## What Is The Logarithm of a Negative Number?

Learning of imaginary numbers, things created to be the square roots of negative numbers, inspired me. It probably inspires anyone who’s the sort of person who’d become a mathematician. The trick was great. I wondered could I do it? Could I find some other useful expansion of the number system?

The square root of a complex-valued number sounded like the obvious way to go, until a little later that week when I learned that’s just some other complex-valued numbers. The next thing I hit on: how about the logarithm of a negative number? Couldn’t that be a useful expansion of numbers?

No. It turns out you can make a sensible logarithm of negative, and complex-valued, numbers using complex-valued numbers. Same with trigonometric and inverse trig functions, tangents and arccosines and all that. There isn’t anything we can do with the normal mathematical operations that needs something bigger than the complex-valued numbers to play with. It’s possible to expand on the complex-valued numbers. We can make quaternions and some more elaborate constructs there. They don’t solve any particular shortcoming in complex-valued numbers, but they’ve got their uses. I never got anywhere near reinventing them. I don’t regret the time spent on that. There’s something useful in trying to invent something even if it fails.

One problem with mathematics — with all intellectual fields, really — is that it’s easy, when teaching, to give the impression that this stuff is the Word of God, built into the nature of the universe and inarguable. It’s so not. The stuff we find interesting and how we describe those things are the results of human thought, attempts to say what is interesting about a thing and what is useful. And what best approximates our ideas of what we would like to know. So I was happy to see this come across my Twitter feed:

The links to a 12-page paper by Deepak Bal, Leibniz, Bernoulli, and the Logarithms of Negative Numbers. It’s a review of how the idea of a logarithm of a negative number got developed over the course of the 18th century. And what great minds, like Gottfried Leibniz and John (I) Bernoulli argued about as they find problems with the implications of what they were doing. (There were a lot of Bernoullis doing great mathematics, and even multiple John Bernoullis. The (I) is among the ways we keep them sorted out.) It’s worth a read, I think, even if you’re not all that versed in how to calculate logarithms. (but if you’d like to be better-versed, here’s the tail end of some thoughts about that.) The process of how a good idea like this comes to be is worth knowing.

Also: it turns out there’s not “the” logarithm of a complex-valued number. There’s infinitely many logarithms. But they’re a family, all strikingly similar, so we can pick one that’s convenient and just use that. Ask if you’re really interested.

## Reading the Comics, January 21, 2017: Homework Edition

Now to close out what Comic Strip Master Command sent my way through last Saturday. And I’m glad I’ve shifted to a regular schedule for these. They ordered a mass of comics with mathematical themes for Sunday and Monday this current week.

Karen Montague-Reyes’s Clear Blue Water rerun for the 17th describes trick-or-treating as “logarithmic”. The intention is to say that the difficulty in wrangling kids from house to house grows incredibly fast as the number of kids increases. Fair enough, but should it be “logarithmic” or “exponential”? Because the logarithm grows slowly as the number you take the logarithm of grows. It grows all the slower the bigger the number gets. The exponential of a number, though, that grows faster and faster still as the number underlying it grows. So is this mistaken?

I say no. It depends what the logarithm is, and is of. If the number of kids is the logarithm of the difficulty of hauling them around, then the intent and the mathematics are in perfect alignment. Five kids are (let’s say) ten times harder to deal with than four kids. Sensible and, from what I can tell of packs of kids, correct.

Rick Detorie’s One Big Happy for the 17th of January, 2017. The section was about how the appearance and trappings of wealth matter for more than the actual substance of wealth so everyone’s really up to speed in the course.

Rick Detorie’s One Big Happy for the 17th is a resisting-the-word-problem joke. There’s probably some warning that could be drawn about this in how to write story problems. It’s hard to foresee all the reasonable confounding factors that might get a student to the wrong answer, or to see a problem that isn’t meant to be there.

Bill Holbrook’s On The Fastrack for the 19th continues Fi’s story of considering leaving Fastrack Inc, and finding a non-competition clause that’s of appropriate comical absurdity. As an auditor there’s not even a chance Fi could do without numbers. Were she a pure mathematician … yeah, no. There’s fields of mathematics in which numbers aren’t all that important. But we never do without them entirely. Even if we exclude cases where a number is just used as an index, for which Roman numerals would be almost as good as regular numerals. If nothing else numbers would keep sneaking in by way of polynomials.

Bill Holbrook’s On The Fastrack for the 19th of January, 2017. I feel like someone could write a convoluted story that lets someone do mathematics while avoiding any actual use of any numbers, and that it would probably be Greg Egan who did it.

Dave Whamond’s Reality Check for the 19th breaks our long dry spell without pie chart jokes.

Mort Walker and Dik Browne’s Vintage Hi and Lois for the 27th of July, 1959 uses calculus as stand-in for what college is all about. Lois’s particular example is about a second derivative. Suppose we have a function named ‘y’ and that depends on a variable named ‘x’. Probably it’s a function with domain and range both real numbers. If complex numbers were involved then the variable would more likely be called ‘z’. The first derivative of a function is about how fast its values change with small changes in the variable. The second derivative is about how fast the values of the first derivative change with small changes in the variable.

Mort Walker and Dik Browne’s Vintage Hi and Lois for the 27th of July, 1959. Fortunately Lois discovered the other way to avoid college costs: simply freeze the ages of your children where they are now, so they never face student loans. It’s an appealing plan until you imagine being Trixie.

The ‘d’ in this equation is more of an instruction than it is a number, which is why it’s a mistake to just divide those out. Instead of writing it as $\frac{d^2 y}{dx^2}$ it’s permitted, and common, to write it as $\frac{d^2}{dx^2} y$. This means the same thing. I like that because, to me at least, it more clearly suggests “do this thing (take the second derivative) to the function we call ‘y’.” That’s a matter of style and what the author thinks needs emphasis.

There are infinitely many possible functions y that would make the equation $\frac{d^2 y}{dx^2} = 6x - 2$ true. They all belong to one family, though. They all look like $y(x) = \frac{1}{6} 6 x^3 - \frac{1}{2} 2 x^2 + C x + D$, where ‘C’ and ‘D’ are some fixed numbers. There’s no way to know, from what Lois has given, what those numbers should be. It might be that the context of the problem gives information to use to say what those numbers should be. It might be that the problem doesn’t care what those numbers should be. Impossible to say without the context.

• #### Joshua K. 6:26 am on Monday, 30 January, 2017 Permalink | Reply

Why is the function in the Hi & Lois discussion stated as y(x) = (1/6)6x^3 – (1/2)2x^2 + Cx +D? Why not just y(x) = x^3 – x^2 + Cx + D?

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• #### Joseph Nebus 5:43 pm on Friday, 3 February, 2017 Permalink | Reply

Good question! I actually put a fair bit of thought into this. If I were doing the problem myself I’d have cut right to x^3 – x^2 + Cx + D. But I thought there’s a number of people reading this for whom calculus is a perfect mystery and I thought that if I put an intermediate step it might help spot the pattern at work, that the coefficients in front of the x^3 and x^2 terms don’t vanish without cause.

