Polynomial upper bounding by $2^x$. True or false?

Question

I have the following proposition and I am not sure whether it is true or not. Don't necessarily need a proof.

$$ \forall n\in \mathbb{N}: \text{ as }x\to\infty , a_0 + a_1x + a_2x^2 + ... + a_nx^n \le 2^x $$ where $a_0,a_1,a_2,...,a_n$ are constants.

Basically, for any polynomial, does $2^x$ eventually pass it? Thanks.

You can also write this as $a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n \in O \left( 2^x \right)$. It's known as Landau notation, or sometimes "Big Oh" notation. — Matt Groff, Apr 24 '20 at 04:58

Fareed Abi Farraj · Answer 1 · 2019-05-01T03:41:14.897

0

It is true, $2^x=e^{\ln {2^x}}=e^{x.\ln 2}$ and the exponential function is always greater than any polynomial as $x$ goes to positive infinity.

edited May 01 '19 at 03:41

answered May 01 '19 at 03:28

Fareed Abi Farraj

This feels a little like kicking the can down the road. It just moves the question to one in terms of $e^x$ instead and completely neglects that there is a proof for that too. – PrincessEev May 01 '19 at 03:30
I'd like to see another proof for the question, but this is all what came into my head so far and I thought it would help, and also Colin doesn't really seem that he wants a proof for this one – Fareed Abi Farraj May 01 '19 at 03:34
1

$e^{x \ln2}\geq (x\ln2)^{n+1}/(n+1)!$ for positive $x$. – Jens Schwaiger May 01 '19 at 03:39

1 Answers1