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First of all, recall the complex definitions of the two trigonometric functions:

$$ \sin(x) = \frac{e^{ix}-e^{-ix}}{2i}, \space \cos(x) = \frac{e^{ix}+e^{-ix}}{2}. $$

Now, the formula for $\sin(2x)$ can be derived from these two. But so can it be extended by multipying each new result by $ 2\cos(2^{b}x) $:

$$ 2\sin x \cos x = \sin(2x), \space 4\sin x \cos x \cos(2x) = \sin(4x), \space ... $$ and so on. Therefore $$ 2^{b}\sin x \prod_{n=0}^{b-1}\cos(2^{n}x) = \sin(2^{b}x). $$ However, it might be possible - and this is the step that could be a fallacy - to rewrite the product so as to generalize the outcome to some real numbers. $$ 2^{b}\sin\ x \prod_{n=0}^{\infty}\frac{\cos(2^{n}x)}{\cos(2^{n}2^{b}x)} = \sin(2^{b}x) => \prod_{n=0}^{\infty}\frac{\cos(2^{n}x)}{\cos(2^{n}bx)} = \frac{\sin(bx)}{b\sin x}.$$

So far, is this true? Pick x = 1 and notice this is false for b = $\frac{\pi}{4}$. However, for x = 1 and b = $\pi$ or x = 1 and b = 1 the product goes to appropriate values.

Can I now deduce from this the formula for the reciprocal of the product?

$$ \prod_{n=0}^{\infty}\frac{\cos(2^{n}bx)}{\cos(2^{n}x)} = \frac{b\sin x}{\sin(bx)} $$

Instead of starting the derivation with sinx, I could have started it with $\frac{1}{\sin x}$ and then continued multiplying the denominator. Hence the invalidity of the formula for $\frac{\pi}{2^{n}}, \space n > 1 \space and \space n \in \mathbb{N}, \space$ cannot be explained by the first product's singularities, right?

Is the idea anyhow correct, at least for some numbers?

jjagmath
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Artur Wiadrowski
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    If something is false for $b=\pi/4$, then it's false. If it's true for some other values of $b$, that's a coincidence. – Gerry Myerson Aug 29 '21 at 12:59
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    One can still ask for which values of $b$ it does work. $\sum_0^\infty b^{i}= \frac1{1-b}$ fails for $b=2$, but it works for $b=\frac12$, and that is not “a coincidence”. – MJD Aug 29 '21 at 13:23
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    There is an error in what you wrote, even for the $2^b$ case, as the term in the product does not converge to $1$, so the product does converge. Perhaps you meant to conjecture: $$ \prod_{n=1}^{\infty}\frac{\cos(2^{-n}bx)}{\cos(2^{-n}x)} = \frac{\sin(bx)} {b\sin x}$$ This is true in the case $b=2^k$ with $k\in\mathbb{N}$, by your argument. More generally the left hand side converges to a well defined function as long as $x$ is not an integer multiple of $\pi$. Numerically the left and right hand sides give identical functions in $x$, for all random values of $b$ that I tried. – tkf Aug 31 '21 at 16:20
  • I could have rewritten the product in your way. Thank you! However, is the derivation rigorous enough to count as a proof? Is the version from the top also correct? – Artur Wiadrowski Aug 31 '21 at 18:12
  • I have added a proof of my version. I do not believe your version is correct, or even well defined. For an infinite product to converge to a non-zero value, a minimal necessary condition is that the terms of the product converge to $1$. The terms in your product appear to go all over the place! This is not a proof that your product does not converge, but it seems very unlikely. – tkf Aug 31 '21 at 18:29
  • Having plotted it I noticed the possible divergence. The reason that I call it that "possible" is because one can also see the product oscillating near the correct outcome, which might have a meaning. – Artur Wiadrowski Aug 31 '21 at 18:32
  • To call it oscillation may be wrong. But perhaps the average value of the product after each iteration approaching the correct result is closer to what I intended to say. – Artur Wiadrowski Aug 31 '21 at 18:34
  • Thanks. For your product to converge $${\rm min}\left( \left[\left|2^n\frac{x(b-1)}{2\pi}\right|\right], \left[\left|2^n\frac{x(b+1)}{2\pi}\right|\right]\right)\to 0$$ This will only happen when $\left|\frac{x(b-1)}{2\pi}\right|$ or $\left|\frac{x(b+1)}{2\pi}\right|$ have finite binary expansion. – tkf Aug 31 '21 at 19:11
  • Actually, not necessarily finite, but still very specific. – tkf Aug 31 '21 at 19:20

2 Answers2

1

As I mentioned in comments, I think the conjecture you intended was actually

$$ \prod_{n=1}^{\infty}\frac{\cos(2^{-n}bx)}{\cos(2^{-n}x)} = \frac{\sin(bx)} {b\sin x}.$$

This follows immediately from the following identity:

$$ \prod_{n=1}^{\infty}{\cos(2^{-n}t)} = \frac{\sin t}t,\qquad (1) $$

as we may simply combine the cases $t=bx$ and $t=x$ to get: \begin{eqnarray*} \prod_{n=1}^{\infty}\frac{\cos(2^{-n}bx)}{\cos(2^{-n}x)} &=& \frac{\prod_{n=1}^{\infty}{\cos(2^{-n}bx)}}{\prod_{n=1}^{\infty}{\cos(2^{-n}x)}}\\ \\&=&\frac{\frac{\sin(bx)}{bx}}{\frac{\sin x}x} \\&=&\frac{\sin(bx)} {b\sin x} \end{eqnarray*}

The proof of $(1)$ is similar to your original argument:

For $m\in \mathbb{N}$ we have: $$\frac{\sin(2^{-m}t)}{2^{-m}t}\prod_{n=1}^m \cos(2^{-n}t)=\frac{\sin t}t.$$

Letting $m\to \infty$ gives us $(1)$.

This argument seems to have been given many times on this site: see for example here or here or here or here.

I thought the way you arrived at the conjecture was very clever (+1) and reminded me of Euler's construction of the Gamma function as an infinite product.

tkf
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note that: $$\sin(2x)=\frac{e^{2ix}-e^{-2ix}}{2i}=\frac{(e^{ix}+e^{-ix})(e^{ix}-e^{-ix})}{2i}=2\frac{e^{ix}-e^{-ix}}{2i}\frac{e^{ix}+e^{-ix}}{2}=2\sin(x)\cos(x)$$ Try this expansion with your hypothesis and see what you get

Henry Lee
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