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Sat, 01 Aug 2020

Here's another recent Math Stack Exchange answer I'm pleased with.

I know this question has been asked many times and there is good information out there which has clarified a lot for me but I still do not understand how the addition and multiplication tables for !!GF(4)!! is constructed?

I've seen [links] but none explicity explain the construction and I'm too new to be told "its an extension of !!GF(2)!!"

The only “reasonable” answer here is “get an undergraduate abstract algebra text and read the chapter on finite fields”. Because come on, you can't expect some random stranger to appear and write up a detailed but short explanation at your exact level of knowledge.

But sometimes Internet Magic Lightning strikes and that's what you do get! And OP set themselves up to be struck by magic lightning, because you can't get a detailed but short explanation at your exact level of knowledge if you don't provide a detailed but short explanation of your exact level of knowledge — and this person did just that. They understand finite fields of prime order, but not how to construct the extension fields. No problem, I can explain that!

I had special fun writing this answer because I just love constructing extensions of finite fields. (Previously: [1] [2])

For any given !!n!!, there is at most one field with !!n!! elements: only one, if !!n!! is a power of a prime number (!!2, 3, 2^2, 5, 7, 2^3, 3^2, 11, 13, \ldots!!) and none otherwise (!!6, 10, 12, 14\ldots!!). This field with !!n!! elements is written as !!\Bbb F_n!! or as !!GF(n)!!.

Suppose we want to construct !!\Bbb F_n!! where !!n=p^k!!. When !!k=1!!, this is easy-peasy: take the !!n!! elements to be the integers !!0, 1, 2\ldots p-1!!, and the addition and multiplication are done modulo !!n!!.

When !!k>1!! it is more interesting. One possible construction goes like this:

The elements of !!\Bbb F_{p^k}!! are the polynomials $$a_{k-1}x^{k-1} + a_{k-2}x^{k-2} + \ldots + a_1x+a_0$$ where the coefficients !!a_i!! are elements of !!\Bbb F_p!!. That is, the coefficients are just integers in !!{0, 1, \ldots p-1}!!, but with the understanding that the addition and multiplication will be done modulo !!p!!. Note that there are !!p^k!! of these polynomials in total.
Addition of polynomials is done exactly as usual: combine like terms, but remember that the the coefficients are added modulo !!p!! because they are elements of !!\Bbb F_p!!.
Multiplication is more interesting:

a. Pick an irreducible polynomial !!P!! of degree !!k!!. “Irreducible” means that it does not factor into a product of smaller polynomials. How to actually locate an irreducible polynomial is an interesting question; here we will mostly ignore it.

b. To multiply two elements, multiply them normally, remembering that the coefficients are in !!\Bbb F_p!!. Divide the product by !!P!! and keep the remainder. Since !!P!! has degree !!k!!, the remainder must have degree at most !!k-1!!, and this is your answer.

Now we will see an example: we will construct !!\Bbb F_{2^2}!!. Here !!k=2!! and !!p=2!!. The elements will be polynomials of degree at most 1, with coefficients in !!\Bbb F_2!!. There are four elements: !!0x+0, 0x+1, 1x+0, !! and !!1x+1!!. As usual we will write these as !!0, 1, x, x+1!!. This will not be misleading.

Addition is straightforward: combine like terms, remembering that !!1+1=0!! because the coefficients are in !!\Bbb F_2!!:

$$\begin{array}{c|cccc} + & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 1 & x & x+1 \\ 1 & 1 & 0 & x+1 & x \\ x & x & x+1 & 0 & 1 \\ x+1 & x+1 & x & 1 & 0 \end{array} $$

The multiplication as always is more interesting. We need to find an irreducible polynomial !!P!!. It so happens that !!P=x^2+x+1!! is the only one that works. (If you didn't know this, you could find out easily: a reducible polynomial of degree 2 factors into two linear factors. So the reducible polynomials are !!x^2, x·(x+1) = x^2+x!!, and !!(x+1)^2 = x^2+2x+1 = x^2+1!!. That leaves only !!x^2+x+1!!.)

To multiply two polynomials, we multiply them normally, then divide by !!x^2+x+1!! and keep the remainder. For example, what is !!(x+1)(x+1)!!? It's !!x^2+2x+1 = x^2 + 1!!. There is a theorem from elementary algebra (the “division theorem”) that we can find a unique quotient !!Q!! and remainder !!R!!, with the degree of !!R!! less than 2, such that !!PQ+R = x^2+1!!. In this case, !!Q=1, R=x!! works. (You should check this.) Since !!R=x!! this is our answer: !!(x+1)(x+1) = x!!.

Let's try !!x·x = x^2!!. We want !!PQ+R = x^2!!, and it happens that !!Q=1, R=x+1!! works. So !!x·x = x+1!!.

