What is the inverse of simply composited elementary functions?
$begingroup$
$A$ be an elementary function, algebraic over $mathbb{C}$,
$f_1$ and $f_2$ are bijective elementary functions with elementary inverses,
$Fcolon zmapsto A(f_1(f_2(z)))$ be a bijective elementary function.
What is the inverse $F^{-1}$ of $F$?
See Wikipedia: Elementary function, consider the definition of the elementary functions of Liouville and Ritt.
Ritt wrote in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90: "That every $F(z)$ of this type has an elementary inverse is obvious."
Is this really obvious? If all mentioned functions are bijective, $F^{-1}=f_2^{-1}circ f_1^{-1}circ A^{-1}$. But only injectivity of $f_2$ and surjectivity of $A$ follow from the bijectivity of $F$. Therefore $A$ may be non-injective.
Bijectivity of $A$ can be defined by restriction and corestriction. But are this new function and its inverse (a partial inverse of the original $A$) still elementary functions? They may have different domain and/or codomain in comparison to the original elementary functions. It depends on the answer to the question "Are all restrictions of an elementary function also elementary functions?".
And what if $A$ is not bijective: can its bijective restriction to the codomain of $f_1circ f_2$ represented by a $single$ bijective elementary function?
algebra-precalculus analysis closed-form
asked Dec 22 '18 at 19:40
IV_IV_
1,446525
$endgroup$
add a comment |
$begingroup$
$A$ be an elementary function, algebraic over $mathbb{C}$,
$f_1$ and $f_2$ are bijective elementary functions with elementary inverses,
$Fcolon zmapsto A(f_1(f_2(z)))$ be a bijective elementary function.
What is the inverse $F^{-1}$ of $F$?
See Wikipedia: Elementary function, consider the definition of the elementary functions of Liouville and Ritt.
Ritt wrote in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90: "That every $F(z)$ of this type has an elementary inverse is obvious."
Is this really obvious? If all mentioned functions are bijective, $F^{-1}=f_2^{-1}circ f_1^{-1}circ A^{-1}$. But only injectivity of $f_2$ and surjectivity of $A$ follow from the bijectivity of $F$. Therefore $A$ may be non-injective.
Bijectivity of $A$ can be defined by restriction and corestriction. But are this new function and its inverse (a partial inverse of the original $A$) still elementary functions? They may have different domain and/or codomain in comparison to the original elementary functions. It depends on the answer to the question "Are all restrictions of an elementary function also elementary functions?".
And what if $A$ is not bijective: can its bijective restriction to the codomain of $f_1circ f_2$ represented by a $single$ bijective elementary function?
algebra-precalculus analysis closed-form
asked Dec 22 '18 at 19:40
IV_IV_
1,446525
$endgroup$
1
$begingroup$
I notice this comment is made on p. 68, and the most of the paper between this assertion and the top of p. 73 involves making precise what an elementary function is, so maybe all the domain issues you're worried about are handled by the nearly 4 pages describing what an elementary function is. For what it's worth, essentially the same statement, That a function of this type has an elementary inverse is obvious, is made on p. 57 of his 1948 book Integration in Finite Terms.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:00
$begingroup$
Possibly the paper Implicitly elementary integrals by Robert H. Risch (1976) could be of use. It mentions a retraction by Ritt for a claim he made in a 1923 paper (but not the 1925 paper you cite). However, Ritt's 1925 paper is discussed in Risch's 1979 paper Algebraic properties of the elementary functions of analysis (bottom of p. 744 to top of p. 745), and this 1979 paper also seems to discuss invertibility issues quite a bit.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:09
add a comment |
$begingroup$
$A$ be an elementary function, algebraic over $mathbb{C}$,
$f_1$ and $f_2$ are bijective elementary functions with elementary inverses,
$Fcolon zmapsto A(f_1(f_2(z)))$ be a bijective elementary function.
What is the inverse $F^{-1}$ of $F$?
See Wikipedia: Elementary function, consider the definition of the elementary functions of Liouville and Ritt.
