Wednesday, January 29, 2014

Word of the Day: Kapuris

Speakers of any language tend to find certain words to be inexplicably colorful, and few Yiddish words are more evocative than kaporis, or as we Galicianers would say, kapuris.  Literally this word describes the ritual on Yom Kippur when a chicken is sacrificed for your sins; and I’ve never seen it myself, but I understand that done properly, the ceremony involves swinging the still-living chicken in the air over your head. Even in the old country, this must have seemed ridiculous, and so the word took on a sarcastic meaning which is illustrated by the following story:

A Jew is brought before a Russian judge charged with stealing a chicken. Since the Jew doesn’t speak Russian, he is accompanied by a translator. The Judge asks (in Russian): “Did you steal the chicken?” and the translator repeats the question in Yiddish.

The Jew answers with outrage: “Ich hâb geganvet a chicken?”

The translator in turn reports thusly: “I stole a chicken”.

The judge asks “Why did you steal the chicken?” which is duly translated. The Jew replies, with increasing irritation: “Ich hâb gedarft a chicken???”

Which is reported by the translator as: “I needed a chicken.”

The Judge wants to get to the bottom of this, so he asks, “Why did you need a chicken?”

The Jew loses his cool and shouts in anger “Ich hâb es gedarft auf kapuris!!!”. Which means that he needed it like a hole in the head.

But the translator reports to the judge the literal meaning: “I needed it for a religious ritual.”

Some younger readers might not fully appreciate just how outrageously funny this joke would have been in their grandparents day This is really a trilingual joke, with the judge’s questions being told in Russian and translator’s version being told in English, and the Jew’s mixing of the American “chicken” in his Yiddish responses being a humorous allusion to the way Yiddish was becoming Americanized in the New World.  “Only in America”, and only for a few brief decades, did we have an audience that could have fully appreciated theis marvellous interplay between all three languages. The punch line in its legal formalism is essentially English. It evokes incredibly sinister references of the blood libel accusations that were still fresh in our collective consciousness as a result of the Beyliss trial, which I wrote about last week. I heard my father tell this joke many years ago; he’s normally a very good joke-teller, but he broke down in uncontrollable laughter in when he tried to say “for a religious ritual”.

The word kapuris is funny to us, in my opinion, for one other reason: it is a ka- word. It seems to me that English speakers hear foreign words as being particularly colorful when they start with ka. From the Germans we have kaputt, which means….well, kaputt. From the Italian we have capisce, which is accompanied by all kinds of Mafia associations. The Japanese have it everywhere, with karate, karaoke, and kamikaze being words that are highly representative of certain aspects of their culture and which the English-speaking world has been quick to import.

I can’t leave off the story of ka-words without telling you about one that I learned from my Chinese wife, who belongs to an ethnic group from Southern China known variously as Teo-Chiu or Chiu-Chow. When two Chiu-Chow people meet abroad, they recognize each other as kakinang, which is a word essentially equivalent to our landsman, the Russian zemlyak, and the Italian paesano. The website of the Overseas Chiu-Chow Assoaciation is called It’s a very cool word.

But I digress. I’m going to close off with one more kapuris joke. This one takes place in a small town in Russia, where the governor had just died. Although he was a notorious anti-semite, the local officials came to the Jewish gravestone-maker who was known as the most skillful engraver in the gubernia. The commisioned him with the task of making the governor’s tombstone, and warned him that they expected his finest work.  The engraver was in a quandary: how was he to lavish his best efforts on an enemy of his people? After some thought, he set himself to work, and the tombstone was duly delivered and installed. At the unveiling, the crowd gasped with wonder at the moving inscription on the tombstone: 


But the engraver’s fellow Jews were dismayed. “We understand you had to do a good job, that you had no choice. But for such a soyneh-Yisroel did you have to create something so beautiful?”

The engraver smiled and waved them off while shaking his head. “Read it again in Yiddish”, he told them.

They looked again at the tombstone and one of them translated out loud: “A schöene, réine kapuris.” (A sheyne, reyne kapuris).

Monday, January 20, 2014

About Those Gaussian Envelopes

I mentioned the other day how I used a Gaussian envelope to numerically calculate the sum of that divergent Ramanujan series,


I thought maybe I should tell you how I came up with this technique. It wasn't a discreet series, it was an integral I had to deal with. It came up when I was working on the problem of the crystal radio: what is the maximum theoretical power you can absorb from a plane wave using a small antenna?

The key to solving this problem is to realize you don't need to know (hardly) anything about antennas. Everything that happens comes as a result of wave interactions. You have an incoming plane wave, which causes currents to flow in your antenna. As a result, your antenna radiates spherically. (Of course its a modified spherical pattern because of the donut characteristic of the dipole field. But that's a minor detail.) The point is, looking at the system from above, you have the incoming waves from the left and the outgoing spherical waves from your antenna like so:

(I talked about this a couple of years ago in this blogpost.) You can see that the little antenna is scattering power in all directions, except in the shaded region you have the possiblility (if you can get the phase just right) to actually remove power from the incident wave. This can only happen effectively within the shaded region when the phase difference between the two systems is less than 180 degrees (or 1/2*lambda).

