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:

<br /> k = \pi + \pi\ell<br />

<br /> \tanh{(k)} \approx \pi\ell<br />

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:

<br /> k = \pi + \pi\ell<br />

<br /> \tanh{(k)} \approx \pi\ell<br />

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.