My 2018 Mathematics A To Z: e


I’m back to requests! Today’s comes from commenter Dina Yagodich. I don’t know whether Yagodich has a web site, YouTube channel, or other mathematics-discussion site, but am happy to pass along word if I hear of one.

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e.

Let me start by explaining integral calculus in two paragraphs. One of the things done in it is finding a `definite integral’. This is itself a function. The definite integral has as its domain the combination of a function, plus some boundaries, and its range is numbers. Real numbers, if nobody tells you otherwise. Complex-valued numbers, if someone says it’s complex-valued numbers. Yes, it could have some other range. But if someone wants you to do that they’re obliged to set warning flares around the problem and precede and follow it with flag-bearers. And you get at least double pay for the hazardous work. The function that gets definite-integrated has its own domain and range. The boundaries of the definite integral have to be within the domain of the integrated function.

For real-valued functions this definite integral has a great physical interpretation. A real-valued function means the domain and range are both real numbers. You see a lot of these. Call the function ‘f’, please. Call its independent variable ‘x’ and its dependent variable ‘y’. Using Euclidean coordinates, or as normal people call it “graph paper”, draw the points that make true the equation “y = f(x)”. Then draw in the x-axis, that is, the points where “y = 0”. The boundaries of the definite integral are going to be two values of ‘x’, a lower and an upper bound. Call that lower bound ‘a’ and the upper bound ‘b’. And heck, call that a “left boundary” and a “right boundary”, because … I mean, look at them. Draw the vertical line at “x = a” and the vertical line at “x = b”. If ‘f(x)’ is always a positive number, then there’s a shape bounded below by “y = 0”, on the left by “x = a”, on the right by “x = b”, and above by “y = f(x)”. And the definite integral is the area of that enclosed space. If ‘f(x)’ is sometimes zero, then there’s several segments, but their combined area is the definite integral. If ‘f(x)’ is sometimes below zero, then there’s several segments. The definite integral is the sum of the areas of parts above “y = 0” minus the area of the parts below “y = 0”.

(Why say “left boundary” instead of “lower boundary”? Taste, pretty much. But I look at the words “lower boundary” and think about the lower edge, that is, the line where “y = 0” here. And “upper boundary” makes sense as a way to describe the curve where “y = f(x)” as well as “x = b”. I’m confusing enough without making the simple stuff ambiguous.)

Don’t try to pass your thesis defense on this alone. But it’s what you need to understand ‘e’. Start out with the function ‘f’, which has domain of the positive real numbers and range of the positive real numbers. For every ‘x’ in the domain, ‘f(x)’ is the reciprocal, one divided by x. This is a shape you probably know well. It’s a hyperbola. Its asymptotes are the x-axis and the y-axis. It’s a nice gentle curve. Its plot passes through such famous points as (1, 1), (2, 1/2), (1/3, 3), and pairs like that. (10, 1/10) and (1/100, 100) too. ‘f(x)’ is always positive on this domain. Use as left boundary the line “x = 1”. And then — let’s think about different right boundaries.

If the right boundary is close to the left boundary, then this area is tiny. If it’s at, like, “x = 1.1” then the area can’t be more than 0.1. (It’s less than that. If you don’t see why that’s so, fit a rectangle of height 1 and width 0.1 around this curve and these boundaries. See?) But if the right boundary is farther out, this area is more. It’s getting bigger if the right boundary is “x = 2” or “x = 3”. It can get bigger yet. Give me any positive number you like. I can find a right boundary so the area inside this is bigger than your number.

Is there a right boundary where the area is exactly 1? … Well, it’s hard to see how there couldn’t be. If a quantity (“area between x = 1 and x = b”) changes from less than one to greater than one, it’s got to pass through 1, right? … Yes, it does, provided some technical points are true, and in this case they are. So that’s nice.

And there is. It’s a number (settle down, I see you quivering with excitement back there, waiting for me to unveil this) a slight bit more than 2.718. It’s a neat number. Carry it out a couple more digits and it turns out to be 2.718281828. So it looks like a great candidate to memorize. It’s not. It’s an irrational number. The digits go off without repeating or falling into obvious patterns after that. It’s a transcendental number, which has to do with polynomials. Nobody knows whether it’s a normal number, because remember, a normal number is just any real number that you never heard of. To be a normal number, every finite string of digits has to appear in the decimal expansion, just as often as every other string of digits of the same length. We can show by clever counting arguments that roughly every number is normal. Trick is it’s hard to show that any particular number is.

