# Evaluating $f(x) f(x/2) f(x/4) f(x/8) \cdots$

Let $f : \mathbb R \to \mathbb R$ be a given function with $\lvert f(x) \rvert \le 1$ and $f(0) = 1$. Is there a nice simplified expression for \begin{align}F(x) &= f(x) f(x/2) f(x/4) f(x/8) \cdots \\ &= \prod_{i=0}^\infty f(x/2^i)?\end{align} If there isn't a general solution, as seems likely, can anything useful be said about the case when $f(x) = \operatorname{sinc} x = \frac{1}{x} \sin x$?

This question arose when idly wondering about the limit of convolving an infinite number of dyadically(?) scaled versions of a kernel $g$ together. Taking the Fourier transform of $g(x) * 2g(2x) * 4g(4x) * 8g(8x) * \cdots$ yields the above expression.

-
Please add tags as appropriate; I couldn't think of any others. – Rahul Jan 9 '11 at 8:45
f(x) = sin x is the only case I know where the answer is nice. It's a nice exercise. I guess something like f(x) = e^x as well. – Qiaochu Yuan Jan 9 '11 at 13:44
@Qiaochu Yuan: If sin(x) is nice, then I would have to conclude (based on the method given below) that sinc(x) will also work out. But perhaps you meant sinc(x) since sin(0) = 0. – hardmath Jan 9 '11 at 18:31
Oh, this reminds me, I went ahead and calculated a related product a few months ago: deoxygerbe.wordpress.com/2010/10/13/… – deoxygerbe Jan 9 '11 at 23:52
@Qiaochu: I guess you mean f(x) = cos x, for which F(x) = sinc x. It's fun to visualize that in terms of convolutions. – Rahul Jan 9 '11 at 23:54

Assuming analyticity of $f(x)$ it seems more tractable to work with the form involving sums after taking logarithms:

$$\log F(x) = \sum_{i=0}^\infty \log f(x/2^i)$$

Now if $\log f(x)$ has convergent power series expansion in a neighborhood of zero:

$$\log f(x) = \sum_{k=1}^\infty a_k x^k$$

Note that since $f(0) = 1$ the constant term $a_0$ of $\log f(x)$ is zero and thus omitted.

Then:

$$\log F(x) = \sum_{k=1}^\infty a_k (\sum_{i=0}^\infty 2^{-ik}) x^k$$

Note that the inner summations are just $\sum_{i=0}^\infty 2^{-ik} = 1/(1 - 2^{-k})$ and thus uniformly bounded by $2$.

-
Thanks, that looks like a good way to go about it. Unfortunately there doesn't seem to be a nice expression for the power series expansion of log f(x) in terms of that of f(x). – Rahul Jan 9 '11 at 23:54
@Rahul Narain: Apparently Sage (via Maxima) can produce the power series expansion of $\log \operatorname{sinc}(x)$. Let me play around with it and see what comes out. – hardmath Jan 10 '11 at 1:27
@RahulNarain Using this approach yields $F\left(x\right)=e^\left(\sum _{n=1}^{\infty}\frac{(-1)^n 2^{4 n-1}B_{2 n} x^n}{\left(4^n-1\right)n(2n)!}\right)$ where $B_{i}$ is the ith Bernoulli Number – kram1032 Dec 15 '12 at 13:22
$x^n$ above should be $x^{2n}$. My bad. – kram1032 Dec 15 '12 at 13:31
This approach can not hold beyond the first zero of $F\left(x\right)$ though, can it? Because the series has to become $-\infty$ at that that point to satisfy it. And beyond that point, it would have to become complex to go negative, which it definitely does... (see my graph of it here ) – kram1032 Dec 15 '12 at 13:54

For what it's worth, I found a partial answer to the thought that motivated my original question. When $f(x) = \operatorname{sinc}(x)$, it is the Fourier transform of a rectangular function, $g(t) = 1/2$ for $t \in [-1,1]$ and $g(t) = 0$ otherwise (or some scaling thereof, depending on convention). The infinite product $f(x) f(x/2) f(x/4) f(x/8) \cdots$ corresponds to the convolution which I wrote as $g(t) * 2g(2t) * 4g(4t) * 8g(8t) * \cdots$ in a small abuse of notation. Each term in the latter convolution is the probability distribution function of a uniformly distributed random variable, and the convolution is the pdf of their sum. This distribution is given by (some scaling of) the Fabius function: a compactly supported, infinitely differentiable function that is not analytic anywhere in its support. So the original infinite product falls off exponentially fast but probably cannot be expressed in any nice form.

-

I've uploaded here plots of some approximation of limit distributions which are defined by inverse Fourier transform: $$\mathrm{h}_a(x):=\mathcal{F^{-1}}\biggl[\prod\limits_{k=1}^{\infty}\mathrm{sinc}(t\cdot a^{-k})\biggr](x).$$

-