When we consider the Taylor Series expansion of $f(x)=b^x$ for some $b \in \mathbb{R}$, we see that $$b^x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}x^n.$$ We can substitute $x$ for $b^x$ to find that $$ b^{b^{x}} = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{xn}.$$
Now, let's say that we want to find a function that accurately describes how we should raise $b$ to $b$'th power $x$ times. We say that $b^{(b)}_x$ says that $b$ should be raised to itself $x$ times, for some $x\in\mathbb{R}$. We can generalize the previous examples, and write $$b^{(b)}_x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-1}.\qquad \text{(1)}$$ (Please note that $b^{(b)n}_{x}$ is not equal to $b^{(bn)}_{x}$.) Using this equation we can also state that $$b^{(b)}_{x-1} = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-2}.\qquad\text{(2)}$$ When we substitute (2) in (1), we find that $$ b^{(b)}_x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}\Big( 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-2} \Big).$$ From now on, we write $ \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!} = A$, because then the next equations look more clear. We could go on substituting in this manner $*$, until we arive at $b^{(b)}_{(x-(x-1))}=b$. We then see that the equality $$ b^{(b)}_x = 1 + \sum A \Big( 1 + \sum A \Big(\cdots \Big(1 + \sum_{n=1}^{\infty} \frac{(\log(b^b))^n}{n!}\Big)\Big)\Big) $$ holds when we iterate $f(x) = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}$ an $x-2$ amount of times, for real $x$.
Question 1: Is this possible?
Question 2: If so, how is this done? How do you describe the formula that precisely defines how $f(x)$ looks like after being iterated $x-2$ times?
Question 3: If such a formula is found, would it imply that a nice way of finding the fourth hyper operator (or "tetration") is found for real numbers?
Thanks,
Max
$*$EDIT: When we proceed in this manner when $x$ is not an integer, we will not find $x=1$. We should be able to iterate the function a real amount of times to find $x=1$.