Yesterday while browsing the web, I discovered an interesting equation: x^{x^{x^{\dots}}}=2 Readers, please don’t rush to scroll down; why not try solving it yourself first?
Actually, it is quite simple. The expression in the exponent of x^{x^{x^{\dots}}}=2 is also 2, so the original equation becomes x^2 = 2, which gives x = \sqrt{2}.
Simple, right? ^_^ But wait, there’s more! Now please solve this equation: x^{x^{x^{\dots}}}=4.
Following the same method, you will find: x^4 = 4, which also gives x = \sqrt{2}. This leaves one completely baffled—does \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}} equal 2 or 4?
In fact, this value should be equal to 2. Let’s look at it using a very informal method: to find the upper bound of this number (this term might not be quite appropriate), \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}} \leq \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots^2}}}=2. Thus, \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}} cannot be greater than 2.
It seems that once infinity is involved, all sorts of strange things can happen. Even such a basic method can produce “extraneous roots”...
Additionally, readers might want to try using the above method to find 2^{2^{2^{\dots}}}: let 2^{2^{2^{\dots}}}=k, then 2^k = k. This equation actually has no solution!!! — Wait, there is one solution: k \to \infty, which means 2^{2^{2^{\dots}}} diverges.
So what is the condition for convergence? From the method above, it is not hard to conclude: when x = \sqrt[n]{n}, f(x) can take a definite value (but not necessarily equal to n). This is the condition for convergence. \sqrt[n]{n} reaches its maximum value when n=2. Thus, the range for which x^{x^{x^{\dots}}} converges is x \in [0, \sqrt{2}].
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