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From dabbling with p-adics, I remember mentions of $\infty $ being the 'infinite prime'. So hence something like: $\frac{x^\infty -1}{x - 1} = \displaystyle \prod_{\substack{d| p_\infty \\ d \neq 1}} \Phi_{d}(x) = \Phi_{p_\infty}(x) = \sum_{n=0}^{\infty -1} x^n$ would perhaps make some abstract sense in some formal power series something-or-other. However, if we instead claim that $\substack{n | \infty \\ n \in \mathbb{N}}$ we get $\frac{x^\infty -1}{x - 1} = \displaystyle \prod_{\substack{n=2}}^\infty \Phi_{n}(x)$, which would be appealing (though to my knowledge, simply untrue) if we work with p-adics etc. For reals, with for instance $x = \frac{1}{100} \in \mathbb{R}$, the RHS seems to hover around the 'expected asymptote' value of $\frac{100}{99}$ with increasing variance as $n$ increases. For the 2-adics, setting $x=2$ and taking the logarithm gives an infinite sum of infinite sums, with each $\log(\Phi_{n}(x))$ producing a sum starting with a 2-adic valuation given by the first non-zero power of $x$ in that polynomial and thence growing, and there are infinitely many of them with $x^1$ (e.g. all primes), so the sums would not converge (unless there could be cancellation in some space? dunno).

So after all these incoherent babble, I was wondering what is known about the convergence of the product of all cyclotomic polynomials with or without $\Phi_1(x)$ -- and whether the first or second representation of $\frac{x^\infty -1}{x - 1}$ make any sense.

Thanks!

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    $\begingroup$ $$\prod_{m=0}^\infty \Phi_{2^m}(x) =\lim_{n \to \infty} \prod_{m=0}^n \Phi_{2^m}(x) = \lim_{n \to \infty} x^{2^n}-1 = -1 \qquad \text{for} \ |x|< 1$$ $\endgroup$ – reuns Jun 11 at 20:13
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The infinite products $\prod_{n=1}^\infty \Phi_n(x)$ and $\prod_{n=2}^\infty \Phi_n(x)$ diverge for all $x \ne 0$. In order to converge, you would certainly need $\Phi_n(x) \to 1$ as $n \to \infty$, but if $n$ is prime $\Phi_n(x) = 1 + x + x^2 + \ldots + x^{n-1} = 1 + x + O(x^2)$.

Note that for an infinite product "converge" means "converge to a nonzero limit".

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