1}(x))$ Or $f:\tanh(x) \to \tanh(x+1)$ Is A Rational Function ...

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Learn more about Teams $f(x)=\tanh(1+\tanh^{-1}(x))$ or $f:\tanh(x) \to \tanh(x+1)$ is a rational function? Ask Question Asked 11 years, 7 months ago Modified 11 years, 7 months ago Viewed 346 times 4 $\begingroup$

This is (again) more a recreational/incidental question.

Playing with iteration of functions I considered the function $$ f(x) = \tanh(1+\tanh^{-1}(x)) \tag1$$ such that $$ f : \tanh(x) \to \tanh(x+1) \tag 2$$ Pari/GP is kind enough to provide the first few coefficients of the Taylor-series of $f(x)$ numerically. $$ f(x) \sim 0.7615941559557649 + 0.4199743416140261 x - 0.3198500042246123 x^2 \\ + 0.2435958939998914 x^3 - 0.1855212092851373 x^4 + 0.1412918687974069 x^5 \\ - 0.1076070615601738 x^6 + 0.08195290922380060 x^7 - 0.06241485672841984 x^8 \\ + O(x^9) $$

Looking at the coefficients it seemed to me that they give just an alternating geometric series with quotient $q=-\tanh(1)$ and a scaling factor $a = \frac 1q - q$ such that -by that numerical heuristic- the power-series of $f(x)$ is $$f(x) \underset{\text{guessed}}{=} -q + a \sum_{k=1}^\infty q^k x^k \tag 3$$ which reduces then to the rational function $$ f(x) \underset{\text{guessed}}{=} { a \over 1-x\cdot q }- \frac 1q \tag 4$$

I'm surprised that this results in such a simple function - how would a proof for the algebraic identity (4) look like?

• sequences-and-series
• functions

Share Cite Follow edited May 7, 2013 at 20:12 Gottfried Helms asked May 7, 2013 at 19:59 Gottfried HelmsGottfried Helms 35.2k33 gold badges6868 silver badges145145 bronze badges $\endgroup$ Add a comment |

3 Answers 3

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$$\tanh(x+y) = \frac{\sinh(x+y)}{\cosh(x+y)} = \frac{\tanh x + \tanh y }{1 + \tanh x \tanh y}$$

$$ \tanh(1 + x) = \frac{\tanh 1 + \tanh x}{1 + \tanh 1 \tanh x} $$

$$ f(\tanh x) = \tanh(1+x) \implies f(x) = \frac{\tanh 1 + x}{1 + x \tanh 1 } $$

Share Cite Follow answered May 7, 2013 at 20:18 user14972user14972 $\endgroup$ 0 Add a comment | 3 $\begingroup$

Here, we use the alternative expressions

$$\mathrm{arctanh}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right) \quad \text{and} \quad \tanh(x)=\frac{1-e^{-2x}}{1+e^{-2x}}.$$

The first is valid in $\vert x \vert < 1$, and the latter valid in all of $\mathbb{C}$. Algbraic manipulation now gives $$\tanh(1+\mathrm{arctanh}(x))=\frac{x+e^2 (x+1)-1}{-x+e^2 (x+1)+1}=\frac{x(e^2+1)+e^2-1}{x(e^2-1)+e^2+1},$$ at least for $\vert x \vert <1$. This result may be analytically continued, of course.

Share Cite Follow answered May 7, 2013 at 20:15 awwalkerawwalker 6,95411 gold badge2323 silver badges3333 bronze badges $\endgroup$ Add a comment | 2 $\begingroup$

If it is true for the ordinary tangent it is true for the hyperbolic tangent (due to $\tan(ix)=\pm i \tanh(x)$ (for some choice of the sign that I don't need to remember in order to answer the question).

And $\tan(x+a)$ is indeed a rational function of $\tan x$ and $\tan a$, by the familiar addition formula for tangent.

This way we can foresee that a rational function expression will exist, and a bit of care about the signs will pin down what the formula is.

Share Cite Follow answered May 7, 2013 at 20:48 zyxzyx 35.8k44 gold badges5050 silver badges109109 bronze badges $\endgroup$ 1

• $\begingroup$ (+1) That dualism to the tangent-function via the focusing on complex values instead of the reals is a nice idea and I think I should keep such an option somewhere in my mind for further application... Thanks! $\endgroup$ – Gottfried Helms Commented May 8, 2013 at 5:39

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