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Let $ \left(x_n\right)_{n = 1}^\infty $ be a sequence of real numbers defined by the initial value $ x_1 = e^x $ for some $ x \in \mathbb{R} $ and the relationship $ x_{n+1} = e^{x^{x_n}} $, such that $ x_2 = e^{x^{e^{x}}} $ and so forth.

Hence, find all $ x $ such that $ \lim\limits_{n \to \infty} x_n = 2 $

So we should be taking the root of both sides, would that be the approach to solve this problem?

Jon Claus
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user82528
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3 Answers3

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If you define $f_x(y) = y^{e^x}$, then you are asking for a real number $x$ such that $$ \lim_{n \to \infty} f_x^n(e^x) = 2, $$ where $f_x^n(e^x)$ means $f_x(f_x(\dots(f_x(e^x))\dots))$ and the function is applied $n$ times. Note that if such a number existed, then we would have $$ a \overset{def}{=} \lim_{n \to \infty} f_x^n(e^x) = \lim_{n \to \infty} f_x^{n+1}(e^x) = f_x \left( \lim_{n \to \infty} f_x^n(e^x) \right) = f_x(a), $$ because $f_x$ is a continuous function, hence $a = a^{e^x}$ and then either $e^x = 1$, which means $x = 0$ (and this is not a solution to your problem), or $e^x \neq 1$ and $a^{e^x - 1} = 0$, hence $a = 0 \neq 2$ (which is again not a solution to your problem). In any case there are no solutions.

The right question to ask is if you iterate the other way around : that is, if you define $g_x(y) = (e^x)^y$ and you consider the equation $$ \lim_{n \to \infty} g_x^n(y) = 2. $$ This has been answered in this solution of mine.

Hope that helps,

Community
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Patrick Da Silva
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lab bhattacharjee had mentioned that $e^{x^2}=2$, thus after taking the natural logarithms of both sides we get:

$$x^2=ln\ 2\implies x=\sqrt{ln\ 2}$$

  • This doesn't appear to quite match lab bhattacharjee's comment above. One should really have $2 = (e^x)^y = ( e^x )^2 = e^{2x}$ (you are raising all of $e^x$ to the power of his $y$, not just the $x$ in the exponent). – user642796 Jun 15 '13 at 07:00
  • I was thinking that since $(((e^x)^{e^x})^{e^x})^{\cdots}=2\implies e^{x^2}=2$ –  Jun 15 '13 at 07:03
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    Again, what you are raising to the power of $y$ matters. This is basically a more complicated version of the ol' $( 2^3)^2 = 8^2 = 64$, but $2^{3^2} = 2^9 = 512$. – user642796 Jun 15 '13 at 07:20
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Take $e^{x}=y$ therefore $y^{y^{y^{y^{...}}}}=2$ and $\log_{2}(y^{y^{y^{y^{...}}}}) = y^{y^{y^{y^{...}}}}\log_{2}(y)=1$ leading us to $2\log_{2}(y)=1$ and $ y= \sqrt{2} \to x=\frac{\ln(2)}{2}$.

Daniel Margolis
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