Exercise 3.5

(a)

(b)

Let $s\to \infty$, we have $e^{-2s}\to 0$, thus $\tanh <!--\; FUNCTION\; APPLICATION\; -->(s)\to 1$. Similarly, when $s\to -\infty$, $\tanh <!--\; FUNCTION\; APPLICATION\; -->(s)\to -1$.

It’s also easy to see that $\tanh <!--\; FUNCTION\; APPLICATION\; -->(s)1$ and $\tanh <!--\; FUNCTION\; APPLICATION\; -->(s)-1$. So $-1$ and $1$ are hard thresholds for $\tanh <!--\; FUNCTION\; APPLICATION\; -->(s)$ when $\left|s\right|$ is large.

When $\left|s\right|$ is small, consider Taylor expansion of $e^s=1+s+\frac{s^2}{2}+O(s^3)$, then $e^{-s}=1-s+\frac{s^2}{2}-O(s^3)$, we have $\tanh <!--\; FUNCTION\; APPLICATION\; -->(s)=\frac{e^s-e^{-s}}{e^s+e^{-s}}=\frac{2s+2O(s^3)}{2+s^2}\approx s$.

So $\tanh <!--\; FUNCTION\; APPLICATION\; -->(s)$ is approximately linear when $\left|s\right|$ is small. There’s no threshold in this case.