8
$\begingroup$

I have been plotting functions of the form $$\left(\sum_{i=-n}^{i=n} \tanh(x - i)\right)-2x$$ which, over the region around zero, look a lot like sine waves. For example, here's the function for $n=10$.

enter image description here

Empirically, it seems to be something approximating $0.00065 \sin(2 \pi x)$. Beyond about $n=10$ it is very stable over the region $x\in[-5,5]$, as you might expect from $\tanh(x)$ being approximately $\pm 1$ for large $\pm x$. I have found plots of the series and $0.00065 \sin(2 \pi x)$ are indistinguishable. Here's the same plot zoomed out with this sine wave in orange:

enter image description here

Is this actually tending towards a sine wave? and if so, what is the exact value of the constant $k$ in

$$\lim_{n \to \infty}\left(\sum_{i=-n}^{i=n} \tanh(x - i)\right)-2x = k \sin(2 \pi x)$$

Update

I've calculated the specific values of this function symbolically with Mathematica, and evaluated it with $N[.,100]$ function at x=1/4, 5/4, 401/4 and 4001/4 which gives

   1/4: 0.0006499727271926147575932558139650291715585570399687299381104388139877502424025845538796044013975974138
   5/4: 0.0006499727271926147575932558139650291715585570399687299381104388139877502424025845538796044013975974138
 401/4: 0.0006499727271926147575932558139650291715585570399687299381104388139877502424025845538796044013975974138
4001/4: 0.0006499727271926147575932558139650291715585570399687299381104388139877502424025845538796044013975974138

I'm not 100% sure this is not a numerical issue, but I also don't see any reason why it should be 0.00065 exactly either.

$\endgroup$
6
$\begingroup$

We have the following expansion:

$$ \lim_{n\to\infty} \sum_{k=-n}^{n} \tanh(x+k) = 2x + \sum_{k=1}^{\infty} \frac{2\pi}{\sinh(\pi^2 k)} \sin(2\pi k x). \tag{*} $$

So the resulting function is not a pure sine function, but since the coefficients decay exponentially fast, the first term is predominant. This can also be glimpsed from the following numerical value:

$$ \frac{2\pi}{\sinh(\pi^2)} \approx 0.000649972728931478984525468377615\cdots, $$


Derivation of the identity $\text{(*)}$.

  1. Assume for a moment that not only the LHS of $\text{(*)}$ converges, but it is also differentiable and can be computed by interchanging the order of limiting operators:

    \begin{align*} \frac{d}{dx}\lim_{n\to\infty} \sum_{k=-n}^{n} \tanh(x+k) &= \lim_{n\to\infty} \sum_{k=-n}^{n} \frac{d}{dx}\tanh(x+k) \\ &= \sum_{k=-\infty}^{\infty} \frac{1}{\cosh^2(x+k)}. \end{align*}

    (We will justify this later, but at this point this is just a technical issue.) The summand of the last sum has exponential decay as $k\to\infty$, so we can use the Poisson summation formula to obtain

    $$ \sum_{k=-\infty}^{\infty} \frac{1}{\cosh^2(x+k)} = \sum_{k=-\infty}^{\infty} f(k) = \sum_{k=-\infty}^{\infty} \hat{f}(k). $$

    To evaluate the last sum, we appeal to the following well-known result

    $$ \hat{f}(\xi) = e^{2\pi i \xi x} \frac{2\pi^2 \xi}{\sinh(\pi^2 \xi)}, \qquad \hat{f}(0) = 2. $$

    Plugging this back, we have

    $$ \frac{d}{dx}\lim_{n\to\infty} \sum_{k=-n}^{n} \tanh(x+k) = 2 + \sum_{k \neq 0} \frac{4\pi^2 k}{\sinh(\pi^2 k)} e^{2\pi i k x}. $$

    Integrating both sides and using the fact that the LHS of $\text{(*)}$ vanishes at $x = 0$, we have

    \begin{align*} \lim_{n\to\infty} \sum_{k=-n}^{n} \tanh(x+k) &= 2x + \sum_{k \neq 0} \frac{2\pi}{i \sinh(\pi^2 k)} e^{2\pi i k x} \\ &= 2x + \sum_{k=1}^{\infty} \frac{2\pi}{\sinh(\pi^2 k)} \sin(2\pi k x) \end{align*}

    as expected.

  2. We justify that the limit and differentiation can be exchanged. This is in fact not hard:

    \begin{align*} \sum_{k=-n}^{n} \tanh(x+k) &= \frac{1}{2} \sum_{k=-n}^{n} [\tanh(k+x) - \tanh(k-x)] \\ &= \frac{1}{2} \sum_{k=-n}^{n} \int_{-x}^{x} \frac{du}{\cosh^2(u+k)} \\ &= \frac{1}{2} \int_{-x}^{x} \left( \sum_{k=-n}^{n} \frac{1}{\cosh^2(u+k)} \right) \, du. \end{align*}

    Since the integrand of the last integral converges uniformly on $\mathbb{R}$ to an even function, it follows that

    \begin{align*} \lim_{n\to\infty}\sum_{k=-n}^{n} \tanh(x+k) &= \frac{1}{2} \int_{-x}^{x} \left( \sum_{k=-\infty}^{\infty} \frac{1}{\cosh^2(u+k)} \right) \, du \\ &= \int_{0}^{x} \left( \sum_{k=-\infty}^{\infty} \frac{1}{\cosh^2(u+k)} \right) \, du. \end{align*}

    This is precisely what we needed to justify in the previous computation.

$\endgroup$
  • $\begingroup$ Ah, Fourier transform... obvious in hindsight. Thanks very much, this problem was really bugging me. $\endgroup$ – Lucas Aug 27 '17 at 23:12

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.