I strongly recommend that you calculate the multiplication table yourself. But here it is if you want to check:

$$\begin{array}{c|cccc} · & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & x & x+1 \\ x & 0 & x & x+1 & 1 \\ x+1 & 0 & x+1 & 1 & x \end{array} $$

To calculate the unique field !!\Bbb F_{2^3}!! of order 8, you let the elements be the 8 second-degree polynomials !!0, 1, x, \ldots, x^2+x, x^2+x+1!! and instead of reducing by !!x^2+x+1!!, you reduce by !!x^3+x+1!!. (Not by !!x^3+x^2+x+1!!, because that factors as !!(x^2+1)(x+1)!!.) To calculate the unique field !!\Bbb F_{3^2}!! of order 27, you start with the 27 third-degree polynomials with coefficients in !!{0,1,2}!!, and you reduce by !!x^3+2x+1!! (I think).

The special notation !!\Bbb F_p[x]!! means the ring of all polynomials with coefficients from !!\Bbb F_p!!. !!\langle P \rangle!! means the ring of all multiples of polynomial !!P!!. (A ring is a set with an addition, subtraction, and multiplication defined.)

When we write !!\Bbb F_p[x] / \langle P\rangle!! we are constructing a thing called a “quotient” structure. This is a generalization of the process that turns the ordinary integers !!\Bbb Z!! into the modular-arithmetic integers we have been calling !!\Bbb F_p!!. To construct !!\Bbb F_p!!, we start with !!\Bbb Z!! and then agree that two elements of !!\Bbb Z!! will be considered equivalent if they differ by a multiple of !!p!!.

To get !!\Bbb F_p[x] / \langle P \rangle!! we start with !!\Bbb F_p[x]!!, and then agree that elements of !!\Bbb F_p[x]!! will be considered equivalent if they differ by a multiple of !!P!!. The division theorem guarantees that of all the equivalent polynomials in a class, exactly one of them will have degree less than that of !!P!!, and that is the one we choose as a representative of its class and write into the multiplication table. This is what we are doing when we “divide by !!P!! and keep the remainder”.

A particularly important example of this construction is !!\Bbb R[x] / \langle x^2 + 1\rangle!!. That is, we take the set of polynomials with real coefficients, but we consider two polynomials equivalent if they differ by a multiple of !!x^2 + 1!!. By the division theorem, each polynomial is then equivalent to some first-degree polynomial !!ax+b!!.

Let's multiply $$(ax+b)(cx+d).$$ As usual we obtain $$acx^2 + (ad+bc)x + bd.$$ From this we can subtract !!ac(x^2 + 1)!! to obtain the equivalent first-degree polynomial $$(ad+bc) x + (bd-ac).$$

Now recall that in the complex numbers, !!(b+ai)(d + ci) = (bd-ac) + (ad+bc)i!!. We have just constructed the complex numbers,with the polynomial !!x!! playing the role of !!i!!.

[ Note to self: maybe write a separate article about what makes this a good answer, and how it is structured. ]

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Fri, 31 Jul 2020

What does it mean to expand a function “in powers of x-1”?

A recent Math Stack Excahnge post was asked to expand the function !!e^{2x}!! in powers of !!(x-1)!! and was confused about what that meant, and what the point of it was. I wrote an answer I liked, which I am reproducing here.

You asked:

I don't understand what are we doing in this whole process

which is a fair question. I didn't understand this either when I first learned it. But it's important for practical engineering reasons as well as for theoretical mathematical ones.

Before we go on, let's see that your proposal is the wrong answer to this question, because it is the correct answer, but to a different question. You suggested: $$e^{2x}\approx1+2\left(x-1\right)+2\left(x-1\right)^2+\frac{4}{3}\left(x-1\right)^3$$

Taking !!x=1!! we get !!e^2 \approx 1!!, which is just wrong, since actually !!e^2\approx 7.39!!. As a comment pointed out, the series you have above is for !!e^{2(x-1)}!!. But we wanted a series that adds up to !!e^{2x}!!.

As you know, the Maclaurin series works here:

$$e^{2x} \approx 1+2x+2x^2+\frac{4}{3}x^3$$

so why don't we just use it? Let's try !!x=1!!. We get $$e^2\approx 1 + 2 + 2 + \frac43$$

This adds to !!6+\frac13!!, but the correct answer is actually around !!7.39!! as we saw before. That is not a very accurate approximation. Maybe we need more terms? Let's try ten:

$$e^{2x} \approx 1+2x+2x^2+\frac{4}{3}x^3 + \ldots + \frac{8}{2835}x^9$$

If we do this we get !!7.3887!!, which isn't too far off. But it was a lot of work! And we find that as !!x!! gets farther away from zero, the series above gets less and less accurate. For example, take !!x=3.1!!, the formula with four terms gives us !!66.14!!, which is dead wrong. Even if we use ten terms, we get !!444.3!!, which is still way off. The right answer is actually !!492.7!!.

What do we do about this? Just add more terms? That could be a lot of work and it might not get us where we need to go. (Some Maclaurin series just stop working at all too far from zero, and no amount of terms will make them work.) Instead we use a different technique.

