Can a non-invertible function be inverted by returning a set of all possible solutions?How should I understand $f^-1(E):=xin A:f(x)in E$?Show that each composite function $f_i circ f_j$ is one of the given functionsThe inverse function of f(x)=ln(x)/x.Find the derivative of $f^-1(x)$ at $x=2$ if $f(x)=x^2 + x + ln x$Inverse function of $x + x^q$ with rational $q$“Class” of functions whose inverse, where defined, is the same “class”Inverse derivative of a functionFinding inverse function of a function with multiple variableTerm for a function that handles all possible inputs?How to show $12^a cdot 18^b$ is injective
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Can a non-invertible function be inverted by returning a set of all possible solutions?
How should I understand $f^-1(E):=xin A:f(x)in E$?Show that each composite function $f_i circ f_j$ is one of the given functionsThe inverse function of f(x)=ln(x)/x.Find the derivative of $f^-1(x)$ at $x=2$ if $f(x)=x^2 + x + ln x$Inverse function of $x + x^q$ with rational $q$“Class” of functions whose inverse, where defined, is the same “class”Inverse derivative of a functionFinding inverse function of a function with multiple variableTerm for a function that handles all possible inputs?How to show $12^a cdot 18^b$ is injective
$begingroup$
Is there a concept of inverting a non-invertible function by returning a set of the possible solutions?
For example:
$g(x) = x^2$
Would it be possible to create an inverse function $f(y)$ where, for example:
$f(4) = -2,2$
(I'm pretty sure I'm going about this wrong, but I'm still learning so I don't know *how* I'm wrong)
functions inverse-function
New contributor
$endgroup$
add a comment |
$begingroup$
Is there a concept of inverting a non-invertible function by returning a set of the possible solutions?
For example:
$g(x) = x^2$
Would it be possible to create an inverse function $f(y)$ where, for example:
$f(4) = -2,2$
(I'm pretty sure I'm going about this wrong, but I'm still learning so I don't know *how* I'm wrong)
functions inverse-function
New contributor
$endgroup$
3
$begingroup$
If you define $f$ as a function from $mathbbR$ to the $textitpower set$ of $mathbbR$, then yes, that would be a function. However, $g circ f$ does not equal the identity anymore, which is a property you'd want/expect from an inverse.
$endgroup$
– Alexander Geldhof
9 hours ago
add a comment |
$begingroup$
Is there a concept of inverting a non-invertible function by returning a set of the possible solutions?
For example:
$g(x) = x^2$
Would it be possible to create an inverse function $f(y)$ where, for example:
$f(4) = -2,2$
(I'm pretty sure I'm going about this wrong, but I'm still learning so I don't know *how* I'm wrong)
functions inverse-function
New contributor
$endgroup$
Is there a concept of inverting a non-invertible function by returning a set of the possible solutions?
For example:
$g(x) = x^2$
Would it be possible to create an inverse function $f(y)$ where, for example:
$f(4) = -2,2$
(I'm pretty sure I'm going about this wrong, but I'm still learning so I don't know *how* I'm wrong)
functions inverse-function
functions inverse-function
New contributor
New contributor
edited 7 hours ago
Asaf Karagila♦
312k33446780
312k33446780
New contributor
asked 9 hours ago
Code SlingerCode Slinger
1164
1164
New contributor
New contributor
3
$begingroup$
If you define $f$ as a function from $mathbbR$ to the $textitpower set$ of $mathbbR$, then yes, that would be a function. However, $g circ f$ does not equal the identity anymore, which is a property you'd want/expect from an inverse.
$endgroup$
– Alexander Geldhof
9 hours ago
add a comment |
3
$begingroup$
If you define $f$ as a function from $mathbbR$ to the $textitpower set$ of $mathbbR$, then yes, that would be a function. However, $g circ f$ does not equal the identity anymore, which is a property you'd want/expect from an inverse.
$endgroup$
– Alexander Geldhof
9 hours ago
3
3
$begingroup$
If you define $f$ as a function from $mathbbR$ to the $textitpower set$ of $mathbbR$, then yes, that would be a function. However, $g circ f$ does not equal the identity anymore, which is a property you'd want/expect from an inverse.