Ritt wrote in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90: "That every $F(z)$ of this type has an elementary inverse is obvious."
Is this really obvious? If all mentioned functions are bijective, $F^{-1}=f_2^{-1}circ f_1^{-1}circ A^{-1}$. But only injectivity of $f_2$ and surjectivity of $A$ follow from the bijectivity of $F$. Therefore $A$ may be non-injective.
Bijectivity of $A$ can be defined by restriction and corestriction. But are this new function and its inverse (a partial inverse of the original $A$) still elementary functions? They may have different domain and/or codomain in comparison to the original elementary functions. It depends on the answer to the question "Are all restrictions of an elementary function also elementary functions?".
And what if $A$ is not bijective: can its bijective restriction to the codomain of $f_1circ f_2$ represented by a $single$ bijective elementary function?
algebra-precalculus analysis closed-form
asked Dec 22 '18 at 19:40
IV_IV_
1,446525
$endgroup$
$A$ be an elementary function, algebraic over $mathbb{C}$,
$f_1$ and $f_2$ are bijective elementary functions with elementary inverses,
$Fcolon zmapsto A(f_1(f_2(z)))$ be a bijective elementary function.
What is the inverse $F^{-1}$ of $F$?
See Wikipedia: Elementary function, consider the definition of the elementary functions of Liouville and Ritt.
Ritt wrote in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90: "That every $F(z)$ of this type has an elementary inverse is obvious."
Is this really obvious? If all mentioned functions are bijective, $F^{-1}=f_2^{-1}circ f_1^{-1}circ A^{-1}$. But only injectivity of $f_2$ and surjectivity of $A$ follow from the bijectivity of $F$. Therefore $A$ may be non-injective.
Bijectivity of $A$ can be defined by restriction and corestriction. But are this new function and its inverse (a partial inverse of the original $A$) still elementary functions? They may have different domain and/or codomain in comparison to the original elementary functions. It depends on the answer to the question "Are all restrictions of an elementary function also elementary functions?".
And what if $A$ is not bijective: can its bijective restriction to the codomain of $f_1circ f_2$ represented by a $single$ bijective elementary function?
algebra-precalculus analysis closed-form
algebra-precalculus analysis closed-form
asked Dec 22 '18 at 19:40
IV_IV_
1,446525
asked Dec 22 '18 at 19:40
IV_IV_
1,446525
edited Dec 23 '18 at 20:39
IV_
asked Dec 22 '18 at 19:40
IV_IV_
1,446525
asked Dec 22 '18 at 19:40
IV_IV_
1,446525
asked Dec 22 '18 at 19:40
IV_IV_
1,446525
1,446525
1
$begingroup$
I notice this comment is made on p. 68, and the most of the paper between this assertion and the top of p. 73 involves making precise what an elementary function is, so maybe all the domain issues you're worried about are handled by the nearly 4 pages describing what an elementary function is. For what it's worth, essentially the same statement, That a function of this type has an elementary inverse is obvious, is made on p. 57 of his 1948 book Integration in Finite Terms.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:00
$begingroup$
Possibly the paper Implicitly elementary integrals by Robert H. Risch (1976) could be of use. It mentions a retraction by Ritt for a claim he made in a 1923 paper (but not the 1925 paper you cite). However, Ritt's 1925 paper is discussed in Risch's 1979 paper Algebraic properties of the elementary functions of analysis (bottom of p. 744 to top of p. 745), and this 1979 paper also seems to discuss invertibility issues quite a bit.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:09
add a comment |
1
$begingroup$
I notice this comment is made on p. 68, and the most of the paper between this assertion and the top of p. 73 involves making precise what an elementary function is, so maybe all the domain issues you're worried about are handled by the nearly 4 pages describing what an elementary function is. For what it's worth, essentially the same statement, That a function of this type has an elementary inverse is obvious, is made on p. 57 of his 1948 book Integration in Finite Terms.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:00
$begingroup$