Expanding the Taylor series “around !!x=a!!” gets us a different series, one that works best when !!x!! is close to !!a!! instead of when !!x!! is close to zero. Your homework is to expand it around !!x=1!!, and I don't want to give away the answer, so I'll do a different example. We'll expand !!e^{2x}!! around !!x=3!!. The general formula is $$e^{2x} \approx \sum \frac{f^{(i)}(3)}{i!} (x-3)^i\tag{$\star$}\ \qquad \text{(when $x$ is close to $3$)}$$

The !!f^{(i)}(x)!! is the !!i!!'th derivative of !! e^{2x}!! , which is !!2^ie^{2x}!!, so the first few terms of the series above are:

$$\begin{eqnarray} e^{2x} & \approx& e^6 + \frac{2e^6}1 (x-3) + \frac{4e^6}{2}(x-3)^2 + \frac{8e^6}{6}(x-3)^3\\ & = & e^6\left(1+ 2(x-3) + 2(x-3)^2 + \frac34(x-3)^3\right)\\ & & \qquad \text{(when $x$ is close to $3$)} \end{eqnarray} $$

The first thing to notice here is that when !!x!! is exactly !!3!!, this series is perfectly correct; we get !!e^6 = e^6!! exactly, even when we add up only the first term, and ignore the rest. That's a kind of useless answer because we already knew that !!e^6 = e^6!!. But that's not what this series is for. The whole point of this series is to tell us how different !!e^{2x}!! is from !!e^6!! when !!x!! is close to, but not equal to !!3!!.

Let's see what it does at !!x=3.1!!. With only four terms we get $$\begin{eqnarray} e^{6.2} & \approx& e^6(1 + 2(0.1) + 2(0.1)^2 + \frac34(0.1)^3)\\ & = & e^6 \cdot 1.22075 \\ & \approx & 492.486 \end{eqnarray}$$

which is very close to the correct answer, which is !!492.7!!. And that's with only four terms. Even if we didn't know an exact value for !!e^6!!, we could find out that !!e^{6.2}!! is about !!22.075\%!! larger, with hardly any calculation.

Why did this work so well? If you look at the expression !!(\star)!! you can see: The terms of the series all have factors of the form !!(x-3)^i!!. When !!x=3.1!!, these are !!(0.1)^i!!, which becomes very small very quickly as !!i!! increases. Because the later terms of the series are very small, they don't affect the final sum, and if we leave them out, we won't mess up the answer too much. So the series works well, producing accurate results from only a few terms, when !!x!! is close to !!3!!.

But in the Maclaurin series, which is around !!x=0!!, those !!(x-3)^i!! terms are !!x^i!! terms intead, and when !!x=3.1!!, they are not small, they're very large! They get bigger as !!i!! increases, and very quickly. (The !! i! !! in the denominator wins, eventually, but that doesn't happen for many terms.) If we leave out these many large terms, we get the wrong results.

The short answer to your question is:

Maclaurin series are only good for calculating functions when !!x!! is close to !!0!!, and become inaccurate as !!x!! moves away from zero. But a Taylor series around !!a!! has its “center” near !!a!! and is most accurate when !!x!! is close to !!a!!.

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Fri, 18 Dec 2015

Math.SE report 2015-08

I only posted three answers in August, but two of them were interesting.

In why this !!\sigma\pi\sigma^{-1}!! keeps apearing in my group theory book? (cycle decomposition) the querent asked about the “conjugation” operation that keeps cropping up in group theory. Why is it important? I sympathize with this; it wasn't adequately explained when I took group theory, and I had to figure it out a long time later. Unfortunately I don't think I picked the right example to explain it, so I am going to try again now.

Consider the eight symmetries of the square. They are of five types:
1. Rotation clockwise or counterclockwise by 90°.
2. Rotation by 180°.
3. Horizontal or vertical reflection
4. Diagonal reflection
5. The trivial (identity) symmetry
What is meant when I say that a horizontal and a vertical reflection are of the same ‘type’? Informally, it is that the horizontal reflection looks just like the vertical reflection, if you turn your head ninety degrees. We can formalize this by observing that if we rotate the square 90°, then give it a horizontal flip, then rotate it back, the effect is exactly to give it a vertical flip. In notation, we might represent the horizontal flip by !!H!!, the vertical flip by !!V!!, the clockwise rotation by !!\rho!!, and the counterclockwise rotation by !!\rho^{-1}!!; then we have

$$ \rho H \rho^{-1} = V$$

and similarly

$$ \rho V \rho^{-1} = H.$$

Vertical flips do not look like diagonal flips—the diagonal flip leaves two of the corners in the same place, and the vertical flip does not—and indeed there is no analogous formula with !!H!! replaced with one of the diagonal flips. However, if !!D_1!! and !!D_2!! are the two diagonal flips, then we do have

$$ \rho D_1 \rho^{-1} = D_2.$$

In general, When !!a!! and !!b!! are two symmetries, and there is some symmetry !!x!! for which

$$xax^{-1} = b$$

we say that !!a!! is conjugate to !!b!!. One can show that conjugacy is an equivalence relation, which means that the symmetries of any object can be divided into separate “conjugacy classes” such that two symmetries are conjugate if and only if they are in the same class. For the square, the conjugacy classes are the five I listed earlier.