The really interesting thing is that since power goes as the square of the wave amplitude, you have two things going on. As the antenna current grows, the re-radiated power grows quadratically. But at the same time, within the shadow region, the absorbed power grows linearly. Because if the incident field is 1 and the re-radiated field is dx, the resultant power is (1-dx)^2 where 2dx is the absorbed power (and you ignore the term in (dx)^2).

You know that if you have two quantities and one of them is growing linearly and the other is growing quadratically, that at first the linear term is going to dominate, but then the quadratic term is going to catch up. And in between there is an optimum. That optimum represents the maximum power you can absorb with a small antenna...and it has nothing to do with the size of the antenna. It's strictly a result of the interacting field patterns.

How do you achieve the optimum conditions? First, you have to get the phase relationship just right. That's called "tuning the antenna" and you do it with a variable inductor. Then, you have to get just the right current. That's called "impedance matching". We won't get into that today.

The point all depends on comparing the re-radiated power to the absorbed power. The spherical power isn't too hard to the worst case, we can look up the standard antenna formulas. But what about the absorbed do we do that?

For starters, let's realize that if we back off far enough away from the antenna....100 wavelengths, or even 1000 wavelengths...then we can use the small-angle approximations like sin(x) = x. I know you math guys hate that approximation, as I did the first time I saw it in Grade 11...but in physics you learn to love it. But that's a whole nother story.

You'll notice that "shadow region" we're working with is a paraboloid of revolution. That comes straight from the focus-directrix definition of a parabola as the locus of all points such that the difference of the two distances is a constant. And we said that within the first lobe, we can get destructive interference. But outside that lobe, we get positive interference...the waves re-inforce. So that works against us...and so on and so forth to infinity, the interference pattern alternately giving us rings of positive and negative interference. 

Also notice that the rings are not the same width. Because of the geometry, they get closer together as you get farther from the axis of symmetry.  (It goes as sqrt(r)...that comes straight from putting successive integer values into the focus-directrix definition of the parabola.) Here then is a picture of the positive/negative interference pattern as though you projected it on a screen 1000 wavelengths behind the antenna:
Without getting bogged down in the details, this basically meant that to figure out the power absorbed by the antenna, I had to integrate a function of the form (again, taking advantage of the small-angle approximations)

It's helpful to know what these functions look like. Maybe you can imagine them, but here is the cos function, courtesy of Excel:

How do you integrate this from zero to infinity? By remembering the physics. Wave interference is a funny thing. If you shine two flashlight beams across each other, then there is all kinds of crazy interference patters going on where they beams cross. If you followed the power flow in detail, its going here and there and everywhere. But the overall result is that Beam A ends up where it was going, and Beam B goes its own way too. Like they never crossed. And essentially, that's the situation we have here. The crazy inteference peaks that get bigger and bigger...those represent power flowing from the incident plane wave and the small receiving antenna. But once you diverge from the axis of symmetry, there really can't be much going on that is of any physical significance. All those oscillating powers have to average out to zero. So let's just force them to do that. And the obvious way to do that mathematically is to multiply the integrand by a gently-decreasing Gaussian, something like this:
As you let k become arbitrarily large (so the envelope gets wider and wider) this integral should converge to a constant value...and in fact it does. I don't remember which one of them comes out non-zero...the sine or the cosine. Ultimately that just tells you the relative phase of the wave pattern. At some point I ran it in Excel and came up with a value. But actually, as I look at it now, I'm thinking that you can do this one exactly with integration by parts (with the obvious u=r^2 substitution). When I was doing this calculation, I think all I really wanted was a ballpark estimate of the power, so I just made the arbitrary assumption that interference was 100% inside the first nodal line (the shadow zone) and negligible outside. That turned out to be not too far off.

And that, my friends, is how I came up with the idea for using a Gaussian Envelope to tame ill-behaved integrals. Necessity was the mother of invention. 

So how much power can you absorb with a crystal radio? According to the calculation, as it turns out, a rather huge amount of power. Remember, the size of the antenna is irrelevant: all you need is to have a perfectly tuned and perfectly matched receiver. In that case, you can absorb a quantity of power equivalent to everything flowing through an area approximately one-half wavelength in diameter. So for a typical AM radio station, that would be all the power flowing through a cross-section of 

(200 m)x(200m)=40,000 sq. meters

Unfortunately it is not so easy to demonstrate this in practise. The biggest problem is the resistance of copper, which becomes huge in comparison to the other parameters. You might be able to do something with all superconducting components, but even then I'm not quite sure, because there's also the parasitic resistance of the ground which comes into the picture.

