# Plot and study of a function numerically challenging

my question is simple and I'm a little bit lost how to proceed. I explain :

Let $$f : x \mapsto 1 - \tanh\left(512x^3\right)$$, I'd like to study $$g\left(x\right) = \frac{f\left(2.5x\right)}{f\left(x\right)}$$

For example, for $$x=0.8$$, I would like to know the value of $$f\left(0.8\right)$$, $$f\left(2\right)$$ and overall, $$g\left(0.8\right)=\frac{f\left(2\right)}{f\left(0.8\right)}$$. However every software I tried give me $$f\left(0.8\right)=f\left(2\right)=0$$ and then the ratio does not exist. I'd like to know if there is a simple way to plot $$g$$, or at least see what range of values this function will take ?

$$\tanh(x) = 1 - 2 \exp(-2x) + O(\exp(-4x))\ \text{as}\ x \to \infty$$ so \eqalign{f(x) &\sim 2 \exp(-1024 x^3)\cr g(x) &\sim \exp(-14976 x^3)} In particular $$g(0.8) \approx \exp(-7667.712) \approx 9.015 \times 10^{-3331}$$

$$1-\tanh t$$ can be written as:

$$\begin{array}{rcl}1-\tanh t&=&1-\frac{e^t-e^{-t}}{e^t+e^{-t}}\\ &=&\frac{e^t+e^{-t}-e^t+e^{-t}}{e^t+e^{-t}}\\&=&\frac{2e^{-t}}{e^t+e^{-t}}\\&=&\frac{2}{e^{2t}-1}\end{array}$$

which should work on your calculator when you now use $$t=512x^3$$. It may still give a number very close to $$0$$, though. If $$e^{2t}$$ is a lot bigger than $$1$$, then you can ignore the "-1" term and get:

$$\begin{array}{rcl}\cdots&\approx&\frac{2}{e^{2t}}\\&=&2e^{-2t}\end{array}$$

which gives $$f(x)=1-\tanh(512x^3)\approx 2e^{-1024x^3}$$.

The next step is $$g(x)$$:

$$\begin{array}{rcl}g(x)&=&\frac{f(2.5x)}{f(x)}\\&=&\frac{\frac{2}{e^{2\cdot 512(2.5x)^3}-1}}{\frac{2}{e^{2\cdot 512x^3}-1}}\\&=&\frac{e^{1024x^3}-1}{e^{16000x^3}-1}\end{array}$$

If $$x$$ is big enough so that you can pretty much ignore the "-1" term, you can simplify:

$$\begin{array}{rcl}\cdots&\approx &\frac{e^{1024x^3}}{e^{16000x^3}}\\&=&e^{-14976x^3}\end{array}$$