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I'm trying to figure out this infinite product: $ f ( x ) = \prod_{ m = 1 }^{ \infty } { \cos (\frac{ x }{ 4 ^ { m }}) } $

I did the same trick as in the $2^m$ version to get this functional equation: $ f(x) \times f(2x) = \frac{ \sin ( x )}{ x } $

And now im really stuck, I'm not even sure if there's only one solution for this equation. I thought convolution might work but that was messy..

My main question is how to solve this infinite product, but anything about this equation will be great too.

Thanks!

What i did was: $ f(x)= \prod_{ m = 1 }^{ \infty } { \cos (\frac{ x }{ 4 ^ { m }}) } = \lim_{N \to \infty } \prod_{ m = 1 }^{ N } { \cos (\frac{ x }{ 4 ^ { m }}) } = \lim_{N \to \infty } \prod_{ m = 1 }^{ N } { \frac{ \sin (\frac{ x }{ 4 ^ { m - 1 }})}{ 4 \times \sin (\frac{ x }{ 4 ^ { m }}) \times \cos (\frac{ 2x }{ 4 ^ { m }})} } = \lim_{N \to \infty } \frac{ \sin ( x )}{ 4 ^ { N } \times \sin (\frac{ x }{ 4 ^ { N }}) \times \prod_{ m = 1 }^{ N } { \cos (\frac{ 2x }{ 4 ^ { m }}) } } = \frac{ \sin ( x )}{ x } \lim_{N \to \infty } \frac{ x }{ 4 ^ { N } \times \sin (\frac{ x }{ 4 ^ { N }})} \lim_{N \to \infty } \frac{ 1 }{ \prod_{ m = 1 }^{ N } { \cos (\frac{ 2x }{ 4 ^ { m }}) } } = \frac{ \sin ( x )}{ x } \times 1 \times \frac{ 1 }{ f \left( 2x \right) } $

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    $\begingroup$ @metamorphy. Terrific or Terrible ? $\endgroup$ Sep 8, 2022 at 6:24

2 Answers 2

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Comments. Probabilistic viewpoint:
Let $X_n, n=1,2,3,\dots$ be IID random variables with $\mathbb P(X_n = 1)=\mathbb P(X_n = -1)=\frac12$. Fix $r \in (0,1)$.Consider the rancom variable $$ Z_r = \sum_{n=1}^\infty r^n X_n . $$ Then your function is $f_{1/4}$, where $$ f_r ( x ) = \prod_{ m = 1 }^{ \infty } \cos \left(r^n x\right) $$ is the characteristic function of $Z_{r}$, that is: $$ \mathbb E\left[\exp\big(ixZ_{r}\big)\right] = f_r(x) . $$ In the case $r=1/2$, note that $Z_{1/2}$ is uniformly distributed on $[-1,1]$, so $$ f_{1/2}(x) = \frac12\int_{-1}^1 e^{ixz}\,dz = \frac{\sin x}{x} . $$ But for the case in this question, $r=1/4$, the random variable $Z_{1/4}$ has a singular distribution, concentrated on a Cantor set. (A fractal measure.) And $f_{1/4}$ is the characteristic function of that fractal measure.


But of course "known" functions do not have continuous singular Fourier transforms; so it is unlikely that $f_{1/4}$ can be written in terms of standard special functions.

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  • $\begingroup$ There is a classic question of Erdõs asking about whether the distribution of $Z_r$ is singular or absolutely continuous. $0<r<\frac12$ singular of course. $r=\frac12$ absolutely continuous. But what about $\frac12 < r < 1$? There are known values of $r$ in each camp. Considerable progress has been made, which I will not review here. $\endgroup$
    – GEdgar
    Sep 8, 2022 at 10:14
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Not more exciting than @metamorphy

Using my favored $1,400$ years old approximation $$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad \text{for} \qquad t\in \left(-\frac \pi 2,\frac \pi 2\right)$$

Using the q-Pochhammer symbol

$$P_n=\prod_{ m = 1 }^{ n} { \cos \left(\frac{x}{4^m}\right)}\sim(-1)^{n+1}\, 4^{n+1}\,\frac{\pi ^2+x^2}{\pi ^2-4x^2}\,\frac{\left(\frac{\pi ^2}{4 x^2};16\right){}_{n+1}}{\left(-\frac{\pi ^2}{x^2};16\right){}_{n+1}}$$

Let $ x=\frac \pi k$ to make $$P_n\sim (-1)^{n+1}\, 4^{n+1}\,\frac{k^2+1}{k^2-4}\,\frac{\left(\frac{k^2}{4};16\right){}_{n+1}}{\left(-k^2;16\right){}_{n+1}}$$

Computing $P_{1000}$ for various $k$ $$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 3 & 0.963283 & 0.963720 \\ 4 & 0.979268 & 0.979525 \\ 5 & 0.986708 & 0.986876 \\ 6 & 0.990761 & 0.990879 \\ 7 & 0.993208 & 0.993295 \\ 8 & 0.994798 & 0.994865 \\ 9 & 0.995889 & 0.995942 \\ 10 & 0.996669 & 0.996712 \\ 11 & 0.997247 & 0.997283 \\ 12 & 0.997686 & 0.997716 \end{array} \right)$$

Edit

Using $$\log (\cos (t))=\sum_{n=1}^\infty (-1)^n \frac{4^{n-1} } {(2n)! } (E_{2 n-1}(1)-E_{2 n-1}(0))\,\, t^{2n}$$ with $t=\frac x {4^m}$

$$P_n=1-\frac{x^2}{30}+\frac{7 x^4}{30600}-\frac{59 x^6}{83538000}+\frac{54121 x^8}{43797302640000}-\frac{21679 x^{10}}{15676653564000000}+\frac{7892306381 x^{12}}{7364296445064839280000000}-\frac{3966051541499 x^{14}}{6515855851628919146551200000000}++O\left(x ^{16}\right)$$

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