How Does the Tanh(x) Approximation Relate to Small Angles?

[wil3]

Hello! So I was reading a paper in which I came across the following:

$k = \pi + \pi\ell$

$\tanh{(k)} \approx \pi\ell$

where "l" is very small. What on Earth is the origin of this approximation? I'm sure it's very simple, but I can't seem to derive it from the angle-sum and small angle approximations for sinh and cosh.

Thanks very much for any help!

 

[Robert1986]

I don't know for sure, but I'd try the power series for $\tanh$. And also $\pi +\pi l \simeq \pi$ or linear apporximation.

 

[Mentallic]

Hello! So I was reading a paper in which I came across the following:

$k = \pi + \pi\ell$

$\tanh{(k)} \approx \pi\ell$

where "l" is very small. What on Earth is the origin of this approximation? I'm sure it's very simple, but I can't seem to derive it from the angle-sum and small angle approximations for sinh and cosh.

Thanks very much for any help!

Unless I'm missing something we have $$\lim_{\ell \to 0}\tanh(k)=\tanh(\pi)\neq 0$$ so how could this approximation possibly be true since $$\lim_{\ell \to 0}\pi \ell = 0$$

 

[Robert1986]

Unless I'm missing something for small $\ell$ we have $\tanh(k)\approx \tanh(\pi)\neq 0$ so how could this approximation possibly be true since for small $\ell$, $\pi \ell \approx 0$.

Yeah, I was just thinking about that - I tried to do a linear approximation and it didn't work. When I first suggested this, I tried to quickly do it in my head, and got it a bit confused and made an error. I'm not sure if OP made a typo.

If he meant $\tan$ then I think it is right:

$$y = \sec^2(\pi)(x-\pi) + \tan(\pi) = x - \pi$$

Now if we let $x = \pi +l \pi$ we have the correct linear approximation and $\tan(k) \simeq l \pi$.

 

[Mentallic]

Yeah, I was just thinking about that - I tried to do a linear approximation and it didn't work. When I first suggested this, I tried to quickly do it in my head, and got it a bit confused and made an error. I'm not sure if OP made a typo.

If he meant $\tan$ then I think it is right:

$$y = \sec^2(\pi)(x-\pi) + \tan(\pi) = x - \pi$$

Now if we let $x = \pi +l \pi$ we have the correct linear approximation and $\tan(k) \simeq l \pi$.

Yeah for tan it works. They're both equivalent at $\ell = 0$ and their first derivatives match too.

 

[wil3]

OH! So I correctly quoted the paper, but the paper itself was incorrect. The author definitely meant tan()-- he switched between two equations. The previous equations had cosh(.) and sinh(.), so I didn't catch the error:

Check out 7b in this paper if you're curious where this is from:
http://www.ctsystemes.com/zeland/publi/00982223.pdf

thanks very much guys!

 

[Mentallic]

OH! So I correctly quoted the paper, but the paper itself was incorrect. The author definitely meant tan()-- he switched between two equations. The previous equations had cosh(.) and sinh(.), so I didn't catch the error:

Check out 7b in this paper if you're curious where this is from:
http://www.ctsystemes.com/zeland/publi/00982223.pdf

thanks very much guys!

At least he plugged the approximation into the formula correctly
An honest typo since the Author kept switching between tan and tanh.