## 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.

## Machines That Give You Logarithms

As I’ve laid out the tools that the Harvard IBM Automatic Sequence-Controlled Calculator would use to work out a common logarithm, now I can show how this computer of the 1940s and 1950s would do it. The goal, remember, is to compute logarithms to a desired accuracy, using computers that haven’t got abundant memory, and as quickly as possible. As quickly as possible means, roughly, avoiding multiplication (which takes time) and doing as few divisions as can possibly be done (divisions take forever).

As a reminder, the tools we have are:

1. We can work out at least some logarithms ahead of time and look them up as needed.
2. The natural logarithm of a number close to 1 is $log_e\left(1 + h\right) = h - \frac12h^2 + \frac13h^3 - \frac14h^4 + \frac15h^5 - \cdots$.
3. If we know a number’s natural logarithm (base e), then we can get its common logarithm (base 10): multiply the natural logarithm by the common logarithm of e, which is about 0.43429.
4. Whether the natural or the common logarithm (or any other logarithm you might like) $\log\left(a\cdot b\cdot c \cdot d \cdots \right) = \log(a) + \log(b) + \log(c) + \log(d) + \cdots$

Now we’ll put this to work. The first step is which logarithms to work out ahead of time. Since we’re dealing with common logarithms, we only need to be able to work out the logarithms for numbers between 1 and 10: the common logarithm of, say, 47.2286 is one plus the logarithm of 4.72286, and the common logarithm of 0.472286 is minus two plus the logarithm of 4.72286. So we’ll start by working out the logarithms of 1, 2, 3, 4, 5, 6, 7, 8, and 9, and storing them in what, in 1944, was still a pretty tiny block of memory. The original computer using this could store 72 numbers at a time, remember, though to 23 decimal digits.

So let’s say we want to know the logarithm of 47.2286. We have to divide this by 10 in order to get the number 4.72286, which is between 1 and 10, so we’ll need to add one to whatever we get for the logarithm of 4.72286 is. (And, yes, we want to avoid doing divisions, but dividing by 10 is a special case. The Automatic Sequence-Controlled Calculator stored numbers, if I am not grossly misunderstanding things, in base ten, and so dividing or multiplying by ten was as fast for it as moving the decimal point is for us. Modern computers, using binary arithmetic, find it as fast to divide or multiply by powers of two, even though division in general is a relatively sluggish thing.)

We haven’t worked out what the logarithm of 4.72286 is. And we don’t have a formula that’s good for that. But: 4.72286 is equal to 4 times 1.1807, and therefore the logarithm of 4.72286 is going to be the logarithm of 4 plus the logarithm of 1.1807. We worked out the logarithm of 4 ahead of time (it’s about 0.60206, if you’re curious).

We can use the infinite series formula to get the natural logarithm of 1.1807 to as many digits as we like. The natural logarithm of 1.1807 will be about $0.1807 - \frac12 0.1807^2 + \frac13 0.1807^3 - \frac14 0.1807^4 + \frac15 0.1807^5 - \cdots$ or 0.16613. Multiply this by the logarithm of e (about 0.43429) and we have a common logarithm of about 0.07214. (We have an error estimate, too: we’ve got the natural logarithm of 1.1807 within a margin of error of $\frac16 0.1807^6$, or about 0.000 0058, which, multiplied by the logarithm of e, corresponds to a margin of error for the common logarithm of about 0.000 0025.

Therefore: the logarithm of 47.2286 is about 1 plus 0.60206 plus 0.07214, which is 1.6742. And it is, too; we’ve done very well at getting the number just right considering how little work we really did.

Although … that infinite series formula. That requires a fair number of multiplications, at least eight as I figure it, however you look at it, and those are sluggish. It also properly speaking requires divisions, although you could easily write your code so that instead of dividing by 4 (say) you multiply by 0.25 instead. For this particular example number of 47.2286 we didn’t need very many terms in the series to get four decimal digits of accuracy, but maybe we got lucky and some other number would have required dozens of multiplications. Can we make this process, on average, faster?

