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I recently wrote a program to do what was described in HAKMEM 123. Copy-pasting verbatim...

Consider the image of the circle $|z| = 1$ under the function

$$f(z) = \sum_n \frac{z^{2^{n}}}{2^{n}}$$

This is physically analogous to a series of clock hands placed end to end. The first hand rotates around the center (0,0) at some rate. the next hand is half as long and rotates around the end of the first hand at twice this rate. The third hand rotates around the end of the second at four times this rate; etc. It would seem that the end of the "last" hand (really there are infinitely many) would sweep through space very fast, tracing out an (infinitely) long curve in the time the first hand rotates once. The hands shrink, however, because of the $2^n$ in the denominator. Thus it is unclear whether the curve's arc length is really infinite.

Was this problem ever solved? Is the curve's arc length infinite or finite? My gut feeling says that it's infinite, but I don't really know how to go about proving it.

(By the way, tagging help would be appreciated.)

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This is a lacunary function. Since $\lim2^{n+1}/2^n>1$, the circle of convergence - which is $|z|=1$ - is a cut (thm 1)‌​, which means that the function isn't holomorphic on the boundary and also has at least one singularity. I think that means the arclength doesn't exist but I'm not sure.. –  anon Oct 6 '11 at 7:08
    
@anon: Fixed :) –  Zev Chonoles Oct 6 '11 at 7:57
    
This is what Mathematica thinks is the image of the unit disk after being transformed by your $f(z)$. Nasty-looking boundary, eh? –  J. M. Oct 6 '11 at 8:38
    
@J.M. Can you kindly share the code snippet responsible for that sort of plots ? I've been wanting something like that... –  M. Alaggan Oct 6 '11 at 8:40
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@M.Alaggan: I actually had to cheat a bit and used a circle with radius $1-10^{-3}$, since the series becomes progressively more difficult to evaluate as one nears the boundary. In any event... f[z_?NumericQ] := NSum[z^2^k/2^k, {k, 0, Infinity}, Method -> "WynnEpsilon", WorkingPrecision -> Precision[z]] /; Precision[z] < Infinity; ParametricPlot[ Through[{Re, Im}[f[r Exp[I t]]]], {r, 0, 1 - 10^-3}, {t, 0, 2 Pi}, PerformanceGoal -> "Quality", PlotPoints -> 100, PlotStyle -> None] –  J. M. Oct 6 '11 at 8:53

1 Answer 1

up vote 5 down vote accepted

Let $f_n(t) = \sum_{k=0}^n 2^{-k} e^{2 \pi i 2^k t}$ be a parametric representation of the $n$'th partial sum of your curve and $v_n(t) = f_n'(t) = 2 \pi i \sum_{k=0}^n e^{2 \pi i 2^k t}$ its velocity. Now by the Salem-Zygmund theorem on lacunary trigonometric series, if $U$ is a uniform random variable on $[0,1]$, as $n \to \infty$, $\text{Re}(v_n(U))/\sqrt{n}$ converges in distribution to a normal random variable with mean 0 and variance $2 \pi^2$. As a consequence, the arc length $\int_0^1 |v_n(t)|\ dt = E [|v_n(U)|] \to \infty$ as $n \to \infty$.

EDIT: I could also have used the more elementary fact that for any Sidon set $E$ there is a positive constant $B$ such that for all $E$-spectral trigonometric polynomials $g$, $\|g\|_1 \le B \|g\|_2$. But I liked the idea that there is a "central limit theorem" here.

EDIT: As joriki noted, this in itself doesn't imply that the arc length of the infinite sum is infinite. However, we can obtain that as follows. Let $f(t) = \sum_{k=0}^\infty 2^{-k} e^{2 \pi i 2^k t}$. The arc length of a parametric curve $g: [0,1] \to \mathbb C$ is the total variation $V(g)$. Suppose $V(f) < \infty$. Now $$f(t) - f_n(t) = \sum_{k=n+1}^\infty 2^{-k} e^{2 \pi i 2^k t} = 2^{-n-1} f(2^{n+1} t)$$ so that $V(f - f_n) = V(f)$. But since $V$ is subadditive, $V(f_n) \le V(f) + V(f_n - f) = 2 V(f)$, contradicting the fact that $V(f_n) \to \infty$.

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Rigorously speaking, how does the arc length of the partial sums going to infinity imply that the limit curve isn't rectifiable? –  joriki Oct 6 '11 at 9:08
    
In fact it doesn't, at least not for general sequences of curves. For instance, a sequence of curves that wrap $n^2$ times around concentric circles of radius $1/n$ has the constant curve at the centre as its limit, but the arc lengths diverge. –  joriki Oct 6 '11 at 11:17
    
@joriki: Good point. I had actually thought about that last night, but it was very late, so I just added that part now. –  Robert Israel Oct 6 '11 at 15:42

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