In Wikipedia, I found the symmetry equation for a sigmoid function as $g(x) + g(-x) = 1$, where $g(x) = 1/1 + \exp(-x)$.
As per the property stated above, $g(x)$ becomes a symmetric function. But, algebraically, a function is said to be symmetric if and only if $g(-x) = g(x)$ or $g(-x) = -g(x)$ for even and odd symmetric respectively.
Can somebody clarify how sigmoid function is symmetric?
Thank you!
In the algebraic sense, symmetric means what you have stated: $g(x)=-g(x)$ or $g(x)=-g(-x)$. In the geometric sense, something is symmetric if it looks the same after a particular sequence of flips, rotations, and translations. The plot of the sigmoid function has point symmetry: if you rotate it 180 degrees around the point $(0,0.5)$ then it looks the same as it originally did.
We have that $f(x) = g(x) - \frac{1}{2}$ is an odd function, equivalently the graph of $y=g(x)$ is rotationally symmetric about the point $(0, \frac{1}{2})$. Beyond that, it's not fair to say the sigmoid function is symmetric.