This conjugacy thing is important for telling when two symmetries are group-theoretically “the same”, and have the same group-theoretic properties. For example, the fact that the horizontal and vertical flips move all four vertices, while the diagonal flips do not. Another example is that a horizontal flip is self-inverse (if you do it again, it cancels itself out), but a 90° rotation is not (you have to do it four times before it cancels out.) But the horizontal flip shares all its properties with the vertical flip, because it is the same if you just turn your head.

Identifying this sameness makes certain kinds of arguments much simpler. For example, in counting squares, I wanted to count the number of ways of coloring the faces of a cube, and instead of dealing with the 24 symmetries of the cube, I only needed to deal with their 5 conjugacy classes.

The example I gave in my math.se answer was maybe less perspicuous. I considered the symmetries of a sphere, and talked about how two rotations of the sphere by 17° are conjugate, regardless of what axis one rotates around. I thought of the square at the end, and threw it in, but I wish I had started with it.
How to convert a decimal to a fraction easily? was the month's big winner. OP wanted to know how to take a decimal like !!0.3760683761!! and discover that it can be written as !!\frac{44}{117}!!. The right answer to this is of course to use continued fraction theory, but I did not want to write a long treatise on continued fractions, so I stripped down the theory to obtain an algorithm that is slower, but much easier to understand.

The algorithm is just binary search, but with a twist. If you are looking for a fraction for !!x!!, and you know !!\frac ab < x < \frac cd!!, then you construct the mediant !!\frac{a+c}{b+d}!! and compare it with !!x!!. This gives you a smaller interval in which to search for !!x!!, and the reason you use the mediant instead of using !!\frac12\left(\frac ab + \frac cd\right)!! as usual is that if you use the mediant you are guaranteed to exactly nail all the best rational approximations of !!x!!. This is the algorithm I described a few years ago in your age as a fraction, again; there the binary search proceeds down the branches of the Stern-Brocot tree to find a fraction close to !!0.368!!.

I did ask a question this month: I was looking for a simpler version of the dogbone space construction. The dogbone space is a very peculiar counterexample of general topology, originally constructed by R.H. Bing. I mentioned it here in 2007, and said, at the time:

[The paper] is on my desk, but I have not read this yet, and I may never.

I did try to read it, but I did not try very hard, and I did not understand it. So my question this month was if there was a simpler example of the same type. I did not receive an answer, just a followup comment that no, there is no such example.

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Sun, 16 Aug 2015

Math.SE report 2015-07

My overall SE posting volume was down this month, and not only did I post relatively few interesting items, I've already written a whole article about the most interesting one. So this will be a short report.

I already wrote up Building a box from smaller boxes on the blog here. But maybe I have a couple of extra remarks. First, the other guy's proposed solution is awful. It's long and complicated, which is forgivable if it had answered the question, but it doesn't. And the key point is “blah blah blah therefore code a solver which visits all configurations of the search space”. Well heck, if this post had just been one sentence that ended with “code a solver which visits all configurations of the search space” I would not have any complaints about that.

As an undergraduate I once gave a talk on this topic. One of my examples was the problem of packing 31 dominoes into a chessboard from which two squares have been deleted. There is a simple combinatorial argument why this is impossible if the two deleted squares are the same color, say if they are opposite corners: each domino must cover one square of each color. But if you don't take time to think about the combinatorial argument you could waste a lot of time on computer search learning that there is no solution in that case, and completely miss the deeper understanding that it brings you. So this has been on my mind for a long time.
I wrote a few posts this month where I thought I gave good hints. In How to scale an unit vector !!u!! in such way that !!a u\cdot u=1!! where !!a!! is a scalar I think I did a good job identifying the original author's confusion; he was conflating his original unit vector !!u!! and the scaled, leading him to write !!au\cdot u=1!!. This is sure to lead to confusion. So I led him to the point of writing !!a(bv)\cdot(bv)=1!! and let him take it from there. The other proposed solution is much more rote and mechanical. (“Divide this by that…”)

In Find numbers !!\overline{abcd}!! so that !!\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}!! the OP got stuck partway through and I specifically addressed the stuckness; other people solved the problem from the beginning. I think that's the way to go, if the original proposal was never going to work, especially if you stop and say why it was never going to work, but this time OP's original suggestion was perfectly good and she just didn't know how to get to the next step. By the way, the notation !!\overline{abcd}!! here means the number !!1000a+100b+10c+d!!.

In Help finding the limit of this series !!\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots!! it would have been really easy to say “use the formula” or to analyze the series de novo, but I think I almost hit the nail on the head here: it's just like !!1+\frac12 + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots!!, which I bet OP already knows, except a little different. But I pointed out the wrong difference: I observed that the first sequence is one-fourth the second one (which it is) but it would have been simpler to observe that it's just the second one without the !!1+\frac12!!. I had to review it just now to give the simpler explanation, but I sure wish I'd thought of it at the time. Nobody else pointed it out either. Best of all, would have been to mention both methods. If you can notice both of them you can solve the problem without the advance knowledge of the value of !!1+\frac12+\frac14+\ldots!!, because you have !!4S = 1+\frac12 + S!! and then solve for !!S!!.

