Today’s the first day of a new month, which means the exponential sum of the day will be simpler than usual. The exponential sum of the day plots the partial sums of
where m, d, and y are the month, day, and (two-digit) year. The n/d term is simply n, and integer, when d = 1 and so it has no effect because exp(2πn) = 1. Here’s today’s sum, the plot formed by the partial sums above.
It’s possible in principle to work out exactly each of the points are on the curve. In order to do so, you’d need to compute
Since we can use the Pythagorean theorem to compute sine from cosine and vice versa, we only need to know how to compute the sine of 5°.
OK, so how do we do that? The sine and cosine of 45° and of 30° are well known, and we can use the trig identity for sin(A – B) to find that the sine of 15° is (√3 – 1)/ 2√2. So how do we get from the sine of 15° to the sine of 5°? We can borrow a trick from medieval astronomers. They did something similar to our work above, then solved the trig identity
as a cubic equation in sin θ to find the sine of 1/3 of an angle with known sine.
If we ask Mathematica to solve this equation for us
Solve[3 x -4 x^3 == Sin[15 Degree], x]
we get three roots, and it’s the middle one that we’re after. (We can evaluate the roots numerically and tell that the middle one has the right magnitude.) So Mathematica says that the sine of 5° is
Unfortunately, this is the end of the line for Mathematica. We know that the expression above must be real, but Mathematica is unable to find its real part. I tried running
FullSimplify on this expression, and aborted evaluation after it ran for 20 minutes or so. If anybody wants to pick up where Mathematica left off, let me know your solution. It may be possible to continue with Mathematica by giving the software some help, or it may be easier to continue with pencil and paper.
Update: Based on the comments below, it seems this is about as far as we can go. This was a surprise to me. We’re solving a cubic equation, less than 5th degree, so the roots are expressible in terms of radicals. But that doesn’t mean we can eliminate the appearance of the i‘s, even though we know the imaginary part is zero.