So let me do another definite integral. Set the left boundary to this “x = 2.718281828(etc)”. Set the right boundary a little more than that. The enclosed area is less than 1. Set the right boundary way off to the right. The enclosed area is more than 1. What right boundary makes the enclosed area ‘1’ again? … Well, that will be at about “x = 7.389”. That is, at the square of 2.718281828(etc).

Repeat this. Set the left boundary at “x = (2.718281828etc)2”. Where does the right boundary have to be so the enclosed area is 1? … Did you guess “x = (2.718281828etc)3”? Yeah, of course. You know my rhetorical tricks. What do you want to guess the area is between, oh, “x = (2.718281828etc)3” and “x = (2.718281828etc)5”? (Notice I put a ‘5’ in the superscript there.)

Now, relationships like this will happen with other functions, and with other left- and right-boundaries. But if you want it to work with a function whose rule is as simple as “f(x) = 1 / x”, and areas of 1, then you’re going to end up noticing this 2.718281828(etc). It stands out. It’s worthy of a name.

Which is why this 2.718281828(etc) is a number you’ve heard of. It’s named ‘e’. Leonhard Euler, whom you will remember as having written or proved the fundamental theorem for every area of mathematics ever, gave it that name. He used it first when writing for his own work. Then (in November 1731) in a letter to Christian Goldbach. Finally (in 1763) in his textbook Mechanica. Everyone went along with him because Euler knew how to write about stuff, and how to pick symbols that worked for stuff.

Once you know ‘e’ is there, you start to see it everywhere. In Western mathematics it seems to have been first noticed by Jacob (I) Bernoulli, who noticed it in toy compound interest problems. (Given this, I’d imagine it has to have been noticed by the people who did finance. But I am ignorant of the history of financial calculations. Writers of the kind of pop-mathematics history I read don’t notice them either.) Bernoulli and Pierre Raymond de Montmort noticed the reciprocal of ‘e’ turning up in what we’ve come to call the ‘hat check problem’. A large number of guests all check one hat each. The person checking hats has no idea who anybody is. What is the chance that nobody gets their correct hat back? … That chance is the reciprocal of ‘e’. The number’s about 0.368. In a connected but not identical problem, suppose something has one chance in some number ‘N’ of happening each attempt. And it’s given ‘N’ attempts given for it to happen. What’s the chance that it doesn’t happen? The bigger ‘N’ gets, the closer the chance it doesn’t happen gets to the reciprocal of ‘e’.

It comes up in peculiar ways. In high school or freshman calculus you see it defined as what you get if you take \left(1 + \frac{1}{x}\right)^x for ever-larger real numbers ‘x’. (This is the toy-compound-interest problem Bernoulli found.) But you can find the number other ways. You can calculate it — if you have the stamina — by working out the value of

1 + 1 + \frac12\left( 1 + \frac13\left( 1 + \frac14\left( 1 + \frac15\left( 1 + \cdots \right)\right)\right)\right)

There’s a simpler way to write that. There always is. Take all the nonnegative whole numbers — 0, 1, 2, 3, 4, and so on. Take their factorials. That’s 1, 1, 2, 6, 24, and so on. Take the reciprocals of all those. That’s … 1, 1, one-half, one-sixth, one-twenty-fourth, and so on. Add them all together. That’s ‘e’.

This ‘e’ turns up all the time. Any system whose rate of growth depends on its current value has an ‘e’ lurking in its description. That’s true if it declines, too, as long as the decline depends on its current value. It gets stranger. Cross ‘e’ with complex-valued numbers and you get, not just growth or decay, but oscillations. And many problems that are hard to solve to start with become doable, even simple, if you rewrite them as growths and decays and oscillations. Through ‘e’ problems too hard to do become problems of polynomials, or even simpler things.

Simple problems become that too. That property about the area underneath “f(x) = 1/x” between “x = 1” and “x = b” makes ‘e’ such a natural base for logarithms that we call it the base for natural logarithms. Logarithms let us replace multiplication with addition, and division with subtraction, easier work. They change exponentiation problems to multiplication, again easier. It’s a strange touch, a wondrous one.

There are some numbers interesting enough to attract books about them. π, obviously. 0. The base of imaginary numbers, \imath , has a couple. I only know one pop-mathematics treatment of ‘e’, Eli Maor’s e: The Story Of A Number. I believe there’s room for more.