Possibly the paper Implicitly elementary integrals by Robert H. Risch (1976) could be of use. It mentions a retraction by Ritt for a claim he made in a 1923 paper (but not the 1925 paper you cite). However, Ritt's 1925 paper is discussed in Risch's 1979 paper Algebraic properties of the elementary functions of analysis (bottom of p. 744 to top of p. 745), and this 1979 paper also seems to discuss invertibility issues quite a bit.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:09
1
1
$begingroup$
I notice this comment is made on p. 68, and the most of the paper between this assertion and the top of p. 73 involves making precise what an elementary function is, so maybe all the domain issues you're worried about are handled by the nearly 4 pages describing what an elementary function is. For what it's worth, essentially the same statement, That a function of this type has an elementary inverse is obvious, is made on p. 57 of his 1948 book Integration in Finite Terms.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:00
$begingroup$
I notice this comment is made on p. 68, and the most of the paper between this assertion and the top of p. 73 involves making precise what an elementary function is, so maybe all the domain issues you're worried about are handled by the nearly 4 pages describing what an elementary function is. For what it's worth, essentially the same statement, That a function of this type has an elementary inverse is obvious, is made on p. 57 of his 1948 book Integration in Finite Terms.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:00
$begingroup$
Possibly the paper Implicitly elementary integrals by Robert H. Risch (1976) could be of use. It mentions a retraction by Ritt for a claim he made in a 1923 paper (but not the 1925 paper you cite). However, Ritt's 1925 paper is discussed in Risch's 1979 paper Algebraic properties of the elementary functions of analysis (bottom of p. 744 to top of p. 745), and this 1979 paper also seems to discuss invertibility issues quite a bit.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:09
$begingroup$
Possibly the paper Implicitly elementary integrals by Robert H. Risch (1976) could be of use. It mentions a retraction by Ritt for a claim he made in a 1923 paper (but not the 1925 paper you cite). However, Ritt's 1925 paper is discussed in Risch's 1979 paper Algebraic properties of the elementary functions of analysis (bottom of p. 744 to top of p. 745), and this 1979 paper also seems to discuss invertibility issues quite a bit.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:09
add a comment |
1 Answer
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$begingroup$
If $f_1$ and $f_2$ and $Acirc f_1 circ f_2$ are all bijective $Xto X$ then necessarily $A$ will be bijective too.
If $A$ fails to be either injective or surjective, this failure would be inherited by $Acirc f_1 circ f_2$. This is obvious for surjectivity; for injectivity it is a consequence of $f_2$ and $f_2$ both being surjective.
answered Dec 23 '18 at 20:45
Henning MakholmHenning Makholm
241k17308547
$endgroup$
$begingroup$
The following question remains. Usually, $A$ is the restriction of an algebraic function $A_0colon zmapsto A(z)$ to the image of $f_1$ o $f_2$. $A_0$ doesn't have to be bijective. How is guaranteed that the inverses of each restriction $A$ of $A_0$ can be represented by a $single$ branch of an algebraic function?
$endgroup$
– IV_
Dec 23 '18 at 21:54
$begingroup$
@IV_: Since you have explicitly stipulated that $f_1$ and $f_2$ are bijective, the image of $f_1circ f_2$ is the entire $mathbb C$.
$endgroup$
– Henning Makholm
Dec 23 '18 at 21:59
$begingroup$
Ah, I see now: The inverse of $A$ is a bijective unary algebraic function and therefore an elementary function - regardless of the multivalued inverse of $A_0$.
$endgroup$
– IV_
Dec 23 '18 at 22:17
add a comment |
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$begingroup$
If $f_1$ and $f_2$ and $Acirc f_1 circ f_2$ are all bijective $Xto X$ then necessarily $A$ will be bijective too.
If $A$ fails to be either injective or surjective, this failure would be inherited by $Acirc f_1 circ f_2$. This is obvious for surjectivity; for injectivity it is a consequence of $f_2$ and $f_2$ both being surjective.
answered Dec 23 '18 at 20:45
Henning MakholmHenning Makholm
241k17308547
$endgroup$
$begingroup$
The following question remains. Usually, $A$ is the restriction of an algebraic function $A_0colon zmapsto A(z)$ to the image of $f_1$ o $f_2$. $A_0$ doesn't have to be bijective. How is guaranteed that the inverses of each restriction $A$ of $A_0$ can be represented by a $single$ branch of an algebraic function?
