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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?

Michael Hardy
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Jacob Wakem
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  • I think I misunderstood what you meant when you wrote f^o(infinity). If you can explain what you wanted, someone can help you format it, or you can see this formatting tutorial for help. – MJD Jun 20 '14 at 06:34
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    The recurrence $a_n=\sqrt{a+a_{n-1}}$ defines a finite nested expression (counting $n$ radicals) and it remains evaluated "inside out" (you need to evaluate $a_{n-1}$ before $a_n$). What's wrong with that ? Compare to another recurrence like $a_n=\frac12(a_{n-1}+\frac a{a_{n-1}})$. –  Jun 20 '14 at 06:50
  • @MJD i meant the infinitieth composition of f; thus, the infinity should be up where the o is. – Jacob Wakem Jun 20 '14 at 15:29
  • @JacobWakem, has my edit corrected the notation to reflect your intentions? – goblin GONE Jun 20 '14 at 17:07
  • @user18921 yes, thank you. Sorry, I tried but I didn't know how. I think that is clear now. – Jacob Wakem Jun 20 '14 at 18:04
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    What exactly is $f^{\infty}$ here? I'd generally think you'd take it as $f^{\infty}(x)=\lim_{n\rightarrow\infty}f^n(x)$, but you haven't said. Also, what's wrong with defining $a_1=x$ for some arbitrary $x$ and then making a sequence $a_{n+1}=\int_{a}^{a_n}f(x),dx$? This is in line with both definitions you point out. – Milo Brandt May 25 '16 at 03:51
  • Maybe this question and the links in there tackle some issues: Backward from infinity? – Han de Bruijn May 25 '16 at 14:18

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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.

Though the infinite summation may looks simpler since we can "see" its innermost part, we still cannot sum it infinitely many times!

The real problem is not that we cannot compute from outside in, it is that we can only perform the operations finitely many times.

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).

BigbearZzz
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    +1 for including computability in the explanation, in the "real life" a definition that doesn't tell you how to compute the definiendum is like the scholastic concept of how many angels fit in the tip of a needle. – Way Too Simple May 26 '16 at 20:22
  • @moonshine don't be so trivial. there is no "real life" . – Jacob Wakem May 27 '16 at 02:55
  • @JacobWakem So you're still around. Do you mind answering my question on how you intended to compute "$f^{\infty}(d)$"? It'd be a nice step toward my better understanding of your question. – BigbearZzz May 27 '16 at 03:14
  • @BigbearZzz f(f(infinity(d)))=f(infinity)d – Jacob Wakem May 27 '16 at 07:08
  • @JacobWakem I still don't understand your proposal. So $f^{\infty}(d)$ is a fixed point of $f$? What if $f$ has more than 1 fixed point? – BigbearZzz May 27 '16 at 07:43
  • If f has more than one fixed point, the limit does not exist. – Jacob Wakem May 27 '16 at 08:10
  • @JacobWakem By your definition $\sqrt{a+\sqrt{a+\cdots}}$ has no meaning then, since $f$ has 2 fixed points. – BigbearZzz May 27 '16 at 08:20
  • @BigbearZzz But one of them (the $\frac{1}{2} \left(\sqrt{4 a+1}+1\right)$ one) make the function $f$ imaginary. – user202729 May 27 '16 at 09:50
  • @BigbearZzz Actually I was mistaken. If f has more than one fixed point, the limit would still exist! This is because f(infinity)(d) != f(infinity(s)) for some s. – Jacob Wakem Jun 07 '16 at 15:42