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Given $x\in \Re$, $a \in \Re$ where $-1 \le x \le 1$ and $0 \le a \le 4$, is it possible to reduce the following expression:

$\tanh(a \tanh^{-1}(x))$

E.g. to some kind of polynomium?

I know that if $a$ is an integer then the following holds:

$\tanh(0 \tanh^{-1}(x)) = 0$

$\tanh(1 \tanh^{-1}(x)) = x$

$\tanh(2 \tanh^{-1}(x)) = 2 x / (1 + x^2)$

$\tanh(3 \tanh^{-1}(x)) = (3 x + x^3) / (1 + 3 x^2)$

$\tanh(4 \tanh^{-1}(x)) = (4 x + 4 x^3) / (1 + 6 x^2 + x^4)$

Any suggestions?

Nis Wegmann
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    Interesting, "Polynomium" is apparently Danish for polynomial: https://da.wikipedia.org/wiki/Polynomium ... perhaps we want brøkdel af polynomier – James S. Cook Dec 22 '15 at 14:40
  • It's definitely not a polynomial, unless $a$ is an integer. (And even then not, but I believe it is a rational function...) – Thomas Andrews Dec 22 '15 at 15:00
  • By the way, the pattern of the coefficients is $\cfrac{\binom a1x+\binom a3x^3+\binom a5x^5+\cdots}{\binom a0+\binom a2x^2+\binom a4x^4+\cdots}$. –  Dec 22 '15 at 15:42

1 Answers1

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$$\tanh^{-1}{x} = \frac12 \log{\left (\frac{1+x}{1-x} \right )}$$

$$\tanh{(a \tanh^{-1} x)} = \frac{e^{(2 a \tanh^{-1} x)}-1}{e^{(2 a \tanh^{-1} x)}+1} = \frac{(1+x)^a-(1-x)^a}{(1+x)^a+(1-x)^a}$$

Ron Gordon
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  • Great! Can the same trick be used for $\tanh(a + b \tanh^{-1}(x))$? – Nis Wegmann Dec 22 '15 at 14:37
  • For that maybe this will help: $$\tanh(a+y)=\frac{\sinh (a) \cosh (y)}{\sinh (a) \sinh (y)+\cosh (a) \cosh (y)}+\frac{\cosh (a) \sinh (y)}{\sinh (a) \sinh (y)+\cosh (a) \cosh (y)}.$$ Let $y=b\tanh^{-1}(x)$ and try to expand. – pshmath0 Dec 22 '15 at 14:38
  • @NisWegmann: I would simply use the standard expression for the tanh of a sum: $$\tanh{(a+b \tanh^{-1}{x})} = \frac{\tanh{a} +\tanh{(b \tanh^{-1}{x})}}{1+\tanh{a} \tanh{(b \tanh^{-1}{x})}}$$ I trust you can do the algebra. – Ron Gordon Dec 22 '15 at 14:41
  • That's even better. – pshmath0 Dec 22 '15 at 14:41
  • @RonGordon: Thanks! I'll try that. – Nis Wegmann Dec 22 '15 at 14:43