But oddly enough, nature provides us with an example of the perfectly tuned and lossless antenna in the form of...the atom. Yes, an atom can behave as a tiny antenna, and in this case the equations really work. The absorption radius of a hydrogen atom in visible light is on the order of the wavelength of light, which represents a cross-sectional area around 10 million times the physical cross-section of a hydrogen atom.

When you take physics in high school, they tell you that it's impossible to explain the photo-electric effect using the wave theory of light because the energy in a wave isn't concentrated enough to knock an electron out of an atom. They even "prove" this by multiplying the power density of the light wave by the cross-sectional area of an atom. Therefore, they claim, light must be made of particles...the so-called "photons".

The people who make this argument (and they include physicists at the very highest level, who should know better) have never done the crystal radio calculation. The cross-section of the atom, like the cross-section of the radio antenna, has nothing to do with it. The same argument that supposedly debunks the wave theory of light also "proves" that a crystal radio can't work without a battery.

But it does.

POST-SCRIPT: Maybe I'm getting a little senile, but it turns out I went through most of these calculations two years ago, here and here. Either way,  it's still stuff you won't find anywhere else.

Sunday, January 19, 2014

The Casimir Effect: Ramanujan Revisited

There's been a small flurry of hits (a few dozen actually) on one of my posts from almost three   four  years ago, when I talked about how you can use one of Ramanujan's weird infinite sums to calculate the Casimir effect. There seem to be two unrelated reasons for this new activity: a math video by a guy named Brady Haran that got reposted on Slate Magazine and got an awful lot of hits...and an ongoing discussion on on the same topic, where my new correspondent Balarka Sen mentioned my post. It's quite a coincidence because I do almost no math on this blog, and Balarka doesn't do any physics...and the only two real math topics I've ever done...the quintic equation and that Ramanujan series....turn out to be huge topics of interest for Balarka.

The topic of the internet video is the series

1 + 2 + 3 + 4 + 5 ....

And I have to say I've never understood how that's supposed to work. My interest was in the alternating series:

1 - 2 + 3 - 4 + 5....

and that's the one I used to calculate the Casimir Effect. Or at least that's what I thought. When I look it over again, (especially after watching that guy's video) it's almost arguable that I really did use the first series.

Except there's a difference the way I do it. I actually have a physically motivated reason for being able to shuffle those series around the way I do. And I'm not sure I made it perfectly clear in my original post. So I think I want to go over it again.

The idea is based on a fact of quantum field theory that I have to admit I don't really understand: that a standing-wave mode of the electromagnetic field cannot be totally at rest: it must have a minimum energy of one-half quantum. It's a very small amount of energy, but the problem is there is an infinite number of modes. So does that mean space is filled with an infinite amount of energy? Hard to say. Because in the case of an infinite universe, each of those mode enegies is spread over an infinite volume. What's the local energy density? Hard to say.

But we can calculate the case of a finite box. We're going to take our "box" to be other words, a parallel-plate capacitor...because that will simplify counting of the modes. Le'ts say the plates are one millimeter apart. We have modes with one standing wave, two, three, etc...with corresponding frequencies of 1, 2, 3 etc. The mode energies are proportional to the mode frequencies, so we get a total energy of

1 + 2 + 3 + 4 + 5.....

which looks a lot like infinity. (I'm basically letting Plank's Constant = 1, in case you were wondering).

Now let's do something funny. Let's compress the capacitor and see how much the energy increases. The funny thing here is that each of the modes deforms by the time the plates are twice as close together, the energy is:

2 + 4 + 6 + 8 + 10....

which looks like twice as much energy. But wait....

The whole problem with these infinite sums is that we assume the presence of infinitely high frequencies. That's a little crazy. Are there actually gamma rays between those capacitor plates...and not just gamma rays, but super-ultra high frequency gamma rays beyond anything imaginable? It doesn't make sense.

Maybe we should only be counting up to a specific frequency cutoff. Lets see how our sums work then. For the capacitor at 1 mm, we have:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8.....; and,counting only up to the same frequency we get

     2    +     4     +   6     +   8..... for the compressed capacitor at 0.5 mm!

If you count it this way, the compressed capacitor has actually less energy that the one you started with. How can that be? Well... maybe it means that unlike a cylinder full of air, which resists being compressed, the parallel plate capacitor has a negative pressure...the plates attract. That would be the Casimir Effect.

But can we calculate it? We're still looking at those monotonically increasing infinite series, which I find intractable. But there's a trick. I can handle the alternating infinite series. Because I assume in physics that the stuff at ultra-high frequencies which is wildly fluctuating between positive and negative must logically just cancel out. It's not hard to verify this numerically. Take that series

1 - 2 + 3 - 4 + 5....

and put it into Excel, and then put a very gradual Gaussian envelope over it, to gently supress the ultra-high frequencies. It's not hard to verify that the value tends to 0.25, once your Gaussian is wide enough, and it stabilizes there as your Gaussian tends to infinite width. So how can we apply this to our capacitor?

