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I've recently taken interest in infinite products, and I'm having trouble with a proof I found in this PDF file: "Infinite Products and Elementary Functions":

An intermediate step in finding an infinite product to represent $\sin(x)$ is given as follows:

$$\sin x=x\lim_{n\to \infty}\sum_{k=0}^{(n-1)/2}(-1)^k\binom{n}{2k+1}\frac{x^{2k}}{n^{2k+1}} \tag{2.8} $$ Of all the stages in Euler’s procedure, which, as a whole, represents a real work of art, the next stage is perhaps the most critical and decisive. Factoring the polynomial in (2.8) into the trigonometric form: $$\sin x=x\lim_{n \to \infty}\prod_{k=1}^{(n-1)/2}\left(1-\dfrac{(1+\cos(2k\pi/n))x^2}{(1-\cos(2k\pi/n))n^2}\right)$$

I've tried a proof by induction, but could only manage to check a couple of iterations, that hinted to a pattern that looks like this: $$\prod_{k=1}^{\frac{n-1}{2}} (1-a_{k,n}\frac{x^2}{n^2})$$ but this is a dead end since I couldn't find a trigonometric expression of $a_{k,n}$. I can't figure out where the trigonometric functions came from in the right hand side of the second equation.

My last attempt was to turn the product into a sum using the natural logarithm, but again this complicated things further.

The author seems to hint at an elementary factorization, but I'm starting to have some doubts. Is there really an elementary approach? Is there some identity I'm missing? Any answer, or ("Socratic") hint is appreciated.

entrelac
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    $$\dfrac{1+\cos\theta}{1-\cos\theta}=\cot^2\frac{\theta}2$$ and try this method. – Bumblebee Jul 29 '15 at 08:23
  • @Nilan I am aware of this identity, but in the PDF it's the next step in the proof (under the guise of $1/\tan^2(k\pi /n)$). I need to turn the polynomial in $(2.8)$ into trig. form. But I'm checking out the link right now. – entrelac Jul 29 '15 at 08:26

1 Answers1

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If we don't insist on writing it with the $\dfrac{1+\cos\theta}{1-\cos\theta}$ form, a factorisation is easy to be had. We have

$$\sum_{k = 0}^{\left\lfloor \frac{n-1}{2}\right\rfloor} (-1)^k \binom{n}{2k+1} \frac{x^{2k}}{n^{2k+1}} = \frac{\bigl(1 + i\frac{x}{n}\bigr)^n- \bigl(1 - i\frac{x}{n}\bigr)^n}{2ix}.$$

Writing

$$1 \pm i \frac{x}{n} = \sqrt{1 + \frac{x^2}{n^2}}\cdot \exp \Biggl(\pm i\arctan \frac{x}{n}\biggr)$$

we see that

$$\frac{\bigl(1 + i\frac{x}{n}\bigr)^n- \bigl(1 - i\frac{x}{n}\bigr)^n}{2ix} = \biggl(1+\frac{x^2}{n^2}\biggr)^{n/2}\cdot \frac{\sin\bigl(n\arctan \frac{x}{n}\bigr)}{x},$$

so the zeros of the polynomial are

$$\pm n\tan \frac{k\pi}{n},\quad 1 \leqslant k \leqslant \biggl\lfloor\frac{n-1}{2}\biggr\rfloor.$$

Grouping the zeros in pairs of negatives, we obtain the factorisation

$$\sum_{k = 0}^{\left\lfloor \frac{n-1}{2}\right\rfloor} (-1)^k \binom{n}{2k+1} \frac{x^{2k}}{n^{2k+1}} = \prod_{k = 1}^{\left\lfloor \frac{n-1}{2}\right\rfloor}\biggl( 1 - \frac{x^2}{n^2\tan^2 \frac{k\pi}{n}}\biggr),$$

since the constant term of the polynomial is $1$.

I unfortunately don't see a natural way that leads to the form

$$\prod_{k = 1}^{\left\lfloor \frac{n-1}{2}\right\rfloor} \biggl( 1 - \frac{(1+\cos (2k\pi/n))x^2}{(1 - \cos (2k\pi/n))n^2}\biggr).$$

Daniel Fischer
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  • This is great! Just out of curiousity, if you don't mind me asking: what led you to think of the first step? Was the $\binom{n}{2k+1}$ enough to make you think of the binomial expansion? Or was there some other intuition? – entrelac Jul 31 '15 at 20:02
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    The binomial coefficient together with the corresponding powers next to it. Pulling the $x$ in, we have $(-1)^k\binom{n}{2k+1}\bigl(\frac{x}{n}\bigr)^{2k+1}$, and then writing $(-1)^k = i^{2k} = \frac{1}{i}\cdot i^{2k+1}$ is often enough useful to give it a try. – Daniel Fischer Jul 31 '15 at 20:08
  • I just noticed something: suppose we multiply the expression containing the sine by some function that has no roots ($e^x$ is an obvious example), wouldn't it lead by the same procedure to the same product ? In other words, the product can correspond to multiple functions. – entrelac Jul 31 '15 at 20:44
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    If we multiply it by $e^x$ or something of that ilk, we no longer have a polynomial. We still have a product representation with the same product for the zeros, if we multiply with a "nice enough" function, but we also get another factor of the form $e^{g(x)}$. If we have a polynomial, that factor is a constant (equivalently, $g(x)$ is constant), and if the constant term of the polynomial is $1$, the constant is $1$. See e.g. the Weierstraß product theorem for the general theory. – Daniel Fischer Jul 31 '15 at 20:55
  • How do the zeros of the polynomial are

    $$\pm n\tan \frac{k\pi}{n}$$? Can you please explain?

     –  Oct 07 '17 at 03:03
  • @N.Maneesh Since $1 + \frac{x^2}{n^2}$ is strictly positive, the polynomial evaluates to $0$ if (and only if) $\sin \bigl(n\arctan \frac{x}{n}\bigr) = 0$, and $x\neq 0$ (the latter because we divide by $x$, removing that zero of the sine). Now $\sin \varphi = 0 \iff \varphi = k\pi$ for some $k \in \mathbb{Z}$. And $n \arctan \frac{x}{n} = k\pi \iff \frac{x}{n} = \tan \frac{k\pi}{n}$. The constraint $x\neq 0$, and $\lvert \arctan \psi\rvert < \frac{\pi}{2}$ (I'm using the principal branch of $\arctan$) yield the condition $1 \leqslant \lvert k\rvert < \frac{n}{2}$. – Daniel Fischer Oct 07 '17 at 11:21