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I have found this interesting answer about increasing the linear range of the hyperbolic tangent function.

Now, I am looking for a proof (or at least have a reference from literature if it shows up to be too convoluted) that the following function becomes near linear for high $\gamma$ values.

$$ f(x,\gamma)=\frac{2}{\left(1+e^{-2x\gamma^{-1}}\right)}-1 $$

Something that seems to make this harder is that:

$$ \lim\limits_{\gamma \to \inf}f(x,\gamma)=0 $$

Is is possible to demonstrate the linear behavior near zero in a clear way?

dawid
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2 Answers2

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First I'm going to make a minor change to the proposed function and use the following one: $$ f_\gamma(x) = \gamma\left(\frac{2}{1 + e^{-2x\gamma^{-1}}} - 1\right) $$ I added the additional $\gamma$ so that the function converges to the line $y = x$.

Now to prove that $f_\gamma(x)$ converges to the line with increasing $\gamma$. We can simply show that the function converges to the line at every point. Let's choose $x_0$ as a fixed point and show the limiting behavior: $$ \begin{align} \lim_{\gamma\to\inf} f_\gamma(x_0) & = \lim_{\gamma\to\inf} \gamma\left(\frac{2}{1 + e^{-2x_0\gamma^{-1}}} - 1\right) \\ &=\lim_{\gamma\to\inf} \frac{\frac{2}{1 + e^{-2x_0\gamma^{-1}}} - 1}{\gamma^{-1}} \\ &\stackrel{\text{l'H}}{=}\lim_{\gamma\to\inf} \frac{-\frac{2(2x_0\gamma^{-2})e^{-2x_0\gamma^{-1}}}{(1 + e^{-2x_0\gamma^{-1}})^2}}{-\gamma^{-2}} \\ &=\lim_{\gamma\to\inf}\frac{4x_0e^{-2x_0\gamma^{-1}}}{(1 + e^{-2x_0\gamma^{-1}})^2} \\ &=\frac{4x_0}{2^2} \\ &= x_0 \end{align} $$ Thus $f_\gamma(x) = x$ as $\gamma$ goes to infinity.

Lukáš J.
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Since $x \ll \gamma$, you may consider a (first-order) Taylor expansion of $f$ around $x = 0$ : $$ \begin{align} f_\gamma(x) &= \frac{2}{1 + e^{-2x/\gamma}} - 1 \\ &= \frac{2}{1 + (1 - 2x/\gamma + \mathcal{O}(x^2))} - 1 \\ &= \frac{1}{1 - x/\gamma + \mathcal{O}(x^2)} - 1 \\ &= (1 + \frac{x}{\gamma} + \mathcal{O}(x^2)) - 1 \\ &= \frac{x}{\gamma} + \mathcal{O}(x^2) \end{align} $$

Abezhiko
  • 8,153