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.

Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

5 thoughts on “Machines That Do Something About Logarithms”

1. If you take the terms of the expansion of log(1+x) in pairs things are better:
Take h^3/3-h^4/4 and get h^3*(4-3*h)/12 for example
These are all positive for h<1 and the series will converge much more quickly.
ps my post will soon be with you

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1. Pairing things up looks nice, although I’m not sure it saves work. It might make for more numerically stable calculations but I haven’t tested that. (To explain: calculations on the computer naturally include a bit of error, because, basically, what we might want to write as 1/3 we write on the computer as 0.333 and that’s a tiny bit different. Sometimes doing calculations in one order will magnify that little difference between what you want and what you actually compute. Sometimes changing the order will mean those little differences stay little, and that’s a stable computation.)

I’m looking forward to your post and appreciate the writing. Thank you.

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