2014年12月5日 · 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 ...

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 …

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2015年6月3日 · $\begingroup$ 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 ...

I am a high school student and am trying to find the taylor expansion of $\tanh(x)$ in terms of a summation form. I have gotten this far, and am aware it might get complicated very quickly. If …

$$ \int (1-\tanh x)dx = \int dx-\int \tanh x dx $$ and $$ \int \tanh x dx=\int \frac{\sinh x}{\cosh x} dx=\ln(\cosh(x)) +C $$ is a standard integral, easy calculated with the substitution $$ \cosh x= …

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

2018年1月29日 · Derivative polynomial of the hyperbolic tangent function. It is known that $$ \tan z=\operatorname{i}\tanh(\operatorname{i}z). $$ So, from the derivative polynomial of the …

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