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I'm considering a parametrized sigmoid function such as the following logistic:
$$f(x)=\frac{e^{a+bx}}{1+e^{a+bx}}$$
And I'm interested only in the interval $\displaystyle x >= 0$ and $\displaystyle x < x_{max}$ (with a given $x_{max}$).

Two tiny questions regarding its properties:

1) Is the sum of sigmoid functions always a sigmoid function itself?

2) Is the log function (i.e. $\displaystyle f(x) = log_{a}(bx^{c})$ or similar) a particular case of a sigmoid function? In other words, can I pick parameters for the logistic function above so that it behaves like a log function in the given interval?

Thanks.

Mino
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2 Answers2

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I am assuming that a sigmoid function is bounded, and its graph has exactly one inflection point (among other properties).

For (1), then, the answer is no. If $f(x)=\frac{e^{3x}}{1+e^{3x}}$ and $g(x)=\frac{e^{2x-5}}{1+e^{2x-5}}$, then $h(x)=f(x)+g(x)$ can be seen to have two points of inflection, so h(x) is not sigmoid.

For (2), your function is unbounded on $0\le x \le x_{\mbox{max}}$, but sigmoid functions are bounded. So these functions have decidedly different behavior. Though, it does depend somewhat on what you mean by "like".

Matthew Conroy
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  • Thanks for the answer to (1). Regarding the answer to (2), i am only interested in the interval $[0, x_{max}]$. In this interval, is it possible to define a "parametrized" log function that corresponds to the sigmoid function defined above? – Mino Nov 09 '10 at 21:37
  • What do you mean by a parametrized log function? Are you looking for a function which is identical to the sigmoid function, or one which is an approximation (i.e., what do you mean by correspond)? – Matthew Conroy Nov 09 '10 at 22:22
  • I'm looking for something that is identical to the sigmoid function in the given interval. – Mino Nov 09 '10 at 23:52
  • Again: what do you mean by a parametrized log function? – Matthew Conroy Nov 10 '10 at 06:08
  • Also, your function, equivalent to $d \ln x +e$ as Ross explained, is unbounded on $0 \le x \le x_{\mbox{max}}$. Your sigmoid functions are not unbounded. Thus sigmoid functions and your log functions cannot be identical. – Matthew Conroy Nov 10 '10 at 06:11
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It is incorrect to speak of "the sigmoid function". The Wikipedia link shows that there are many to choose from. For (2), note that $\displaystyle f(x) = \log_{a}(bx^{c})=c\log_{a}{x}+\log_{a}{b}=d\ln{x}+e$ by absorbing the constants into the change of base of log. This is definitely not sigmoidal.

Ross Millikan
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