Simplest exponential sum

Today‘s exponential sum curve is simply a triangle.

2018_09_06

But yesterday‘s curve was more complex

2018_09_05

and tomorrow‘s curve will be more complex as well.

2018_09_07

Why is today’s curve so simple? The vertices of the curves are the partial sums of the series

\sum_{n=0}^N \exp\left( 2\pi i \left( \frac{n}{m} + \frac{n^2}{d} + \frac{n^3}{y} \right ) \right )

where m is the month, d is the day, and y is the last two digits of the year. Typically the sums are too complicated to work out explicitly by hand, but today’s sum is fairly simple. We have m = 9, d = 6, and y = 18. The three fractions add to (2n + 3n² + n³)/18. Reduced mod 18, the numerators are

0, 6, 6, 6, 12, 12, 12, 0, 0, 0, 6, 6, 6, 12, 12, 12, 0, 0

The repetition in the terms of the sum leads to the straight lines in the plot. The terms in the exponential sum only take on three values, the three cube roots of 1. These three roots are 1, a, and b where

a = exp(2πi/3) = -1/2 + i√3/2

and

b = exp(-2πi/3) = -1/2 – i√3/2

is the complex conjugate of a.

Using Mathematica we have

    Table[ Exp[2 Pi I (2 n + 3 n^2 + n^3)/18], {n, 0, 17}] 
        /. {Exp[2 Pi I/3] -> a, Exp[-2 Pi I/3] -> b}

which produces

1, a, a, a, b, b, b, 1, 1, 1, a, a, a, b, b, b, 1, 1

When we take the partial sums, we get four points in a straight line because they differ by a:

1, 1 + a, 1 + 2a, 1 + 3a

 then three points in a straight line because they differ by b:

(1 + 3a), (1 + 3a) + b, (1 + 3a) + 2b, (1 + 3a) + 3b

and so forth.

 

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