I have read about nested radicals like $$\sqrt{a+\sqrt{a+\cdots}},$$ and they define the expression as the limit of sequence defined by $a_1=\sqrt a$ and $a_n=\sqrt{a+a_{n-1}}$. Why instead isn't it defined by $f(f(\cdots=x$ iff $f^\infty(d)=x$ for all $d$ in the domain of $f$? This seems to conform more to intuition and how we evaluate finitely nested expressions (from the inside out, not outside in). Furthermore, it would let us consider things like the $$\int_a^{\int_a^\cdots} f(x)\;dx.$$ Is there another more general definition out there for infinite nestings that I just can't find?
2026-03-25 14:04:16.1774447456
Definition of Infinite Nesting?
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The reason is more subtle that you think. Suppose that we play along with your definition of $$ f(f(...=x \quad\text{iff}\quad f^{\infty}(d)=x $$ for some $d\in \mathcal D(f)$, whatever that means. Think about this for a second, how do you begin to evaluate $f^{\infty}(d)$? Sure, we can evaluate $f^{n}(d)$ for any value of $n\in\Bbb N$ "from inside out", but to begin evaluating $f^{\infty}(d)$?
In itself, the symbol $$ \sqrt{a+\sqrt{a+\cdots}} $$ could very well has no meaning since, as you might have noticed, the expression cannot be computed from "inside out". Unlike $$ \frac 1a+\frac 1{2a}+\frac 1{4a}+\cdots\ $$ where we tend to parse it as $$ ...\left(...\left(\left(\frac 1a+\frac 1{2a}\right)+\frac 1{4a}\right)+\cdots\right. $$ and the "innermost" part can be seen, the expression $\sqrt{a+\sqrt{a+\cdots}}$ is parsed as $$ \sqrt{a+\left( \sqrt{a+\left( \cdots \right)} \right)} $$ and we can only see its outermost part.
Once we understand the problem, we can begin contemplating a way to remedy it. Over centuries, mathematician came out with an ingenious method, that is for the expressions of those kind, we simplify it to the finite case and let $n\to\infty$. That is taken to be the definition of the expression.
This way, the problem of "inside out" or "outside in" also go away since for the case where operations are perform a finite number of times, we can always find the innermost part.
Considering you question concerning $$ \int_a^{\int_a^\cdots} f(x)\;dx\ , $$ I don't think there's a sensible way to interpret that. If we let $$ G(a,b):=\int_a^b f(x)dx\ , $$ then $$ G\left(a,G(a,b)\right):=\int_a^{\int_a^b f(x)dx} f(x)dx\ . $$ From this we conclude that $b$ can be arbitrary so your expression, as I interpret it, doesn't give a unique value (if it converges at all).