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My motivation to ask this question stems from this question.

The ellipsis notation is used in two ways:

  1. For finite number of terms: When you want to save space and time, you may omit the terms between the first few terms and the last term by using an ellipsis, provided that the few terms have clearly established the pattern/sequence that each of these terms must have.

  2. For infinite number of terms: When you have established a pattern/sequence with the first few terms, you use ellipsis after those few terms to mean "and so on" for the rest of the infinite terms.


I want to discuss particularly the second case.

For example; when you have an infinite sum or product, it is understood that:

  • $a_1+a_2 + a_3+ \ldots := \lim\limits_{n \to ∞} \left(a_1+a_2 + a_3+ \ldots+ a_n \right)$
  • $a_1 \cdot a_2 \cdot \ldots := \lim\limits_{n \to ∞} \left(a_1 \cdot a_2 \cdot \ldots a_n\right)$

Now consider another example. If I write down something like this: $$\sqrt{a_1 + \sqrt{a_2 + \sqrt{ \ldots }}}$$ Any reasonable person would interpret and/or define this as: $$ \lim\limits_{n\to∞} \sqrt{a_1 + \sqrt{a_2 + \sqrt{ \ldots +\sqrt{a_n}}}}$$

Or here's one more: $$ f_1\left(f_2\left(f_3\left( \ldots (x) \right) \right) \right):= \lim\limits_{n \to ∞} f_1\left(f_2\left(f_3\left( \ldots f_n(x) \right) \right) \right) $$


However these are only examples of how one would define or give meaning to an infinite expression which would otherwise be a meaningless garbage.

My question: Is there a way to define "ellipsis" in infinite expressions using limits (or any other means) in general (I only list down examples), to give it a rigourous meaning?

If not, then is it not possible to do something like this for all kinds of infinite expressions? Some counterexamples would help in that case.


Edit 2: After some thought, what I was sort of trying to ask is if something like this could be done.

Let $(x_n)$ be an infinite sequence of real numbers. We need an infinite sequence to make any sense out of the ellipsis notation, otherwise I assume it would be ambiguous.

Let $E$ be an infinite expression in $x_i$'s for each $i$. Note that, expression is once again, an informal word. But in this context, I would like it to mean "some combination of symbols that have some meaning in $\mathbb{R}$"

So, $E=E(x_1,x_2,x_3, \ldots, x_n, \ldots)$, this is an infinite expression. And to make any sense out of $E$, I'd define $E(x_1,x_2, \ldots):= \lim\limits_{n \to ∞} E(x_1,x_2, \ldots, x_n)$. How's this for a change? Now one could also replace the set $\mathbb{R}$ with some arbitrary set $X$ if they like in which case my sequence would then be in $X$. Does this make any sense or have I just gone complete nuts?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Jul 3, 2021 at 10:48

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I believe you can eliminate the ellipsis "..." if you use recursive definitions with conditional expressions such as the following.


$$\sqrt{a_1+\sqrt{a_2+\sqrt{\ldots}}}=\lim\limits_{n\to\infty}\sqrt{a_1+\sqrt{a_2+\sqrt{\ldots+\sqrt{a_n}}}}=\lim\limits_{n\to\infty}b_{1,n}\tag{1}$$

$$b_{i,n}=\sqrt{a_i+\left\{\begin{array}{cc} b_{i+1,n} & i<n \\ 0 & \text{True} \\ \end{array}\right.}=\sqrt{\left\{\begin{array}{cc} a_i+b_{i+1,n} & i<n \\ a_n & \text{True} \\ \end{array}\right.}\tag{2}$$


$$f_1\left(f_2\left(f_3\left(\ldots(x)\right)\right)\right)=\lim\limits_{n\to\infty}f_1\left(f_2\left(f_3\left(\ldots f_n(x)\right)\right)\right)=\lim\limits_{n\to\infty}g_{1,n}\tag{3}$$

$$g_{i,n}=f_i\left(\left\{\begin{array}{cc} g_{i+1,n} & i<n \\ x & \text{True} \\ \end{array}\right.\right)\tag{4}$$


Formulas (2) and (4) above can be written in a more standard functional form as follows which is the way they'd actually be implemented in Mathematica.


$$b(i,n)=\sqrt{a(i)+\left\{\begin{array}{cc} b(i+1,n) & i<n \\ 0 & \text{True} \\ \end{array}\right.}=\sqrt{\left\{\begin{array}{cc} a(i)+b(i+1,n) & i<n \\ a(n) & \text{True} \\ \end{array}\right.}\tag{5}$$

$$g(i,n)=f_i\left(\left\{\begin{array}{cc} g(i+1,n) & i<n \\ x & \text{True} \\ \end{array}\right.\right)\tag{6}$$

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  • $\begingroup$ This is a very helpful answer, I'll upvote and mark it later probably. But let me first try to figure out and get comfortable with the notations you've used. One question: Can you replace ellipsis in every kind of infinite expression using recursive definition? And two, would be kind enough to tell me what could go wrong or is wrong with my edit 2 (if at all, it is wrong). $\endgroup$
    – William
    Jul 3, 2021 at 7:45
  • $\begingroup$ @William You don't always need to use a recursive definition. For example: $a_1+a_2+a_3+\ldots=\lim\limits_{n\to\infty}\left(a_1+a_2+a_3+\ldots+a_n\right)=\lim\limits_{n\to\infty}\sum\limits_{k=1}^n a_n$. But my point is for more complicated recursive functions you can replace the ellipsis "..." with a a recursive piecewise conditional expression which is mathematically precise (see reference.wolfram.com/language/ref/Piecewise.html). It seems to me your edit 2 involves a function of multiple variables versus the recursive functions which you originally used as examples. $\endgroup$ Jul 3, 2021 at 15:53
  • $\begingroup$ @William In my comment above, $\sum\limits_{k=1}^n a_n$ should have been $\sum\limits_{k=1}^n a_k$. $\endgroup$ Jul 3, 2021 at 16:32
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    $\begingroup$ @William With respect to your edit 2, it seems to me you need to define some examples of the expression E, but I suspect you'll just end up with expressions similar to your previous non-recursive and recursive ellipsis "..." examples. $\endgroup$ Jul 3, 2021 at 21:51
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    $\begingroup$ @William As I said Mathematica evaluates the expression associated with the first true condition, and the value True satisfies this condition for the default expression. I copied the Piecewise function with the Latex option from Mathematica and it automatically inserts the True condition into the generated Latex. Mathematica supports the Boolean values True and False (see reference.wolfram.com/language/ref/True.html). $\endgroup$ Jul 4, 2021 at 21:51

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