Exploring the Boundaries of a Recursive Sequence

In order to find the value of $x$, we can represent it as the limit of an infinite sequence $a_n$ such that $x=lim_{n rightarrow infty} a_n$ and $x=exp(-x)$. This limit of infinite sequence remains unchanged even after one step of recursion. However, the sequence is not monotonic, as can be observed from numerical experiments, and hence proving its convergence is tricky.
One way to solve this is by using the fact that exponential function is continuous. Therefore, $lim_{ntoinfty}a_n=lim_{ntoinfty}a_{n+1}=lim_{ntoinfty}e^{-a_n}=e^{-lim_{ntoinfty}a_n}$, which gives us the value of $x$ satisfying $x=e^{-x}$, i.e., $c$. To establish the existence of $c$, we can use induction to show that $1geq a_ngeq 0$.


Supposing that the value of $c$ in the interval $(0,1)$ satisfies the equation $e^{-x}=x$.

And give the following sequence




What is the method to demonstrate that the sequence

sequence converge
converges to the limit $c$ as $n$ approaches infinity?

I attempted to establish a distinct sequence labeled as $$b_n=a_n-c$$ and verify its validity using the ratio test, but my efforts were fruitless.

Solution 1:

The function $f(x) = mathrm{e}^{-x}$ is a contraction over the interval
relevant range
$(0, 1]$ which implies that if started within this range, the sequence $x_{n+1} = f(x_n)$ will converge.

Solution 2:

Instead of using
recursive sequence
, it is simpler to express $e^y$ as $exp(y)$.

The value of $a_n$ is obtained by applying the exponential function to the number $-1$ repeatedly, a total of $n$ times.

Then you can write:

The equation can be written as $x=displaystylelim_{n to e^{-x}}n$.

The limit of an infinite sequence will not be altered by a single recursion step.

The rest is clear.

When it comes to demonstrating the positivity and boundedness of the sequence

limit exists
, it can be shown that each term in the sequence satisfies both conditions.

$$0 leq a_n leq 1$$

The numerical experiment with values of $a_3$, $a_4$, $a_5$, and $a_6$ (e.g., $a_3=0.500$, $a_4=0.606$, $a_5=0.545$, $a_6=0.580$) shows that the sequence is not monotonic. Therefore, dealing with
proof of convergence
can be complicated.

Solution 3:

Provided that the exponential function is continuous, then…

The limit of $a_n$ as $n$ approaches infinity is equal to the limit of $a_{n+1}$ as $n$ approaches infinity. Additionally, the limit of $e^{-a_n}$ as $n$ approaches infinity is equal to $e^{-lim_{ntoinfty}a_n}$.

The solution of $x=e^{-x}$, denoted as $c$, is the limit of this sequence.

To establish its existence, we can use induction to show that $1geq a_ngeq 0$. Then, we can assert that there is a subsequence with decreasing odd terms and a subsequence with increasing even terms. For $k=1$, it’s clear that $
a_{2k-1}geq a_{2k+1}
$. Similarly, we can show that $
a_{2k}leq a_{2k+2}
$ by noting that $1geq a_n$ and $e$ is monotonic. In particular, we have $
a_2=e^{-1}leq a_4=e^{-a_3}.

In the event that $a_{2k-1}$ is greater than or equal to $a_{2k+1}$ and $a_{2k}$ is less than or equal to $a_{2k+2}$ for some value of $k$, then…

If $e^{-a_{2k}}$ is greater than or equal to $e^{-a_{2k+2}}$, then $a_{2k+1}$ is greater than or equal to $a_{2k+3}$. Similarly, if $e^{-a_{2k+1}}$ is less than or equal to $e^{-a_{2k+3}}$, then $a_{2k+2}$ is less than or equal to $a_{2k+4}$.

By applying the monotone convergence theorem, we can deduce that both $lim_{ktoinfty}a_{2_k}=l_0$ and $lim_{ktoinfty}a_{2k-1}=l_1$ are in existence. It can be observed that the odd terms are always greater than the even terms. This can be easily verified for $k=1$. Assuming that this is true for some $k$, i.e., $a_{2k-1}geq a_{2k}$, we can proceed further.

By applying induction, we can prove that $a_{2k}leq a_{2k+2}leq a_{2k+1}leq a_{2k-1}$. Additionally, it is important to note that the supremum and infimum for the corresponding sequences are $l_0$ and $l_1$, respectively. Thus, we can conclude that $l_1geq l_0$. The proof also involves showing that $a_{2k+1}=e^{-a_{2k}}geq e^{-a_{2k-1}}=a_{2k}implies e^{-a_{2k}}=a_{2k+1}geq e^{a_{2k+1}}=a_{2k+2}$. This completes the induction.

Through induction, it can be proven that $a_{2k+1}=a_{2k}^{1/e}geq a_{2k+1}^{1/e}=a_{2k+2}$. As we take the limit of both sides, we obtain $l_0^{1/e}geq l_1^{1/e}$ and $l_0geq l_1$, which leads to $l_0=l_1=l$, causing $a_kto l$.

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