Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The functions $\tanh x$ and $\arctan x$ have a similar graph. Is there a formula to transform $\tanh x$ to $\arctan x$?

share|improve this question
    
Is an identity like $\mathrm{tanh}(ix)=i\mathrm{tan}(x)$ what you're looking for? –  Sid Raval Dec 18 '11 at 16:35
add comment

6 Answers 6

If we let $f$ and $g$ be defined by

$$\begin{align} f&:\mathbb{C}\to\mathbb{C}\\ &z\mapsto -i\tanh(z)\\ g&:\mathbb{R}\to\mathbb{C}\\ &:x\mapsto i\arctan(x) \end{align}$$

Then $f$ and $g$ are semi-inverses, in the sense that $f\circ g=$ is the identity map.

Another way to say the same thing: If we consider multiplication by $i$ as a function represented by $\mu_i$, then $$\mu_i^{-1}\circ\tanh\circ \mu_i\circ\arctan=\mathrm{id}_{\mathbb{R}}$$

So there is this relationship between the functions, along with all the others mentioned in other answers.

share|improve this answer
add comment

There is no formula that "transforms $\tanh x$ to $\arctan x$" even though the graphs of the two functions look apparently similar: Both functions are odd, are defined on all of ${\mathbb R}$, and for $x>0$ are monotonically increasing to some finite limit. But that's where the similarities end: In particular

$${\pi\over2}-\arctan(x)\sim{1\over x}\ ,\qquad 1-\tanh(x)\sim 2e^{-2x}$$

when $x\to\infty$. This shows that the asymptotic behavior of the two functions is completely different.

As indicated in other answers, $\tan$ and $\tanh$ are related to the function $\exp$ whereas $\arctan$ and ${\rm artanh}$ are related to the function $\log$, whereby the transition from trigonometric functions to hyperbolic ones lives in the complex domain.

share|improve this answer
add comment

If you define $f(t) = \arctan(\text{arctanh}(t))$, you'll have $f(\tanh(x)) = \arctan(x)$. $f$ is analytic in $|t|<1$, with Maclaurin series $f(t) = t + \frac{t^5}{15} + \frac{t^7}{45} + \frac{64}{2835} t^9 + \frac{71}{4725} x^{11}+ \ldots$

share|improve this answer
add comment

There is the identity $\arcsin(\tanh x) = \arctan(\sinh x)$, if that's what you're looking for. See Gudermannian function on Wikipedia.

share|improve this answer
    
...and the Gudermannian is yet another fine example of a sigmoidal function... –  J. M. Dec 19 '11 at 2:23
add comment

They are only similar around the origin because their Taylor series agree up to fourth order, that is,

$\tanh(x) = x -x^3/3 + 2x^5/15 + O(x^7)$

$\arctan(x) = x - x^3/3 + x^5/5 + O(x^7)$

I don't believe there is any deeper reason why these look alike since one is an inverse trigonometric function while the other is a hyperbolic function (not an inverse).

share|improve this answer
add comment

The two functions resemble each other on the real line, but they're not the only sigmoidal functions around. There are functions as simple as $\dfrac{x}{\sqrt{1+x^2}}$ and as complicated as $\dfrac2{\sqrt \pi}\int_0^x e^{-t^2}\mathrm dt$ that also have the s-shape.

Also, the resemblance ends when we look at their behavior for complex arguments:

arctangent plots

hyperbolic tangent plots

The former possesses branch cuts, while the latter exhibits periodicity on the imaginary axis.

On the other hand, a look at the Maclaurin series for $\tan\tanh\,z$ and $\mathrm{artanh}\arctan\,z$ gives a clue as to why there is some resemblance between $\arctan$ and $\tanh$:

$$\begin{align*} \tan\tanh\,z&=z-\frac{z^5}{15}+\frac{z^7}{45}-\frac{z^9}{2835}-\frac{13 z^{11}}{4725}+\cdots\\ \mathrm{artanh}\arctan\,z&=z+\frac{z^5}{15}-\frac{z^7}{45}+\frac{64 z^9}{2835}-\frac{71 z^{11}}{4725}+\cdots \end{align*}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.