Oh, one little remarkable thing that’s of no use whatsoever. Mathworld’s page about approximations to ‘e’ mentions this. Work out, if you can coax your calculator into letting you do this, the number:

\left(1 + 9^{-(4^{(42)})}\right)^{\left(3^{(2^{85})}\right)}

You know, the way anyone’s calculator will let you raise 2 to the 85th power. And then raise 3 to whatever number that is. Anyway. The digits of this will agree with the digits of ‘e’ for the first 18,457,734,525,360,901,453,873,570 decimal digits. One Richard Sabey found that, by what means I do not know, in 2004. The page linked there includes a bunch of other, no less amazing, approximations to numbers like ‘e’ and π and the Euler-Mascheroni Constant.

The End 2016 Mathematics A To Z: Riemann Sum


I see for the other A To Z I did this year I did something else named for Riemann. So I did. Bernhard Riemann did a lot of work that’s essential to how we see mathematics today. We name all kinds of things for him, and correctly so. Here’s one of his many essential bits of work.

Riemann Sum.

The Riemann Sum is a thing we learn in Intro to Calculus. It’s essential in getting us to definite integrals. We’re introduced to it in functions of a single variable. The functions have a domain that’s an interval of real numbers and a range that’s somewhere in the real numbers. The Riemann Sum — and from it, the integral — is a real number.

We get this number by following a couple steps. The first is we chop the interval up into a bunch of smaller intervals. That chopping-up we call a “partition” because it’s another of those times mathematicians use a word the way people might use the same word. From each one of those chopped-up pieces we pick a representative point. Now with each piece evaluate what the function is for that representative point. Multiply that by the width of the partition it was in. Then take those products for each of those pieces and add them all together. If you’ve done it right you’ve got a number.

You need a couple pieces in place to have “the” Riemann Sum for something. You need a function, which is fair enough. And you need a partitioning of the interval. And you need some representative point for each of the partitions. Change any of them — function, partition, or point — and you may change the sum you get. You expect that for changing the function. Changing the partition? That’s less obvious. But draw some wiggly curvy function on a sheet of paper. Draw a couple of partitions of the horizontal axis. (You’ll probably want to use different colors for different partitions.) That should coax you into it. And you’d probably take it on my word that different representative points give you different sums.

Very different? It’s possible. There’s nothing stopping it from happening. But if the results aren’t very different then we might just have an integrable function. That’s a function that gives us the same Riemann Sum no matter how we pick representative points, as long as we pick partitions that get finer and finer enough. We measure how fine a partition is by how big the widest chopped-up piece is. To be integrable the Riemann Sum for a function has to get to the same number whenever the partition’s size gets small enough and however we pick points inside. We get the lovely quiet paradox in which we add together infinitely many things, each of them infinitesimally tiny, and get a regular old number out of all that work.

We use the Riemann Sum for what we call numerical quadrature. That’s working out integrals on the computer. Or calculator. Or by hand. When we do it by evaluating numbers instead of using analysis. It’s very easy to program. And we can do some tricks based on the Riemann Sum to make the numerical estimate a closer match to the actual integral.

And we use the Riemann Sum to learn how the Riemann Integral works. It’s a blessedly straightforward thing. It appeals to intuition well. It lets us draw all sorts of curves with rectangular boxes overlaying them. It’s so easy to work out the area of a rectangular box. We can imagine adding up these areas without being confused.

We don’t use the Riemann Sum to actually do integrals, though. Numerical approximations to an integral, yes. For the actual integral it’s too hard to use. What makes it hard is you need to evaluate this for every possible partition and every possible pick of representative points. In grad school my analysis professor worked through — once — using this to integrate the number 1. This is the easiest possible thing to integrate and it was barely manageable. He gave a good try at integrating the function ‘f(x) = x’ but admitted he couldn’t do it. None of us could.

When you see the Riemann Sum in an Introduction to Calculus course you see it in simplified form. You get partitions that are very easy to work with. Like, you break the interval up into some number of equally-sized chunks. You get representative points that follow one of a couple good choices. The left end of the partition. The right end of the partition. The middle of the partition.

That’s fine, numerically. If the function is integrable it doesn’t matter what partition or representative points we pick. And it’s fine for learning about whether functions are integrable. If it matters whether you pick left or middle or right ends of the partition then the function isn’t integrable. The instructor can give functions that break integrability based on a given partition or endpoint choice or whatever.

But that isn’t every possible partition and every possible pick of representative points. I suppose it’s possible to work all that out for a couple of really, really simple functions. But it’s so much work. We’re better off using the Riemann Sum to get to formulas about integrals that don’t depend on actually using the Riemann Sum.

So that is the curious position the Riemann Sum has. It is a fundament of integral calculus. It is the way we first define the definite integral. We rely on it to learn what definite integrals are like. We use it all the time numerically. We never use it analytically. It’s too hard. I hope you appreciate the strange beauty of that.