# Smooth $f(x)$ where $f(x) = 1$ for $x = \pm 1, \pm 5, \pm 9, \ldots$ and $f(x) = -1$ for $x = \pm 3, \pm 7, \pm 11$

Question: Find smooth $f(x)$ where $f(x) = 1$ for $x = \pm 1, \pm 5, \pm 9, \ldots$ and $f(x) = -1$ for $x = \pm 3, \pm 7, \pm 11$

Condition: For the purpose of this question, I define smoothness as no sudden changes in the function and its fourier transform is finite in (band)width. For sync, the FT is rect function which has a finite width. I know that a rigorous definition of smoothness is lot more complicated - but please go with my notion for now.

Consider the top two curves in the picture below. The first one is the $\operatorname{sinc}(x) = \dfrac{\sin(\frac{\pi x}{2})}{(\frac{\pi x}{2})}$ function. It is smooth and has finite width in FT domain. It has same values for $\pm1$ and it alternates sign for $\pm3, \pm5$ and so on. But its value does not remain constant going from $\pm1$ to $\pm5$ to $\pm7,\ldots$ as per my Question. So this function doesn't satisfy my requirement.

The second one is a sine function multiplied by sign function. It satisfies the values I need, but since there is an abrupt change at $0$, its fourier transform is infinite in width. Hence this too doesn't satisfy my requirement.

I am looking for functions that are (only for visual example) like in the 3 bottom figures (that I have hand drawn with no particular mathematical accuracy). I am not particular about how the function should look like in between $-1$ to $+1$. It could be totally different looking than what I have shown, as long as it meets the requirements stated in my question (and condition).

• Do piecewise functions satisfy your definition of smoothness? If yes, then it is easy to create such a function from the sin variant you have already tried. – shardulc says Reinstate Monica Jun 24 '17 at 1:02
• No, because piecewise functions tend to have infinite width in Fourier/frequency domain. – Srini Jun 24 '17 at 1:17

If piecewise functions are allowed, consider $$f(x) = \begin{cases} \operatorname{sgn}(x)\sin(\pi x/2) & : |x| \ge 1\\ \sin^2(\pi x/2) & : |x| < 1 \end{cases}$$

• Piecewise functions are not allowed because they don't have finite width in FT domain as per my requirement. Wherever sign function or heaviside functions are present, their FT usually are infinite in width. This definitely is the case with your answer. – Srini Jun 24 '17 at 1:20