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 ...

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 …

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 …

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^{ …

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, …

2018年2月26日 · @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 …

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

$\begingroup$ 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, …

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.

2019年4月9日 · Given a signed 16-bit word (with 8 bits of fraction length), what's the easiest way to implement the tanh function with reasonable accuracy? Keep in mind that the target is an …