$endgroup$
– IV_
Dec 23 '18 at 21:54
$begingroup$
@IV_: Since you have explicitly stipulated that $f_1$ and $f_2$ are bijective, the image of $f_1circ f_2$ is the entire $mathbb C$.
$endgroup$
– Henning Makholm
Dec 23 '18 at 21:59
$begingroup$
Ah, I see now: The inverse of $A$ is a bijective unary algebraic function and therefore an elementary function - regardless of the multivalued inverse of $A_0$.
$endgroup$
– IV_
Dec 23 '18 at 22:17
add a comment |
$begingroup$
If $f_1$ and $f_2$ and $Acirc f_1 circ f_2$ are all bijective $Xto X$ then necessarily $A$ will be bijective too.
If $A$ fails to be either injective or surjective, this failure would be inherited by $Acirc f_1 circ f_2$. This is obvious for surjectivity; for injectivity it is a consequence of $f_2$ and $f_2$ both being surjective.
answered Dec 23 '18 at 20:45
Henning MakholmHenning Makholm
241k17308547
$endgroup$
$begingroup$
The following question remains. Usually, $A$ is the restriction of an algebraic function $A_0colon zmapsto A(z)$ to the image of $f_1$ o $f_2$. $A_0$ doesn't have to be bijective. How is guaranteed that the inverses of each restriction $A$ of $A_0$ can be represented by a $single$ branch of an algebraic function?
$endgroup$
– IV_
Dec 23 '18 at 21:54
$begingroup$
@IV_: Since you have explicitly stipulated that $f_1$ and $f_2$ are bijective, the image of $f_1circ f_2$ is the entire $mathbb C$.
$endgroup$
– Henning Makholm
Dec 23 '18 at 21:59
$begingroup$
Ah, I see now: The inverse of $A$ is a bijective unary algebraic function and therefore an elementary function - regardless of the multivalued inverse of $A_0$.
$endgroup$
– IV_
Dec 23 '18 at 22:17
add a comment |
$begingroup$
If $f_1$ and $f_2$ and $Acirc f_1 circ f_2$ are all bijective $Xto X$ then necessarily $A$ will be bijective too.
If $A$ fails to be either injective or surjective, this failure would be inherited by $Acirc f_1 circ f_2$. This is obvious for surjectivity; for injectivity it is a consequence of $f_2$ and $f_2$ both being surjective.
answered Dec 23 '18 at 20:45
Henning MakholmHenning Makholm
241k17308547
$endgroup$
If $f_1$ and $f_2$ and $Acirc f_1 circ f_2$ are all bijective $Xto X$ then necessarily $A$ will be bijective too.
If $A$ fails to be either injective or surjective, this failure would be inherited by $Acirc f_1 circ f_2$. This is obvious for surjectivity; for injectivity it is a consequence of $f_2$ and $f_2$ both being surjective.
answered Dec 23 '18 at 20:45
Henning MakholmHenning Makholm
241k17308547
answered Dec 23 '18 at 20:45
Henning MakholmHenning Makholm
241k17308547
answered Dec 23 '18 at 20:45
Henning MakholmHenning Makholm
241k17308547
answered Dec 23 '18 at 20:45
Henning MakholmHenning Makholm
241k17308547
241k17308547
$begingroup$
The following question remains. Usually, $A$ is the restriction of an algebraic function $A_0colon zmapsto A(z)$ to the image of $f_1$ o $f_2$. $A_0$ doesn't have to be bijective. How is guaranteed that the inverses of each restriction $A$ of $A_0$ can be represented by a $single$ branch of an algebraic function?
$endgroup$
– IV_
Dec 23 '18 at 21:54
$begingroup$
@IV_: Since you have explicitly stipulated that $f_1$ and $f_2$ are bijective, the image of $f_1circ f_2$ is the entire $mathbb C$.
