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. — 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. — 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. — Lord_Farin (talk) 20:07, 14 May 2014 (UTC)

Real counterpart

There has arisen a need for a real-valued counterpart for this page, which basically establishes the result needed in place of Exponential Dominates Polynomial, i.e.:

$\lim_{x \to \infty} \dfrac {P(x)}{\exp x} = 0$

A proof based on L'Hôpital was extracted from Expectation of Exponential Distribution:

The exponential function tends to infinity and therefore the limit is indeterminate ($\frac \infty \infty$) and satisfies L'Hôpital's Rule, Corollary 2. This allows us to take the derivatives of the numerator and denominator and achieve the same limit. This changes the limit to:

$\lim_{u \to \infty} {\frac 1 {\exp u}}$

This limit tends to zero.

Inclusion of this result will yet again improve PW by a small margin :). — Lord_Farin (talk) 14:24, 20 September 2014 (UTC)

Do we have a result for $P_n \to +\infty$ or $-\infty$ for any nonconstant polynomial? --GFauxPas (talk) 13:17, 21 September 2014 (UTC)

I think we don't. At least I couldn't find it. — Lord_Farin (talk) 13:22, 21 September 2014 (UTC)

It should be easy to craft -- I've seen a proof that uses at its heart the result that if $n > m$ then $a_1 x^n$ dominates $a_2 x^m$ (or whatever the language is) whatever $a_1$ and $a_2$ are. --prime mover (talk) 14:25, 21 September 2014 (UTC)

Interestingly enough, we don't have a LHR up for limits at infinity. --GFauxPas (talk) 20:20, 21 September 2014 (UTC)

What's a LHR? I think I might have suffered a stroke, because I can't understand the simplest things at the moment. --prime mover (talk) 20:45, 21 September 2014 (UTC)

L'Hôpital's Rule, sorry. --GFauxPas (talk) 20:46, 21 September 2014 (UTC)