That said, I probably screwed up by writing them as 1/6 and 1/2. That looks too much like I’m just dividing by what the coefficients are. If I had taken more time to think out the post I should have written 1/(23) and 1/(12). This might’ve given a slightly better chance at connecting the powers of x and the fractions in the denominator. I’m not sure how much help that would give, since I didn’t describe how to take antiderivatives here. But I think it’d be a better presentation and I should remember that in future situations like that.

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## Reading the Comics, October 29, 2016: Rerun Comics Edition

There were a couple of rerun comics in this week’s roundup, so I’ll go with that theme. And I’ll put in one more appeal for subjects for my End of 2016 Mathematics A To Z. Have a mathematics term you’d like to see me go on about? Just ask! Much of the alphabet is still available.

John Kovaleski’s Bo Nanas rerun the 24th is about probability. There’s something wondrous and strange that happens when we talk about the probability of things like birth days. They are, if they’re in the past, determined and fixed things. The current day is also a known, determined, fixed thing. But we do mean something when we say there’s a 1-in-365 (or 366, or 365.25 if you like) chance of today being your birthday. It seems to me this is probability based on ignorance. If you don’t know when my birthday is then your best guess is to suppose there’s a one-in-365 (or so) chance that it’s today. But I know when my birthday is; to me, with this information, the chance today is my birthday is either 0 or 1. But what are the chances that today is a day when the chance it’s my birthday is 1? At this point I realize I need much more training in the philosophy of mathematics, and the philosophy of probability. If someone is aware of a good introductory book about it, or a web site or blog that goes into these problems in a way a lay reader will understand, I’d love to hear of it.

I’ve featured this installment of Poor Richard’s Almanac before. I’ll surely feature it again. I like Richard Thompson’s sense of humor. The first panel mentions non-Euclidean geometry, using the connotation that it does have. Non-Euclidean geometries are treated as these magic things — more, these sinister magic things — that defy all reason. They can’t defy reason, of course. And at least some of them are even sensible if we imagine we’re drawing things on the surface of the Earth, or at least the surface of a balloon. (There are non-Euclidean geometries that don’t look like surfaces of spheres.) They don’t work exactly like the geometry of stuff we draw on paper, or the way we fit things in rooms. But they’re not magic, not most of them.

Stephen Bentley’s Herb and Jamaal for the 25th I believe is a rerun. I admit I’m not certain, but it feels like one. (Bentley runs a lot of unannounced reruns.) Anyway I’m refreshed to see a teacher giving a student permission to count on fingers if that’s what she needs to work out the problem. Sometimes we have to fall back on the non-elegant ways to get comfortable with a method.

Dave Whamond’s Reality Check for the 25th name-drops Einstein and one of the three equations that has any pop-culture currency.

Guy Gilchrist’s Today’s Dogg for the 27th is your basic mathematical-symbols joke. We need a certain number of these.

Berkeley Breathed’s Bloom County for the 28th is another rerun, from 1981. And it’s been featured here before too. As mentioned then, Milo is using calculus and logarithms correctly in his rather needless insult of Freida. 10,000 is a constant number, and as mentioned a few weeks back its derivative must be zero. Ten to the power of zero is 1. The log of 10, if we’re using logarithms base ten, is also 1. There are many kinds of logarithms but back in 1981, the default if someone said “log” would be the logarithm base ten. Today the default is more muddled; a normal person would mean the base-ten logarithm by “log”. A mathematician might mean the natural logarithm, base ‘e’, by “log”. But why would a normal person mention logarithms at all anymore?

Jef Mallett’s Frazz for the 28th is mostly a bit of wordplay on evens and odds. It’s marginal, but I do want to point out some comics that aren’t reruns in this batch.

## A Leap Day 2016 Mathematics A To Z: Kullbach-Leibler Divergence

Today’s mathematics glossary term is another one requested by Jacob Kanev. Kaven, I learned last time, has got a blog, “Some Unconsidered Trifles”, for those interested in having more things to read. Kanev’s request this time was a term new to me. But learning things I didn’t expect to consider is part of the fun of this dance.

## Kullback-Leibler Divergence.

The Kullback-Leibler Divergence comes to us from information theory. It’s also known as “information divergence” or “relative entropy”. Entropy is by now a familiar friend. We got to know it through, among other things, the “How interesting is a basketball tournament?” question. In this context, entropy is a measure of how surprising it would be to know which of several possible outcomes happens. A sure thing has an entropy of zero; there’s no potential surprise in it. If there are two equally likely outcomes, then the entropy is 1. If there are four equally likely outcomes, then the entropy is 2. If there are four possible outcomes, but one is very likely and the other three mediocre, the entropy might be low, say, 0.5 or so. It’s mostly but not perfectly predictable.

Suppose we have a set of possible outcomes for something. (Pick anything you like. It could be the outcomes of a basketball tournament. It could be how much a favored stock rises or falls over the day. It could be how long your ride into work takes. As long as there are different possible outcomes, we have something workable.) If we have a probability, a measure of how likely each of the different outcomes is, then we have a probability distribution. More likely things have probabilities closer to 1. Less likely things have probabilities closer to 0. No probability is less than zero or more than 1. All the probabilities added together sum up to 1. (These are the rules which make something a probability distribution, not just a bunch of numbers we had in the junk drawer.)

The Kullback-Leibler Divergence describes how similar two probability distributions are to one another. Let me call one of these probability distributions p. I’ll call the other one q. We have some number of possible outcomes, and we’ll use k as an index for them. pk is how likely, in distribution p, that outcome number k is. qk is how likely, in distribution q, that outcome number k is.

To calculate this divergence, we work out, for each k, the number pk times the logarithm of pk divided by qk. Here the logarithm is base two. Calculate all this for every one of the possible outcomes, and add it together. This will be some number that’s at least zero, but it might be larger.

The closer that distribution p and distribution q are to each other, the smaller this number is. If they’re exactly the same, this number will be zero. The less that distribution p and distribution q are like each other, the bigger this number is.

And that’s all good fun, but, why bother with it? And at least one answer I can give is that it lets us measure how good a model of something is.

Suppose we think we have an explanation for how something varies. We can say how likely it is we think there’ll be each of the possible different outcomes. This gives us a probability distribution which let’s call q. We can compare that to actual data. Watch whatever it is for a while, and measure how often each of the different possible outcomes actually does happen. This gives us a probability distribution which let’s call p.

If our model is a good one, then the Kullback-Leibler Divergence between p and q will be small. If our model’s a lousy one, then this divergence will be large. If we have a couple different models, we can see which ones make for smaller divergences and which ones make for larger divergences. Probably we’ll want smaller divergences.

Here you might ask: why do we need a model? Isn’t the actual data the best model we might have? It’s a fair question. But no, real data is kind of lousy. It’s all messy. It’s complicated. We get extraneous little bits of nonsense clogging it up. And the next batch of results is going to be different from the old ones anyway, because real data always varies.