And here’s one way to do it. Besides working out the common logarithms for the whole numbers 1 through 9, also work out the common logarithms for 1.1, 1.2, 1.3, 1.4, et cetera up to 1.9. And then …

We started with 47.2286. Divide by 10 (a free bit of work) and we have 4.72286. Divide 4.72286 is 4 times 1.180715. And 1.180715 is equal to 1.1 — the whole number and the first digit past the decimal — times 1.07337. That is, 47.2286 is 10 times 4 times 1.1 times 1.07337. And so the logarithm of 47.2286 is the logarithm of 10 plus the logarithm of 4 plus the logarithm of 1.1 plus the logarithm of 1.07337. We are almost certainly going to need fewer terms in the infinite series to get the logarithm of 1.07337 than we need for 1.180715 and so, at the cost of one more division, we probably save a good number of multiplications.

The common logarithm of 1.1 is about 0.041393. So the logarithm of 10 (1) plus the logarithm of 4 (0.60206) plus the logarithm of 1.1 (0.041393) is 1.6435, which falls a little short of the actual logarithm we’d wanted, about 1.6742, but two or three terms in the infinite series should be enough to make that up.

Or we could work out a few more common logarithms ahead of time: those for 1.01, 1.02, 1.03, and so on up to Our original 47.2286 divided by 10 is 4.72286. Divide that by the first number, 4, and you get 1.180715. Divide 1.180715 by 1.1, the first two digits, and you get 1.07337. Divide 1.07337 by 1.07, the first three digits, and you get 1.003156. So 47.2286 is 10 times 4 times 1.1 times 1.07 times 1.003156. So the common logarithm of 47.2286 is the logarithm of 10 (1) plus the logarithm of 4 (0.60206) plus the logarithm of 1.1 (0.041393) plus the logarithm of 1.07 (about 0.02938) plus the logarithm of 1.003156 (to be determined). Even ignoring the to-be-determined part that adds up to 1.6728, which is a little short of the 1.6742 we want but is doing pretty good considering we’ve reduced the whole problem to three divisions, looking stuff up, and four additions.

If we go a tiny bit farther, and also have worked out ahead of time the logarithms for 1.001, 1.002, 1.003, and so on out to 1.009, and do the same process all over again, then we get some better accuracy and quite cheaply yet: 47.2286 divided by 10 is 4.72286. 4.72286 divided by 4 is 1.180715. 1.180715 divided by 1.1 is 1.07337. 1.07337 divided by 1.07 is 1.003156. 1.003156 divided by 1.003 is 1.0001558.

So the logarithm of 47.2286 is the logarithm of 10 (1) plus the logarithm of 4 (0.60206) plus the logarithm of 1.1 (0.041393) plus the logarithm of 1.07 (0.029383) plus the logarithm of 1.003 (0.001301) plus the logarithm of 1.001558 (to be determined). Leaving aside the to-be-determined part, that adds up to 1.6741.

And the to-be-determined part is great: if we used just a single term in this series, the margin for error would be, at most, 0.000 000 0052, which is probably small enough for practical purposes. The first term in the to-be-determined part is awfully easy to calculate, too: it’s just 1.0001558 – 1, that is, 0.0001558. Add that and we have an approximate logarithm of 1.6742, which is dead on.

And I do mean dead on: work out more decimal places of the logarithm based on this summation and you get 1.674 205 077 226 78. That’s no more than five billionths away from the correct logarithm for the original 47.2286. And it required doing four divisions, one multiplication, and five additions. It’s difficult to picture getting such good precision with less work.

Of course, that’s done in part by having stockpiled a lot of hard work ahead of time: we need to know the logarithms of 1, 1.1, 1.01, 1.001, and then 2, 1.2, 1.02, 1.002, and so on. That’s 36 numbers altogether and there are many ways to work out logarithms. But people have already done that work, and we can use that work to make the problems we want to do considerably easier.

But there’s the process. Work out ahead of time logarithms for 1, 1.1, 1.01, 1.001, and so on, to whatever the limits of your patience. Then take the number whose logarithm you want and divide (or multiply) by ten until you get your working number into the range of 1 through 10. Divide out the first digit, which will be a whole number from 1 through 9. Divide out the first two digits, which will be something from 1.1 to 1.9. Divide out the first three digits, something from 1.01 to 1.09. Divide out the first four digits, something from 1.001 to 1.009. And so on. Then add up the logarithms of the power of ten you divided or multiplied by with the logarithm of the first divisor and the second divisor and third divisor and fourth divisor, until you run out of divisors. And then — if you haven’t already got the answer as accurately as you need — work out as many terms in the infinite series as you need; probably, it won’t be very many. Add that to your total. And you are, amazingly, done.

