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With only the formal definition of $$f(x) = \exp(-x)= \sum \frac{(-x)^n}{n!}$$ how can we show that $$\lim_{x\to \infty} f(x)=0?$$

I am looking for a proof that would not use the identity $\exp(x)\exp(-x)=1$ in order to find a strategy for similar problems (i.e. limits of functions defined as power series $\sum a_n x^n$ where $a_n$ does not have a fixed sign beginning at any $n_0$).

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  • $\begingroup$ Use the fact that $f'(x) = -f(x)$. see my answer. $\endgroup$
    – Paramanand Singh
    Aug 12, 2015 at 6:09
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    $\begingroup$ Perhaps he wants to find a method that will work on Bessel functions like $$\sum\frac{(-1)^n x^n}{(n!)^2}$$ $\endgroup$
    – GEdgar
    Aug 13, 2015 at 1:12

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I don't think that this is possible.

The proof has to be very specific to the given series because of the following: Consider $g(x) = \sum a_n x^n$ with $a_n = \frac{1}{n!}$ for all $n \neq n_0$ and $a_n = \frac{1}{n!} + \delta$ for some $\delta > 0$ and some fixed even integer $n_0 \geq 2$.

We then have

$$ g(x) = e^x + \delta x^{n_0} \xrightarrow[x \to -\infty]{} \infty, $$ since the first summand vanishes and the second one goes to $\infty$.

This shows that just chaning a single coefficient with an arbitrarily small perturbation makes the claim invalid.

BTW: The same argument works for odd $n_0$, but in this case, we get convergence to $-\infty$ instead of $\infty$.

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    $\begingroup$ It doesn't mean it's impossible, it just means it's hard. :P $\endgroup$ Aug 11, 2015 at 14:19
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    $\begingroup$ @columbus8myhw: Yes, but the OP stated that he wanted a proof which would allow him to solve the same question for more general series. The above argument shows that it will at least be very difficult to adapt the proof for the $e^x$ series to more general series. Nevertheless, the question is very interesting. $\endgroup$
    – PhoemueX
    Aug 11, 2015 at 14:24
  • $\begingroup$ Ah.${{}{{}}{{}{{}}}}$ $\endgroup$ Aug 11, 2015 at 14:24
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From the power series we can obtain $f'(x) = -f(x)$. Also $f(0) = 1$. Using these we will show that $f(x) > 0$ for all $x$.

Let's suppose that there is a number $a$ such that $f(a) = 0$ and consider the function $$g(x) = f(a - x)f(x - a)$$ then its derivative is given by \begin{align} g'(x) &= -f'(a - x)f(x - a) + f(a - x)f'(x - a)\notag\\ &= f(a - x)f(x - a) - f(a - x)f(x - a)\notag\\ &= 0\notag \end{align} so that $g(x)$ is a constant and therefore $1 = g(a) = g(0) = 0$ and we get a contradiction. Hence $f(x) \neq 0$ for all $x$. Since $f(0) = 1$ it follows by continuity that that $f(x) > 0$ for all $x$.

From $f'(x) = -f(x) < 0$ we see that $f(x)$ is strictly decreasing and since it is bounded below by $0$, the limit $\lim_{x \to \infty}f(x) = L$ exists. Hence $\lim_{x \to \infty}f'(x) = -L$ also exists. Now by mean value theorem we have $$f(x + 1) - f(x) = f'(\xi)\tag{1}$$ where $x < \xi < x + 1$. Taking limit as $x \to \infty$ on both sides of $(1)$ we get $L - L = -L$ so that $L = 0$. Hence $\lim_{x \to \infty}f(x) = 0$.

Note: As noted by PhoemueX in his answer, there is no general theorem to prove that a power series tends to $0$ as $x \to \infty$ based on certain criteria for its coefficients. Each power series needs to be analyzed properly in order to determine its behavior near $\pm\infty$. Here I managed to get a nice property of $f(x)$ namely $f'(x) = -f(x)$ from the series representation which is very useful in the analysis given in my answer.

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One approach would be to show, via the power series than $f(N) = f(1)^N$ focusing on integer $N > 1$ so your power series manipulations are manageable. Then establish that $f(1) < 1$. It's easy from here to show that $f(N)$ converges to $0$ in the limit of large $N$. Now you must prove that $f(x) \leq f([x])$ for any real $x$ where $[x]$ is the greatest integer not exceeding $x$.

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    $\begingroup$ Yeah, but that seems a little silly, because he specified he didn't want to use $e^te^{-t}=1$. Using $f(N)=f(1)^N$ isn't quite equivalent to that, but it's exactly the same sort of thing... $\endgroup$ Aug 11, 2015 at 16:50
  • $\begingroup$ Not quite sure I understand your point. $\endgroup$
    – josh
    Aug 12, 2015 at 1:29
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I think the correct strategy in other situations is to look for something analogous to $\exp(t)\exp(-t)=1$.

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Another obstruction is that the proof has to work with $x$ replaced with $ax$ for positive $a$, which changes the $n$'th coefficient by a factor of $a^n$.

A more accessible question is to find nice conditions on coefficients of a formal power series $1 + \sum a_n x^n$ so that the logarithm or the inverse of the series have negative coefficients.

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