Prove a function is exponential decaying after a finite time

Question

Consider a function $f(t) = te^{-t}$ and $t > 0$, obviously, it is not exponentially decaying. However, the derivative of $f(t)$ is less than $0$ for $t > 1$, and from plot, $f(t)$ is exponentially decreasing after $t > 1$.

Is there any paper considering such function and proving that $f(t)$ is exponentially decaying after $t > 1$, or can someone prove it? Thanks!!

Actually it is for some more general case, like f(t) = t^20e^{-t}, can we say this function is exponentially decreasing after t > 20?

Answer 1

For all $x\ge0$,

$$xe^{-x}<e^{-x/2}.$$

This does not deserve a paper.

Answer 2

It should be easily provable that for any polynomial $x\to P(x)$ : $$P(x)e^{-x}$$ will be exponentially decaying for large enough $x$.

It's derivative is $$P'(x)e^{-x} - P(x)e^{-x} = e^{-x}(P'(x)-P(x))$$

Only has zeros where $P'(x)-P(x) = 0$.

Since this is a polynomial of same degree as $P(x)$ by the fundamental theorem of algebra it will have at most degree of $P(x)$ number of real zeros.

Now we have proven monotocity for all $x$ above some largest $x$, let us call it $x_0$. What remains is to prove limit at infinity is $0$.

This you can probably finish, I suppose.

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Added for rigour: This question assumes a couple of things:

1. Our function is differentiable. This is ensured by a theorem saying that the product of two differentiable functions is differentiable.

2. The product theorem of differentiable functions $$(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$$

3. A differentiable function which has no zeros in the derivative on an interval is strictly monotonic on this interval.

4. The function has a limit 0 as $t\to \infty$. Limits are usually treated before derivatives are introduced in course literature. One of the main reasons for this being, that differentiation in such literature often is defined as a limit:

$$f'(x) = \lim_{h\to 0} \left\{ \frac{f(x+h)-f(x)}{h} \right\}$$

Answer 3

To prove decay or increase you should consider the derivative with respect to your desired parameter, here $t$, $$df(t)/dt = e^{-t} - te^{-t}$$ so simply, when $t>1$ $$df(t)/dt < 0 $$ Now if you were living in the 1667 A.C, that could have been a paper and you famous. However, these days you can easily find what you have discovered in any textbook and thus no one would publish your discovery! Simply because there is no novelty in this work.