Furthermore, one of the purposes of a model is to be simpler than reality. A model should do away with complications so that it is easier to analyze, easier to make predictions with, and easier to teach than the reality is. But a model mustn’t be so simple that it can’t represent important aspects of the thing we want to study.

The Kullback-Leibler Divergence is a tool that we can use to quantify how much better one model or another fits our data. It also lets us quantify how much of the grit of reality we lose in our model. And this is at least some of the use of this quantity.

• #### howardat58 5:20 pm on Wednesday, 23 March, 2016 Permalink | Reply

****** This makes sense !

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• #### Joseph Nebus 3:16 am on Thursday, 24 March, 2016 Permalink | Reply

Thank you kindly. I hope people who specialize in information theory think I’m not impossibly far off.

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• #### howardat58 4:57 pm on Thursday, 24 March, 2016 Permalink | Reply

I read a paper a while ago by this chap who investigated the speed of evolution from an information theoretical viewpoint. It’s worth a read. I made a copy. Just as well, as recently I searched “speed of evolution” and got a load of people who rubbished that paper, my guess is that they didn’t understand it. besides, the conclusions were clearly unacceptable. But that’s science.

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• #### Joseph Nebus 5:06 pm on Saturday, 26 March, 2016 Permalink | Reply

I’m interested by this paper. But I am so weak in information theory, and incredibly weak in biology, that I probably wouldn’t get much of interest from it. If lucky I might know what most of the key symbols mean …

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## A Leap Day 2016 Mathematics A To Z: Isomorphism

Gillian B made the request that’s today’s A To Z word. I’d said it would be challenging. Many have been, so far. But I set up some of the work with “homomorphism” last time. As with “homomorphism” it’s a word that appears in several fields and about different kinds of mathematical structure. As with homomorphism, I’ll try describing what it is for groups. They seem least challenging to the imagination.

## Isomorphism.

An isomorphism is a kind of homomorphism. And a homomorphism is a kind of thing we do with groups. A group is a mathematical construct made up of two things. One is a set of things. The other is an operation, like addition, where we take two of the things and get one of the things in the set. I think that’s as far as we need to go in this chain of defining things.

A homomorphism is a mapping, or if you like the word better, a function. The homomorphism matches everything in a group to the things in a group. It might be the same group; it might be a different group. What makes it a homomorphism is that it preserves addition.

I gave an example last time, with groups I called G and H. G had as its set the whole numbers 0 through 3 and as operation addition modulo 4. H had as its set the whole numbers 0 through 7 and as operation addition modulo 8. And I defined a homomorphism φ which took a number in G and matched it the number in H which was twice that. Then for any a and b which were in G’s set, φ(a + b) was equal to φ(a) + φ(b).

We can have all kinds of homomorphisms. For example, imagine my new φ1. It takes whatever you start with in G and maps it to the 0 inside H. φ1(1) = 0, φ1(2) = 0, φ1(3) = 0, φ1(0) = 0. It’s a legitimate homomorphism. Seems like it’s wasting a lot of what’s in H, though.

An isomorphism doesn’t waste anything that’s in H. It’s a homomorphism in which everything in G’s set matches to exactly one thing in H’s, and vice-versa. That is, it’s both a homomorphism and a bijection, to use one of the terms from the Summer 2015 A To Z. The key to remembering this is the “iso” prefix. It comes from the Greek “isos”, meaning “equal”. You can often understand an isomorphism from group G to group H showing how they’re the same thing. They might be represented differently, but they’re equivalent in the lights you use.

I can’t make an isomorphism between the G and the H I started with. Their sets are different sizes. There’s no matching everything in H’s set to everything in G’s set without some duplication. But we can make other examples.

For instance, let me start with a new group G. It’s got as its set the positive real numbers. And it has as its operation ordinary multiplication, the kind you always do. And I want a new group H. It’s got as its set all the real numbers, positive and negative. It has as its operation ordinary addition, the kind you always do.

For an isomorphism φ, take the number x that’s in G’s set. Match it to the number that’s the logarithm of x, found in H’s set. This is a one-to-one pairing: if the logarithm of x equals the logarithm of y, then x has to equal y. And it covers everything: all the positive real numbers have a logarithm, somewhere in the positive or negative real numbers.

And this is a homomorphism. Take any x and y that are in G’s set. Their “addition”, the group operation, is to multiply them together. So “x + y”, in G, gives us the number xy. (I know, I know. But trust me.) φ(x + y) is equal to log(xy), which equals log(x) + log(y), which is the same number as φ(x) + φ(y). There’s a way to see the postive real numbers being multiplied together as equivalent to all the real numbers being added together.

You might figure that the positive real numbers and all the real numbers aren’t very different-looking things. Perhaps so. Here’s another example I like, drawn from Wikipedia’s entry on Isomorphism. It has as sets things that don’t seem to have anything to do with one another.

Let me have another brand-new group G. It has as its set the whole numbers 0, 1, 2, 3, 4, and 5. Its operation is addition modulo 6. So 2 + 2 is 4, while 2 + 3 is 5, and 2 + 4 is 0, and 2 + 5 is 1, and so on. You get the pattern, I hope.

The brand-new group H, now, that has a more complicated-looking set. Its set is ordered pairs of whole numbers, which I’ll represent as (a, b). Here ‘a’ may be either 0 or 1. ‘b’ may be 0, 1, or 2. To describe its addition rule, let me say we have the elements (a, b) and (c, d). Find their sum first by adding together a and c, modulo 2. So 0 + 0 is 0, 1 + 0 is 1, 0 + 1 is 1, and 1 + 1 is 0. That result is the first number in the pair. The second number we find by adding together b and d, modulo 3. So 1 + 0 is 1, and 1 + 1 is 2, and 1 + 2 is 0, and so on.

So, for example, (0, 1) plus (1, 1) will be (1, 2). But (0, 1) plus (1, 2) will be (1, 0). (1, 2) plus (1, 0) will be (0, 2). (1, 2) plus (1, 2) will be (0, 1). And so on.

The isomorphism matches up things in G to things in H this way:

In G φ(G), in H
0 (0, 0)
1 (1, 1)
2 (0, 2)
3 (1, 0)
4 (0, 1)
5 (1, 2)

I recommend playing with this a while. Pick any pair of numbers x and y that you like from G. And check their matching ordered pairs φ(x) and φ(y) in H. φ(x + y) is the same thing as φ(x) + φ(y) even though the things in G’s set don’t look anything like the things in H’s.

Isomorphisms exist for other structures. The idea extends the way homomorphisms do. A ring, for example, has two operations which we think of as addition and multiplication. An isomorphism matches two rings in ways that preserve the addition and multiplication, and which match everything in the first ring’s set to everything in the second ring’s set, one-to-one. The idea of the isomorphism is that two different things can be paired up so that they look, and work, remarkably like one another.

