Talk:Limit at Infinity of Polynomial over Complex Exponential

I'm not sure what the problem is with $e^{iy}$ being constant ... --GFauxPas (talk) 13:12, 14 May 2014 (UTC)


 * If $y = 0$ then $e^{iy}$ = 1. If $y = \pi$ then $e^{iy} = -1$. $1 \ne -1$. $e^{iy}$ is not constant. --prime mover (talk) 13:18, 14 May 2014 (UTC)


 * But we're taking the limit as $x \to +\infty$, we're not concerned with $y$? --GFauxPas (talk) 13:37, 14 May 2014 (UTC)


 * That's not the issue. I can see what you're trying to do, and I know where you're going with it. But the fact is that $e^{iy}$ is not constant. Your result works, but your reasoning on how to get there seems flawed to me.


 * My recommendation is to consider applying the modulus. --prime mover (talk) 14:57, 14 May 2014 (UTC)

I think I've sufficiently addressed the issues. &mdash; Lord_Farin (talk) 18:43, 14 May 2014 (UTC)


 * "Constant with respect to $x$"? Not a concept I've seen before -- I believe it probably needs explaining. I thinbk I know what it means, but it's not obvious. --prime mover (talk) 18:49, 14 May 2014 (UTC)


 * Well, we can consider this an intermediate proof. The modulus approach is more elegant and universal. I just don't feel like typing it up right now. &mdash; Lord_Farin (talk) 19:01, 14 May 2014 (UTC)


 * Not sure if this is the modulus approach you were thinking of, but is this better? --GFauxPas (talk) 20:02, 14 May 2014 (UTC)


 * Yes, that's about it. Technically Exponential Dominates Polynomial should be adjusted to $x \in \R$ instead of the implicit $n \in \N$, but it's a brainless exercise. &mdash; Lord_Farin (talk) 20:07, 14 May 2014 (UTC)