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Background (though not necessary for the mechanics of what I'm asking): According to Lapidus' $\textit{Fractal Geometry, Complex Dimensions and Zeta Functions}$, the Cantor String consists of lengths $3^{-n}$ with multiplicity $2^n$ and the geometric zeta function of this string is $$\zeta (w) = \frac{1}{1-2\cdot 3^{-w}}$$ where $w \in \mathbb{C}$. The complex dimensions of the Cantor String are obtained by finding the poles of $\zeta (s)$.

As such we want to solve $2 \cdot 3^{-w} = 1$ where $w \in \mathbb{C}$ and that the values of $w$ are $\{D + inp\ : n \in \mathbb{Z}\}$ where $D=\log_{3}{2}$ and $p=\frac{2 \pi}{\log 3}$. How do they arrive at this? I have tried looking up the definition of complex logs and such but that is not very helpful thus far.

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  • $\begingroup$ Just checking, for that $p$ term, is the denominator just $3$, or perhaps $ln3$? $\endgroup$ – imranfat Jun 18 at 0:54
  • $\begingroup$ My apologies. It is log 3. $\endgroup$ – Salazar_3854708 Jun 18 at 1:00
  • $\begingroup$ I have a pretty good guess about where this question comes from, and why you have tagged it fractals, but your question, as it is written, doesn't provide any of that context. What motivates this question? Where does it come from? Why have tagged it fractals? When you say that you have "looked up the definition of complex logs", where did you look it up? What part of the definition is unclear? $\endgroup$ – Xander Henderson Jun 18 at 1:27
  • $\begingroup$ @XanderHenderson it is edited. $\endgroup$ – Salazar_3854708 Jun 18 at 1:49
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Key Idea

The complex exponential function $s \mapsto \mathrm{e}^s$ is periodic along lines the lines $\{ i\tau : \tau \in \mathbb{R}\}$. This implies that equations of the form $$ \mathrm{e}^{s} = \alpha$$ do not generally have unique solutions. In particular \begin{align} 1 - 2\cdot 3^{-w} = 0 &\iff \mathrm{e}^{-w \log(3)} = \frac{1}{2} = \mathrm{e}^{-\log(2)} \\ &\iff -w \log(3) = -\log(2) + i2k\pi \end{align} for some integer $k$. Solving for $w$ gives $$ w = \frac{\log(2)}{\log(3)} + i \frac{2k\pi}{\log(3)}. $$

In Detail

The complex exponential function has periodic behaviour: if $s = \sigma + i\tau$ (where $\sigma$ and $\tau$ are real), then $$ \mathrm{e}^{s} = \mathrm{e}^{\sigma} \bigl( \cos(\tau) + i\sin(\tau) \bigr), \tag{*}$$ where $\sigma\mapsto\mathrm{e}^{\sigma}$ is the real exponential function, and $\cos$ and $\sin$ are the real sine and cosine functions. There are a number of ways that one can obtain this identity[1], but, for the purposes of this answer, I will take formula (*) as the definition.

Note that both the sine and cosine functions are $2\pi$ periodic. That is, $$ \cos(\tau + 2k\pi) = \cos(\tau) \qquad\text{and}\qquad \sin(\tau + 2k\pi) = \sin(\tau) $$ for all real $\tau$ and all integers $k$. This implies that the complex exponential function is $2\pi$-periodic along vertical lines in the complex plane. That is, $$ \mathrm{e}^{s + i2k\pi} = \mathrm{e}^{s} $$ for any complex $s$ and any integer $k$. Something slightly stronger is true: if $s \ne t + i2k\pi$ for some integer $k$, then $$ \mathrm{e}^{s} \ne \mathrm{e}^{t}. $$ In other words, the exponential function is injective along horizontal strips in the complex plane.

This observation is the tool required to solve the problem. Using properties of the complex exponential and logarithm functions (i.e. they are inverses of each other, on the appropriate domains), $$ 0 = 1 - 2\cdot 3^{-w} = 1 - 2\mathrm{e}^{\log(3^{-w})} = 1 - 2\mathrm{e}^{-w\log(3)}, $$ where $\log$ denotes any chosen branch of the complex logarithm. Isolate the exponential term to get $$ \mathrm{e}^{-w \log(3)} = \frac{1}{2} = \mathrm{e}^{\log(1/2)}, $$ where $1/2 = \mathrm{e}^{\log(1/2)}$ is a property of the real (rather than complex) logarithm. Applying the observation that the exponential is injective on horizontal strips, this implies that that the equation is solved if and only if $$ -w \log(3) = \log\left(\frac{1}{2}\right) + i2k\pi, $$ where $k$ is any integer. Isolating $w$ gives the result.


[1] Common techniques for obtaining (*) include

  • define the various functions involved from their power series, then deduce the identity;
  • define the various functions involved as solutions to systems of differential equations, then use properties of those systems to deduce the identity;[2]
  • just take the formula as the definition;
  • and so on. Taking (*) as the definition greatly simplifies the remaining exposition.

[2] In the context of fractal strings, it may be a good idea to familiarize yourself with the idea that the exponential, sine, and cosine functions are the solutions to certain sets of differential equations. One of the goals of Lapidus and van Frankenhuijsen's book is to use the geometric and spectral zeta functions to tie the geometry of a string to the spectrum of the Laplace operator (with Dirichlet conditions) on that string. As such, having a solid understanding of differential operators on these kinds of domains may be helpful.

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  • $\begingroup$ Thank you very much Xander! $\endgroup$ – Salazar_3854708 Jun 18 at 3:21
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A key ingredient is that $ln(a+bi)=ln(r) + arg(a+bi)$ where $r$ is the absolute value of $a+bi$. With this in mind, your exponential equation can be written as $3^w=2$ and this can be converted into $w(ln3)=ln2+2ni\pi$. The "addition" of $2ni\pi$ is the complex "multiple" that is part of the solution set. (You got to have seen this somewhere in your course). Dividing by $ln3$ gives the final solution. Using the Change of base formula you can rewrite $ln2/ln3$ if needed

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  • $\begingroup$ But when you are taking logs, you get $\ln 3$ on the LHS and $\ln 2$ on the RHS. The complex number $w$ isn't being 'logged' at all. The logs are being taken of real numbers so how come we are getting such an expression on the RHS? $\endgroup$ – Salazar_3854708 Jun 18 at 1:52
  • $\begingroup$ No, when you take the $ln$ of $2$, in complex terms, the absolute value of $ln2$ is $ln2$ and the argument is $0$ because $ln2$ can be written as $ln2+0i$, and then the multiples (2pi)i get tacked on. $\endgroup$ – imranfat Jun 18 at 2:03
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    $\begingroup$ This is now clear. Thank you! $\endgroup$ – Salazar_3854708 Jun 18 at 2:16
  • $\begingroup$ I would accept Xander's answer. You may not even get an answer as clear as that in a college classroom. $\endgroup$ – imranfat Jun 18 at 15:32

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