Mixing
Is the meaning of *indeterminate* in the context of polynomial theory the same as in the context of, say,...
$begingroup$
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 $xtosum_{i=0}^{n}a_{i}x^{i}$ 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 $a_{i}=b_{i}left(i=0,dots,nright).$
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
$$fleft(xright)=sum_{i=0}^{n}alpha_{i}x^{i}$$
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 $fleft(xright)=sum_{i=0}^{n}alpha_{i}x^{i}$?
elementary-number-theory polynomials definition indeterminate-forms
asked Jan 25 at 2:00
Steven HattonSteven Hatton
979422
$endgroup$
|
show 4 more comments
$begingroup$
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 $xtosum_{i=0}^{n}a_{i}x^{i}$ 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 $a_{i}=b_{i}left(i=0,dots,nright).$
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
$$fleft(xright)=sum_{i=0}^{n}alpha_{i}x^{i}$$
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 $fleft(xright)=sum_{i=0}^{n}alpha_{i}x^{i}$?
elementary-number-theory polynomials definition indeterminate-forms
asked Jan 25 at 2:00
Steven HattonSteven Hatton
979422
$endgroup$
3
$begingroup$
Do you have any reason to believe that there might be any relationship between the two? There are many (highly) overloaded terms in mathematics.
$endgroup$
– Bill Dubuque
Jan 25 at 2:38
3
$begingroup$
Please explain why you believe that "they both seem to arise as limits of ratios".
$endgroup$
– Bill Dubuque
Jan 25 at 2:57
2
$begingroup$
The notion of "indeterminate" applies over any ring but "limit of rations" does not. So how can they be related?
$endgroup$
– Bill Dubuque
Jan 25 at 4:01
2
$begingroup$
The short answer to your title is "no".
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:01
2
$begingroup$
@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.
$endgroup$
– Bill Dubuque
Jan 25 at 21:31
|
show 4 more comments
$begingroup$
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 $xtosum_{i=0}^{n}a_{i}x^{i}$ 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 $a_{i}=b_{i}left(i=0,dots,nright).$
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
$$fleft(xright)=sum_{i=0}^{n}alpha_{i}x^{i}$$
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 $fleft(xright)=sum_{i=0}^{n}alpha_{i}x^{i}$?
elementary-number-theory polynomials definition indeterminate-forms
asked Jan 25 at 2:00
Steven HattonSteven Hatton
979422
$endgroup$
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 $xtosum_{i=0}^{n}a_{i}x^{i}$ 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 $a_{i}=b_{i}left(i=0,dots,nright).$
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
$$fleft(xright)=sum_{i=0}^{n}alpha_{i}x^{i}$$
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 $fleft(xright)=sum_{i=0}^{n}alpha_{i}x^{i}$?
elementary-number-theory polynomials definition indeterminate-forms
elementary-number-theory polynomials definition indeterminate-forms
asked Jan 25 at 2:00
Steven HattonSteven Hatton
979422
asked Jan 25 at 2:00
Steven HattonSteven Hatton
979422
asked Jan 25 at 2:00
Steven HattonSteven Hatton
979422
asked Jan 25 at 2:00
Steven HattonSteven Hatton
979422
asked Jan 25 at 2:00
Steven HattonSteven Hatton
979422
979422
3
$begingroup$
Do you have any reason to believe that there might be any relationship between the two? There are many (highly) overloaded terms in mathematics.
$endgroup$
– Bill Dubuque
Jan 25 at 2:38
3
$begingroup$
Please explain why you believe that "they both seem to arise as limits of ratios".
$endgroup$
– Bill Dubuque
Jan 25 at 2:57
2
$begingroup$
The notion of "indeterminate" applies over any ring but "limit of rations" does not. So how can they be related?
$endgroup$
– Bill Dubuque
Jan 25 at 4:01
2
$begingroup$
The short answer to your title is "no".
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:01
2
$begingroup$
@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.
$endgroup$
– Bill Dubuque
Jan 25 at 21:31
|
show 4 more comments
3
$begingroup$
Do you have any reason to believe that there might be any relationship between the two? There are many (highly) overloaded terms in mathematics.
$endgroup$
– Bill Dubuque
Jan 25 at 2:38
3
$begingroup$
Please explain why you believe that "they both seem to arise as limits of ratios".
$endgroup$
– Bill Dubuque
Jan 25 at 2:57
2
$begingroup$
The notion of "indeterminate" applies over any ring but "limit of rations" does not. So how can they be related?
$endgroup$
– Bill Dubuque
Jan 25 at 4:01
2
$begingroup$
The short answer to your title is "no".
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:01
2
$begingroup$
@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.
$endgroup$
– Bill Dubuque
Jan 25 at 21:31
3
3
$begingroup$
Do you have any reason to believe that there might be any relationship between the two? There are many (highly) overloaded terms in mathematics.
$endgroup$
– Bill Dubuque
Jan 25 at 2:38
$begingroup$
Do you have any reason to believe that there might be any relationship between the two? There are many (highly) overloaded terms in mathematics.
$endgroup$
– Bill Dubuque
Jan 25 at 2:38
3
3
$begingroup$
Please explain why you believe that "they both seem to arise as limits of ratios".
$endgroup$
– Bill Dubuque
Jan 25 at 2:57
$begingroup$
Please explain why you believe that "they both seem to arise as limits of ratios".
$endgroup$
– Bill Dubuque
Jan 25 at 2:57
2
2
$begingroup$
The notion of "indeterminate" applies over any ring but "limit of rations" does not. So how can they be related?
$endgroup$
– Bill Dubuque
Jan 25 at 4:01
$begingroup$
The notion of "indeterminate" applies over any ring but "limit of rations" does not. So how can they be related?
$endgroup$
– Bill Dubuque
Jan 25 at 4:01
2
2
$begingroup$
The short answer to your title is "no".
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:01
$begingroup$
The short answer to your title is "no".
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:01
2
2
$begingroup$
@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.
$endgroup$
– Bill Dubuque
Jan 25 at 21:31
$begingroup$
@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.
$endgroup$
– Bill Dubuque
Jan 25 at 21:31
|
show 4 more comments
1 Answer
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oldest
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$begingroup$
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.
answered Jan 27 at 5:28
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
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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.
answered Jan 27 at 5:28
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 27 at 5:28
Eric WofseyEric Wofsey
190k14216347
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 27 at 5:28
Eric WofseyEric Wofsey
190k14216347
$endgroup$
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.
answered Jan 27 at 5:28
Eric WofseyEric Wofsey
190k14216347
edited Jan 27 at 5:33
answered Jan 27 at 5:28
Eric WofseyEric Wofsey
190k14216347
answered Jan 27 at 5:28
Eric WofseyEric Wofsey
190k14216347
answered Jan 27 at 5:28
Eric WofseyEric Wofsey
190k14216347
190k14216347
add a comment |
add a comment |
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3
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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.
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– Bill Dubuque
Jan 25 at 2:38
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Please explain why you believe that "they both seem to arise as limits of ratios".
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– Bill Dubuque
Jan 25 at 2:57
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The notion of "indeterminate" applies over any ring but "limit of rations" does not. So how can they be related?
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– Bill Dubuque
Jan 25 at 4:01
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The short answer to your title is "no".
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– Lord Shark the Unknown
Jan 25 at 4:01
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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.
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– Bill Dubuque
Jan 25 at 21:31