I figured out I can make it work by analyzing pressures instead of energies. The mode pressures in the capacitor are proportional to the energies:

1 + 2 + 3 + 4 + 5....

but when you bring the plates twice as close together, the mode pressures don't double: they quadruple. Because pressure is energy per unit volume, and you now have twice the energy in half the volume. So comparing the pressures:

1.0 mm capacitor:   1 + 2  +  3  +  4  +  5  +  6  +  7  +  8....
0.5 mm capacitor:         4       +     8        +    12     +    16....

(Notice I've lined them up so the frequencies agree.) We still don't know what the pressure is between the plates, but oddly enough, we can now calculate the change in pressure when we went from d=1 to d=0.5  .  It's the difference between the two series, which is just the alternating series we've been talking about:

dP = 1 - 2 + 3 - 4 + 5....= 0.25

Now...the people who want the monotonic series to add up to -1/12, they say: look what you've done: you've just said S - 4S = 0.25, and they solve for S and get -1/12. I don't know about that. I can't justify it physically. But it turns out I'm going to get the same result, with my own brand of physical reasoning.

If the change in pressure of the capacitor was 1/4 in compressing from 1 to 0.5, then what do you think it would have been expanding going from 1 to 2 mm? It's not hard to verify by dimensional analysis that it would have been exactly a quarter as much, or 1/16. And in going from 2 to 4?...1/64. You can see that in expanding the plates from 1 millimeter to infinity, the pressure increased by:

1/16  +  1/64   +   1/256.....   = 1/12

But the there can't be any pressure between the plates when they're infinitely far apart, can there? So we...wait for it...renormalize. We say that the pressure at infinity is zero, and the pressure at 1 millimeter separation is...-1/12.

Now as for the people who say I could have got that result from the get-go just by using Ramanujan's result for divergent series...I disagree. Yes, superficially you could write as I did, from the mode pressures, the sum:

1 + 2 + 3 + 4 + 5....= -1/12

But that's wrong. There is infinite pressure between the plates. This calculation neglects the ever-so-slightly-larger infinite pressure which is also found outside the plates. It's the difference between those infinities that adds up to -1/12. And that's why the plates experience a force drawing them together.

Saturday, January 18, 2014

It's gotta be S5

I think I've got it.

The problem was to show for the general fifth-degree equation that the Galois Group of the resolvent sextic was just S5, the same as the Galois group of the quintic. I already talked about how you could generate the coefficients of the sextic by calculating the elementary symmetric polynomials of Dummits Functions. Which brought us to the question of: what is the Galois Group>

What I did is I wrote out all six Dummit functions in A,B,C,D and E (in a kind of shorthand, actually).
The "standard form" of Dummit's function is

1.  AA(BE + CD) + BB(CA + DE) + CC(DB + EA)....

which is obviously based on the rotation ABCDE. Starting from this one, there are exactly five other "rotations" which work, and these are all of them:

1.  ABCDE  (= ACEBD)
2.  ACBED  (= ABDCE)
3.  AECBD        etc.

The terms in brackets are included to show that the alternate forms generate the same Dummit Function.

Then I took all possible 5-cycles on A,B,C,D and E and examined what there effect was on the Dummit Functions. (EDIT: Let me explain this a little better. First I'll shift A to B, B to C, etc. Obviously the first one is just going to map to a rotation of itself, which means for the sake of the associated Dummit Function, it goes nowhere. But the second one, ACBED, gets transformed to BDCAE...which is the same as No. 6 (or a reverse image, which comes to the same thing. Likewise you can plot where they all go. The first "orbit" is (26435). Then I tries a different five-cycle...actually, I chose a cycle of the 2nd function, because I wanted it to map to itself. The other five mapped to each other in a second five-cycle. I've listed all six five-cycles below:


And then of course you get the images of these under self-compounding...24 5-cycles altogether. (Just like S5.) The point is they didn't mushroom out into every possible 5-cycle in S6. So let's see what happens when you start multiplying them together? At first I started getting  3-cycles...actually, direct products of 3-cycles, like these:


And then I started getting new 5-cycles. But it was easy to verify that each new one was already within the orbits of my original 6. So far so good.

I was starting to think I was getting only even permutations...that would be interesting. Could the Galois Group of the resolvent be A5? I had already 24 5-cycles, and a whole bunch of 3-cycles, so unless I had some odd permutations, it looked like I might be heading there. So I went back to Dummit's Functions and just tried swapping A and B. And I got:


That's an odd permutation, so it can't be A5. I've got 5-cycles, 3-cycles, and odd permutations...if it's not S5, then God help me. 

