Linger

Dolores O’Riordan, singer in the Cranberries, died today.  In the fall of 1993 I was living in an apartment by myself for the first time, the Baltimorean on N. Charles Street.  I was devoting myself full-time to being a writer and kind of hating it.  I didn’t know anyone in Baltimore and the people in my program were mostly older than me and socially inaccessible and I was lonely.   The apartment was always too hot.  I ate spaghetti with jar sauce for dinner by myself and listened to “Linger.”  It’s still the sound of loneliness to me, after all these years.

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Farblandia

The job fell to me of giving an overview talk about Benson Farb’s entire career at his birthday conference last fall.  Hard task!  I didn’t cover nearly everything but I think I gave a decent impression of what Farbisme is all about.

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Wanlin Li, “Vanishing of hyperelliptic L-functions at the central point”

My Ph.D. student Wanlin Li has posted her first paper!  And it’s very cool.  Here’s the idea.  If chi is a real quadratic Dirichlet character, there’s no reason the special value L(1/2,chi) should vanish; the functional equation doesn’t enforce it, there’s no group whose rank is supposed to be the order of vanishing, etc.  And there’s an old conjecture of Chowla which says the special value never vanishes.  On the very useful principle that what needn’t happen doesn’t happen.

Alexandra Florea (last seen on the blog here)  gave a great seminar here last year about quadratic L-functions over function fields, which gave Wanlin the idea of thinking about Chowla’s conjecture in that setting.  And something interesting developed — it turns out that Chowla’s conjecture is totally false!  OK, well, maybe not totally false.  Let’s put it this way.  If you count quadratic extensions of F_q(t) up to conductor N, Wanlin shows that at least c N^a of the corresponding L-functions vanish at the center of the critical strip.  The exponent a is either 1/2,1/3, or 1/5, depending on q.  But it is never 1.  Which is to say that Wanlin’s theorem leaves open the possibility that o(N) of the first N hyperelliptic L-functions vanishes at the critical point.  In other words, a density form of Chowla’s conjecture over function fields might still be true — in fact, I’d guess it probably is.

The main idea is to use some algebraic geometry.  To say an L-function vanishes at 1/2 is to say some Frobenius eigenvalue which has to have absolute value q^{1/2} is actually equal to q^{1/2}.  In turn, this is telling you that the hyperelliptic curve over F_q whose L-function you’re studying has a map to some fixed elliptic curve.  Well, that’s something you can make happen by physically writing down equations!  Of course you also need a lower bound for the number of distinct quadratic extensions of F_q(t) that arise this way; this is the most delicate part.

I think it’s very interesting to wonder what the truth of the matter is.  I hope I’ll be back in a few months to tell you what new things Wanlin has discovered about it!

A rejection from Gordon Lish

At some point I’m going to go through my gigantic file of rejection letters from the mid-1990s, when I was trying to be a fiction writer, and write a post about them, but for now, here’s the one I have from Gordon Lish, from 27 June 1996, when he was editing The Quarterly:

Excellent Jordan — you only impress me further with your force. I
would take the cut if I liked it. I don’t — but I do admire the man

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How I won the talent show

So I don’t want to brag or anything but I won the neighborhood talent show tonight.  Look, here’s my trophy:

My act was “The Great Squarerootio.”  I drank beer and computed square roots in my head.  So I thought it might be fun to explain how to do this!  Old hat for my mathematician readers, but this is elementary enough for everyone.

Here’s how it works.  Somebody in the audience said “752.”  First of all, you need to know your squares.  I think I know them all up to 36^2 = 1296.  In this case, I can see that 752 is between 27^2 = 729 and 28^2 = 784, a little closer to 729.  So my first estimate is 27.  This is not too bad:  27^2 = 729 is 23 away from 752, so my error is 23.

Now here’s the rule that makes things go:

new estimate = old estimate + error/(2*old estimate).

In this case, the error is 23 and my old estimate is 27, so I should calculate 23/2*27; well, 23 is just a bit less than 27, between 10 and 20% less, so 23/2*27 is a bit less than 1/2, but should be bigger than 0.4.

So our new estimate is “27.4 something.”

Actual value of 27 + 23/54: 27.4259…

so I probably would have gotten pretty close on the hundredth place if I’d taken more care with estimating the fraction.  Of course, another way to get even closer is to take 27.4 as your “old estimate” and repeat the process!  But then I’d have to know the square of 27.4, which I don’t by heart, and computing it mentally would give me some trouble.

Why does this work?  The two-word answer is “Newton’s method.”  But let me give a more elementary answer.

Suppose x is our original estimate, and n is the number we’re trying to find the square root of, and e is the error; that is, n = x^2 + e.

