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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:

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

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

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

to

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

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    @DaniloAraújoSilva : It doesn't work as well. The difference is visible. BTW, you're missing a factor of $x$ in your infinite product.2012-04-19

2 Answers 2

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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. Th$x$.2012-04-18
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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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    Spelli$n$g: It's Jim Belk.2012-04-19