A differentiable approximation of modulus?

Question

I'm trying to find a differentiable approximation of the "fract" function, which returns the fractional portion of a real number.

$y = x-\lfloor x\rfloor$

I have something that works "ok", that I got by adapting a bandlimited saw wave.

$y=0.5-\frac{sin(2\pi x)+sin(4\pi x)/2+sin(6\pi x)/3+sin(8\pi x)/4+sin(10\pi x)/5}{\pi}$

I can add more harmonics to make the band limited saw wave closer to the actual "fract" function, but for my usage case, all these trig function calls are getting pretty expensive.

I was curious, are there other (better quality / lower computational complexity) ways to differentiably approximate this function?

Answer 1

A function like

$$x-\frac{x^n}{1+x^n}$$

well approximates $x-\lfloor x \rfloor$ on the interval $[0, 2]$ for large $n$. If we take it on the interval $\left[{1 \over 2}, {3 \over 2}\right]$ and periodically repeat it, we get a nice almost differentiable approximation. $x^n$ can be efficiently calculated using the binary exponentiation algorithm (thus it would be handy if $n=2^k$).

NB: I said almost differentiable, since in the boundary points the derivatives on different sides are $\approx 1 - \frac{n}{2^{n-1}}$ and $1-\frac{n}{(3/2)^{n+1}}$, so, if $n$ is suitably large, the derivative is $\approx 1$ from practical perspective.

A live example on desmos.

Answer 2

Assume $x\to f(x)$ is the discontinous function you want to get smoother/more regular. Then, for example the local mean value integral $$F(x) = \frac{1}{2\Delta_x}\int_{x-\Delta_x}^{x+\Delta_x}f(\varphi)d\varphi \hspace{1cm} \text{(local averaging)}$$ will be differentiable for any $\Delta_x\in \mathbb R^+$ (why?) You can estimate this as a discrete sum (low pass filter) or you can calculate an explicit expression for it analytically as a continuous time convolution since $f(x)$ is so nice in this example. A slightly smoother and more complicated one is if we iterate it: $$F_2(x) = \frac{1}{2\Delta_x}\int_{x-\Delta_x}^{x+\Delta_x}F(\varphi)d\varphi \hspace{1cm} \text{(linear interpolation)}$$

Answer 3

Let $\{ x \}$ mean $x - \lfloor x \rfloor$. Given your goal of having the derivative be cheaply computable by computer, you should use a piecewise defined function:

$$ f(x) = \begin{cases} \{ x \} & \epsilon \leq \{ x \} \leq 1 - \epsilon \\ g(x) & 0 \leq \{ x \} < \epsilon \\ h(x) & 1 - \epsilon < \{ x \} < 1 \end{cases} $$

where $g$ and $h$ are any easily computed functions (e.g. quadratic polynomials) that satisfy the conditions listed below, and $\epsilon$ is a small positive number.

The point being that for most values of $x$, you have $f'(x) = 1$ so there is very little work in computing the derivative.

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The conditions for $f$ to be differentiable are:

• $g(0) = h(1)$
• $g(\epsilon) = \epsilon$
• $h(1-\epsilon) = 1 - \epsilon$
• $g'(0) = h'(1)$
• $g'(\epsilon) = 1$
• $h'(1 - \epsilon) = 1$

For symmetry, you probably want $h(x) = 1 - g(1-x)$. Then the conditions reduce to

• $g(0) = 1/2$
• $g(\epsilon) = \epsilon$
• $g'(\epsilon) = 1$

Answer 4

Here is a differentiable one. Just repeat $f$ every unit interval.

$f(x) = x + \dfrac{(\frac12-x)c(1+c)}{(x+c)(1-x+c)}$ for every $x \in [0,1]$, with parameter $c \to 0^+$.

The gradient also tends to $1$ at the midpoint as $c \to 0^+$.

The main advantage is that no exponentiation is needed; just pure arithmetic.

Example with $c=0.001$.