One of the common uses of isomorphisms is describing the evolution of systems. We often like to look at how some physical system develops from different starting conditions. If you make a little variation in how things start, does this produce a small change in how it develops, or does it produce a big change? How big? And the description of how time changes the system is, often, an isomorphism.

Isomorphisms also appear when we study the structures of groups. They turn up naturally when we look at things called “normal subgroups”. The name alone gives you a good idea what a “subgroup” is. “Normal”, well, that’ll be another essay.

• #### Gillian B 10:27 pm on Friday, 18 March, 2016 Permalink | Reply

Yayay!

I chose that, of all things, from an old Dr Who episode in “The Pyramids of Mars”. Sutek (old Egyptian god) wants to use the TARDIS himself, but the Doctor tells him it’s isomorphic – and my mother yelled from the kitchen “I KNOW WHAT THAT MEANS!” (she was about halfway through her maths degree at the time). So thank you! I’m going to pass this on to her, for the memories.

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• #### Gillian B 2:14 am on Monday, 21 March, 2016 Permalink | Reply

Allow me to reprint an email I received today:

From:
“Liz Richards”

To:
reynardo

Subject:
Re: Isomorphism

Thank you, thank you, the you. I’ve printed out the isomorphic page.

Love

Mum

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• #### Joseph Nebus 2:48 am on Thursday, 24 March, 2016 Permalink | Reply

Aw, goodness. That’s wonderful. Thank you.

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## The Set Tour, Part 10: Lots of Spheres

The next exhibit on the Set Tour here builds on a couple of the previous ones. First is the set Sn, that is, the surface of a hypersphere in n+1 dimensions. Second is Bn, the ball — the interior — of a hypersphere in n dimensions. Yeah, it bugs me too that Sn isn’t the surface of Bn. But it’d be too much work to change things now. The third has lurked implicitly since all the way back to Rn, a set of n real numbers for which the ordering of the numbers matters. (That is, that the set of numbers 2, 3 probably means something different than the set 3, 2.) And fourth is a bit of writing we picked up with matrices. The selection is also dubiously relevant to my own thesis from back in the day.

## Sn x m and Bn x m

Here ‘n’ and ‘m’ are whole numbers, and I’m not saying which ones because I don’t need to tie myself down. Just as with Rn and with matrices this is a whole family of sets. Each different pair of n and m gives us a different set Sn x m or Bn x m, but they’ll all look quite similar.

The multiplication symbol here is a kind of multiplication, just as it was in matrices. That kind is called a “direct product”. What we mean by Sn x m is that we have a collection of items. We have the number m of them. Each one of those items is in Sn. That’s the surface of the hypersphere in n+1 dimensions. And we want to keep track of the order of things; we can’t swap items around and suppose they mean the same thing.

So suppose I write S2 x 7. This is an ordered collection of seven items, every one of which is on the surface of a three-dimensional sphere. That is, it’s the location of seven spots on the surface of the Earth. S2 x 8 offers similar prospects for talking about the location of eight spots.

With that written out, you should have a guess what Bn x m means. Your guess is correct. It’s a collection of m things, each of them within the interior of the n-dimensional ball.

Now the dubious relevance to my thesis. My problem was modeling a specific layer of planetary atmospheres. The model used for this was to pretend the atmosphere was made up of some large number of vortices, of whirlpools. Just like you see in the water when you slide your hand through the water and watch the little whirlpools behind you. The winds could be worked out as the sum of the winds produced by all these little vortices.

In the model, each of these vortices was confined to a single distance from the center of the planet. That’s close enough to true for planetary atmospheres. A layer in the atmosphere is not thick at all, compared to the planet. So every one of these vortices could be represented as a point in S2, the surface of a three-dimensional sphere. There would be some large number of these points. Most of my work used a nice round 256 points. So my model of a planetary atmosphere represented the system as a point in the domain S2 x 256. I was particularly interested in the energy of this set of 256 vortices. That was a function which had, as its domain, S2 x 256, and as range, the real numbers R.

But the connection to my actual work is dubious. I was doing numerical work, for the most part. I don’t think my advisor or I ever wrote S2 x 256 or anything like that when working out what I ought to do, much less what I actually did. Had I done a more analytic thesis I’d surely have needed to name this set. But I didn’t. It was lurking there behind my work nevertheless.

The energy of this system of vortices looked a lot like the potential energy for a bunch of planets attracting each other gravitationally, or like point charges repelling each other electrically. We work it out by looking at each pair of vortices. Work out the potential energy of those two vortices being that strong and that far apart. We call that a pairwise interaction. Then add up all the pairwise interactions. That’s it. [1] The pairwise interaction is stronger as each vortex is stronger; it gets weaker as the vortices get farther apart.

In gravity or electricity problems the strength falls off as the reciprocal of the distance between points. In vortices, the strength falls off as minus one times the logarithm of the distance between points. That’s a difference, and it meant that a lot of analytical results known for electric charges didn’t apply to my problem exactly. That was all right. I didn’t need many. But it does mean that I was fibbing up above, when I said I was working with S2 x 256. Pause a moment. Do you see what the fib was?

I’ll put what would otherwise be a footnote here so folks have a harder time reading right through to the answer.

[1] Physics majors may be saying something like: “wait, I see how this would be the potential energy of these 256 vortices, but where’s the kinetic energy?” The answer is, there is none. It’s all potential energy. The dynamics of point vortices are weird. I didn’t have enough grounding in mechanics when I went into them.

That’s all to the footnote.

Here’s where the fib comes in. If I’m really picking sets of vortices from all of the set S2 x 256, then, can two of them be in the exact same place? Sure they can. Why couldn’t they? For precedent, consider R3. In the three-dimensional vectors I can have the first and third numbers “overlap” and have the same value: (1, 2, 1) is a perfectly good vector. Why would that be different for an ordered set of points on the surface of the sphere? Why can’t vortex 1 and vortex 3 happen to have the same value in S2?

The problem is if two vortices were in the exact same position then the energy would be infinitely large. That’s not unique to vortices. It would be true for masses and gravity, or electric charges, if they were brought perfectly on top of each other. Infinitely large energies are a problem. We really don’t want to deal with them.

We could deal with this by pretending it doesn’t happen. Imagine if you dropped 256 poker chips across the whole surface of the Earth. Would you expect any two to be on top of each other? Would you expect two to be exactly and perfectly on top of each other, neither one even slightly overhanging the other? That’s so unlikely you could safely ignore it, for the same reason you could ignore the chance you’ll toss a coin and have it come up tails 56 times in a row.

And if you were interested in modeling the vortices moving it would be incredibly unlikely to have one vortex collide with another. They’d circle around each other, very fast, almost certainly. So ignoring the problem is defensible in this case.

Or we could be proper and responsible and say, “no overlaps” and “no collisions”. We would define some set that represents “all the possible overlaps and arrangements that give us a collision”. Then we’d say we’re looking at S2 x 256 except for those. I don’t think there’s a standard convention for “all the possible overlaps and collisions”, but Ω is a reasonable choice. Then our domain would be S2 x 256 \ Ω. The backslash means “except for the stuff after this”. This might seem unsatisfying. We don’t explicitly say what combinations we’re excluding. But go ahead and try listing all the combinations that would produce trouble. Try something simple, like S2 x 4. This is why we hide all the complicated stuff under a couple ordinary sentences.

