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If you are given any arbitrary function f(x), how is it possible to make it circular with a period of n, writing the function explicitly?

The function would be: $$f(x_{2\space periodic})= \begin{equation} \begin{cases} &f(x), x<n\\ &f(x-n),n<x<2n\\ &f(x-2n), 2n< x < 3n\\ &... \end{cases} \end{equation} $$

But to write it explicitly in terms of x is a bit more difficult. Using the Fourier Transform would work, but is not something I want to employ (for one given that I am working on a computer problem). Are there any simple ways to make function repeat itself, especially if the domain is limited?

Dole
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    When you say "working on a computer problem", do you mean you're programming? Because what you've written is perfectly easy to turn into an algorithm. (If your language/data type supports it and does the right thing, the modulo operator a%b giving the remainder of $a$ when divided by $b$ basically does all this for you as $f(x)=f(x%n)$) – Milo Brandt Aug 16 '16 at 04:19

2 Answers2

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Suppose a function $f$ is defined on an interval $[a,b)$ and we want it to create a periodic version of this function $f_p$ that loops over the values of the function on $[a,b)$. Define $f_p(x) = f(x)$ for $x \in [a,b)$ and recursively define $f_p(x) = f_p(x-(b-a))$. So for example, at $x=b$, we expect to get the value of $f$ at $a$ and $$f_p(b) = f_p(b-(b-a)) =f_p(a) =f(a).$$

benguin
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  • If you want something more explicit, if you have a function $f$ that's defined on $[0,a)$, then you can define the periodic version of $f$ as $f_p(x) = f(x \mod a).$ It shouldn't be too much work to generalize this to more general intervals. – benguin Aug 16 '16 at 03:41
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Let $D$ be the domain of $f.$ Assuming that $D$ is bounded and not empty, let $c=\inf D$ and $d=\sup D.$ Let $n>d-c.$ First extend the domain to the interval $[c,c+n)$ by assigning an arbitrary value (e.g. $0$) to $f(x)$ for any $x\in [c,c+n)$ \ $D.$

Let $[x]$ denote the largest integer not exceeding $x.$ For any $x$ let $x'=x-n [(x-c)/n].$ Then $x'\in [c,c+n)$ and $(x'-x)/n\in \mathbb Z.$ So let $f(x)=f(x').$

DanielWainfleet
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