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 ?

Thanks in advance


2 Answers 2


$$\tanh(x) = 1 - 2 \exp(-2x) + O(\exp(-4x))\ \text{as}\ x \to \infty$$ 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:


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:

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


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