I start with the fact that the zeros of $\sin x$ are $ n\pi$, $n\in\mathbb{Z}$. Therefore, it should be possible to express it as an infinite product: $\sin x = x (x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\cdots$ $ = x\prod_{n=1}^\infty (x^2 - n^2\pi^2) $ From here I cannot factor out $\pi^2$ as there are infinite terms and $\pi^\infty$ doesn't look reasonable. If I take out $n^2$ then I have $x(n!)^2 \prod_{n=1}^\infty (\frac{x^2}{n^2} - \pi^2)$ But this isn't complete. Infact, this looks wrong (n is defined inside the product afterall). What further manipulations would enable me to get the correct product: $\sin x = x\prod(1-\frac{x^2}{n^2\pi^2})$
Trying to get the infinite product for $\sin x$
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trigonometry
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0many complex analysis books have discussions of infinite products (including the weierestrass product product formula and usually the specific examples of the sine and gamma functions) – 2011-08-26
1 Answers
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The infinite product representation for the sine function is $\sin(\pi x)=\pi x\prod_{1}^\infty \left(1-\frac{x^2}{n^2}\right).$ So in the post, $\sin x$ should be replaced by $\sin(\pi x)$. Then the issue raised in the post disappears.
To prove the result, one needs quite a bit more function theory than the informal type of reasoning about zeros. Note that for example $e^z$ has no zeros, but $e^z$ is not equal to the empty product.
Your informal attack on the infinite product is not very distant from that of Euler. That is very good company to be in!
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0The factor $\pi x$ in the expansion of $\sin \pi x$ can be seen as a consequence of $\displaystyle\underset{x\rightarrow 0}{\lim }\displaystyle\frac{\sin \pi x}{x}=\pi$. – 2011-08-26