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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?

Geek
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  • Prove what, exactly? What do you mean by "exponentially decaying", precisely? – MPW Mar 13 '20 at 15:59
  • For t > 1, f(t) < ae^(-bt), where a and b are constants. – Geek Mar 13 '20 at 16:01
  • You can see my answer, it is valid for any polynomial you can multiply $e^{-t}$ with. What you need to find is the largest real zero to the polynomial $P'(x)-P(x)$ – mathreadler Mar 13 '20 at 16:31

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For all $x\ge0$,

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

This does not deserve a paper.

  • Thanks. This is a key step for a proof in my paper. I am actually considering a more general case like x^{20}e^{-x}, can we say this function exponentially decaying for x > 20? – Geek Mar 13 '20 at 16:19
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    It works for any polynomial. –  Mar 13 '20 at 16:51
  • Yes definitely not publishable as a paper. This is more what you would expect to be an example, an exercise or maybe exam question in some bachelor level's first calculus course. – mathreadler Mar 13 '20 at 19:17
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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.

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:

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

mathreadler
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  • Thanks for your reply. I know this method, the function in my question is a simpler case of the model you provided, but I am wondering this method is rigorous or not, so that is why I seek for some paper or book to refer to. – Geek Mar 13 '20 at 16:46
  • @Geek : Something like this should exist as examples or exercises in calculus textbooks for Bachelors' first year students. I am afraid I don't have any examples of such books since my own course literature from this part of my studies was not in the English language. – mathreadler Mar 13 '20 at 16:55
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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.

William The conqueror
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  • Thanks for you reply. I have modified my question. I need to rigorously prove it is exponential decreasing after t > 2 (or t > 20 for my modified version), not simply monotonic decreasing. – Geek Mar 13 '20 at 16:38