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We know $\sin(x)=0$ has solutions $0,\pm\pi,\pm2\pi,\pm3\pi,\dots$.

So $\sin(x)$, if interpreted as a polynomial, could be written as:

$a_0x^0+a_1x^1+a_2x^2+\cdots$ and we know this polynomial too:


So, the question is, is it possible to transform the factored form of $\sin(x)$:

$$\sin(x)=a x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\dots$$


$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\ ?$$

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In $\TeX$ you can code $\pm2\pi$ as \pm2\pi. (If you want $\pm x$ you need a space between \pm and x so that it doesn't look as if "pmx" is the control sequence.) (I editied accordingly.) Also, notice that when it's a binary rather than unary operator you get more space between $\pm$ and $x$, thus: $y \pm x$ (just as with "$+$" and "$-$"). – Michael Hardy Apr 18 '12 at 2:46
I wrote a blog post a few years ago that partially addresses this question:… – Jim Belk Apr 18 '12 at 2:48
@JimBelk, nice post. Thx. – GarouDan Apr 18 '12 at 16:00
@MichaelHardy, thx about the clue. I'll remember^^ but is interesting that $^+_{-}$ works too. – GarouDan Apr 18 '12 at 16:01
@DaniloAraújoSilva : It doesn't work as well. The difference is visible. BTW, you're missing a factor of $x$ in your infinite product. – Michael Hardy Apr 19 '12 at 12:48
up vote 10 down vote accepted

The answer is yes, you can factor $\sin(z)$ into a product of zeros. The general theory behind this is Weierstrass factorization. For your example,

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

In fact, Euler famously used an unrigorously derived form of this identity to solve the Basel problem. I say "unrigorously" here because, while one can show that the function can be written as a product over it's zeros, it's the outside term (1 in this case, infront of the first 'z') that takes work to derive. For example, if we had the function $e^z\sin(z)$, then there would be an $e^z$ factor on the outside of the product. Since $e^z$ doesn't have any zeros, you cannot break it down into such a product, so you just tack it on as a factor. More difficult functions have even more intricate product representations but the general rule of thumb is that the function factors into a product over zeroes times something that looks like $e^{g(s)}$.

An interesting consequence of this is that it's not necessarly possible to directly transform such a product into the infinite polynomial corresponding to the function. One can however, write down a correspondence between products and sums of the zeroes and the polynomial coefficients. This comes from Vieta's formulas which is precisely what Euler used to show $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$.

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Great answer Sam. This formula really works. Fine. This Euler beautiful result I already know, but this Weierstrauss factorization is new to me. Thx. – GarouDan Apr 18 '12 at 16:02

The proposal is: $$\sin x = a(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots$$

The standard result, already posted by Sam, is in effect $$ \sin x = \frac{x-\pi}{\pi}\cdot \frac{x+\pi}{\pi} \cdot \frac{x - 2\pi}{2\pi}\cdot \frac{x+2\pi}{2\pi}\cdot\frac{x-3\pi}{3\pi}\cdot\frac{x+3\pi}{3\pi} \cdots $$ So the coefficient "$a$" in front of the whole thing is more . . . . interesting . . . than might be initally guessed. Might Euler have considered $$ a = \frac{1}{\pi^2}\cdot\frac{1}{(2\pi)^2}\cdot\frac{1}{(3\pi)^2}\cdots $$ to be some sort of "infinitely small number"? Might it actually be fruitful in some way to think of it that way?

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I think you are probably right. It's pointed out again here:… – Alex R. Apr 18 '12 at 3:52
Well, the formula works so I think yes. Jim Belt use this in his post. – GarouDan Apr 18 '12 at 16:04
Very interesting posting by Jim Belt. – Michael Hardy Apr 19 '12 at 4:37
Spelling: It's Jim Belk. – Michael Hardy Apr 19 '12 at 12:51

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