We want to know how much we should change x in order to get a better estimate.  So let’s say we modify x by d.  Then

$(x+d)^2 = x^2 + 2xd + d^2$

Now let’s say we’ve wisely chosen x to be the closest integer to the square root of n, so d should be pretty small, less than 1 at any rate; then d^2 is really small.  So small I will decide to ignore it, and estimate

$(x+d)^2 = x^2 + 2xd$.

What we want is

$(x+d)^2 = n = x^2 +e$.

For this to be the case, we need e = 2xd, which tells us we should take d = e/2x; that’s our rule above!

Here’s another way to think about it.  We’re trying to compute 752^(1/2), right?  And 752 is 729+23.  So what is (729+23)^(1/2)?

If you remember the binomial theorem from Algebra 2, you might remember that it tells you how to compute (x+y)^n for any n.  Well, any positive whole number n, right?  That’s the thing!  No!  Any n!  It works for fractions too!  Your Algebra 2 teacher may have concealed this awesome fact from you!  I will not be so tight-lipped.

The binomial theorem says

$(x+y)^n = x^n + {n \choose 1} x^{n-1} y + {n \choose 2} x^{n-2} y^2 + \ldots +$

Plugging in x=729,y=23,n=1/2 we get

$(729+23)^{1/2} = 729^{1/2} + {1/2 \choose 1} 729^{-1/2} \cdot 23 + {1/2 \choose 2} 729^{-3/2} \cdot 23^2 + \ldots$

Now 729^{1/2} we know; that’s 27.  What is the binomial coefficient 1/2 choose 1?  Is it the number of ways to choose 1 item out of 1/2 choices?  No, because that makes no sense.  Rather:  by “n choose 1” we just mean n, by “n choose 2” we just mean (1/2)n(n-1), etc.

So we get

$(729+23)^{1/2} = 27 + (1/2) \cdot 23 / 27 + (-1/8) \cdot 23^2 / 27^3 + \ldots$

And look, that second term is 23/54 again!  So the Great Squarerootio algorithm is really just “use the first two terms in the binomial expansion.”

To sum up:  if you know your squares up to 1000, and you can estimate fractions reasonably fast, you can get a pretty good mental approximation to the square root of any three-digit number.  Even while drinking beer!  You might even be able to beat out cute kids in your neighborhood and win a big honking cup!

Sex Has Thrown A Bomb Into Business

This article, written in 1927 by the psychoanalyst Smith Ely Jeliffe (a dude) has a take on workplace sexism that is, to me, startlingly contemporary.

Why Men Fail

That’s the book I picked up off the shelf while working in Memorial Library today.  It’s an book of essays by psychiatrists about failure and suboptimal function, published in 1936.  In the introduction I find:

We see what a heavy toll disorders of the mind exact from human happiness when we realize that of all the beds in all the hospitals throughout the United States one in every two is for mental disease; in other words, there are as many beds for mental ailments as for all other ailments put together.

That’s startling to me!  Can it really have been so?  What’s the proportion now?

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Landlord rights and Wisconsin home rule follies

The era of small government remains over in Wisconsin, as the state legislature continues to chew away at municipal self-governance.  This time:  cities are prohibited from requiring regular inspections of rental properties.

Just to remind you again what the Wisconsin Constitution says on this point:

Cities and villages organized pursuant to state law may determine their local affairs and government, subject only to this constitution and to such enactments of the legislature of statewide concern as with uniformity shall affect every city or every village.

Over the years, the state has accorded to itself the power to declare just about anything a city might do “of statewide concern,” rendering the Home Rule Amendment essentially null.  The statewide effect of Beloit requiring landlords to subject their rental properties to safety inspections every once in a while seems pretty minor to me.  I guess that’s why I’m not on the Wisconsin Supreme Court.

And yes, I get that there’s lots of interpretation of the Commerce Clause that runs roughly along the same lines.  And yes, I get that a strong interpretation of home rule would keep states from invalidating discriminatory municipal ordinances unless they ran afoul of federal law.  But these judges say they’re pure custodians of the Constitutional text.  It gets up my nose when they act as if it doesn’t exist.

Previous blog post where I complain at length about previous SC-WI home rule jurisprudence.

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Are Alabama’s House seats gerrymandered?

This map has a lot of people saying so:

Here’s what I think:  Alabama’s House maps might well be gerrymandered, but the Moore-Jones numbers aren’t very strong evidence.

First of all, about that weirdly shaped District 7 where so many Democrats live.  That’s a majority-minority district.  The Voting Rights Act requires the creation of some such districts, and that provision has increased the representation of racial minorities in Congress.  But most people agree they hurt Democrats overall.  You might be able to draw a district map for Alabama, Republican though it is, with two districts where Democrats have a chance instead of one.  But you’d also increase the likelihood of Alabama sending an all-white delegation.