In Visualization of Rhombus made of Radii and Chords it seemed that OP just needed to see a diagram (“I really really don't see how two circles can form a rhombus?”), so I drew one.

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Sun, 19 Jul 2015

Math.SE report 2015-04

[ Notice: I originally published this report at the wrong URL. I moved it so that I could publish the June 2015 report at that URL instead. If you're seeing this for the second time, you might want to read the June article instead. ]

A lot of the stuff I've written in the past couple of years has been on Mathematics StackExchange. Some of it is pretty mundane, but some is interesting. I thought I might have a little meta-discussion in the blog and see how that goes. These are the noteworthy posts I made in April 2015.

Languages and their relation : help is pretty mundane, but interesting for one reason: OP was confused about a statement in a textbook, and provided a reference, which OPs don't always do. The text used the symbol !!\subset_\ne!!. OP had interpreted it as meaning !!\not\subseteq!!, but I think what was meant was !!\subsetneq!!.

I dug up a copy of the text and groveled over it looking for the explanation of !!\subset_\ne!!, which is not standard. There was none that I could find. The book even had a section with a glossary of notation, which didn't mention !!\subset_\ne!!. Math professors can be assholes sometimes.
Is there an operation that takes !!a^b!! and !!a^c!!, and returns !!a^{bc}!! is more interesting. First off, why is this even a reasonable question? Why should there be such an operation? But note that there is an operation that takes !!a^b!! and !!a^c!! and returns !!a^{b+c}!!, namely, multiplication, so it's plausible that the operation that OP wants might also exist.

But it's easy to see that there is no operation that takes !!a^b!! and !!a^c!! and returns !!a^{bc}!!: just observe that although !!4^2=2^4!!, the putative operation (call it !!f!!) should take !!f(2^4, 2^4)!! and yield !!2^{4\cdot4} = 2^{16} = 65536!!, but it should also take !!f(4^2, 4^2)!! and yield !!4^{2\cdot2} = 2^4 = 256!!. So the operation is not well-defined. And you can take this even further: !!2^4!! can be written as !!e^{4\log 2}!!, so !!f!! should also take !!f(e^{2\log 4}, e^{2\log 4})!! and yield !!e^{4(\log 4)^2} \approx 2180.37!!.

They key point is that the representation of a number, or even an integer, in the form !!a^b!! is not unique. (Jargon: "exponentiation is not injective".) You can raise !!a^b!!, but having done so you cannot look at the result and know what !!a!! and !!b!! were, which is what !!f!! needs to do.

But if !!f!! can't do it, how can multiplication do it when it multiplies !!a^b!! and !!a^c!! and gets !!a^{b+c}!!? Does it somehow know what !!a!! is? No, it turns out that it doesn't need !!a!! in this case. There is something magical going on there, ultimately related to the fact that if some quantity is increasing by a factor of !!x!! every !!t!! units of time, then there is some !!t_2!! for which it is exactly doubling every !!t_2!! units of time. Because of this there is a marvelous group homomophism $$\log : \langle \Bbb R^+, \times\rangle \to \langle \Bbb R ,+\rangle$$ which can change multiplication into addition without knowing what the base numbers are.

In that thread I had a brief argument with someone who thinks that operators apply to expressions rather than to numbers. Well, you can say this, but it makes the question trivial: you can certainly have an "operator" that takes expressions !!a^b!! and !!a^c!! and yields the expression !!a^{bc}!!. You just can't expect to apply it to numbers, such as !!16!! and !!16!!, because those numbers are not expressions in the form !!a^b!!. I remembered the argument going on longer than it did; I originally ended this paragraph with a lament that I wasted more than two comments on this guy, but looking at the record, it seems that I didn't. Good work, Mr. Dominus.
how 1/0.5 is equal to 2? wants a simple explanation. Very likely OP is a primary school student. The question reminds me of a similar question, asking why the long division algorithm is the way it is. Each of these is a failure of education to explain what division is actually doing. The long division answer is that long division is an optimization for repeated subtraction; to divide !!450\div 3!! you want to know how many shares of three cookies each you can get from !!450!! cookies. Long division is simply a notation for keeping track of removing !!100!! shares, leaving !!150!! cookies, then !!5\cdot 10!! further shares, leaving none.

In this question there was a similar answer. !!1/0.5!! is !!2!! because if you have one cookie, and want to give each kid a share of !!0.5!! cookies, you can get out two shares. Simple enough.

I like division examples that involve giving cookies to kids, because cookies are easy to focus on, and because the motivation for equal shares is intuitively understood by everyone who has kids, or who has been one.

