function $e^x$ as quotient of two polynomials

Question

Show that the function $e^x$ can not be expressed as a quotient of two polynomials. A solution could be this:

Suppose $e^x=\frac{P(x)}{Q(x)}$ then $\quad \frac{e^x}{P(x)} = \frac{1}{Q(x)}$

We know that $e^x \quad$ tends to infinity faster than any power of $x$. So if we do

$\lim\limits_{x\to \infty}\frac{e^x}{P(x)} = \infty $ and

$\lim\limits_{x\to \infty}\frac{1}{Q(x)} = 0 $

we get a contradiction.

Do you think is it a good solution?

and .... could anyone suggest a better solution?

Answer 1

It depends on what you are allowed to assume, but if you are allowed to use standard differentiation rules - then a somewhat nice proof might be the following:

Assume, seeking a contradiction, that there exists polynomials $P$ and $Q$, such that $e^x = \frac{P(x)}{Q(x)}$.

Then by differenting both sides, we get $e^x = \frac{P'(x) Q(x) - P(x)Q'(x)}{Q(x)^2}$ by the quotient rule.

Therefore $\frac{P(x)}{Q(x)} = \frac{P'(x) Q(x) - P(x)Q'(x)}{Q(x)^2}$.

Rearranging both sides gives us $P(x) Q(x) = P'(x) Q(x) - P(x)Q'(x)$.

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Now you have two polynomials on the left and the right. You should be able to reach a contradiction by considering the degree of the polynomials on each side.

Answer 2

If you want a more formal reasoning, you can use the derivative. Let $n,m$ be the degrees of the polynomials $P_n(x),Q_m(x)$, respectively: $$e^x=\frac{P_n(x)}{Q_m(x)}$$ Then, $e^x\,Q(x)=P(x)$ and you can differentiate both sides to obtain: $$e^x\,Q_m(x)+e^x\,Q'_{m-1}(x)=P'_{n-1}(x)$$ so in the left-hand side you have a new polynomial of the same degree as $Q_m$, $R_m(x)=Q_m(x)+Q'_{m-1}(x)$ and: $$e^x\,R_m(x)=P'_{n-1}(x)$$ Observe that the degree of the polynomial in the right-hand side decreases.

Repeating this process $n$ times, you finally obtain a polynomial $S_m(x)$ such that: $$e^x\,S_m(x)=0$$ which is a contradiction.

Answer 3

For any two polynmials $P(x)$ and $Q(x)$, the limits$$\lim_{x\to+\infty}\frac{P(x)}{Q(x)}\qquad\text{and}\qquad\lim_{x\to-\infty}\frac{P(x)}{Q(x)}$$are identical in $\mathbb{R}\cup\{\pm\infty\}$. But$$\lim_{x\to+\infty}e^x=+\infty\qquad\text{and}\qquad\lim_{x\to-\infty}e^x=0.$$