It’s not hard to describe “no overlaps” mathematically. (You would say something like “vortex number j and vortex number k are not at the same position”, with maybe a rider of “unless j and k are the same number”. Or you’d put it in symbols that mean the same thing.) “No collisions” is harder. For gravity or electric charge problems we can describe at least some of them. And I realize now I’m not sure if there is an easy way to describe vortices that collide. I have difficulty imagining how they might, since vortices that are close to one another are pushing each other sideways quite intently. I don’t think that I can say they can’t, though. Not without more thought.

• #### sheldonk2014 3:52 pm on Thursday, 10 December, 2015 Permalink | Reply

If I’m getting your draft correctly letters can be put into mathematical equations
Way cool Joseph

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• #### Joseph Nebus 11:31 pm on Saturday, 12 December, 2015 Permalink | Reply

Really you can put anything into equations. They’re just shorthand ways of writing down interesting and true ideas. We’ll often use letters to stand in for numbers if we know some things about the number but not which one it is, or if we don’t care which number it is. It’s convenient to have a way to refer to a number without pinning it down to being “eight” or something like that.

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• #### Joseph Nebus 7:28 pm on Saturday, 28 March, 2015 Permalink | Reply Tags: basketball ( 15 ), bits, Claude Shannon ( 2 ), entropy ( 29 ), information content ( 3 ), information theory ( 18 ), logarithms, March madness ( 8 ), memory ( 3 ), tournaments ( 2 )

When I wrote last weekend’s piece about how interesting a basketball tournament was, I let some terms slide without definition, mostly so I could explain what ideas I wanted to use and how they should relate. My love, for example, read the article and looked up and asked what exactly I meant by “interesting”, in the attempt to measure how interesting a set of games might be, even if the reasoning that brought me to a 63-game tournament having an interest level of 63 seemed to satisfy.

When I spoke about something being interesting, what I had meant was that it’s something whose outcome I would like to know. In mathematical terms this “something whose outcome I would like to know” is often termed an experiment’ to be performed or, even better, a message’ that presumably I wil receive; and the outcome is the “information” of that experiment or message. And information is, in this context, something you do not know but would like to.

So the information content of a foregone conclusion is low, or at least very low, because you already know what the result is going to be, or are pretty close to knowing. The information content of something you can’t predict is high, because you would like to know it but there’s no (accurately) guessing what it might be.

This seems like a straightforward idea of what information should mean, and it’s a very fruitful one; the field of “information theory” and a great deal of modern communication theory is based on them. This doesn’t mean there aren’t some curious philosophical implications, though; for example, technically speaking, this seems to imply that anything you already know is by definition not information, and therefore learning something destroys the information it had. This seems impish, at least. Claude Shannon, who’s largely responsible for information theory as we now know it, was renowned for jokes; I recall a Time Life science-series book mentioning how he had built a complex-looking contraption which, turned on, would churn to life, make a hand poke out of its innards, and turn itself off, which makes me smile to imagine. Still, this definition of information is a useful one, so maybe I’m imagining a prank where there’s not one intended.

And something I hadn’t brought up, but which was hanging awkwardly loose, last time was: granted that the outcome of a single game might have an interest level, or an information content, of 1 unit, what’s the unit? If we have units of mass and length and temperature and spiciness of chili sauce, don’t we have a unit of how informative something is?

We have. If we measure how interesting something is — how much information there is in its result — using base-two logarithms the way we did last time, then the unit of information is a bit. That is the same bit that somehow goes into bytes, which go on your computer into kilobytes and megabytes and gigabytes, and onto your hard drive or USB stick as somehow slightly fewer gigabytes than the label on the box says. A bit is, in this sense, the amount of information it takes to distinguish between two equally likely outcomes. Whether that’s a piece of information in a computer’s memory, where a 0 or a 1 is a priori equally likely, or whether it’s the outcome of a basketball game between two evenly matched teams, it’s the same quantity of information to have.

So a March Madness-style tournament has an information content of 63 bits, if all you’re interested in is which teams win. You could communicate the outcome of the whole string of matches by indicating whether the “home” team wins or loses for each of the 63 distinct games. You could do it with 63 flashes of light, or a string of dots and dashes on a telegraph, or checked boxes on a largely empty piece of graphing paper, coins arranged tails-up or heads-up, or chunks of memory on a USB stick. We’re quantifying how much of the message is independent of the medium.

## How Interesting Is A Basketball Tournament?

Yes, I can hear people snarking, “not even the tiniest bit”. These are people who think calling all athletic contests “sportsball” is still a fresh and witty insult. No matter; what I mean to talk about applies to anything where there are multiple possible outcomes. If you would rather talk about how interesting the results of some elections are, or whether the stock market rises or falls, whether your preferred web browser gains or loses market share, whatever, read it as that instead. The work is all the same.

Mark Litzler’s Joe Vanilla for the 14th of March, 2015. We all make events interesting in our own ways.

To talk about quantifying how interesting the outcome of a game (election, trading day, whatever) means we have to think about what “interesting” qualitatively means. A sure thing, a result that’s bound to happen, is not at all interesting, since we know going in that it’s the result. A result that’s nearly sure but not guaranteed is at least a bit interesting, since after all, it might not happen. An extremely unlikely result would be extremely interesting, if it could happen.

• #### Angie Mc 9:52 pm on Friday, 20 March, 2015 Permalink | Reply

I’ll hop in at the sweet 16. I’m be in charge of cheering and you can be in charge of the stats :D Super cool post, Joseph!

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• #### Joseph Nebus 8:36 pm on Sunday, 22 March, 2015 Permalink | Reply

Thanks. I hope to be interesting with this sort of thing. I actually do have a follow-up in the works.

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• #### Angie Mc 6:45 pm on Tuesday, 24 March, 2015 Permalink | Reply

Great! I caught the Notre Dame vs Butler game…WOW! Excellence right to the end :D

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• #### Joseph Nebus 3:40 am on Friday, 27 March, 2015 Permalink | Reply

Oh, good, good. Uhm … I called the outcome of that right, if I’m reading my brackets correctly. That seems to have turned out all right.

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• #### Garfield Hug 6:17 am on Saturday, 21 March, 2015 Permalink | Reply

Math as in stats, calculus or algebra has been my weakest link. Still I managed to do Bs for math subjects in university. I am enjoying the way you use math to define things and just want to say thanks for making math not so complicated. This is your passion in spreading math in your topics of your blog 😊 Thanks for the teaching and I am learning☺

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• #### Joseph Nebus 8:36 pm on Sunday, 22 March, 2015 Permalink | Reply

Thank you so. I’m glad you’re enjoying. I do want to share the fun I have in mathematics.