I know there are guys that would do a lot cleaner job of this than me, but there it is.

The Galois Group of the Resolvent

Okay, I thought I was done but I'm not. We generated a sixth-degree equation from the general quintic. People say "that's why you can't solve the quintic...because the resolvent you generate is of higher order than the equation you started with."

That's not the reason you can't solve the quintic. There are solvable sextic equations...maybe our resolvent is one of them? The question is: what is the Galois Group of the resolvent sextic?

I'm thinking it has to be S5. Because there are absolutely no permutations of the roots of the sextic that aren't generated by permutations of the roots of the quintic. And furthermore...every single permutation of the roots of the quintic affects some of the roots of the sextic. Does that make it a one-to-one correspondence?

Because if it doesn't....if there are two different elements of S5 that give the same permutation of the resolvents...then the order of the Galois group is less than 120...which makes it 60 or less. And unless it's A5, then all bets are may indeed have a solvable sextic. So the problem is to show explicitly just what is the Galois Group of the sextic. I'm thinking it has to be S5. But I'm having trouble constructing it.

The Galois Group in question comes from the reshuffling of the six resolvents functions, so it if obviously at most a subgroup of S6. And it's easy to find S5 as a subgroup of S6...just take one letter and hold it fixed while the others get shuffled around. But that's not what the resolvent functions do. They all move under permutation, and they do so in a way that is somehow...symmetric.

Furthermore, there are 5-cycles in the Galois group...permutations that, when repeated 5 times, take us back to where we started. There is only one way to make a five-cycle is S6...hold one element fixed and let the others move. So without loss of generality, we must have

(1 2 3 4 5)

as a permutation, and also:

(2 3 4 5 6)
(1 3 4 5 6)

....except we can't pick and choose all of them willy-nilly, because there are 120 5-cycles of this kind in S6 and that's just too many! We need exactly 24, because that's how many there are in we have to pick them carefully. It's not that easy.

So that's what I'm working on: to find (and explicitly describe!) the group of permutations of six letters which is isomorphic to S5, and show that this is the Galois Group of the resolvent sextic. 

How to Generate Dummit's Resolvents

I said I was going to give this topic a rest, but it turns out I've got one more thing to say. Last time, in the comments, I said you could generate Dummit's Resolvents for a fifth-degree equation by taking a Lagrange Resolvent to the fifth power, then taking the other three Lagrange Resolvents you get from the first one by permuting the fifth roots of unity, taking the fifth power of those and adding all the resulting 480 terms togethers. You collect all the terms symmetric and throw them away, and what's left is Dummit's Resolvent.

And it almost works...but not quite. Lagrange's Resolvent is linear in the roots, so all the terms you get by taking fifth powers are fifth-order in the roots. But Dummit's Resolvents are made up of terms that are fourth-order in the roots:

AA(BE + CD) + BB(CA + DE).... etc.

And so my method can't work exactly. But it's close. The leftover terms when you follow my system look a lot like Dummit's:

AAA(BE + CD) + BBB(CA + DE)...etc; and also,

A(BBEE + CCDD) + B(CCAA + DDEE)....etc.

Since for the sake of Galois Theory we are only interested in the permutation properties of the resolvents, we don't need to keep the whole term...obviously the terms in A^cubed etc. are going to have permute the same way as Dummit's terms in A-squared. (But differently if you reduced it to linear in A, because you need to distinguish the terms outside the brackets from the ones inside!)

It's not even that hard to see why Dummit's resolvents "work", if you think about it. It's an awful lot of work to take the fifth power of all those Lagrange Resolvents and add them together: but if you take a step backwards and look at the big picture, you can see there are only a handful of different forms that the resulting terms can take:

AAAAA + BBBBB + CCCCC....etc; (five terms of this type)

AAAAB + BBBBC....etc; (twenty terms of this type)

AAABB etc. (twenty terms)

AAABC etc. (thirty terms)*

AABBC etc. (thirty terms)*

AABCD etc. (twenty terms)

There are only six different types of terms that you get; and only the fourth and fifth (I've marked them with an asterik) survive the symmetrization. And the reason is kind of funny. The terms in AAAAA etc. are obviously symmetric, so they add up to an integer value which you can throw away. The terms in AAAAB etc....well, when you take the fifth powers, these all have a fifth root of unity stuck to them. (Try it if you don't believe me.) There are twenty terms...five of them have a multiplier of w, five of them w^2, etc. You know the sum of the fifth roots of unity is -1; so when you add all twenty terms together, you get:

-(AAAAB + BBBBC +.....+ CCCCA +...etc)

which is also symmetric, so you throw them away.