$endgroup$
– Henning Makholm
Dec 23 '18 at 21:59
$begingroup$
Ah, I see now: The inverse of $A$ is a bijective unary algebraic function and therefore an elementary function - regardless of the multivalued inverse of $A_0$.
$endgroup$
– IV_
Dec 23 '18 at 22:17
add a comment |
$begingroup$
The following question remains. Usually, $A$ is the restriction of an algebraic function $A_0colon zmapsto A(z)$ to the image of $f_1$ o $f_2$. $A_0$ doesn't have to be bijective. How is guaranteed that the inverses of each restriction $A$ of $A_0$ can be represented by a $single$ branch of an algebraic function?
$endgroup$
– IV_
Dec 23 '18 at 21:54
$begingroup$
@IV_: Since you have explicitly stipulated that $f_1$ and $f_2$ are bijective, the image of $f_1circ f_2$ is the entire $mathbb C$.
$endgroup$
– Henning Makholm
Dec 23 '18 at 21:59
$begingroup$
Ah, I see now: The inverse of $A$ is a bijective unary algebraic function and therefore an elementary function - regardless of the multivalued inverse of $A_0$.
$endgroup$
– IV_
Dec 23 '18 at 22:17
$begingroup$
The following question remains. Usually, $A$ is the restriction of an algebraic function $A_0colon zmapsto A(z)$ to the image of $f_1$ o $f_2$. $A_0$ doesn't have to be bijective. How is guaranteed that the inverses of each restriction $A$ of $A_0$ can be represented by a $single$ branch of an algebraic function?
$endgroup$
– IV_
Dec 23 '18 at 21:54
$begingroup$
The following question remains. Usually, $A$ is the restriction of an algebraic function $A_0colon zmapsto A(z)$ to the image of $f_1$ o $f_2$. $A_0$ doesn't have to be bijective. How is guaranteed that the inverses of each restriction $A$ of $A_0$ can be represented by a $single$ branch of an algebraic function?
$endgroup$
– IV_
Dec 23 '18 at 21:54
$begingroup$
@IV_: Since you have explicitly stipulated that $f_1$ and $f_2$ are bijective, the image of $f_1circ f_2$ is the entire $mathbb C$.
$endgroup$
– Henning Makholm
Dec 23 '18 at 21:59
$begingroup$
@IV_: Since you have explicitly stipulated that $f_1$ and $f_2$ are bijective, the image of $f_1circ f_2$ is the entire $mathbb C$.
$endgroup$
– Henning Makholm
Dec 23 '18 at 21:59
$begingroup$
Ah, I see now: The inverse of $A$ is a bijective unary algebraic function and therefore an elementary function - regardless of the multivalued inverse of $A_0$.
$endgroup$
– IV_
Dec 23 '18 at 22:17
$begingroup$
Ah, I see now: The inverse of $A$ is a bijective unary algebraic function and therefore an elementary function - regardless of the multivalued inverse of $A_0$.
$endgroup$
– IV_
Dec 23 '18 at 22:17
add a comment |
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$begingroup$
I notice this comment is made on p. 68, and the most of the paper between this assertion and the top of p. 73 involves making precise what an elementary function is, so maybe all the domain issues you're worried about are handled by the nearly 4 pages describing what an elementary function is. For what it's worth, essentially the same statement, That a function of this type has an elementary inverse is obvious, is made on p. 57 of his 1948 book Integration in Finite Terms.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:00
$begingroup$
Possibly the paper Implicitly elementary integrals by Robert H. Risch (1976) could be of use. It mentions a retraction by Ritt for a claim he made in a 1923 paper (but not the 1925 paper you cite). However, Ritt's 1925 paper is discussed in Risch's 1979 paper Algebraic properties of the elementary functions of analysis (bottom of p. 744 to top of p. 745), and this 1979 paper also seems to discuss invertibility issues quite a bit.
$endgroup$
– Dave L. Renfro
Dec 22 '18 at 21:09