$endgroup$
– Alexander Geldhof
9 hours ago
$begingroup$
If you define $f$ as a function from $mathbbR$ to the $textitpower set$ of $mathbbR$, then yes, that would be a function. However, $g circ f$ does not equal the identity anymore, which is a property you'd want/expect from an inverse.
$endgroup$
– Alexander Geldhof
9 hours ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.
It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.
Finally, one could also view them simply as relations with a full domain.
$endgroup$
$begingroup$
Multivalued functions were the missing piece - thanks!
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
It is common to use the notation $f^-1(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set $x : f(x) in A $. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^-1(x)$ instead of $f^-1(x)$. So if in your case the context is clear, it is fine to write $g^-1(4) = 2, -2$.
Also there are functions that are multivalued by default like the complex logarithm for example.
$endgroup$
$begingroup$
So $g^-1(4)$ could be the same as $g^-1(4)$ in a certain context? Would that make $g^-1$ a multivalued function?
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.
What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.
$endgroup$
add a comment |
$begingroup$
To give a function is the same to give the sets where the function is constant: $cmapsto f^-1(c)$. This would be a map $F:mathbbRto power(mathbbR)$. The function $F$ is almost injective (since $f^-1(c)cap f^-1(c')=emptyset$ for $cneq c'$). It might not be injective if $f^-1(c)=f^-1(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:
$$F: Image(f)to power(mathbbR)$$
is injective.
This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.
$endgroup$
add a comment |
$begingroup$
We usually want functions to only take one value in value space.
If there are several different possible ones, we call them branches of a function.
For example $f^-1(t) = -sqrtt$ is one such branch that is inverse to $f(t) = t^2$
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
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$begingroup$
This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.
It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.
Finally, one could also view them simply as relations with a full domain.
$endgroup$
$begingroup$
Multivalued functions were the missing piece - thanks!
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.
It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.
Finally, one could also view them simply as relations with a full domain.
$endgroup$
$begingroup$
Multivalued functions were the missing piece - thanks!
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.
It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.
Finally, one could also view them simply as relations with a full domain.
$endgroup$
This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.
It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.
Finally, one could also view them simply as relations with a full domain.
edited 9 hours ago
answered 9 hours ago
Theo BenditTheo Bendit
23.6k12359
23.6k12359
$begingroup$
Multivalued functions were the missing piece - thanks!
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
Multivalued functions were the missing piece - thanks!
$endgroup$
– Code Slinger
5 hours ago
$begingroup$
Multivalued functions were the missing piece - thanks!
$endgroup$
– Code Slinger
5 hours ago
$begingroup$
Multivalued functions were the missing piece - thanks!
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
It is common to use the notation $f^-1(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set $x : f(x) in A $. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^-1(x)$ instead of $f^-1(x)$. So if in your case the context is clear, it is fine to write $g^-1(4) = 2, -2$.
Also there are functions that are multivalued by default like the complex logarithm for example.
$endgroup$
$begingroup$
So $g^-1(4)$ could be the same as $g^-1(4)$ in a certain context? Would that make $g^-1$ a multivalued function?
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
It is common to use the notation $f^-1(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set $x : f(x) in A $. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^-1(x)$ instead of $f^-1(x)$. So if in your case the context is clear, it is fine to write $g^-1(4) = 2, -2$.
Also there are functions that are multivalued by default like the complex logarithm for example.
$endgroup$
$begingroup$
So $g^-1(4)$ could be the same as $g^-1(4)$ in a certain context? Would that make $g^-1$ a multivalued function?
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
It is common to use the notation $f^-1(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set $x : f(x) in A $. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^-1(x)$ instead of $f^-1(x)$. So if in your case the context is clear, it is fine to write $g^-1(4) = 2, -2$.
Also there are functions that are multivalued by default like the complex logarithm for example.
$endgroup$
It is common to use the notation $f^-1(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set $x : f(x) in A $. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^-1(x)$ instead of $f^-1(x)$. So if in your case the context is clear, it is fine to write $g^-1(4) = 2, -2$.
Also there are functions that are multivalued by default like the complex logarithm for example.
answered 9 hours ago
LionCoderLionCoder
677315
677315
$begingroup$
So $g^-1(4)$ could be the same as $g^-1(4)$ in a certain context? Would that make $g^-1$ a multivalued function?
