Is there a power series expansion of the Signum function?

Question

I would like to define a linear operator as the sign of a another operator. But to use it I would need to expand it out as a power series.

Roughly, I'm wondering if something like this

\begin{eqnarray} \mathrm{Sgn}\left(\hat{O}\right)=\sum_{n=0}^{\infty}C_{n}\hat{O}^{n} \end{eqnarray}

exists?

What are you trying to do? Are you trying to make up an operator like $\operatorname{Sgn}(\frac d{dx})$ where, perhaps, $\operatorname{Sgn}(\frac d{dx})e^{ax}=\operatorname{Sgn}(a)e^{ax}$ or something? — Akiva Weinberger, Jun 09 '15 at 16:43

score 5 · Accepted Answer · answered Jun 09 '15 at 13:56

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A power series that converges for a certain $x_0$ converges absolutely for all $x$ with $|x|<|x_0|$ and then the limit function is necessarily continuous. Since the sign function is not continuous, no such series exists.

answered Jun 09 '15 at 13:56

Hagen von Eitzen

374,180

More precisely, continuity follows from the fact there is total convergence (or convergence of the norms) in any closed interval $[-\delta,\delta]$ for positive $\delta$ less than the radius of convergence. – egreg Jun 09 '15 at 13:59
Well that makes life trickier. Thanks, though. – quantum_loser Jun 09 '15 at 14:01

score 1 · Answer 2 · answered Jun 09 '15 at 22:17

1

As explained in the other post by Hagen von Eitzen, this is not possible due to the discontinuity of the Sign function at $x = 0$. Having said that, you are obviously free to circumvent this difficulty by representing the Sign function as the limit of a continuous sigmoidal function. For example, $f(x) = tanh(x/ \epsilon)$ will do fine. This function can be expanded in a power series without difficulty, and in the limit of $\epsilon$ to zero the function is equal to the Sign function.

answered Jun 09 '15 at 22:17

M. Wind

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This is potentially quite helpful. Thank you. – quantum_loser Jun 10 '15 at 17:14
By the way, you mean ArcTan, correct? – quantum_loser Jun 11 '15 at 09:08
Never mind, both work. – quantum_loser Jun 11 '15 at 09:30
There are several choices for the sigmoidal function. I would probably choose $tanh(x/ \epsilon)$, but $arctan(x/ \epsilon)$ should also work fine. Then there is $x/ \sqrt (x^2 + \epsilon ^2)$. And the error-function. Obviously they all have different series expansions. It is up to you to make a choice. Good luck with your project ! – M. Wind Jun 11 '15 at 18:58

Is there a power series expansion of the Signum function?

2 Answers2