And the same thing happens whith the third and sixth terms. But the terms I've marked off are a little different. For them, you get all the terms with with a multiplier of a fifth roots of unity: but you also get real terms. You collect the terms with the fifth roots, and they add up to a negative term...the other terms add up positive. You get:

-(AAABC + AAACD +.... + CCCDA +...) + (AAABE + AAACD....)*

You get all thirty terms, but some of them are negative and some of them are positive....the ones I've marked with an asterik. You know how to symmetrize this! It's just

-(all thirty terms) + 2*(AAABE + AAACD)

You throw away the symmetric terms and what's left are the (modified) Dummits Resolvents.

And it's easy to make a sixth degree equation out of them. Remember, there were 120 possible Lagrange resolvents based on the order you choose the roots of the fifth: but when you take fifth powers, this reduces to 24 different ones because some of them are just rotations of others. Then I grouped them according to ones that are related via legal permutations of the roots of unity....this gave me six groups of four. Each group of four generates, by the reduction I've shown above, a bunch of symmetric term (which you throw away) plus a leftover group of ten terms which have the form of Dummit's Resolvent.

It's not hard to convince yourself that while Dummits Resolvents are not symmetric in A,B,C,D and E, their generalized sums and products certainly are. It's a bit of work but before long you can derive therfore integer values for all the elementary symmetric polynomials in the conjugate Dummit Functions.

And those are the coefficients of the sextic equation which is called the "Resolvent Equation" of your original quintic equation.

Wednesday, January 15, 2014

Let's give Galois a rest

I don't think I'll have that much more to say about Galois theory. But I think it's been a pretty good run. I don't think there's anywhere on the Internet where you'll find a better explanation of why you can't solve the quintic than what we've posted here over the last couple of months. At least, you won't find a more understandable explanation.

It's not that I'm smarter than the other people in the game. It's that I'm playing a different game than they are. In academia, the big thing in math is to be rigorous and abstract. What they value is exactness and economy, and if you ask them why something works that way, they say: which line of the proof do you not understand? I don't think that's what math is about.

The difference is most clear when you consider my emphasis on the question: what are the functions on five letters which map to each other under any permutation of those five letters? (That's the question Balarka answered last month on stackexchange which led to our correspondence in the last several blogposts.) Mathematicians don't ask this question. In doing Galois theory, they ask about the existence of normal subgroups. I would venture to say that the great majority of them don't even realize that the significance of those normal subgroups is precisely that they are associated with my "functions that map to each other under permutation." And by the's not a one-to-one association. I didn't realize it at the time, but the function Balarka gave me, "Dummit's Function", is associated with the trivial subgroup consisting of exactly one element...the identity subgroup. So for S5, the resulting quotient group is just S5 itself. What it means is you won't find Dummit's Functions if you're looking for a normal subgroup to generate them.

What makes me think mathematicians don't look at things this way? I think I'm on pretty solid ground here. I would say that the fact that Dummit's Functions weren't even discovered until 1991 (!) is pretty good evidence that this way of thinking is outside the mainstream.

The point of all this is that the unsolvability of the quintic equation makes (almost!) intuitive sense if you play my game and look for functions of five letters that map to each other under permutation...but it becomes totally opaque if you abstract away all the concrete manifestations of the group action, and restrict yourself to talking about towers of normal subgroups. Even the well-known definition of a normal subgroup makes no intuitive's a subgroup for which "every left coset is also a right coset". What are we supposed to make of that? Mathematicians do that kind of thing all the time...they define two numbers p and q as being relatively prime if "there exists a and b such that ab-pq=1". That's a definition? It's no so harmful in this instance because we all know intuitively what it means for two numbers to be relatively prime, but in group theory, when you cascade one opaque definition like this on top of another, the whole subject rapidly becomes incomprehensible.

So at the end of the day, I'm pretty satisfied that I've explained what's really going on with the fifth degree equation. There may be a few fine points I haven't nailed down, but I'm not too worried about that. If you want to know why the quintic is unsolvable, I don't think you'll find a better place to start looking for answers than this blog right here.

Having gotten that out of the way, I think I'm going to return to my previous order of business, where I was posting my last-year's articles from the Jewish Post. I have to admit that my physics audience is not necessarily taking to this material with enthusiasm, but it's kind of important for me to get them out there. So let's see what's next...

Tuesday, January 7, 2014

Galois Group C5: Constructing a Polynomial

I don't have any great illusions that I'm doing significant math here, but having started down the road, it's funny where I've ended up. Using recursive relations of the form

A2 = A1^2 -n

and requiring A4 to equal A1, we were able to generate any number of cubic equations with cyclic Galois groups. Of course we always got a cyclic relationship between A1, A2 and A3, but we didn't always get cubic equations with integer coefficients. The surprising thing was that sometimes we did...for n=2, 4, 8 and (apparently) 14.

It wasn't surprising that we got the cyclic relationship for n=2, because that comes straight from the properties nth-cyclotomic polynomial. The sum of any nth root of unity with its conjugate, squared, differs by two from some other nth roots of unity. And specifically for the 7th roots of unity, you get the roots of a cubic by summing the 1st and 6th, the 2nd and 5th, and the 4th and 7th.

