The Swiss Minaret Ban, or why I love America, part 13,207

You may have heard of the controversy over the recent ban on new construction of minarets in Switzerland.  22 of 26 cantons, representing 57.5% of the popular vote, voted for the ban.  Multiculturalists and religious officials were quite upset with the ban.

To me, this ban is interesting in not only what it says about Switzerland as a nation, but also as a nation of law in contrast to the US.

To be perfectly honest, I don’t think I would much care for a minaret in my neighborhood.  This is not due to any distaste for Islam, but rather distaste for so much of my home going to something which will only annoy me.  I imagine this is what many Swiss voters felt.

But what I feel as a single voter is of little consequence.  Rather, it’s how the legal infrastructure is set up to deal with my feelings.  And clearly, in the US we have a huge advantage.  Recall the First Amendment to the Constitution:

Congress shall make no law respecting an establishment of religion, or prohibiting the free exercise thereof…

If we wanted a national ban on minarets, as the Swiss have done, we’d have to amend the Constitution.  Good luck with that.  But, such a national ban shouldn’t be done in the first place.  Minarets should be allowed to be built where there is local approval.  If there is an area chock-a-block full of devout Muslims, then they should get a minaret.  If a few Muslims want to build a minaret in a place that will annoy the local populace, then local regulations will prevent the building.  The system we have is, of course, not perfect: if it were, then the Mormon temple serving the Boston area would not be dominating a residential neighborhood in Belmont.  But, at the very least, we cannot embarrass ourselves as the Swiss apparently have by making a local issue first a national, and then an international, one.

Nonlinear system producing periodic functions

Consider the following system:

\(y’\) = \(-z^3\); \(z’\) = \(y^3\); \(y[0]=1\), \(z[0]=0\).

Show that the solutions \(y=f[x]\) and \(z=g[x]\) are periodic functions, and evaluate the period.


Clearly, \(y^3 y’\) + \(z^3 z’=0\), so that the quantity \(y^4+z^4\) is constant.  Given the boundary conditions, \(y^4+z^4=1\) for all values of \(x\).  When plotted in phase space, this is a closed curve and is therefore representative of a periodic system.

To compute the period \(p\), we begin with the obvious observation that

\(p=\int_{0}^{p}dx=4 \int_{0}^{p/4}dx\).

Note that


so that

\(p=4 \int_{0}^{1}dz(1-z^4)^{-\frac{3}{4}}\)

This integral is doable by using the substitution \(w=1-z^4\) to reveal that


The true nerd will recognize the above integral is in the form of a Beta integral and is thus equal to

\(p\) = \(\frac{\Gamma[\frac{1}{4}]^2}{\sqrt{\pi}}\) \(\approx 7.4163\)