A well-known formula for the logarithm is given by

$$\log x = -\frac{\pi}{AGM(a^2,b^2)}, \qquad x < 1$$

where AGM is the arithmetic-geometric mean, and $a$ and $b$ are given by

$$a = \sum_{k\in\mathbb{Z}}x^{k^2}, \qquad b = \sum_{k\in\mathbb{Z}}x^{(k+1/2)^2};$$

Or, equivalently,

$$a = 1 + 2x + 2x^4 + 2x^9 + 2x^{16} + 2x^{25} + \dots$$ and $$b = 2x^{1/4} + 2x^{9/4} + 2x^{25/4} + 2x^{49/4} + \dots$$

Why does it happen that $a = b$ occurs when $x = 0.8$?


The two values $a(0.8)$ and $b(0.8)$ appear "deceivingly" equal, but they actually are not!

Other "near-miss" values include $$0<\theta_3(0,0.9)-\theta_2(0,0.9)<0.5\times 10^{-39}.$$

Let $f(x)=\theta_3(0,x)-\theta_2(0,x)$. The graph of $f(x)$, for values $0<x<1$ shows a global minimum at $x_*$ near $x=0.9$ (of course $f(x_*)>0$, still) and also a local maximum for some $0.9<x^*<1$. It would be really interesting to figure out these numbers, especially $x_*$. In any case, there are two values of $x$ in the range $0.8<x<1$ for which $$\theta_3'(0,x)=\theta_2'(0,x).$$

By the way, can someone post the graph for $y=f(x)$? It would be a nice documentation for the discussion and analysis here.

  • $\begingroup$ I'm afraid you are misusing the big O notation (though in here it is kinda clear what you mean): if you really meant big O there, then it would just mean that this difference is smaller than $C\cdot 10^{-39}$ for some constant $C$, which of course tells us absolutely nothing. This is more of a nitpick though. $\endgroup$ – Wojowu Oct 8 '16 at 9:01

$$\sum_{k \in \mathbb Z} x^{k^2} = \theta_3(0,x)$$ while $$\sum_{k \in \mathbb Z} x^{(k+1/2)^2} = \theta_2(0,x)$$ where $\theta_2$ and $\theta_3$ are Jacobi theta functions. The difference $$\theta_2(0,0.8) - \theta_3(0,0.8) \approx 9.280378636257491074676461535977 \times 10^{-19}$$ according to Maple.

EDIT: The difference $$\theta_3(0,x) - \theta_2(0,x) = \sum_{j \in \mathbb Z} (-1)^j x^{(j/2)^2} =\theta_3(\pi/2, x^{1/4}) $$ The Poisson summation formula gives us the identity $$ \theta_3(\pi/2, e^{-t^2}) = \frac{\sqrt{\pi}}{t} \theta_2(0, e^{-\pi^2/t^2})$$ and for $t \to 0+$, this goes to $0$ very rapidly: $$\theta_3(\pi/2, e^{-t^2}) \sim \frac{2 \sqrt{\pi}}{t} e^{-\pi^2/(4 t^2)}$$ i.e. $$ \theta_3(0,x) - \theta_2(0,x) \sim \frac{4 \sqrt{\pi}}{\sqrt{\ln(1/x)}} \exp\left(-\frac{\pi^2}{\ln(1/x)}\right) $$ For $x = 0.8$, the right side above is extremely close to the value I gave for $\theta_2(0,0.8) - \theta_3(0,0.8)$ (all digits shown match).

  • 2
    $\begingroup$ The difference is even smaller for 0.9. It looks like my original question was pretty stupid: wolframalpha.com/input/?i=theta2(0,0.9)-theta3(0,0.9) $\endgroup$ – Zazzle Oct 7 '16 at 20:49
  • 2
    $\begingroup$ On the contrary: it is connected to some interesting properties of the theta functions. $\endgroup$ – Robert Israel Oct 7 '16 at 21:29
  • 2
    $\begingroup$ There's something just slightly funny about saying "all digits shown match" when "all digits shown match" was precisely the original problem! :-) $\endgroup$ – wchargin Oct 7 '16 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.