There is a general pedagogical principle that an ounce of examples are worth a pound of theory. My answer here is a good example of that. When you explain the theory, you're telling the student how to understand it. When you give an example, though, if it's the right example, the student can't help but understand it, and when they do they'll understand it in their own way, which is better than if you told them how.
How to read a cycle graph? is interesting because hapless OP is asking for an explanation of a particularly strange diagram from Wikipedia. I'm familiar with the eccentric Wikipedian who drew this, and I was glad that I was around to say "The other stuff in this diagram is nonstandard stuff that the somewhat eccentric author made up. Don't worry if it's not clear; this author is notorious for that."
In Expected number of die tosses to get something less than 5, OP calculated as follows: The first die roll is a winner !!\frac23!! of the time. The second roll is the first winner !!\frac13\cdot\frac23!! of the time. The third roll is the first winner !!\frac13\cdot\frac13\cdot\frac23!! of the time. Summing the series !!\sum_n \frac23\left(\frac13\right)^nn!! we eventually obtain the answer, !!\frac32!!. The accepted answer does it this way also.

But there's a much easier way to solve this problem. What we really want to know is: how many rolls before we expect to have seen one good one? And the answer is: the expected number of winners per die roll is !!\frac23!!, expectations are additive, so the expected number of winners per !!n!! die rolls is !!\frac23n!!, and so we need !!n=\frac32!! rolls to expect one winner. Problem solved!

I first discovered this when I was around fifteen, and wrote about it here a few years ago.

As I've mentioned before, this is one of the best things about mathematics: not that it works, but that you can do it by whatever method that occurs to you and you get the same answer. This is where mathematics pedagogy goes wrong most often: it proscribes that you must get the answer by method X, rather than that you must get the answer by hook or by crook. If the student uses method Y, and it works (and if it is correct) that should be worth full credit.

Bad instructors always say "Well, we need to test to see if the student knows method X." No, we should be testing to see if the student can solve problem P. If we are testing for method X, that is a failure of the test or of the curriculum. Because if method X is useful, it is useful because for some problems, it is the only method that works. It is the instructor's job to find one of these problems and put it on the test. If there is no such problem, then X is useless and it is the instructor's job to omit it from the curriculum. If Y always works, but X is faster, it is the instructor's job to explain this, and then to assign a problem for the test where Y would take more time than is available.

I see now I wrote the same thing in 2006. It bears repeating. I also said it again a couple of years ago on math.se itself in reply to a similar comment by Brian Scott:

If the goal is to teach students how to write proofs by induction, the instructor should damned well come up with problems for which induction is the best approach. And if even then a student comes up with a different approach, the instructor should be pleased. ... The directions should not begin [with "prove by induction"]. I consider it a failure on the part of the instructor if he or she has to specify a technique in order to give students practice in applying it.

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Fri, 19 Jun 2015

Math.SE report 2015-05

A lot of the stuff I've written in the past couple of years has been on math.StackExchange. Some of it is pretty mundane, but some is interesting. My summary of April's interesting posts was well-received, so here are the noteworthy posts I made in May 2015.

What matrix transforms !!(1,0)!! into !!(2,6)!! and tranforms !!(0,1)!! into !!(4,8)!!? was a little funny because the answer is $$\begin{pmatrix}2 & 4 \\ 6 & 8 \end{pmatrix}$$ and yeah, it works exactly like it appears to, there's no trick. But if I just told the guy that, he might feel unnecessarily foolish. I gave him a method for solving the problem and figured that when he saw what answer he came up with, he might learn the thing that the exercise was designed to teach him.
Is a “network topology'” a topological space? is interesting because several people showed up right away to say no, it is an abuse of terminology, and that network topology really has nothing to do with mathematical topology. Most of those comments have since been deleted. My answer was essentially: it is topological, because just as in mathematical topology you care about which computers are connected to which, and not about where any of the computers actually are.

Nobody constructing a token ring network thinks that it has to be a geometrically circular ring. No, it only has to be a topologically circular ring. A square is fine; so is a triangle; topologically they are equivalent, both in networking and in mathematics. The wires can cross, as long as they don't connect at the crossings. But if you use something that isn't topologically a ring, like say a line or a star or a tree, the network doesn't work.

The term “topological” is a little funny. “Topos” means “place” (like in “topography” or “toponym”) but in topology you don't care about places.
Is there a standard term for this generalization of the Euler totient function? was asked by me. I don't include all my answers in these posts, but I think maybe I should have a policy of including all my questions. This one concerned a simple concept from number theory which I was surprised had no name: I wanted !!\phi_k(n)!! to be the number of integers !!m!! that are no larger than !!n!! for which !!\gcd(m,n) = k!!. For !!k=1!! this is the famous Euler totient function, written !!\varphi(n)!!.

But then I realized that the reason it has no name is that it's simply !!\phi_k(n) = \varphi\left(\frac n k\right)!! so there's no need for a name or a special notation.

