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# "Existence Proof" Correction

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.

Note by Ethan Robinett
2 years, 2 months ago

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@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. · 2 years, 2 months ago

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 $$|x|\to \infty$$. · 2 weeks, 2 days ago

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. · 2 years, 2 months ago