want the closer h is to zero, that is, the closer the number whose logarithm we want is to 1.

So most of my reference table is easy enough to make. But there’s a column left out: what is the logarithm of 2? Or 3, or 4, or so on? The infinite-series formula there doesn’t work that far out, and if you give it a try, let’s say with the logarithm of 5, you get a good bit of nonsense, numbers swinging positive and negative and ever-larger.

Of course we’re not limited to formulas; we can think, too. 3, for example, is equal to 1.5 times 2, so the logarithm of 3 is the logarithm of 1.5 2 plus the logarithm of 2, and we have the logarithm of 1.5, and the logarithm of 2 is … OK, that’s a bit of a problem. But if we had the logarithm of 2, we’d be able to work out the logarithm of 4 — it’s just twice that — and we could get to other numbers pretty easily: 5 is, among other things, 2 times 2 times 1.25 so its logarithm is twice the logarithm of 2 plus the logarithm of 1.25. We’d have to work out the logarithm of 1.25, but we can do that by formula. 6 is 2 times 2 times 1.5, and we already had 1.5 worked out. 7 is 2 times 2 times 1.75, and we have a formula for the logarithm of 1.75. 8 is 2 times 2 times 2, so, triple whatever the logarithm of 2 is. 9 is 3 times 3, so, double the logarithm of 3.

We’re not required to do things this way. I just picked some nice, easy ways to factor the whole numbers up to 9, and that didn’t seem to demand doing too much more work. I’d need the logarithms of 1.25 and 1.75, as well as 2, but I can use the formula or, for that matter, work it out using the rest of my table: 1.25 is 1.2 times 1.04 times 1.001 times 1.000602, approximately. But there are infinitely many ways to get 3 by multiplying together numbers between 1 and 2, and we can use any that are convenient.

We do still need the logarithm of 2, but, then, 2 is among other things equal to 1.6 times 1.25, and we’d been planning to work out the logarithm of 1.6 all the time, and 1.25 is useful in getting us to 5 also, so, why not do that?

So in summary we could get logarithms for any numbers we wanted by working out the logarithms for 1.1, 1.2, 1.3, and so on, and 1.01, 1.02, 1.03, et cetera, and 1.001, 1.002, 1.003 and so on, and then 1.25 and 1.75, which lets us work out the logarithms of 2, 3, 4, and so on up to 9.

I haven’t yet worked out, but I am curious about, what the fewest number of “extra” numbers I’d have to calculate are. That is, granted that I have to figure out the logarithms of 1.1, 1.01, 1.001, et cetera anyway. The way I outlined things I have to also work out the logarithms of 1.25 and 1.75 to get all the numbers I need. Is it possible to figure out a cleverer bit of factorization that requires only one extra number be worked out? For that matter, is it possible to need no extra numbers? My instinctive response is to say no, but that’s hardly a proof. I’d be interested to know better.

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