Dec 5, 2014 · An easy way to compute the coefficients of the Taylor series of $\tanh$ is to consider that: $$\cosh(z)=\prod_{n=0}^{+\infty}\left(1+\frac{4z^2}{(2n+1)^2 \pi^2}\right ...

$\begin{align*} \tanh(x) &= x -x^3/3 + 2x^5/15 + O(x^7)\\ \arctan(x) &= x - x^3/3 + x^5/5 + O(x^7) \end{align*}$ 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).

Jun 3, 2015 · $\begingroup$ Neither is a Maclaurin series, which has to look like $\sum a_n x^n$. You can do a formal division, and obtain after some pain a few terms.

Also $0\leq f(x)$, and if $0<x$: $$ (x\tanh x)' = \tanh x + x \operatorname{sech}^2x >0 . $$ It's not hard ...

float fast_tanh(float x){ float x2 = x * x; float a = x * (135135.0f + x2 * (17325.0f + x2 * (378.0f + x2))); float b = 135135.0f + x2 * (62370.0f + x2 * (3150.0f + x2 * 28.0f)); return a / b; } I should add that if x might be large (not relevant for the OP), you would want to clamp the output to +/-1 for x beyond about +/-4.97.

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$\displaystyle\tanh(x)=u,$ $$\implies u^2=\frac{\sinh^2(x)}{\cosh^2(x)}=\frac{\sinh^2(x)}{1+\sinh^2(x)}$$ ...

I used the formula $$\frac{e^x - e^{-x}}{e^x + e^{-x}}$$ I calculated the inverse to find the range, but I got the incorrect answer. Please help me find the domain and range of $\tanh(x)$. calculus

Feb 5, 2022 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.