Meaning (Definition) of "Decays Exponentially"?

Question

When I look this up I get an unambiguous definition that $f(x) $ decays exponentially if it is a function of the form $f(x) = Ae^{-\lambda x}$ for some real $A$ and $\lambda > 0$. This fits my expectation up to a point, but doesn't deal with an exercise I'm doing: "show $x e^{-x}$ decays exponentially as $x \to \infty$". Clearly this is not a function of the form $Ae^{-\lambda x}$.

Might exponential decay mean that there exists some real $A$, $\lambda > 0$, and $x_0$ such that for all $x > x_0$, $|f(x)| < Ae^{-\lambda x}$ ? In that case I can complete the exercise as follows"

Proposition: for $x > 2$, $x < e^{x/2}$. Proof: for $x > 2$, $x^2 > 2x$ so $e^{x/2} = 1 + x/2 + (1/2!).(x/2)^2 + ... > 1 + x/2 + x/2 + ... > x$.

Proposition: for $x > 2$, $|x e^{-x}| < e^{-x/2}$ and therefore decays exponentially. Proof: $ 0 < x e^{-x} < e^{x/2}e^{-x} = e^{-x/2}$.

However, I'm left with some doubts about my interpretation. Does $f(x) = 0$ decay exponentially ? I would think it fails to satisfy the "decay" part. So, should the term refer to a non-zero function with modulus bounded by a diminishing exponential, or even a function which is approximated in some way by a diminishing exponential ?

To take the given exercise further ...

Proposition: for any $\epsilon > 0$ there is $x_0$ such that for all $x > x_0$, $ x < e^{\epsilon x}$. Proof: for any $\epsilon, n$ there is $x_0$ such that $x_0 > ((n+1)!/\epsilon^{n+1})^{1/n}$. Then $(\epsilon x_0)^{n+1}/(n+1)! = x_0.\epsilon ^{n+1}.x_0^n/(n+1)! > x_0 \implies e^{\epsilon x_0} > x_0$

Proposition: $xe^{-x} $ is "approximated" by a decreasing exponential function. Proof: for any $\epsilon > 0$ there is $x_o$ such that for $x > x_0$, $e^{-x} < xe^{-x} < e^{(\epsilon - 1)x}$

I'd appreciate feedback on the meaning of this term.

Actually, to clarify my question following some feedback, it's the definition I'm looking for rather than an interpretation of its meaning.

Answer 1

The meaning of the exponential decay is essentially a decay so strong that it beats the growth of any polynomial: for any $\lambda>0$,

$$\lim_{x\to\infty}P(x)e^{-\lambda x}=0.$$

You could claim that $P(x)e^{-\lambda x}$ is not a "pure" exponential decay. But considering

$$A:=\max_x\left(P(x)e^{-\lambda x/2}\right)$$ we have

$$P(x)e^{-\lambda x}\le Ae^{-\lambda x/2},$$ so that $P(x)e^{-\lambda x}$ actually decreases faster than an exponential of coefficient $\lambda/2$.

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You can feel it by looking at a plot of $\log\left(P(x)e^{-\lambda x}\right)=\log(P(x))-\lambda x$ for various polynomials, which is always asymptotic to a straight line.

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A clean definition is indeed:

There exists some real $A$, $\lambda > 0$, and $x_0$ such that for all $x > x_0$, $|f(x)|\le Ae^{-\lambda x}$.

Or, in the big-O notation, simply

$$f(x)=O\left(e^{-\lambda x}\right).$$

Answer 2

In the field of Partial Differential Equations, we use to say that

a solution $u$ of some differential equation decays exponentially fast (at infinity) if it can be bounded in the following way: $$|u(x)| \leq C \exp (-\alpha |x|)$$ for every $|x| \gg 1$, where $C>0$ and $\alpha>0$ are suitable constants.

This is a very important property for solutions of a PDE (mainly of elliptic type), and it has been widely investigated in the literature. In our dialect, the exponential decay never refers to being equal to an exponential function, since this would (almost) never happen. The case that $u \equiv 0$ may, in principle, happen: anyway, for our purposes this is a trivial and totally clear situation.