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I am faced with explaining to a bunch of people who have taken a course in real analysis, but no course in complex analysis, why $\sin(x)=x\prod_{n\geq1}(1-x^2/\pi^2n^2)$. I vaguely remember being taught a proof of this of the following form: check product converges, and then observe (because sin has simple zeros) that the ratio of the two sides is holomorphic with no zeros or poles; now check that its growth at infinity is not too large (to rule out it being $e^x$ or whatever) and deduce the ratio is a constant, which must be 1". But this "growth" trick uses some complex analysis which they don't know (and in writing this question I realise that I'm no longer sure of it either). Is there any way I can get away with just real analysis? I am happy to use use basic properties of complex numbers and even radius of convergence, but I want to avoid Cauchy's (integral) theorem and beyond.

Gaurav Mishra
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Kevin Buzzard
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  • I don't know about proof, but take the limit as $x\to0$ of $\sin x/x$. –  May 20 '14 at 14:12
  • i wonder if you can show rhs satisfies the second order differential equation $y^{\prime \prime} + y = 0$ and conditions $y(0) = 0$ and $y^\prime(0) = 1?$ – abel May 20 '14 at 14:14
  • @abel: my gut feeling is that any attempt to prove that the RHS satisfies the differential equation will get bogged down in trying to e.g. relate sum 1/n^2 to sum 1/n^4 (when one is checking coefficients of the power series match up) and this is not elementary. – Kevin Buzzard May 20 '14 at 15:00
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    This has already been answered here http://math.stackexchange.com/questions/157372/infinite-product-of-sine-function – Winther May 20 '14 at 15:13

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