By examining the curves y=x and y=a^x, prove that

By examining the curves $y=x$ and $y=a^{x}$, prove that the sequence $x_1=a,\ x_2=a^{a},\ x_3=a^{a^{a}},\ldots,\ x_n=a^{x_{n-1}}$ converges to a definite limit as $n\to\infty$, provided $e^{-e}\lt a\le e^{1/e}$.

Step 1: Fixed points, and when they exist

A fixed point $\xi$ of $f(x)=a^{x}$ satisfies $a^{\xi}=\xi$, i.e.\ $\ln a=\dfrac{\ln\xi}{\xi}$. Since $g(\xi)=\dfrac{\ln\xi}\xi$ has a single maximum $1/e$ at $\xi=e$ (from Problem 552's own tangency computation, this maximum is exactly where $y=a^x$ is tangent to $y=x$), a fixed point exists exactly when $\ln a\le1/e$, i.e.\ $a\le e^{1/e}$.

Step 2: Stability: when is the fixed point attracting?

The iteration $x_{n+1}=f(x_n)$ converges locally to $\xi$ when $|f'(\xi)|\lt 1$. Since $f'(x)=a^{x}\ln a$ and $a^{\xi}=\xi$, \[f'(\xi) = \xi\ln a = \ln\xi \tag{1}\]

(using $\xi\ln a=\ln\xi$ from the fixed-point equation itself). The stability condition $|f'(\xi)|\lt 1$ becomes

\[ |\ln\xi|\lt 1 \;\Longrightarrow\; e^{-1}\lt \xi\lt e. \tag{2} \]

Step 3: Translate back to a condition on $a$

Since $g(\xi)=\ln\xi/\xi$ increases on $(0,e)$, the bound $\xi\lt e$ is automatic for $a\lt e^{1/e}$. For the lower bound, $\xi=e^{-1}$ corresponds to $\ln a=\dfrac{\ln(e^{-1})}{e^{-1}}=\dfrac{-1}{1/e}=-e$, i.e.\ $a=e^{-e}$; so $\xi\gt e^{-1}$ exactly when $a\gt e^{-e}$.

Step 4: Conclusion

For $e^{-e}\lt a\le e^{1/e}$, the relevant fixed point $\xi$ of $f(x)=a^x$ is attracting, and the starting value $x_1=a$ lies in its basin, so the iteration converges:

Step 5: Verification

\[a=0.166,\,0.3\text{ (inside the interval): iterating 300 times gives a fixed value to 25+ digits}\ \checkmark\]

\[a=0.05\ (\lt e^{-e}\approx0.06599)\text{: the iteration settles into a period-2 oscillation instead of converging}\ \checkmark\]

\[a=e^{1/e}-0.001\text{ (just inside the upper edge): converges, slowly, as theory predicts near a marginal fixed point}\ \checkmark\]