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More notes on power series
It seems I wasn't done thinking about this. I pointed out in yesterday's article that, having defined the cosine function as:

     coss = zipWith (*) (cycle [1,0,-1,0]) (map ((1/) . fact) [0..])

     sins = zipWith (*) (cycle [0,1,0,-1]) (map ((1/) . fact) [0..])

     sins = (srt . (add one) . neg . sqr) coss

    solution_of_equation f0 f1 = func
        where func = int f0 (int f1 (neg func))
 
    sins = solution_of_equation 0 1
    coss = solution_of_equation 1 0

Well, that was fun.

One problem with the power series approach is that the answer you get is not usually in a recognizable form. If what you get out is

     [1.0,0.0,-0.5,0.0,0.0416666666666667,0.0,-0.00138888888888889,0.0,2.48015873015873e-05,0.0,...]

     [1.0,1.0,-0.5,0.5,-0.625,0.875,-1.3125,2.0625,-3.3515625,5.5859375,-9.49609375,16.40234375,...]

One thing I was thinking about in the shower is doing Fourier analysis; this should at least identify the functions that are sinusoidal. Suppose that we know (or believe, or hope) that some power series a₁x + a₃x³ + ... actually has the form c₁ sin x + c₂ sin 2x + c₃ sin 3x + ... . Then we can guess the values of the c_i by solving a system of n equations of the form:

$$\sum_{i=1}^n i^kc_i = k!a_k\qquad{\hbox{($k$ from 1 to $n$)}}$$

But what about more general cases? I have no idea. If you have the happy inspiration to square the mystery power series above, you get [1, 2, 0, 0, 0, ...], so it is √(2x+1), but what if you're not so lucky? I wasn't; I solved it by a variation of Gareth McCaughan's method of a few days ago: f·f' is the derivative of f²/2, so integrate both sides of f·f' = 1, getting f²/2 = x + C, and so f = √(2x + C). Only after I had solved the equation this way did I try squaring the power series, and see that it was simple.

I'll keep thinking.

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