So I thought I'd try and generate the 5th-degree equation having a cyclic galois group by doing the same trick on the 11th roots of unity. I had already figured out that I didn't need to be a genius at factoring...I could plot the cyclic relationship in Excel and monitor the 5 elementary symmetric polynomials in the roots to get the coefficients of the equation. Sure enough, I verified that the following equation had its roots in a cyclic relationship:

x^5 +x^4 -4x^3 - 3x^2 +3x +1 = 0

Now, when we did the cubic, we got an extra polynomial "for free"...remember how? Applying the cyclic relationship three times gave us an eigth degree equation...we factored out the trivial quadratic corresponding to the case of the three roots being equal, leaving us a sextic. The cubic generated by the seventh roots of unity factored cleanly into the sextic, giving us a brand new cubic, also cyclic in the roots.

So what about the fifth degree?

Well, when we apply the recursive relationship five times, we get a thirty-second degree equation, which, upon factoring out the trivial quadratic, gives us thirty roots. Thirty?

I started hunting for roots in Excel, and it seemed I was finding all kinds of 5-cycles wherever I looked, and for the most part they didn't give me equations in integer coefficients. Remember, I had Excel displaying the values of the elementary symmetric functions...that's where I got the nice fifth-degree equation I showed you earlier. But most of the 5-cycles I generated corresponded to garbage polynomials with irrational coefficients. Where were they coming from?

Eventually, I figured it out. They came from the 31st roots of unity, and the 33rd roots of unity. When you apply the recursive relationship to the sum of any root and its conjugate, the new power of the root you generate is just double the one you started with. So operating on the eleventh roots of unity, you get the powers:

1, 2, 4, 8, 16 (which is the same as 5) and 10 (which is the same as -1).

So five iterations brings you around to the conjugate of the root you started with. It works the same if its the root itself or the conjugate, because either way your going to be taking sums.

So what about the 31st roots of unity? Well here, the sequence is:

1, 2, 4, 8, 16, 32(=1)

so adding these to their conjugates give you a 5-cycle. But you can also take these alternate choices for the roots:

3, 6, 12, 24, 17, 34(=3)      or     5, 10, 20, 9, 18, 36(=5) 

and in either case, adding them to their conjugates gives you a five-cycle in the roots. None of which, by the way, give you fifth-degree equations in integer coefficients.

So that's counting the ring generated by the eleventh roots of unity, that's four. But we had a thirtieth-degree equation to account for...there should be six rings of roots. That's where the 33rd-degree equation comes in.

The cycle {1, 2, 4, 8, 16, 32(=-1)} gives us five. The cycle {5, 10, 20, 8, 16, 32(=-1)} gives us five more. And the cycle {3, 6, 12, 24, 14, 28(=-3)} is the same cycle we already generated from the 11th degree!

Those are the six five-cycles accounting for all the roots of the 30th-degree equation. Only one of the 5-cycles gives us a 5th-degree in integer coefficients; but in any case, there they are.

I don't know what you want to take away from this discussion, but if you ever had any doubts that group theory applies to the theory of equations, well, those doubts should pretty much be laid to rest.

Saturday, January 4, 2014

Excel To The Rescue

We've been having some fun with cubic equations. The other day we figured out that the relations A^2-4=B, when applied transitively three times in a row, generates the roots of a cubic equation. The trick is that you have to find a starting point whereby, after three iterations, you get back to where you've started. You can see what is happening on this graph:

You can see that starting with A=1.6, we generate B= -1.4 and C= -2.4. Let's look at the elementary symmetric polynomials in A, B and C to see what our cubic equation should look like:

A + B + C = -2.2
AB + BC + CA = -2.72
ABC = 5.376

Now, if we remember what Balarka did the other day, he generated an octic equation by substituing the recursive relationship into itself three times, factored out the trivial quadratic, and then factored the resulting sextic into two terms, one of which was:

x^3 + 2x^2 -3x +5

This is really close to the equations you get from my symmetric polynomials in A, B and C....considering I just drew those lines on the graph by eyeball (no cheating), I'm a bit amazed how well it worked. (Remember the signs alternate when you sub in the coefficients...I told Balarka he had a sign error yesterday, but I was wrong.)

It was actually quite a surprise when Balarka's octic factored cleanly. We had started with the recursive relationship A^2 - 2 = B....going through the same steps, generating the octic, factoring out the quadratic, and factoring the resulting sextic...well, we knew that one had to factor because there was already a known factor that was generated from combining the seventh roots of unity. But we didn't actually expect the arbitrary relationship A^2-4 to also factor.