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• #### elkement 7:18 pm on Tuesday, 24 March, 2015 Permalink | Reply

Excellent post!! Are you perhaps planning to write about entropy?

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• #### Joseph Nebus 3:40 am on Friday, 27 March, 2015 Permalink | Reply

Thank you! I am indeed sidling my way up to that.

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## Echoing “Fourier Echoes Euler”

The above tweet is from the Analysis Fact of The Day feed, which for the 5th had a neat little bit taken from Joseph Fourier’s The Analytic Theory Of Heat, published 1822. Fourier was trying to at least describe the way heat moves through objects, and along the way he developed thing called Fourier series and a field called Fourier Analysis. In this we treat functions — even ones we don’t yet know — as sinusoidal waves, overlapping and interfering with and reinforcing one another.

If we have infinitely many of these waves we can approximate … well, not every function, but surprisingly close to all the functions that might represent real-world affairs, and surprisingly near all the functions we’re interested in anyway. The advantage of representing functions as sums of sinusoidal waves is that sinusoidal waves are very easy to differentiate and integrate, and to add together those differentials and integrals, and that means we can turn problems that are extremely hard into problems that may be longer, but are made up of much easier parts. Since usually it’s better to do something that’s got many easy steps than it is to do something with a few hard ones, Fourier series and Fourier analysis are some of the things you get to know well as you become a mathematician.

The “Fourier Echoes Euler” page linked here shows simply one nice, sweet result that Fourier proved in that major work. It demonstrates what you get if, for absolutely any real number x, you add together $\cos\left(x\right) - \frac12 \cos\left(2x\right) + \frac13 \cos\left(3x\right) - \frac14 \cos\left(4x\right) + \frac15 \cos\left(5x\right) - \cdots$ et cetera. There’s one step in it — “integration by parts” — that you’ll have to remember from freshman calculus, or maybe I’ll get around to explaining that someday, but I would expect most folks reading this far could follow this neat result.

• #### howardat58 12:10 am on Thursday, 6 November, 2014 Permalink | Reply

Not too good ! The marauder link didn’t work.

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• #### Joseph Nebus 5:31 am on Thursday, 6 November, 2014 Permalink | Reply

It doesn’t? I’m surprised, and sorry for the trouble.

Are you referring to the link in the embedded tweet, or to the one that’s in my final paragraph? Both seem to work for me but goodness knows how WordPress shows things differently to me-as-blog-author than it does to other people.

If the raw URL helps any http://www.mathmarauder.com/archives/227 should be it.

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• #### elkement 7:41 pm on Wednesday, 12 November, 2014 Permalink | Reply

Great coincidence – I have just discovered that Excel can do Fourier transforms. I tried to find some hidden periodicities in the daily average ambient temperature, but the FFT has just a single peak, corresponding to 365 days :-)

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• #### Joseph Nebus 2:58 am on Thursday, 13 November, 2014 Permalink | Reply

Oh, nice; I didn’t know Excel did that.

I suppose it’s fair enough to have a strong peak at about 365 days. Next project: what does hourly temperature data tell us?

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## My Math Blog Statistics, October 2014

So now let me go over the mathematics blog statistics for October. I’ll get to listing countries; people like that.

It was a good month in terms of getting people to read: total number of pages viewed was 625, up from 558, and this is the fourth-highest month on record. The number of unique visitors was up too, from 286 in September to 323 in October, and that’s the third-highest since WordPress started giving me those statistics. The views per visitor barely changed, going from 1.95 to 1.93, which I’m comfortable supposing is a statistical tie. I reached 18,507 total page views by the end of October, and maybe I’ll reach that nice round-ish 19,000 by the end of November.

The countries sending me the most visitors were the usual set: the United States with 393, the United Kingdom with 35, and Austria with 23. Curiously, Argentina sent me 20 readers, while Canada plummeted down to a mere nine. Did I say something wrong, up there? On the bright side my Indian readership has grown to nine, which is the kind of trend I like. Sending just a single reader this past month were Albania, Brazil, Denmark, Estonia, Finland, Indonesia, Japan, the Netherlands, Nicaragua, Norway, Poland, Saint Kitts and Nevis, Serbia, Spain, Sweden, Taiwan, Turkey, and the United Arab Emirates. Brazil, Estonia, Finland, the Netherlands, and Sweden were single-reader countries last month, and Finland and Sweden also the month before. I feel embarrassed by the poor growth in my Scandinavian readership, but at least it isn’t dwindling.

The most popular posts in October got a little bit away from the comics posts; the ones most often read were:

There weren’t any really great bits of search term poetry this month, but there were still some evocative queries that brought people to me, among them:

• where did negative numbers come from
• show me how to make a comic stip for rationalnumbers
• desert island logarithm
• herb jamaal math ludwig
• in the figure shown below, Δabc and Δdec are right triangles. if de = 6, ab = 20, and be = 21, what is the area of Δdec?
• origin is the gateway to your entire gaming universe.

That “origin is the gateway” thing has come up before. I stil don’t know what it means. I’m a little scared by it.

• #### elkement 10:58 pm on Saturday, 8 November, 2014 Permalink | Reply

Do you also have some clearly distinguishable all-time-high favorites? On my blog there are a few overly popular articles – and I have no idea what is so special about those…

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• #### Joseph Nebus 9:35 pm on Sunday, 9 November, 2014 Permalink | Reply

I haven’t looked at all-time highs, but will when I have the chance. I suspect that my series of posts about trapezoids would be, since one or more of them turns up every time I look at the popular articles of the month, though. The trapezoids thing must have hit something that a lot of people are searching the web for.

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## Calculus without limits 5: log and exp

I’ve been on a bit of a logarithms kick lately, and I should say I’m not the only one. HowardAt58 has had a good number of articles about it, too, and I wanted to point some out to you. In this particular reblogging he brings a bit of calculus to show why the logarithm of the product of two numbere has to be the sum of the logarithms of the two separate numbers, in a way that’s more rigorous (if you’re comfortable with freshman calculus) than just writing down a couple examples along the lines of how 102 times 103 is equal to 105. (I won’t argue that having a couple specific examples might be better at communicating the point, but there’s a difference between believing something is so and being able to prove that it’s true.)

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The derivative of the log function can be investigated informally, as log(x) is seen as the inverse of the exponential function, written here as exp(x). The exponential function appears naturally from numbers raised to varying powers, but formal definitions of the exponential function are difficult to achieve. For example, what exactly is the meaning of exp(pi) or exp(root(2)).
So we look at the log function:-

View original post

• #### howardat58 8:41 pm on Thursday, 23 October, 2014 Permalink | Reply

Hey thanks, Joseph.
I quite agree with your comment about examples. Yes, examples first, then you know why you are trying to prove something generally.

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• #### Joseph Nebus 4:13 am on Wednesday, 29 October, 2014 Permalink | Reply

Thank you. I’ve wondered sometimes if the whole trick to teaching mathematics is figuring out just enough examples to set up the generalization, and then a couple examples to show why the generalization works. It’s surely got to be more than that, though.

