Compactly supported sigmoid-like function (pursuit of closed-form expression)

Question

Introduction:

This question concerns the existence, uniqueness and most importantly the expression (closed form or otherwise) of a compactly supported sigmoid-like function, whose derivative has the same shape as the function. This function has many practical applications due to its property of being compactly supported while also yielding well-behaving derivatives of high orders.

Description:

Suppose $f$ is a real function that satisfies

1. $f(-1) = -1$, $f(1) = 1$
2. $f$ has continuous derivatives of all orders
3. $f$ has $n$'th derivatives $f^{(n)}(-1) = f^{(n)}(1) = 0$ for any integer $n > 0$
4. $f'\left(\frac{x-1}{2}\right) = f(x)+1$, $x \in [-1,1]$

It can be assumed that $f(x)=-1$ for $x \le -1$ and $f(x)=1$ for $x \geq 1$.

We wish to obtain a closed-form expression of $f$

No successful attempts have yet been made in obtaining any closed or computable form of $f(x)$. Instead we present the function obtained numerically, depicting it with its derivatives (figure 1) and comparing it with closely related functions (figure 2).

The bold line illustrates that $f(x)$ is identical to its first derivative scaled in $x$ (visualization of [4.]).

It is possible to construct various compactly supported sigmoid-like functions by combining a regular sigmoid function such as $erf(x)$ and $tanh(x)$ with an argument that diverges at $x=-1$ and $x=1$. Indeed $f(x)$ is an antiderivative of a class of functions called bump functions with more references documenting their construction. We provide examples of two functions $g(x)$ and $h(x)$ that do not satisfy [4.] comparing them to the numerically obtained $f(x)$

Any suggestions or ideas for obtaining the closed-form expression of $f$ or methods to compute efficiently its function values are appreciated.

Answer 1

Too long for the comments:

I still need to check this with more detail, but I think the idea is right.

Lets assume $f$ is odd.

Set $x=1$ in $f'(\tfrac{x-1}{2})=f(x)+1$ we get $f'(0)=2$.

Deriving $n-1$ times we get $\frac{1}{2^{n-1}}f^{(n)}(\tfrac{x-1}{2}) = f^{(n-1)}(x)$. Setting $x=1$ we get $f^{(n)}(0) = 2^{n-1}f^{(n-1)}(1) = 0$ for $n\ge 2$.

Now set $x=0$ to get $f'(-\tfrac{1}{2})=1$ and $f''(-\tfrac{1}{2})=2 f'(0)=4$ and $f^{(n)}(-\tfrac{1}{2}) = 2^{n-1}f^{(n-1)}(0) = 0$ for $n\ge 3$.

Since $f$ is odd we get $f'(\tfrac{1}{2})=1$ and $f''(\tfrac{1}{2})=-4$ and $f^{(n)}(\tfrac{1}{2}) = 0$ for $n\ge 3$.

Now set $x=\tfrac{1}{2},-\tfrac{1}{2},\tfrac{1}{4},\tfrac{-1}{4},\tfrac{3}{4}\ldots$. All the derivatives will be $0$ from some point on. This means that the radius of convergence of the Taylor series around that points is $0$.

Answer 2

You can consider the first order derivative and call it $w = f'$. It is a compactly supported bell shaped function.

$$ \forall x \in \left[-1, 1 \right], \quad w'(x) =C \left( w(2x+1) - w(2x-1) \right)\tag{*}\label{FuncEq} $$

As such it admits a Fourier transform $\hat{w}$. From Eq.\eqref{FuncEq} \begin{align} 2\pi i \xi \hat{w}(\xi) &= \frac{C}{2} \hat{w}\left(\frac{\xi}{2} \right)\left( e^{\pi i \xi} - e^{-\pi i \xi} \right)\\ &= i C \hat{w}\left(\frac{\xi}{2} \right)\sin(\pi \xi) \end{align} it implies that we must necessarily have $C=2$.

By induction: \begin{align} \hat{w}(\xi) &= \hat{w}\left( \frac{\xi}{2^{N+1}} \right) \prod_{n=0}^N\mathrm{sinc}\left( \frac{ \xi}{2^{n}} \right) \end{align} where $\mathrm{sinc}$ is the cardinal sine defined by: \begin{equation} \mathrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \end{equation} And then \begin{align} \hat{w}(\xi) &= \hat{w}\left( 0 \right) \prod_{n=0}^\infty\mathrm{sinc}\left( \frac{\xi}{2^{n}} \right) \end{align} As \begin{align} \mathrm{sinc}\left( \xi\right) = \prod_{k=1}^\infty \cos\left( \frac{\pi \xi}{2^{k}} \right) \end{align} we also have: \begin{align}\label{eq:what} \hat{w}(\xi) &= \hat{w}\left( 0 \right) \prod_{n=0}^\infty\mathrm{sinc}\left( \frac{\xi}{2^{n}} \right)\\ &=\hat{w}\left( 0 \right) \prod_{n=0}^\infty\prod_{k=1}^\infty \cos\left( \frac{\pi \xi}{2^{k+n}} \right)\\ & = \hat{w}\left( 0 \right) \prod_{h=1}^\infty \cos^{h}\left( \frac{\pi \xi}{2^{h}} \right) \end{align} $w$ being compactly supported in $\left[-1, 1 \right]$ it can be extended into a periodic function of period $2$ and, as it is smooth, $w$ is equal to its Fourier extension on $\left[-1, 1 \right]$. The Poisson summation formula yields: \begin{align} w(x) = \frac{1}{2}\sum_{k=-\infty}^{\infty}\hat{w}\left(\frac{k}{2}\right)e^{\pi i k x} \end{align} That can be simplified further but that is the idea.

More details here: https://papers.ssrn.com/sol3/papers.cfm?abstract_id=4665218