So what about A^2-6? Well, I could have tried graphing it, but then I had a better idea...why not program the relationship into Excel? It's a super-easy spreadsheet literally need only five or six cells. You just keep iterating your seed value until you get the recursion working. And of course, with instant correction, you can easily work it out to four or five decimal places. It's kind of fun.

I tried it first on A^2-4, and verified that the cycle did in fact fact, there were (of course) two cycles, corresponding to the two cubics that we factored out of the sextic. But how could we tell that these were the roots of a cubic equation in integer coefficients? That was easy! I just got Excel to display the values of the sums and products of the roots. At the sime time as the cyclic relationship converged back on itself, you could see the sums and products zeroing in on integer values. We had not only found the roots but also the cubic equation which they came from?

So what about A^2 - 6? (For some reason it seemed reasonable to stick to even values.) This one was different. I found the following triplets with cyclic relationships:

(1.7425, -2.9636, 2.7827)
(2.0547, -1.7782, -2.8380)

but checking the elementary symmetric combinations of these triplets, it was clear that they were not the roots of cubic equations with integer coefficients:

(1.56, -8.56, -14.37)
(-2.56, -4.44, 10.37)

(These triplets came up as integers when we worked with A^2-4). But were the coefficients of the two cubic equations at least something like the roots of a quadratic? Indeed they were. Examining the pair of triplets listed above, you can see by taking sums and products that the coefficients are related as follows:

The x^2 coefficients are given by the relationship x^2 + x - 4 = 0
The x coefficients are given by the relationship x^2 + 13x +38 = 0
The constant coefficients are given by x^2 +4x - 149 = 0

And that's about all I've got to say for now...

Thursday, January 2, 2014

Cubic Equations With Galois Group C3

The other day I identified a cubic equation whose Galois Group, I claimed, was C3. Here is the equation:

x^3 + x^2 - 2x -1 = 0

I promised to show you that the roots of this equation had a cyclic relationship. Let's remember how we got them. We started with the seventh roots of unity, and generated this cubic equation by letting its roots be

A = w^1 + w^6
B = w^2 + w^5
C = w^3 + w^4

It's very easy to see from these relationships that A^2 = B + 2, and similar equations relate B to C and C back to A. Those are the cyclic relationships.

You can also notice that the splitting field of the original equation is just the Q(A), or the rationals extended by appending A. Or B, or just have to append one of the roots, and the resulting field contains all three. That's different from a typical cubic where the three extensions of Q are quite disjoint. It's obvious if you have two imaginary roots and one real root, the field you get by adjoing the real root is not going to include the comples roots. So no simple equation can link one of the roots to the next, as we obtain in the present case. In short, for a cubic to have a Galois group which is cyclic, it pretty much means that all three roots live in the same simple extension, and hence functions of each other.

Does this give us a recipe for constructing cyclic cubics? In the present case, we had the relationship A^2-2=B. What if we say we're going to look for a field where, for example, A^2-4=B? Will we generate another cubic equation with a cyclic relation among the roots?

I'm not totally sure this will work. A funny thing happens if we try...but let's first try with the case where we know it has to work...with A^2 - 2 = B. If the roots have a cyclic relationship, then we ought to also have:

B^2 - 2 = C; and,

C^2 - 2 = A

But this brings us back full circle...we can substitute one equation into another, and eleminate B and C, giving us:

A^8 - 8A^6 + 20A^4 - 16A^2 -A + 2 = 0

It's not a cubic at's an eight-degree equation. But wait...maybe that makes sense. After all, one "trivial" solution to our cyclic relationship between A, B and C is simply that

A^2-2=A (because A = B = C)

So this should theoretically factor out of our eight-degree equation...and indeed it does, leaving us the sixth-degree equation:

A^6 + A^5 - 5A^4 - 3A^3 + 7A^2 + A - 1 = 0

Still not the cubic we're looking for...or is it? Maybe the cubic we already found is actually buried inside this sixth-degree equation. And in fact, we find that it is...on synthetic division, we find that it does indeed go factor through, leaving the residue:

A^3 - 3A + 1 = 0

And there it is...a whole new cubic, generating a different set of A's, B's and C's than the ones we already had, but with the same cyclic relationship amongst themselves. It's not obvious to me that these new ABC's are especially connected to the seventh roots of unity, but I stand to be corrected. For now, all I can say is...there they are. (I actually found them by's not to hard to verify that the numbers (-1.9, 0.35, 1.8) satisfy not only the cubic equation written above, but also the cyclic relationship wherewith they were generated.

So can you start with any similar cyclic relationship and generate a pair of cubic equations with cyclic roots? Sadly, it does not appear so easy. I tried starting with A^2 - 4 = B, generated an eight-degree equation, factored out the quadratic giving me the sixth-degree, but then....I couldn't see any way to factor the sixth degree into two cubics.

And that's about as far as I'm likely to get down this road in the immediate future. Unless anyone out there has any better ideas...