0
$\begingroup$

Assuming that the solution of $e^{-x}=x$ is $c\in (0,1)$

And give the following sequence

$$a_{n+1}=e^{-a_n}$$

$$a_1=1$$

How can i prove that the sequence converge and that the limit is $$\lim_{n \to \infty}a_n=c$$

I tried to define a new sequence $$b_n=a_n-c$$ and to prove it using the ratio test but without success

$\endgroup$
-1
$\begingroup$

Exponential function is continuous,then

$$ \lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}e^{-a_n}=e^{-\lim_{n\to\infty}a_n}. $$

So the limit of this sequence is the solution of $x=e^{-x}$ which is $c$.

To see the existence, first note by induction that $1\geq a_n\geq 0$. Now we claim that subsequence with odd terms decreasing and subsequence with even terms increasing. For $k=1$, $ a_{2k-1}\geq a_{2k+1} $ is immediate and to see $a_{2k}\leq a_{2k+2}$ note that $1\geq a_n$ and $e$ is monotone. Thus, $ a_2=e^{-1}\leq a_4=e^{-a_3}. $

Suppose now for some $k$, $a_{2k-1}\geq a_{2k+1}$ and $a_{2k}\leq a_{2k+2}$. Then

$$ e^{-a_{2k}}\geq e^{-a_{2k+2}}\implies a_{2k+1}\geq a_{2k+3},\\ e^{-a_{2k+1}}\leq e^{-a_{2k+3}}\implies a_{2k+2}\leq a_{2k+4}. $$

This tells us that both $\lim_{k\to\infty}a_{2_k}=l_0$ and $\lim_{k\to\infty}a_{2k-1}=l_1$ exist by monotone convergence theorem. Now we claim that every odd term is bigger than every even term. Trivially for $k=1$, this claim holds. Suppose now that it is true for some $k$, $a_{2k-1}\geq a_{2k}$. Then

$$ a_{2k+1}=e^{-a_{2k}}\geq e^{-a_{2k-1}}=a_{2k}\implies e^{-a_2k}=a_{2k+1}\geq e^{a_{2k+1}}=a_{2k+2} $$ which shows that the claim is true for $k+1$ and induction is completed. So we show that $a_{2k}\leq a_{2k+2}\leq a_{2k+1}\leq a_{2k-1}$. Note also that $l_0$ is supremum and $l_1$ is infimum for corresponding sequences. Then $l_1\geq l_0$.

By induction one can show $a_{2k+1}=a_{2k}^{1/e}\geq a_{2k+1}^{1/e}=a_{2k+2}$. Taking limit on both sides gives us $l_0^{1/e}\geq l_1^{1/e}$ and $l_0\geq l_1$. Thus $l_0=l_1=l$ and $a_k\to l$.

$\endgroup$
1
  • 1
    $\begingroup$ If the limit exists, it must be $c$. Showing the existence of the limit is the less easy task. $\endgroup$ Feb 13 '16 at 21:37
1
$\begingroup$

$f(x) = \mathrm{e}^{-x}$ is a contraction in the relevant range $(0, 1]$, thus $x_{n + 1} = f(x_n)$ converges if started inside that range.

$\endgroup$
2
  • 1
    $\begingroup$ Can you elaborate or provide some reference? I undertand contraction is a kind of coordinate transform, how does it work here? $\endgroup$
    – Yuriy S
    Feb 13 '16 at 22:40
  • 1
    $\begingroup$ @YuriyS Check this document. The relevant theorem is 3.6. You can also search for Banach Fixed Point Theorem. $\endgroup$
    – user64066
    Feb 14 '16 at 12:49
0
$\begingroup$

Write $e^y=\exp(y)$ - it's easier to write recursive sequence:

$$a_n=\exp(-\exp(-\exp(-...\exp(-1)))))$$

Then you can write:

$$x=\lim_{n \rightarrow \infty} a_n=\exp(-x)$$

Since just one step of recursion will not change the limit of infinite sequence.

The rest is clear.

As for proving the limit exists - every term in the sequence is positive, and every term is bounded.

$$0 \leq a_n \leq 1$$

The sequence is actually not monotonic (as can bee seen by numerical experiment $a_3=0.500, a_4=0.606, a_5=0.545, a_6=0.580$ etc), so the proof of convergence is tricky

$\endgroup$
2
  • $\begingroup$ You assume that the limit exist,but how do you prove it exist? $\endgroup$
    – aaadddaaa
    Feb 13 '16 at 21:42
  • $\begingroup$ See the edit I made. We start with 1 and you can see that the sequence is bounded. That's a start, not the whole proof of course $\endgroup$
    – Yuriy S
    Feb 13 '16 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.