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This question is a follow-on to Is "indeterminate" a synonym for "variable" or for "transcendent"? . I have reproduced the quotations and refined some of my original observation regarding them. See the end of this post for the statement of my question. If you only follow one hyperlink, I suggest it be the one to the video.

The following excerpt is from Fundamentals of Mathematics, Volume 1 Foundations of Mathematics: The Real Number System and Algebra, Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle. (BBFSK) Bold text emphasis added.

[W]e have already seen that in general the function $x\to\sum_{i=0}^{n}a_{i}x^{i}$ does not uniquely determine the $a_{i}.$ But for calculation with such expressions as $\sum_{i=0}^{n}a_{i}x^{i}$ it would be very convenient to be able to assume that the coefficients $a_{i}$ are uniquely determined by the values of the expression. This will unquestionably be the case (for an element $x$ with certain properties) if in $R$ or in a suitable extension of $R$ we can find an element $x$ such that an equation $\sum_{i=0}^{n}a_{i}x^{i}=0$ always implies $a_{0}=a_{1}=\dots=a_{n}=0;$ for then we can recognize, as in [a previous section], that $\sum_{i=0}^{n}a_{i}x^{i}=\sum_{i=0}^{n}b_{i}x^{i}$ implies (comparison of coefficients) the equations $a_{i}=b_{i}\left(i=0,\dots,n\right).$ An element $x$ with this property will be called a transcendent over $R.$ If $R$ is the field of rational numbers, then in agreement with the definition in [the subsequent chapter and section on algebraic numbers], any transcendental number may be chosen as a transcendent over $R$ in the present sense. Since a transcendent $x$ cannot satisfy any algebraic equation $\sum_{i=0}^{n}a_{i}x^{i}=0$ with $a_{n}\ne0,$ it cannot be characterized (i.e., determined) by statements involving only $x,$ elements of $R,$ and equality, addition, and multiplication in $R.$ Thus the transcendents are also called indeterminates.${}^{10}$ But a name of this sort must not be allowed to conceal the fact that a transcendent must be a definite element (of an extension ring of $R$) and that the existence of such elements must in every case be proved. As an indeterminate over the field of rational numbers we may choose any transcendental number such as $\mathrm{e}$ or $\pi$.

${}^{10}$In §§2 and 3 the symbol $x$ will almost always denote an indeterminate; more precisely, $x$ is a variable for which only interterminates can be substituted. On the other hand, in §1 the variable $x$ provided it is not bound may be replaced by any of the elements of the ring.

§1 Entire Rational Functions; §2 Polynomials; §3 Use of Inteterminates as a Method of Proof.

This explains why, as discussed here: In Weyl's 'Classical Groups' is this a proper statement about a polynomial vanishing identically? I initially balked at Weyl's presentation.

This is the statement in Weyl's The Classical Groups Their Invariants and Representations which led me to (wrongly?) infer that Weyl meant "indeterminate" and variable to be synonyms [quotation marks "..." in the original]:

A formal expression

$$f\left(x\right)=\sum_{i=0}^{n}\alpha_{i}x^{i}$$

involving the “indeterminate” (or variable) $x$, whose coefficients $\alpha_{i}$ are numbers in a field $k$, is called a $\left(k-\right)\text{polynomial}$ of formal degree $n$.

Although I have not found time to fully investigate what is meant by indeterminate in BBFSK, I realize that an indeterminate is some kind of non-algebraic or transcendent number or element, e.g., an irrational number. When I happened upon this entertaining and insightful video Indeterminate: the hidden power of 0 divided by 0, I remembered that I had already encountered the term indeterminate in mathematics. Indeed, the authors of the chapter on Polynomials in BBFSK gave license to set $0^{0}=1$ in that context.

Now I ask: Is the meaning of indeterminate in the context of polynomial theory the same as in the context of, say, L'Hopital's rule? For example $\frac{0}{0}, \infty^{0}$, etc., discussed in the video. If so, how would one of these be used as an argument $x$ of the polynomial formal expression $f\left(x\right)=\sum_{i=0}^{n}\alpha_{i}x^{i}$?

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    Do you have any reason to believe that there might be any relationship between the two? There are many (highly) overloaded terms in mathematics. – Bill Dubuque Jan 25 '19 at 02:38
  • To begin with, in one case the objects are called "indeterminate" (forms). In the other case they are called "indeterminate" (elements). It is reasonable to suspect the same word might mean the same or a similar thing in both contexts. But, in addition, they both seem to arise as limits of ratios. Furthermore, both categories have a sense of "intractability". – Steven Thomas Hatton Jan 25 '19 at 02:48
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    Please explain why you believe that "they both seem to arise as limits of ratios". – Bill Dubuque Jan 25 '19 at 02:57
  • I believe it because that's how they seem to me. I encountered the term indeterminate in Vol-I, Part B, Chapter 4, Section 2. It provided a forward reference to Vol-I, Part B, Chapter 6, Section 8, which provided a forward reference to Vol-III. By glancing at these, I inferred that indeterminates arise by going form countable to uncountable sets, which suggests taking a limit. – Steven Thomas Hatton Jan 25 '19 at 03:50
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    The notion of "indeterminate" applies over any ring but "limit of rations" does not. So how can they be related? – Bill Dubuque Jan 25 '19 at 04:01
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    The short answer to your title is "no". – Angina Seng Jan 25 '19 at 04:01
  • @BillDubuque I will not be able to fully address your question until I return to this topic after reviewing my notes from the the first two volumes of the Feynman Lectures on Physics, Edwards's Advance Calculus of Several Variables, Misner Thorne and Wheeler's Gravitation, Dirac's General Relativity, Bergamnn's Relativity, Joos's Theoretical Physics, and several other sources. That will take several months. I have reasons for asking my question, and they are good. – Steven Thomas Hatton Jan 25 '19 at 07:12
  • @BillDubuque I should also add that in the case of polynomials with rational number coefficients it is clear that indeterminates described in the quoted text are closely related to the indeterminate forms I asked about. Since I asked about the meaning of the word indeterminate and not about indeterminate forms or indeterminate elements, it appears that in that context, the answer is yes, the the word refers transcendents over the field of rational numbers. Indeterminate elements arise from expressions of indeterminate form. – Steven Thomas Hatton Jan 25 '19 at 21:22
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    @Steven I doubt you'll get a helpful answer unless can explain much more clearly why you believe that these notions are "closely related" This is very far from clear from what you have written above. – Bill Dubuque Jan 25 '19 at 21:31

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No, these are totally unrelated uses of the term "indeterminate". They are each "not determined" in a vague intuitive sense (in rather different ways) but there is no connection beyond that, and certainly no rigorous mathematical connection.

Eric Wofsey
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