Recently I posted a proof stating that there exist functions that can be written as the product of a polynomial and an exponential function such that said function approaches zero as the single variable approaches infinity. I want to apologize, as this proof was incorrect, and I also want to thank Abhishek Sinha for pointing this out, as I probably wouldn't have noticed this. The note has been removed. I noticed no logical error in my proof, but the result was incorrect, which indicates that somewhere, a mistake was made. However, after examining the issue, it is apparent that the opposite is true. We will prove:

There do not exist functions \(f(x)\) such that:

\(f(x) = e^{ax}p(x)\) where \(p(x)\) is a polynomial and \(a \in \mathbb{R}^{+}\)

and \(\displaystyle \lim_{x\to \infty} f(x) \neq \pm \infty\)

With the one crucial exception of \(p(x) = 0\), which is the trivial case.

Proof: Let \(p(x) = \displaystyle \sum_{i=0}^{n} b_ix^i\), where {\(b_i\)} is a set of fixed complex coefficients. Then we have:

\(f(x) = e^{ax}\displaystyle \sum_{i=0}^{n} b_ix^i\)

Now consider \(\displaystyle \lim_{x\to \infty} e^{ax}\displaystyle \sum_{i=0}^{n} b_ix^i\)

If \(\displaystyle \lim_{x\to \infty}\displaystyle \sum_{i=0}^{n} b_ix^i \neq 0\) then

\(\displaystyle \lim_{x\to \infty} f(x) = \pm \infty\) in all cases. Let us consider the case in which:

\(\displaystyle \lim_{x\to \infty}\displaystyle \sum_{i=0}^{n} b_ix^i = 0\)

In this case we have:

\(\displaystyle \lim_{x\to \infty} e^{ax}\displaystyle \sum_{i=0}^{n} b_ix^i = \infty \times 0\)

Which is an indeterminate form. If we rearrange this, we obtain:

\(\displaystyle \lim_{x\to \infty} e^{ax}\displaystyle \sum_{i=0}^{n} b_ix^i = \displaystyle \lim_{x\to \infty} \frac{\displaystyle \sum_{i=0}^{n} b_ix^i }{\frac{1}{e^{ax}}} = \frac{0}{0}\)

Now, this form is conducive to L'Hopital's rule. Notice that:

\(\displaystyle \lim_{x\to \infty} \frac{d^j}{dx^j}(\frac{1}{e^{ax}}) =\pm \infty\) \(\forall\) \(j \geq 0\)

Also notice that:

\(\frac{d^{n+1}}{dx^{n+1}} \displaystyle \sum_{i=0}^{n} b_ix^i = 0\)

Then, if we apply L'Hopital's Rule \(n+1\) times, we arrive at:

\(\displaystyle \lim_{x\to \infty} \frac{\frac{d^{n+1}}{dx^{n+1}}\displaystyle \sum_{i=0}^{n} b_ix^i }{\frac{d^{n+1}}{dx^{n+1}}\frac{1}{e^{ax}}} = \frac{0}{\infty} = 0\)

Then \(\displaystyle \lim_{x\to \infty} f(x) \neq \pm \infty\) if and only if \(\displaystyle \lim_{x\to \infty}\displaystyle \sum_{i=0}^{n} b_ix^i = 0\)

However, let us examine the implications of the above condition.

\(\displaystyle \sum_{i=0}^{n} b_ix^i \) is a polynomial of degree \(n\) and hence can be written as the product:

\(\displaystyle \sum_{i=0}^{n} b_ix^i = \displaystyle \prod_{i=1}^{n} (x-r_i)\)

where {\(r_i\)} is the set of all roots of \(\displaystyle \sum_{i=0}^{n} b_ix^i\).

Then:

\(\displaystyle \lim_{x\to \infty}\displaystyle \sum_{i=0}^{n} b_ix^i = 0\)

\(\Rightarrow\) \(\displaystyle \lim_{x\to \infty}\displaystyle \prod_{i=1}^{n} (x-r_i) =0\)

\(\Rightarrow\) \((\displaystyle \lim_{x\to \infty} x-r_1)(\displaystyle \lim_{x\to \infty} x-r_2) ...(\displaystyle \lim_{x\to \infty} x-r_n) = 0\)

\(\Rightarrow\) \((\infty)^n=0\)

Assuming \(p(x)\) is a polynomial of finite order, this is impossible. Hence, the only condition for converge of \(f(x)\) cannot be satisfied by any polynomial of finite order other than \(p(x)=0\). Then, by contradiction, we conclude:

There do not exist functions \(f(x)\) such that:

\(f(x) = e^{ax}p(x)\) where \(p(x)\) is a polynomial and \(a \in \mathbb{R}^{+}\)

and \(\displaystyle \lim_{x\to \infty} f(x) \neq \pm \infty\)

With again, the only exception being \(p(x) = 0\).

QED

Again, I apologize for the incorrect proof. I didn't mean to mislead anyone.

## Comments

Sort by:

TopNewest@Abhishek Sinha – Ethan Robinett · 2 years, 2 months ago

Log in to reply

– Abhishek Sinha · 2 years, 2 months ago

Now the statement looks fine. The general rule is that you can't kill an exponential growth by a polynomial.Log in to reply

I just want to add a small point; in fact, when \(|x|\to \infty\) the only possibility for a finite degree polynomial \(p(x)\) is \(\lim_{|x|\to \infty}|p(x)|=\infty\). To see this, just assume that \(\exists c\in \mathbb{R}\), such that \(\lim_{|x|\to \infty}|p(x)|=c\), and then consider either of the polynomials \(q(x)=p(x)\pm c\). Then the hypothesis would yield \(\lim_{|x|\to \infty}|q(x)|=0\), which again is impossible, as can be shown (and is shown in the answer) by considering the factorization of the polynomial and taking \(|x|\to \infty\). – Samrat Mukhopadhyay · 2 weeks, 2 days ago

Log in to reply

Can't one simplify this proof by using the rule that the limit of a product of two functions is the same as the product of two functions' limits?

\( e^{ax} \) for positive a is an exponential growth which we know limits to positive infinity as x approaches infinity. Polynomial functions are combinations of its parameters through arithmetic, thus by the rules of limits their limit as x approaches infinity is infinity (because for all x with a non-zero coefficient c, the limit of x as x approaches infinity is infinity). The product of two positive infinities is a positive infinity and hence there exists no function that does not limit to positive infinity if the function is a product of an exponential growth and a polynomial function excluding the zero polynomial. – Ralph Schraven · 2 years, 2 months ago

Log in to reply

– Ethan Robinett · 2 years, 2 months ago

The reason I did it this way is because a while ago, I wrote a proof that suggested the opposite. I wanted to make it clear that said proof was incorrect, because if you read the proof (it's been removed, if you want I can summarize it for you) it was unclear why the actual proof was wrong. All of the logical steps seemed to be correct, but the result was not. Really, the above proof expresses what you just said, as well as the fact that no non-zero polynomial can approach 0 at infinity. So, in summary, yes you can simplify this a lot as you are well aware, but I try to write these things not just for higher-level readers such as yourself, but also for those that are not as experienced with proof-based math so that they may both learn and understand with a little more clarity the reasons for many facts.Log in to reply