As often happens, I found the answer myself shortly after I asked the question. I wonder if the reason for this is that my time to come up with the answer is Poisson-distributed. Then if I set a time threshold for how long I'll work on the problem before asking about it, I am likely to find the answer to almost any question that exceeds the threshold shortly after I exceed the threshold. But if I set the threshold higher, this would still be true, so there is no way to win this particular game. Good feature of this theory: I am off the hook for asking questions I could have answered myself. Bad feature: no real empirical support.
how many ways can you divide 24 people into groups of two? displays a few oddities, and I think I didn't understand what was going on at that time. OP has calculated the first few special cases:

1:1 2:1 3:3 4:3 5:12 6:15

which I think means that there is one way to divide 2 people into groups of 2, 3 ways to divide 4 people, and 15 ways to divide 6 people. This is all correct! But what could the 1:1, 3:3, 5:12 terms mean? You simply can't divide 5 people into groups of 2. Well, maybe OP was counting the extra odd person left over as a sort of group on their own? Then odd values would be correct; I didn't appreciate this at the time.

But having calculated 6 special cases correctly, why can't OP calculate the seventh? Perhaps they were using brute force: the next value is 48, hard to brute-force correctly if you don't have a enough experience with combinatorics.

I tried to suggest a general strategy: look at special cases, and not by brute force, but try to analyze them so that you can come up with a method for solving them. The method is unnecessary for the small cases, where brute force enumeration suffices, but you can use the brute force enumeration to check that the method is working. And then for the larger cases, where brute force is impractical, you use your method.

It seems that OP couldn't understand my method, and when they tried to apply it, got wrong answers. Oh well, you can lead a horse to water, etc.

The other pathology here is:

I think I did what you said and I got 1.585times 10 to the 21

for the !!n=24!! case. The correct answer is $$23\cdot21\cdot19\cdot17\cdot15\cdot13\cdot11\cdot9\cdot7\cdot5\cdot3\cdot1 = 316234143225 \approx 3.16\cdot 10^{11}.$$ OP didn't explain how they got !!1.585\cdot10^{21}!! so there's not much hope of correcting their weird error.

This is someone who probably could have been helped in person, but on the Internet it's hopeless. Their problems are Internet communication problems.
Lambda calculus typing isn't especially noteworthy, but I wrote a fairly detailed explanation of the algorithm that Haskell or SML uses to find the type of an expression, and that might be interesting to someone.
I think Special representation of a number is the standout post of the month. OP speculates that, among numbers of the form !!pq+rs!! (where !!p,q,r,s!! are prime), the choice of !!p,q,r,s!! is unique. That is, the mapping !!\langle p,q,r,s\rangle \to pq+rs!! is reversible.

I was able to guess that this was not the case within a couple of minutes, replied pretty much immediately:

I would bet money against this representation being unique.

I was sure that a simple computer search would find counterexamples. In fact, the smallest is !!11\cdot13 + 19\cdot 29 = 11\cdot 43 + 13\cdot 17 = 694!! which is small enough that you could find it without the computer if you are patient.

The obvious lesson to learn from this is that many elementary conjectures of this type can be easily disproved by a trivial computer search, and I frequently wonder why more amateur mathematicians don't learn enough computer programming to investigate this sort of thing. (I wrote recently on the topic of An ounce of theory is worth a pound of search , and this is an interesting counterpoint to that.)

But the most interesting thing here is how I was able to instantly guess the answer. I explained in some detail in the post. But the basic line of reasoning goes like this.

Additive properties of the primes are always distributed more or less at random unless there is some obvious reason why they can't be. For example, let !!p!! be prime and consider !!2p+1!!. This must have exactly one of the three forms !!3n-1, 3n,!! or !!3n+1!! for some integer !!n!!. It obviously has the form !!3n+1!! almost never (the only exception is !!p=3!!). But of the other two forms there is no obvious reason to prefer one over the other, and indeed of the primes up to 10,000, 611 are of the type !!3n!! and and 616 are of the type !!3n-1!!.

So we should expect the value !!pq+rs!! to be distributed more or less randomly over the set of outputs, because there's no obvious reason why it couldn't be, except for simple stuff, like that it's obviously almost always even.

So we are throwing a bunch of balls at random into bins, and the claim is that no bin should contain more than one ball. For that to happen, there must be vastly more bins than balls. But the bins are numbers, and primes are not at all uncommon among numbers, so the number of bins isn't vastly larger, and there ought to be at least some collisions.

In fact, a more careful analysis, which I wrote up on the site, shows that the number of balls is vastly larger—to have them be roughly the same, you would need primes to be roughly as common as perfect squares, but they are far more abundant than that—so as you take larger and larger primes, the number of collisions increases enormously and it's easy to find twenty or more quadruples of primes that all map to the same result. But I was able to predict this after a couple of minutes of thought, from completely elementary considerations, so I think it's a good example of Lower Mathematics at work.

This is an example of a fairly common pathology of math.se questions: OP makes a conjecture that !!X!! never occurs or that there are no examples with property !!X!!, when actually !!X!! almost always occurs or every example has property !!X!!.

