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I'm trying to solve for $s$ the summation:

$$\sum_{q=2}^\infty \phi(q) \left(\frac{1}{q^2}\right)^s = 1$$

where $\phi(q)$ is Euler's totient function.

The context is trying to calculate the fractal similarity dimension of a synthetic model of recursive ever-smaller kissing circles similar in appearance to parts of the Mandelbrot set.

I have written fairly naive C++ code to bisect $s$ between $0$ and $2$ in double precision, evaluating the left hand side using double precision for each term and accumulating them with higher precision types like double-double and quad-double unevaluated sums (using libqd).

I'm repeatedly doubling the number of terms to see how it behaves, it seems to be slowly increasing, converging to around $s \approx 1.24$ but how can I be sure that it won't increase all the way to $2$, or eventually decrease again, or that rounding errors aren't destroying all hope of an accurate result?

Claude
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1 Answers1

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$\displaystyle\sum_{n=1}^{\infty}\frac{\varphi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}$ from here, thus you're solving $\zeta(2s-1)=2\zeta(2s)$. With PARI/GP, one finds $$s=1.2393753928669801303357436306951677\ldots$$ by issuing solve(s=1.2,1.3,zeta(2*s-1)-2*zeta(2*s)). This is the only solution, since $$\frac{d}{ds}\big(\zeta(2s-1)-2\zeta(2s)\big)=-2\sum_{n=3}^{\infty}\frac{(n-2)\log n}{n^{2s}}<0.$$

metamorphy
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