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
So $g^-1(4)$ could be the same as $g^-1(4)$ in a certain context? Would that make $g^-1$ a multivalued function?
$endgroup$
– Code Slinger
5 hours ago
$begingroup$
So $g^-1(4)$ could be the same as $g^-1(4)$ in a certain context? Would that make $g^-1$ a multivalued function?
$endgroup$
– Code Slinger
5 hours ago
$begingroup$
So $g^-1(4)$ could be the same as $g^-1(4)$ in a certain context? Would that make $g^-1$ a multivalued function?
$endgroup$
– Code Slinger
5 hours ago
add a comment |
$begingroup$
It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.
What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.
$endgroup$
add a comment |
$begingroup$
It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.
What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.
$endgroup$
add a comment |
$begingroup$
It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.
What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.
$endgroup$
It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.
What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.
answered 9 hours ago
Sam SkywalkerSam Skywalker
54913
54913
add a comment |
add a comment |
$begingroup$
To give a function is the same to give the sets where the function is constant: $cmapsto f^-1(c)$. This would be a map $F:mathbbRto power(mathbbR)$. The function $F$ is almost injective (since $f^-1(c)cap f^-1(c')=emptyset$ for $cneq c'$). It might not be injective if $f^-1(c)=f^-1(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:
$$F: Image(f)to power(mathbbR)$$
is injective.
This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.
$endgroup$
add a comment |
$begingroup$
To give a function is the same to give the sets where the function is constant: $cmapsto f^-1(c)$. This would be a map $F:mathbbRto power(mathbbR)$. The function $F$ is almost injective (since $f^-1(c)cap f^-1(c')=emptyset$ for $cneq c'$). It might not be injective if $f^-1(c)=f^-1(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:
$$F: Image(f)to power(mathbbR)$$
is injective.
This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.
$endgroup$
add a comment |
$begingroup$
To give a function is the same to give the sets where the function is constant: $cmapsto f^-1(c)$. This would be a map $F:mathbbRto power(mathbbR)$. The function $F$ is almost injective (since $f^-1(c)cap f^-1(c')=emptyset$ for $cneq c'$). It might not be injective if $f^-1(c)=f^-1(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:
$$F: Image(f)to power(mathbbR)$$
is injective.
This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.
$endgroup$
To give a function is the same to give the sets where the function is constant: $cmapsto f^-1(c)$. This would be a map $F:mathbbRto power(mathbbR)$. The function $F$ is almost injective (since $f^-1(c)cap f^-1(c')=emptyset$ for $cneq c'$). It might not be injective if $f^-1(c)=f^-1(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:
$$F: Image(f)to power(mathbbR)$$
is injective.
This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.
answered 9 hours ago
Hudson LimaHudson Lima
664
664
add a comment |
add a comment |
$begingroup$
We usually want functions to only take one value in value space.
If there are several different possible ones, we call them branches of a function.
For example $f^-1(t) = -sqrtt$ is one such branch that is inverse to $f(t) = t^2$
$endgroup$
add a comment |
$begingroup$
We usually want functions to only take one value in value space.
If there are several different possible ones, we call them branches of a function.
For example $f^-1(t) = -sqrtt$ is one such branch that is inverse to $f(t) = t^2$
$endgroup$
add a comment |
$begingroup$
We usually want functions to only take one value in value space.
If there are several different possible ones, we call them branches of a function.
For example $f^-1(t) = -sqrtt$ is one such branch that is inverse to $f(t) = t^2$
$endgroup$
We usually want functions to only take one value in value space.
If there are several different possible ones, we call them branches of a function.
For example $f^-1(t) = -sqrtt$ is one such branch that is inverse to $f(t) = t^2$
answered 9 hours ago
mathreadlermathreadler
16k72263
16k72263
add a comment |
add a comment |
Code Slinger is a new contributor. Be nice, and check out our Code of Conduct.
Code Slinger is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
If you define $f$ as a function from $mathbbR$ to the $textitpower set$ of $mathbbR$, then yes, that would be a function. However, $g circ f$ does not equal the identity anymore, which is a property you'd want/expect from an inverse.
$endgroup$
– Alexander Geldhof
9 hours ago