I don't know what causes this. Rik Signes speculates that it's just wishful thinking: OP is doing some project where it would be useful to have !!pq+rs!! be unique, so posts in hope that someone will tell them that it is. But there was nothing more to it than baseless hope. Rik might be right.

[ Addendum 20150619: A previous version of this article included the delightful typo “mathemativicians”. ]

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Sun, 14 Jun 2015

Math.SE report 2015-06

[ This page originally held the report for April 2015, which has moved. It now contains the report for June 2015. ]

Is “smarter than” a transitive relationship? concerns a hypothetical "is smarter than" relation with the following paradoxical-seeming property:

most X's are smarter than most Y's, but most Y's are such that it is not the case that most X's are smarter than it.

That is, if !!\mathsf Mx.\Phi(x)!! means that most !!x!! have property !!\Phi!!, then we want both $$\mathsf Mx.\mathsf My.S(x, y)$$ and also $$\mathsf My.\mathsf Mx.\lnot S(x, y).$$

“Most” is a little funny here: what does it mean? But we can pin it down by supposing that there are an infinite number of !!x!!es and !!y!!s, and agreeing that most !!x!! have property !!P!! if there are only a finite number of exceptions. For example, everyone should agree that most positive integers are larger than 7 and that most prime numbers are odd. The jargon word here is that we are saying that a subset contains “most of” the elements of a larger set if it is cofinite.

There is a model of this property, and OP reports that they asked the prof if this was because the "smarter than" relation !!S(x,y)!! could be antitransitive, so that one might have !!S(x,y), S(y,z)!! but also !!S(z,x)!!. The prof said no, it's not because of that, but the OP want so argue that it's that anyway. But no, it's not because of that; there is a model that uses a perfectly simple transitive relation, and the nontransitive thing nothing but a distraction. (The model maps the !!x!!es and !!y!!s onto numbers, and says !!x!! is smarter than !!y!! if its number is bigger.) Despite this OP couldn't give up the idea that the model exists because of intransitive relations. It's funny how sometimes people get stuck on one idea and can't let go of it.
How to generate a random number between 1 and 10 with a six-sided die? was a lot of fun and attracted several very good answers. Top-scoring is Jack D'Aurizio's, which proposes a completely straightforward method: roll once to generate a bit that selects !!N=0!! or !!N=5!!, and then roll again until you get !!M\ne 6!!, and the result is !!N+M!!.

But several other answers were suggested, including two by me, one explaining the general technique of arithmetic coding, which I'll probably refer back to in the future when people ask similar questions. Don't miss NovaDenizen's clever simplification of arithmetic coding, which I want to think about more, or D'Aurizio's suggestion that if you threw the die into a V-shaped trough, it would land with one edge pointing up and thus select a random number from 1 to 12 in a single throw.

Interesting question: Is there an easy-to-remember mapping from edges to numbers from 1–12? Each edge is naturally identified by a pair of distinct integers from 1–6 that do not add to 7.
The oddly-phrased Category theory with objects as logical expressions over !!{\vee,\wedge,\neg}!! and morphisms as? asks if there is a standard way to turn logical expressions into a category, which there is: you put an arrow from !!A\to B!! for each proof that !!A!! implies !!B!!; composition of arrows is concatenation of proofs, and identity arrows are empty proofs. The categorial product, coproduct, and exponential then correspond to !!\land, \lor, !! and !!\to!!.

This got me thinking though. Proofs are properly not lists, they are trees, so it's not entirely clear what the concatenation operation is. For example, suppose proof !!X!! concludes !!A!! at its root and proof !!Y!! assumes !!A!! in more than one leaf. When you concatenate !!X!! and !!Y!! do you join all the !!A!!'s, or what? I really need to study this more. Maybe the Lambek and Scott book talks about it, or maybe the Goldblatt Topoi book, which I actually own. I somehow skipped most of the Cartesian closed category stuff, which is an oversight I ought to correct.
In Why is the Ramsey`s theorem a generalization of the Pigeonhole principle I gave what I thought was a terrific answer, showing how Ramsey's graph theorem and the pigeonhole principle are both special cases of Ramsey's hypergraph theorem. This might be my favorite answer of the month. It got several upvotes, but OP preferred a different answer, with fewer details.

There was a thread a while back about theorems which are generalizations of other theorems in non-obvious ways. I pointed out the Yoneda lemma was a generalization of Cayley's theorem from group theory. I see that nobody mentioned the Ramsey hypergraph theorem being a generalization of the pigeonhole principle, but it's closed now, so it's too late to add it.
In Why does the Deduction Theorem use Union? I explained that the English word and actually has multiple meanings. I know I've seen this discussed in elementary logic texts but I don't remember where.
Finally, Which is the largest power of natural number that can be evaluated by computers? asks if it's possible for a computer to calculate !!7^{120000000000}!!. The answer is yes, but it's nontrivial and you need to use some tricks. You have to use the multiplying-by-squaring trick, and for the squarings you probably want to do the multiplication with DFT. OP was dissatistifed with the answer, and seemed to have some axe to grind, but I couldn't figure out what it was.

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