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 ...

Having stronger gradients: since data is centered around 0, the derivatives are higher. To see this, calculate the derivative of the tanh function and notice that its range (output values) is …

Feb 26, 2018 · @elkout says "The real reason that tanh is preferred compared to sigmoid (...) is that the derivatives of the tanh are larger than the derivatives of the sigmoid." I think this is a …

Assuming the numbers are stored in fixed point with an 8 bit fractional part then the approximation to $\tanh(x)$ should work to the limit implied by the resolution, or for arguments $\tanh^{ …

Jan 29, 2018 · 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 …

Set $ y = \tanh^{-1} t $ and take $\tanh$ to take both sides so we have $$ \tanh y = t .$$

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 …

My maths professor Siegfried Goeldner who got his PhD in mathematics at the Courant Institute at New York University under one of the German refugees from Goetingen, in 1960, …

$\begingroup$ Well, since $\tanh\frac1{x}$ doesn't seem to have a Maclaurin series expansion... have a look at these though. $\endgroup$ – J. M. ain't a mathematician Commented Aug 16, …

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