## Machines That Do Something About Logarithms

I’m going to assume everyone reading this accepts that logarithms are worth computing, and try to describe how Harvard’s IBM Automatic Sequence-Controlled Calculator would work them out.

The first part of this is kind of an observation: the quickest way to give the logarithm of a number is to already know it. Looking it up in a table is way faster than evaluating it, and that’s as true for the computer as for you. Obviously we can’t work out logarithms for every number, what with there being so many of them, but we could work out the logarithms for a reasonable range and to a certain precision and trust that the logarithm of (say) 4.42286 is going to be tolerably close to the logarithm of 4.423 that we worked out ahead of time. Working out a range of, say, 1 to 10 for logarithms base ten is plenty, because that’s all the range we need: the logarithm base ten of 44.2286 is the logarithm base ten of 4.42286 plus one. The logarithm base ten of 0.442286 is the logarithm base ten of 4.42286 minus one. You can guess from that what the logarithm of 4,422.86 is, compared to that of 4.42286.

This is trading computer memory for computational speed, which is often worth doing. But the old Automatic Sequence-Controlled Calculator can’t do that, at least not as easily as we’d like: it had the ability to store 72 numbers, albeit to 23 decimal digits. We can’t just use “worked it out ahead of time”, although we’re not going to abandon that idea either.

The next piece we have is something useful if we want to work out the natural logarithm — the logarithm base e — of a number that’s close to 1. We have a formula that will let us work out this natural logarithm to whatever accuracy we want:

$\log_{e}\left(1 + h\right) = h - \frac12 h^2 + \frac13 h^3 - \frac14 h^4 + \frac15 h^5 - \cdots \mbox{ if } |h| < 1$

In principle, we have to add up infinitely many terms to get the answer right. In practice, we only add up terms until the error — the difference between our sum and the correct answer — is smaller than some acceptable margin. This seems to beg the question, because how can we know how big that error is without knowing what the correct answer is? In fact we don’t know just what the error is, but we do know that the error can’t be any larger than the absolute value of the first term we neglect.

Let me give an example. Suppose we want the natural logarithm of 1.5, which the alert have noticed is equal to 1 + 0.5. Then h is 0.5. If we add together the first five terms of the natural logarithm series, then we have $0.5 - \frac12 0.5^2 + \frac13 0.5^3 - \frac14 0.5^4 + \frac15 0.5^5$ which is approximately 0.40729. If we were to work out the next term in the series, that would be $-\frac16 0.5^6$, which has an absolute value of about 0.0026. So the natural logarithm of 1.5 is 0.40729, plus or minus 0.0026. If we only need the natural logarithm to within 0.0026, that’s good: we’re done.

In fact, the natural logarithm of 1.5 is approximately 0.40547, so our error is closer to 0.00183, but that’s all right. Few people complain that our error is smaller than what we estimated it to be.

If we know what margin of error we’ll tolerate, by the way, then we know how many terms we have to calculate. Suppose we want the natural logarithm of 1.5 accurate to 0.001. Then we have to find the first number n so that $\frac1n 0.5^n < 0.001$; if I'm not mistaken, that's eight. Just how many terms we have to calculate will depend on what h is; the bigger it is — the farther the number is from 1 — the more terms we'll need.

The trouble with this is that it’s only good for working out the natural logarithms of numbers between 0 and 2. (And it’s better the closer the number is to 1.) If you want the natural logarithm of 44.2286, you have to divide out the highest power of e that’s less than it — well, you can fake that by dividing by e repeatedly — and what you get is that it’s e times e times e times 2.202 and we’re stuck there. Not hopelessly, mind you: we could find the logarithm of 1/2.202, which will be minus the logarithm of 2.202, at least, and we can work back to the original number from there. Still, this is a bit of a mess. We can do better.

The third piece we can use is one of the fundamental properties of logarithms. This is true for any base, as long as we use the same base for each logarithm in the equation here, and I’ve mentioned it in passing before:

$\log\left(a\cdot b\cdot c\cdot d \cdots\right) = \log\left(a\right) + \log\left(b\right) + \log\left(c\right) + \log\left(d\right) + \cdots$

That is, if we could factor a number whose logarithm we want into components which we can either look up or we can calculate very quickly, then we know its logarithm is the sum of the logarithms of those components. And this, finally, is how we can work out logarithms quickly and without too much hard work.

