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 ...

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 non-issue. I never seen this being a problem in the literature. If it bothers you that one derivative is smaller than another, you can just scale it.

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 $\tanh^{-1}(\pm[1 - \frac{1}{2^8}]) \approx \pm3.1$.

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 [0,1]. The range of the tanh function is [-1,1] and that of the sigmoid function is …

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, pronounced sinh as /ʃaɪn/, cosh as /kɒʃ/ ("cosh") and tanh as /θæn/, i.e., as shine, cosh and than with a soft th like in theta---the same pronunciation in three countries, in three continents, but 53 years ago.

If I apprximate $ \tanh(x) \sim 1+ \frac{1}{x} $ I know how to compute the integral of course :) integration;

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 tangent function $\tan z$, we can derive the derivative polynomial of …

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 FPGA, so things like multiplications are OK (as long as I can prevent the word-length from growing too fast), but divisions are tricky and I would like to avoid them as ...

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