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## How Richard Feynman Got From The Square Root of 2 to e

I wanted to bring to greater prominence something that might have got lost in comments. Elke Stangl, author of the Theory And Practice Of Trying To Combine Just Anything blog, noticed that among the Richard Feynman Lectures on Physics, and available online, is his derivation of how to discover e — the base of natural logarithms — from playing around.

e is an important number, certainly, but it’s tricky to explain why it’s important; it hasn’t got a catchy definition like pi has, and even the description that most efficiently says why it’s interesting (“the base of the natural logarithm”) sounds perilously close to technobabble. As an explanation for why e should be interesting Feynman’s text isn’t economical — I make it out as something around two thousand words — but it’s a really good explanation since it starts from a good starting point.

That point is: it’s easy to understand what you mean by raising a number, say 10, to a positive integer: 104, for example, is four tens multiplied together. And it doesn’t take much work to extend that to negative numbers: 10-4 is one divided by the product of four tens multiplied together. Fractions aren’t too bad either: 101/2 would be the number which, multiplied by itself, gives you 10. 103/2 would be 101/2 times 101/2 times 101/2; or if you think this is easier (it might be!), the number which, multiplied by itself, gives you 103. But what about the number $10^{\sqrt{2}}$? And if you can work that out, what about the number $10^{\pi}$?

There’s a pretty good, natural way to go about writing that and as Feynman shows you find there’s something special about some particular number pretty close to 2.71828 by doing so.

• #### elkement 7:30 pm on Friday, 17 October, 2014 Permalink | Reply

So there is not an easy way to boil down this section to a very short post, is it? I suppose you would need to resort to Taylor’s expansions, I guess? But Feynman tried to do without explaining too much theoretical concepts upfront – so that’s probably why it takes more lines and one complete example…

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• #### howardat58 7:53 pm on Friday, 17 October, 2014 Permalink | Reply

Taylor series do rather depend on calculus and the fact that d/dx(exp(x)) = exp(x), and he wanted to do it all without calculus.
You can define e as the solution to the equation
ln(x)=1, and with a proper definition of the natural log ( which needs the ideas of calculus) this will work.
Check my recent post on this:

http://howardat58.wordpress.com/

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• #### Joseph Nebus 5:54 pm on Saturday, 18 October, 2014 Permalink | Reply

I don’t know that the section couldn’t be boiled down to something short, actually; I didn’t think to try. Probably it would be possible to get to the conclusion more quickly, but I think at the cost of giving up Feynman’s fairly clear intention to bring the reader there by a series of leading investigatory questions, of getting there the playful way.

It’s a good writing exercise to consider, though, and I might give it a try.

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• #### howardat58 7:55 pm on Friday, 17 October, 2014 Permalink | Reply

Joseph, I read the log bit and the complex number bit. The latter is excellent and should be plastered on the wall of every high school math classroom. Thanks.

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• #### Joseph Nebus 6:01 pm on Saturday, 18 October, 2014 Permalink | Reply

Oh, you’re quite welcome. Glad you could draw something from it.

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## 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.

• #### howardat58 12:54 am on Thursday, 4 September, 2014 Permalink | Reply

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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• #### Joseph Nebus 12:14 am on Friday, 5 September, 2014 Permalink | Reply

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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• #### mommycookforme 2:18 pm on Thursday, 4 September, 2014 Permalink | Reply

Wow! A mathematical genius! Thanks for sharing us, have a wonderful day!:)

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• #### Joseph Nebus 12:11 am on Friday, 5 September, 2014 Permalink | Reply

Aw, hardly a genius, but thank you for thinking so.

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## 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.

• #### Boxing Pythagoras 6:33 pm on Sunday, 24 August, 2014 Permalink | Reply

I definitely need this book, now. Computers have been so naturally ingrained in our lives that people often forget how incredulous the concept once was.

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• #### Joseph Nebus 8:40 pm on Tuesday, 26 August, 2014 Permalink | Reply

Oh, yes, people do forget how fantastical computers were until quite recently. This book goes into a lot of the hardware, though, which I think is great. There’s something wonderful in how a piece of metal can be made to remember a fact, and it’s not really explained anymore in introductions to computers even though that’s become much more standard than memory had been in the 1940s.

I’d picked up the book from the university library. I haven’t checked how available and expensive it is as a used book.

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• #### chrisbrakeshow 3:04 am on Tuesday, 26 August, 2014 Permalink | Reply

This is some pretty heavy stuff, man. Thanks for digging deep!

-john

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• #### Joseph Nebus 8:41 pm on Tuesday, 26 August, 2014 Permalink | Reply

I’m glad you like. There’ll be a follow-up soon.

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• #### chrisbrakeshow 6:25 am on Wednesday, 27 August, 2014 Permalink | Reply

Awesome man, I will keep my eyes peeled and/or glued to the WP Reader!

-john

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• #### howardat58 3:43 am on Wednesday, 27 August, 2014 Permalink | Reply

I am a bit stuck with the text only in the comment section but here goes:
It started with logs-like-products so I wrote the chord slope function for log as
(log(x + x*h) – log(x)/(x*h)
which quickly and to my surprise became
(1/x)*log(1+h)/h
This provided a rationale for the formal definition of log(x) as
integral(1/t) from t=1 to t=x
I then thought to try the standard “add up the rectangles” approach, but with unequal widths, in a geometric progression.
So for 5 intervals and k^5=x the interval points were 1, k, k^2, k^3, k^4 and k^5 (=x)
The sum of the rectangles came to 5*(1 – 1/k) which eventually, via the trapezoidal rule, gave the n th estimate of log(x) as

n(root(x,n) – 1/root(x,n))

where root(x,n) is the nth root of x.

I tried it out with n as a power of 2, so square rooting is the only messy bit, and with n=2^10 I got log(e) = 0.9999…..
That is 4 dp in ten steps
Not brilliant but it works, and I have never seen anything like this before.

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• #### Joseph Nebus 2:22 am on Thursday, 28 August, 2014 Permalink | Reply

I’m still working out to my satisfaction the algebra behind this (in particular the formula n(root(x, n) – 1/root(x, n)) keeps coming out about twice the natural logarithm and I’m not sure just where the two comes in, but have suspicions), but I do believe you’re right about the basic approach. I believe that it’s structurally similar to the “method of exhaustion” that the Ancient Greeks would use to work out the areas underneath a curve, long before there was a calculus to work out these problems. In any case it’s a good scheme.

Of course, the tradeoff for this is that you need to have the n-th root of the number whose logarithm you want. This might not be too bad, especially if you decide to use an n that’s a power of two, though.

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• #### howardat58 5:01 pm on Thursday, 28 August, 2014 Permalink | Reply

Transcription error !
The divisor of 2 was scribbled so small that I missed it.
Yes, it should be divided by 2.
More on a separate post, the lines are getting shorter.

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• #### Joseph Nebus 2:44 am on Friday, 29 August, 2014 Permalink | Reply

I’m glad to have that sorted out.

