On a "simplification" in the infinite product formula for cosine

Question

I am looking for the mistake in the reasoning below, which I just cannot find.

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A book has this nice proof of

$$\sin(x) = x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$

Then the authors proceed to get an infinite product for $\cos(x)$ as

$$\cos(x) = \frac{\sin(2x)}{2\sin(x)} = \prod_{n=1}^\infty \left(1-\frac{4x^2}{(2n-1)^2\pi^2}\right)$$

As this part isn't elaborated, I guess the authors imply that we should just write down infinite products for $\sin(x)$ and $\sin(2x)$ and note that every $n$th element in $\sin(x)$ is also present in $\sin(2x)$ at $2n$, so only odd numbered elements will stay in $sin(2x)$ resulting in the formula.

Now, what gives me a hard time, is when I rewrite $\sin(x)$ as

$$\sin(x) = x\prod_{n=1}^\infty \left(\frac{n^2\pi^2-x^2}{n^2\pi^2}\right)$$

And then do the division

$$\cos(x) = \frac{2x\prod_{n=1}^\infty(\frac{n^2\pi^2-4x^2}{n^2\pi^2})}{2x\prod_{n=1}^\infty(\frac{n^2\pi^2-x^2}{n^2\pi^2})} = \prod_{n=1}^\infty\frac{n^2\pi^2-4x^2}{n^2\pi^2-x^2}$$

and as the following sets {$n, n=1,2,...$} and {$2n, n=1,2,...$} are both countable then there is a bijective transformation from one set to another, meaning that again every $n$th element in the denominator corresponds to $2n$th element (this time multiplied by $4$) in the numerator and we will be left with

$$\prod_{n=1}^\infty[4((2n-1)^2\pi^2-4x)]$$

which (obviously) diverges.

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I really cannot figure out the place where I made the mistake or maybe I used an illegal operation at some point, but as of this moment everything seems fine to me. I have thought maybe I somehow reordered the product, but in this case it converges absolutely, so reordering shouldn't change the limit. I also thought maybe I was wrong about infinite fraction reduction, but then again, seems like the authors use the same technique, also it doesn't contradict any law known to me. At this point I've become a bit paranoid about every operation used in the process, so please forgive me for possible stupid thoughts. Would be much appreciated if somebody pointed out the mistake.

Answer 1

You are considering a convergent product $$p=\prod_{n=1}^\infty\frac{x_n}{y_n}$$ such that, for every $n$, $$x_{2n}=4y_n\tag{$\ast$}$$ Can we use $(\ast)$ to get another formula for $p$? Well, not so easily...

To see why, let us remember that $p$ is defined as the limit of $$p_k=\prod_{n=1}^k\frac{x_n}{y_n}$$ when $k\to\infty$. Now, considering the even indices to simplify the notations and using $(\ast)$, one has $$p_{2k}=\prod_{n=1}^k\left(x_{2n-1}x_{2n}\right)\prod_{n=1}^{2k}\frac1{y_n}=\prod_{n=1}^k\left(x_{2n-1}(4y_{n})\right)\prod_{n=1}^{2k}\frac1{y_n}=\prod_{n=1}^k\left(4x_{2n-1}\right)\prod_{n=k+1}^{2k}\frac1{y_n}$$ hence the simplification you suggest is equivalent to omitting the products $$\prod_{n=k+1}^{2k}\frac1{y_n}$$ Since, in the present case, these products do not converge to $1$ (by far...), forgetting them is illicit.

Answer 2

Even if you crashed with the noticed bijection: noticing that the even members correspond and cancel leaving only the odd members left over to contribute seems the only way out of the polynomial dead-end if you want to arrive the textbook formula. I think the missing link are going from Euler's $$\sin[2x]=\Im[e^{xi+xi}]=2\sin[x]\cos[x]$$ $$\frac{\sin[πx]}{πx} =\prod_{n=1}^\infty\left[1^2-\left[\frac{x}n\right]^2\right]$$ to $$\cos\left[\frac\pi2x\right]=\frac{\sin[πx]}{2\sin[\frac\pi2x]} =\frac{πx\prod_{n=1}^\infty\left[1^2-[\frac{x}{n}]^2\right]}{2\frac\pi2x\prod_{n=1}^\infty\left[1^2-[\frac{x}{2n}]^2\right]} \\=\frac{ [\frac{1-x²}{1}][\frac{4-x²}{4}] [\frac{9-x²}{9}][\frac{16-x²}{16}]\cdots} {[\frac{4-x²}{4}][\frac{16-x²}{16}] [\frac{36-x²}{36}][\frac{64-x²}{64}]\cdots} =\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x}{n}\right]^2 \right]$$$$\sin\left[\fracπ2x\right]=\cos\left[\fracπ2[x-1]\right]$$ yielding an all-rational$$i^x=\exp\left[i\fracπ2x\right] =\cos\left[\frac\pi2x\right]+i\sin\left[\frac\pi2x\right] \\=\left[\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x}{n}\right]^2 \right]\right] +i\left[\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x-1}{n}\right]^2 \right]\right] $$