Alabama, without District 7, is about 78% white, and white people in Alabama are about 85% Republican.  It’s not gerrymandering that Dems don’t have a chance in those six districts under normal conditions; it would happen just about any way you drew the maps.

But the Moore-Jones election was anything but normal conditions!  Did the party draw a map designed to withstand a historic Democratic turnout wave?

I doubt it.  Suppose you wanted to draw a map that would keep your big House majority even if just over half of Alabamian voters chose the Democrat.  You’ve got no chance in AL-7, and in the other 6 districts combined, the Republican is winning by 10 points.  Well, the last thing you’d do is draw an ultra-Republican district like AL-4; that makes the other districts way too close.  You’d take some of those wards and move them over to shore up AL-5, which Moore won by less than half a percent.  You might also try to concentrate the more Democratic parts of AL-1 and AL-2 into one, creating a district you might lose in a wave but leaving the rest of the state so solidly Republican that even an election more disastrous than this one for Republicans would leave five seats in GOP hands.

Outside district 7, Alabama is a very Republican state, even when offered the weakest Republican candidate in recent memory.  That’s the simplest explanation for why Moore finished ahead in districts 1-6, and it’s the one I favor.

Update:  Some people have communicated to me that in their view district 7 is more majority-minority than the Voting Rights Act requires, and that the district was drawn this way on purpose in order to increase Republican margin in the other 6 districts.

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“Worst of the worst maps”: a factual mistake in Gill v. Whitford

The oral arguments in Gill v. Whitford, the Wisconsin gerrymandering case, are now a month behind us.  But there’s a factual error in the state’s case, and I don’t want to let it be forgotten.  Thanks to Mira Bernstein for pointing this issue out to me.

Misha Tseytlin, Wisconsin’s solicitor general, was one of two lawyers arguing that the state’s Republican-drawn legislative boundaries should be allowed to stand.  Tseytlin argued that the metrics that flagged Wisconsin’s maps as drastically skewed in the GOP’s favor were unreliable:

And I think the easiest way to see this is to take a look at a chart that plaintiff’s own expert created, and that’s available on Supplemental Appendix 235. This is plain — plaintiff’s expert studied maps from 30 years, and he identified the 17 worst of the worst maps. What is so striking about that list of 17 is that 10 were neutral draws.  There were court-drawn maps, commission-drawn maps, bipartisan drawn maps, including the immediately prior Wisconsin drawn map.

That’s a strong claim, which jumped out at me when I read the transcripts–10 of the 17 very worst maps, according to the metrics, were drawn by neutral parties!  That really makes it sound like whatever those metrics are measuring, it’s not partisan gerrymandering.

But the claim isn’t true.

(To be clear, I believe Tseytlin made a mistake here, not a deliberate misrepresentation.)

The table he’s referring to is on p.55 of this paper by Simon Jackman, described as follows:

Of these, 17 plans are utterly unambiguous with respect to the sign of the efficiency gap estimates recorded over the life of the plan:

Let me unpack what Jackman’s saying here.  These are the 17 maps where we can be sure the efficiency gap favored the same party, three elections in a row.  You might ask: why wouldn’t we be sure about which side the map favors?  Isn’t the efficiency gap something we can compute precisely?  Not exactly.  The basic efficiency gap formula assumes both parties are running candidates in every district.  If there’s an uncontested race, you have to make your best estimate for what the candidate’s vote shares would have been if there had been candidates of both parties.  So you have an estimate for the efficiency gap, but also some uncertainty.  The more uncontested races, the more uncertain you are about the efficiency gap.

So the maps on this list aren’t the 17 “worst of the worst maps.”  They’re not the ones with the highest efficiency gaps, not the ones most badly gerrymandered by any measure.  They’re the ones in states with so few uncontested races that we can be essentially certain the efficiency gap favored the same party three years running.

Tseytlin’s argument is supposed to make you think that big efficiency gaps are as likely to come from neutral maps as partisan ones.  But that’s not true.  Maps drawn by Democratic legislatures have average efficiency gap favoring Democrats; those by GOP on average favor the GOP; neutral maps are in between, and have smaller efficiency gaps overall.

That’s from p.35 of another Jackman paper.  Note the big change after 2010.  It wasn’t always the case that partisan legislators automatically thumbed the scales strongly in their favor when drawing the maps.  But these days, it kind of is.  Is that because partisanship is worse now?  Or because cheaper, faster computation makes it easier for one-party legislatures to do what they always would have done, if they could?  I can’t say for sure.

Efficiency gap isn’t a perfect measure, and neither side in this case is arguing it should be the single or final arbiter of unconstitutional gerrymandering.  But the idea that efficiency gap flags neutral maps as often as partisan maps is just wrong, and it shouldn’t have been part of the state’s argument before the court.