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Previous two posts:


Update: The first link only verifies continuity on $\mathbb R$, and so continuity cannot be used for complex heights. The limit of ${}^{z+n}a$ as $n\to\infty$ converges to the same value for all $z\in\mathbb C$ since $[\ln({}^\infty a)]^{z+n}\to0$.


Let $D_a$ be the points where the below definition converges.

I believe I have managed to prove that for $a\in(1,e^{1/e})$ and $z\in D_a$,

$${}^za=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a-({}^\infty a-{}^na)[\ln({}^\infty a)]^z)$$

is the unique tetration under the conditions that

  1. ${}^0a=1$

  2. $\displaystyle{}^{z+1}a=a^{({}^za)}$ for all $z\in D_a$

  3. $\displaystyle\lim_{n\to\infty}\frac{{}^\infty a-{}^{n+z}a}{{}^\infty a-{}^na}=[\ln({}^\infty a)]^z$ as a limit over $n\in\mathbb N$ for all $z\in D_a$.

Here is my attempted proof:

Let $b\pm\epsilon$ refer to a value within $\epsilon$ of $b$ for simplicity.

From $(3)$ we know that for all $\epsilon>0$, there exists $N$ s.t. for all $n>N$,

$$\frac{{}^\infty a-{}^{n+z}a}{{}^\infty a-{}^na}=[\ln({}^\infty a)]^z\pm\epsilon$$

By continuity of exponentiation, this can be rewritten in terms of another $\bar\epsilon>0$:

$$\frac{{}^\infty a-{}^{n+z}a}{{}^\infty a-{}^na}=[\ln({}^\infty a)]^{z\pm\bar\epsilon}$$

Solving for ${}^{n+z}a$ gives

$${}^{n+z}a={}^\infty a-({}^\infty a-{}^na)([\ln({}^\infty a)]^{z\pm\bar\epsilon})$$

Logging $n$ times and applying $(2)$ gives

$${}^za=\log_a^{\circ n}({}^\infty a-({}^\infty a-{}^na)([\ln({}^\infty a)]^{z\pm\bar\epsilon}))$$

Taking the limit as $n\to\infty$ gives us

$${}^za=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a-({}^\infty a-{}^na)([\ln({}^\infty a)]^{z\pm\bar\epsilon}))$$

However, from the first link, we know that this limit converges to a continuous function on $D_a$. So for this to be true for all $\bar\epsilon>0$, we must have

$${}^za=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a-({}^\infty a-{}^na)([\ln({}^\infty a)]^z))$$

Note also that the definitions of ${}^\infty a$ and ${}^na$ for natural $n$ in the above are defined by $(1)$ and $(2)$.

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