That actually refocuses me/answers a lot of questions, because it hadn’t really _occurred_ to me that I might cover integrals in the course. But now that I look at the syllabus from this past year, that’s exactly what the last professor did.

The Analysis I course I took in undergrad definitely had linear algebra as a prerequisite. And it covered the definitions of “metric” and “norm”, convergence of sequences in different spaces and norms, and built up at the end to completeness and compactness.

In the second term we mostly spent a lot of time talking about convergence of sequences of functions, but we also talked about Riemann integrals of functions from R to an arbitrary Banach space. (The professor felt there was no reason to talk about Rieman integrals over other domains, since the Lesbesgue integral subsumes all of that). We did the inverse and implicit function theorems, and at the end of the course we defined the Frechet derivative, which I think of as one of the defining moments of my growth as a mathematician.

I don’t think I had to compute a single integral in the entire year.

(It wouldn’t even have occurred to me to count things like Stokes’s theorem as a possible topic for the course; I have that classes with differential geometry or something instead).

—

And this sort of answers my question about why people would do the single-variable version: it lets you do derivatives and integrals. I was imagining a course that was just about convergence and completeness and compactness, and I don’t know that that course benefits much from a restriction to a single variable—especially since most of the interesting convergence properties happen for sequences of functions.

But if people are thinking of the course as “rigorous calculus” rather than “the background you’d need for functional analysis to make sense” then that choice makes a lot more sense.

]]>Econ -003: This is just one example of a specialty condition which complicates the US tax code. There is an EconTalk with a flat-tax* advocate who claims that citizens of a few post-Soviet states (with a new government and a clean slate on which to write the tax code) — I think Estonia among them — can fit their tax form on a postcard.

* This flat tax only kicks in above a threshold, so the poor do pay less. Complicated tax codes abet bureaucracy and give advantage to those who can afford to pay many people to help them flout the intent of the law — like Apple Computer.

Econ 202: Government spending is transparent when done on the spending side and opaque when done through the (non)taxation side. However, lowering taxes shares the advantage with bidding out contracts that the government itself doesn’t have to figure out what to do with the money.

]]>Oh, you probably meant the thing I just posted with a list of relevant House reps. That’s from Alexander Chee’s post:

I show a solution to that problem as follows:

Given a number x and a set S of n positive integers, MINIMUM is the problem of deciding whether x is the minimum of S. We can easily obtain an upper bound of n comparisons: find the minimum in the set and check whether the result is equal to x. Is this the best we can do? Yes, since we can obtain a lower bound of (n – 1) comparisons for the problem of determining the minimum and another obligatory comparison for checking whether that minimum is equal to x. A representation of a set S with n positive integers is a Boolean circuit C, such that C accepts the binary representation of a bit integer i if and only if i is in S. Given a positive integer x and a Boolean circuit C, we define SUCCINCT-MINIMUM as the problem of deciding whether x is the minimum bit integer which accepts C as input. For certain kind of SUCCINCT-MINIMUM instances, the input (x, C) is exponentially more succinct than the cardinality of the set S that represents C. Since we prove that SUCCINCT-MINIMUM is at least as hard as MINIMUM in order to the cardinality of S, then we could not decide every instance of SUCCINCT-MINIMUM in polynomial time. If some instance (x, C) is not in SUCCINCT-MINIMUM, then it would exist a positive integer y such that y < x and C accepts the bit integer y. Since we can evaluate whether C accepts the bit integer y in polynomial time and we have that y is polynomially bounded by x, then we can confirm SUCCINCT-MINIMUM is in coNP. If any single coNP problem cannot be solved in polynomial time, then P is not equal to coNP. Certainly, P = NP implies P = coNP because P is closed under complement, and therefore, we can conclude P is not equal to NP.

You could read the details in the link below…

]]>I’ve asked on Twitter about this — see thread and responses from Elsevier ]]>