Bigger almost-linear zone for tanh(x)

Question

The hyperbolic tangent is almost linear around the origin.

I'm looking for a way to modify the tanh function in order to have a bigger almost-linear zone while keeping the asymptote to y = 1. In other words, the function should 'leave' later the y = x straight line and approach sooner the y = 1 straight line, coming closer to the (1, 1) point. (think about "sooner" and "later" as if x was a time variable).

I'd like this 'linearity control' to be a parameter. If useful, feel free to ignore what happens for x < 0.

Thanks for any suggestion!

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score 3 · Accepted Answer · edited Jul 14 '18 at 16:09

3

One function that works is $\sqrt[a]{tanh(x^a)}$ for a real parameter $a$. The higher the value of $a$, the wider the "linear-looking" area around $x=0$. Unfortunately for negative values of $x$ it is required that $a$ is a rational number in lowest terms, with both an odd numerator and denominator (that is, if you wish to have outputs in the real numbers).

Animation:

edited Jul 14 '18 at 16:09

Oppenede

546

answered Jul 14 '18 at 15:33

Conkerman

86

1

$a$ can be anything if you write $\text{sgn}(x) \sqrt[a]{\tanh \left(\left(x^2\right)^{a/2}\right)}$ – Oppenede Jul 14 '18 at 15:40
That's perfect! Thanks a lot! – Claudio Jul 14 '18 at 16:21

score 0 · Answer 2 · answered Jul 14 '18 at 14:10

You need to increase the value of $a$ in $\tanh(ax)$.

Let $f(x) = \tanh(2x)$. This squishes $\tanh(ax)$ by a factor of $2$. Since $\tanh x \approx 1$ for $x < -1$ and $x > 1$, then the change in the $y$ values won't be too noticeable.

Here is a Desmos graph of the function.

Bigger almost-linear zone for tanh(x)

2 Answers2

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