## Machines That Think About Logarithms

I confess that I picked up Edmund Callis Berkeley’s Giant Brains: Or Machines That Think, originally published 1949, from the library shelf as a source of cheap ironic giggles. After all, what is funnier than an attempt to explain to a popular audience that, wild as it may be to contemplate, electrically-driven machines could “remember” information and follow “programs” of instructions based on different conditions satisfied by that information? There’s a certain amount of that, though not as much as I imagined, and a good amount of descriptions of how the hardware of different partly or fully electrical computing machines of the 1940s worked.

But a good part, and the most interesting part, of the book is about algorithms, the ways to solve complicated problems without demanding too much computing power. This is fun to read because it showcases the ingenuity and creativity required to do useful work. The need for ingenuity will never leave us — we will always want to compute things that are a little beyond our ability — but to see how it’s done for a simple problem is instructive, if for nothing else to learn the kinds of tricks you can do to get the most of your computing resources.

The example that most struck me and which I want to share is from the chapter on the IBM Automatic Sequence-Controlled Calculator, built at Harvard at a cost of “somewhere near 3 or 4 hundred thousand dollars, if we leave out some of the cost of research and development, which would have been done whether or not this particular machine had ever been built”. It started working in April 1944, and wasn’t officially retired until 1959. It could store 72 numbers, each with 23 decimal digits. Like most computers (then and now) it could do addition and subtraction very quickly, in the then-blazing speed of about a third of a second; it could do multiplication tolerably quickly, in about six seconds; and division, rather slowly, in about fifteen seconds.

The process I want to describe is the taking of logarithms, and why logarithms should be interesting to compute takes a little bit of justification, although it’s implicitly there just in how fast calculations get done. Logarithms let one replace the multiplication of numbers with their addition, for a considerable savings in time; better, they let you replace the division of numbers with subtraction. They further let you turn exponentiation and roots into multiplication and division, which is almost always faster to do. Many human senses seem to work on a logarithmic scale, as well: we can tell that one weight is twice as heavy as the other much more reliably than we can tell that one weight is four pounds heavier than the other, or that one light is twice as bright as the other rather than is ten lumens brighter.

What the logarithm of a number is depends on some other, fixed, quantity, known as the base. In principle any positive number will do as base; in practice, these days people mostly only care about base e (which is a little over 2.718), the “natural” logarithm, because it has some nice analytic properties. Back in the day, which includes when this book was written, we also cared about base 10, the “common” logarithm, because we mostly work in base ten. I have heard of people who use base 2, but haven’t seen them myself and must regard them as an urban legend. The other bases are mostly used by people who are writing homework problems for the part of the class dealing with logarithms. To some extent it doesn’t matter what base you use. If you work out the logarithm in one base, you can convert that to the logarithm in another base by a multiplication.

The logarithm of some number in your base is the exponent you have to raise the base to to get your desired number. For example, the logarithm of 100, in base 10, is going to be 2 because 102 is 100, and the logarithm of e1/3 (a touch greater than 1.3956), in base e, is going to be 1/3. To dig deeper in my reserve of in-jokes, the logarithm of 2038, in base 10, is approximately 3.3092, because 103.3092 is just about 2038. The logarithm of e, in base 10, is about 0.4343, and the logarithm of 10, in base e, is about 2.303. Your calculator will verify all that.

All that talk about “approximately” should have given you some hint of the trouble with logarithms. They’re only really easy to compute if you’re looking for whole powers of whatever your base is, and that if your base is 10 or 2 or something else simple like that. If you’re clever and determined you can work out, say, that the logarithm of 2, base 10, has to be close to 0.3. It’s fun to do that, but it’ll involve such reasoning as “two to the tenth power is 1,024, which is very close to ten to the third power, which is 1,000, so therefore the logarithm of two to the tenth power must be about the same as the logarithm of ten to the third power”. That’s clever and fun, but it’s hardly systematic, and it doesn’t get you many digits of accuracy.

So when I pick up this thread I hope to explain one way to produce as many decimal digits of a logarithm as you could want, without asking for too much from your poor Automatic Sequence-Controlled Calculator.