If you’d like, I’d be happy to set your work as a guest post on the main page. WordPress even allows the use of basic LaTeX commands … certainly in main posts; I’m not sure how it does in comments. (I’m afraid to try, given how hard it can be to get right the first time and that there’s no preview and editing of comments that I’ve found.)

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• #### howardat58 3:26 am on Friday, 29 August, 2014 Permalink | Reply

I would like that. Thankyou.
What I use is a very simple program called mathedit , which lets you put the stuff in picture form (media) and I figured out how to modify the display size to fill the width. Have a look at my latest post to see the effect.
If I save any pics to my blogsite you can access them, put them in a post and so on using the url
I was about to start on this, with explanation, anyway, and I have figured out a way of improving the trapezium rule method dramatically.
If you email me at howard_at_58@yahoo.co.uk I can send you stuff directly to check out.
Gracias

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• #### howardat58 3:32 am on Friday, 29 August, 2014 Permalink | Reply

I am putting this as a separate reply as it might then be readable!!

I would like that. Thankyou.
What I use is a very simple program called mathedit , which lets you put the stuff in picture form (media) and I figured out how to modify the display size to fill the width. Have a look at my latest post to see the effect.
If I save any pics to my blogsite you can access them, put them in a post and so on using the url
I was about to start on this, with explanation, anyway, and I have figured out a way of improving the trapezium rule method dramatically.
If you email me at howard_at_58@yahoo.co.uk I can send you stuff directly to check out.
Gracias

Like

• #### howardat58 3:33 am on Friday, 29 August, 2014 Permalink | Reply

That didn’t work!!!!!!!!!!!!!!!!! trying again

I would like that. Thankyou.
What I use is a very simple program called mathedit , which lets you put the stuff in picture form (media) and I figured out how to modify the display size to fill the width. Have a look at my latest post to see the effect.
If I save any pics to my blogsite you can access them, put them in a post and so on using the url
I was about to start on this, with explanation, anyway, and I have figured out a way of improving the trapezium rule method dramatically.
If you email me at howard_at_58@yahoo.co.uk I can send you stuff directly to check out.
Gracias

Like

## Writing About E (Not By Me)

It’s tricky to write about $e$. That is, it’s not a difficult thing to write about, but it’s hard to find the audience for this number. It’s quite important, mathematically, but it hasn’t got an easy-to-understand definition like pi’s “the circumference of a circle divided by its diameter”. E’s most concise definition, I guess, is “the base of the natural logarithm”, which as an explanation to someone who hasn’t done much mathematics is only marginally more enlightening than slapping him with a slice of cold pizza. And it hasn’t got the sort of renown of something like the golden ratio which makes the number sound familiar and even welcoming.

Still, the Mean Green Math blog (“Explaining the whys of mathematics”) has been running a series of essays explaining $e$, by looking at different definitions of the number. The most recent of this has been the twelfth in the series, and they seem to be arranged in chronological order under the category of Algebra II topics, and under the tag of “E” essays, although I can’t promise how long it’ll be before you have to flip through so many “older” page links on the category and tag pages that it’s harder to find that way. If I see a master page collecting all the Definitions Of E essays into one guide I’ll post that.

## Listening To Vermilion Sands

My Beloved is reading J G Ballard’s Vermillion Sands; early in one of the book’s stories is a character wondering if an odd sound comes from one of the musical … let’s call it instruments, one with a 24-octave range. We both thought, wow, that’s a lot of range. Is it a range any instrument could have?

As we weren’t near our computers this turned into a mental arithmetic problem. It’s solvable in principle because, if you know the frequency of one note, then you know the frequency of its counterpart one octave higher (it’s double that), and one octave lower (it’s half that). It’s not solvable, at this point, because we don’t have any information about what the range is supposed to be. So here’s roughly how we worked it out.

The note A above middle C is 440 Hertz, or at least you can use that for tuning ever since the International Standards Organization set that as a tuning standard in 1953. (As with any basically arbitrary standard this particular choice is debatable, although, goodness but this page advocating a 432 Hertz standard for A doesn’t do itself any favors by noting that “440 Hz is the unnatural standard tuning frequency, removed from the symmetry of sacred vibrations and overtones that has declared war on the subconscious mind of Western Man” and, yes, Nikola Tesla and Joseph Goebbels turn up in the article because you might otherwise imagine taking it seriously.) Anyway, it doesn’t matter; 440 is just convenient as it’s a number definitely in hearing range.

So I’m adding the assumption that 440 Hz is probably in the instrument’s range. And I’ll work on the assumption that it’s right in the middle of the range, that is, that we should be able to go down twelve octaves and up twelve octaves, and see if that assumption leads me to any problems. And now I’ve got the problem defined well enough to answer: is 440 divided by two to the twelfth power in human hearing range? Is 440 times two to the twelfth power in range?

I’m not dividing 440 by two a dozen times; I might manage that with pencil and paper but not in my head. But I also don’t need to. Two raised to the tenth power is pretty close to 1,000, as anyone who’s noticed that the common logarithm of two is 0.3 could work out. Remembering a couple approximations like that are key to doing any kind of real mental arithmetic; it’s all about turning the problem you’re interested in into one you can do without writing it down.

Another key to this sort of mental arithmetic is noticing that two to the 12th power is equal to two to the second power (that is, four) times two to the tenth power (approximately 1,000). In algebra class this was fed to you as something like “ax + y = (ax)(a y)”, and it’s the trick that makes logarithms a concept that works.

Getting back to the question, 440 divided by two twelve times over is going to be about 440 divided by 4,000, which is going to be close enough to one-tenth Hertz. There’s no point working it out to any more exact answer, since this is definitely below the range of human hearing; I think the lower bound is usually around ten to thirty Hertz.

Well, no matter; maybe the range of the instrument starts higher up and keeps on going. To see if there’s any room, what’s the frequency of a note twelve octaves above the 440-Hertz A?

That’s going to be 440 Hertz times 4,000, which to make it simpler I’ll say is something more than 400 times 4000. The four times four is easy, and there’s five zeroes in there, so, that suggests an upper range on the high side of 1,600,000 Hertz. Again, I’m not positive the upper limit of human hearing but I’m confident it’s not more than about 30,000 Hertz, and I leave space below for people who know what it is exactly to say. There’s just no fitting 24 octaves into the human hearing range.

So! Was Ballard just putting stuff into his science fiction story without checking whether the numbers make that plausible, if you can imagine a science fiction author doing such a thing?

It’s conceivable. It’s also possible Ballard was trying to establish the character was a pretentious audiophile snob who imagines himself capable of hearing things that no, in fact, can’t be discerned. However, based on the setting … the instruments producing music in this story (and other stories in the book), set in the far future, include singing plants and musical arachnids and other things that indicate not just technology but biology has changed rather considerably. If it’s possible to engineer a lobster that can sing over a 24 octave range, it’s presumably possible to engineer a person who can listen to it.

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