Fourier Coefficient Visualizer (ZIP)

The Python interpreter with everything you need to run the program is available for free download on the web. For an introduction to the discrete Fourier transform, see my exposition on Fourier Pseudorandomness, or check out the detailed but typically excessive article on Wikipedia.

Let me know what you think in the comments!

]]>On Fourier Pseudorandomness (PDF)

Feel free to ask questions, request clarification, pick nits, and otherwise discuss in the comments!

]]>We briefly considered this construction at the 2010 University of Georgia mathematics REU, and in order to clarify some of the details involved, I wrote up a proof which discusses parts of the construction in greater length than the reference material we considered. I believe that the content should be understandable with only modest mathematical training, but if you would like clarification on anything, let me know in the comments! Enjoy.

]]>The Riemann zeta function . . . [Read More]]]>

The Riemann zeta function is most simply described as a series, the infinite sum over n of 1/n^{z} for complex z. This series converges when the real part of z is greater than 1, and diverges when it is less than 1. The convergence properties of the series are less straightforward when Re(z) = 1. Notice, for instance, that substituting z = 1 describes the harmonic series, which diverges, so we see that the series diverges there.

Looking at the zeta function for values of z with real part less than 1 is a bit trickier. Through an analytic technique similar in flavor to integration by parts, a representation can be obtained which is good for Re(z) > 0. From there, it is possible to use a functional equation which relates values of the function with corresponding values reflected across the line Re(z) = 1/2. It turns out that the Riemann zeta function is meromorphic on the entire complex plane with a single simple pole at the point z = 1. For those without background in complex analysis, this means that the function is infinitely differentiable and has a convergent Taylor series at every point except the pole, and if you divide by (z-1), you get a function which has this property on the entire plane.

So the zeta function is, in some sense, a very nice function. In fact, there are deep connections between the zeta function and prime number theory which require some study to appreciate. Again and again, a fundamental question emerges about the function: Where on the complex plane does the function take value zero? It is possible to show that the function is strictly non-zero on the half-plane Re(z) > 1. Using the functional equation, this also shows that the function is non-zero in Re(z) < 0, except at the negative even integers z = -2, -4, -6, ... . The real question concerns zeros that occur in what is known as the "critical strip" 0 ≤ Re(z) ≤ 1. It is known that there are infinitely many of these "non-trivial" zeros, and that they have reflective symmetry across the real axis, and across the line Re(z) = 1/2.

The Riemann hypothesis claims that the non-trivial zeros of the zeta function all lie on the line Re(z) = 1/2. The conjecture, proposed by Bernhard Riemann in 1859, has been attacked by many of the most prominent modern mathematicians, but no proof has been forthcoming. However, no *disproof* has emerged either. Computers have shown that the first 10 trillion zeros in the critical strip, arranged in order of positive imaginary part, satisfy the hypothesis. It is commonly accepted in the mathematical community that the hypothesis is true (hence the name “hypothesis” rather than “conjecture”), but rigorous mathematicians will not be truly satisfied until a proof has been found.

The typical definition for the sum of an arbitrary collection of non-negative summands is to take the supremum of sums over all finite subsets of the summands. It can be shown that if the collection S of positive summands in the sum is uncountable, then the sum must be infinite. Indeed, suppose by way of contradiction that the sum of elements in S is finite. Then the collection S_{n} of summands s ≥ 1/n must be finite for each n–otherwise, the supremum over all finite collections of such summands would be larger than k/n for any k. But then we may write S as the countable union over all n of the S_{n}, and this implies that S is countable, a contradiction with the starting hypothesis. Thus we must conclude that the sum is infinite.

Although a general sum is not defined for complex summands, the argument can be extended to this context to show that partial sums of an uncountable collection of complex numbers may have arbitrarily large magnitude. Certainly, using the same argument as above, an uncountable number of summands must have magnitude at least 1/n for some n. Further, an uncountable number of these must have argument within an interval of length at most π/4. The combination of these two conditions restricts either the imaginary or the real part of the summands, fixing the sign and providing a strictly positive lower bound on the magnitude. Since we have infinitely (uncountably) many such summands, the result follows.

]]>From a topological point of view, this certainly seems to be a reasonable observation. After all, the rationals are dense and have measure zero, so it isn’t surprising that by making the measure positive one can add the condition of openness. But from a more direct perspective, the construction seems to yield a very strange set–a spattering of droplets across the real line which cover very little length but come arbitrarily close to any point. Like an Impressionist painting, one must add a little distance to make out the comprehensible form.

Image by Claude Monet.

]]>To formalize . . . [Read More]]]>

To formalize this problem, consider the party as a complete graph on six vertices, as in the figure to the left. Each vertex represents a single guest, and each edge is colored by either blue, to represent that the connected vertices are acquaintances, or by yellow, to represent that the connected vertices are strangers. Then the result states that there exists a complete subgraph on three vertices with edges all the same color.

To see this, pick a vertex of the graph. Then because there are five edges going from this vertex to other vertices, there must be at least three edges which have the same color. Assume without loss of generality that the three edges are blue, as in the figure to the right. Then if there is a blue edge between any of the three vertices connected to the chosen vertex (vertices 3, 4 and 6 in the figure), then there is a complete same-colored subgraph, and the result holds.

So suppose instead that there are no blue edges between the three nodes. Then all three of the edges must be yellow, as in the figure to the left, and this forms a complete same-colored subgraph of the opposite color. Thus no matter the choice of coloring on the graph, we can find a same-colored complete subgraph on three vertices. As for the dinner party, we *certainly* can find a group of three people with the desired property, if we only care to look.

n^{2} + (n+1)^{2} + … + (n+k)^{2} = (n+k+1)^{2} + … + (n+2k)^{2}.

For example,

0^{2} = 0,

3^{2} + 4^{2} = 5^{2},

10^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2},

21^{2} + 22^{2} + 23^{2} + 24^{2} = 25^{2} . . . [Read More]]]>

n^{2} + (n+1)^{2} + … + (n+k)^{2} = (n+k+1)^{2} + … + (n+2k)^{2}.

For example,

0^{2} = 0,

3^{2} + 4^{2} = 5^{2},

10^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2},

21^{2} + 22^{2} + 23^{2} + 24^{2} = 25^{2} + 26^{2} + 27^{2},

and so on.