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Name for a function whose effect is canceled by another function?
What is the name for a function of a matrix that changes the matrix size?Name for a set of pairs of elements that equalise two functions?Is there a name for the function of a semicircle?What do we call a “function” which is not defined on part of its domain?Is there a name for this function with properties…Is there a name for the function $e^-lvertlog xrvert$?Another name for the trivial model structureIs there a term for this basic effect?Is there a standard name for functions whose fibers are finite on every element in their image?name for a function that composes additively
$begingroup$
I've been saying function $f$ "eclipses" $g$ if $f(g(x)) = f(x)$ for all $x$. For example, if $f(x) = lvert x rvert$ and $g(x) = -x$, then $f$ eclipses $g$.
Is there an established word for this property?
functions terminology
asked 9 hours ago
DoradusDoradus
1836
$endgroup$
add a comment |
$begingroup$
I've been saying function $f$ "eclipses" $g$ if $f(g(x)) = f(x)$ for all $x$. For example, if $f(x) = lvert x rvert$ and $g(x) = -x$, then $f$ eclipses $g$.
Is there an established word for this property?
functions terminology
asked 9 hours ago
DoradusDoradus
1836
$endgroup$
2
$begingroup$
Don't know that there is one. "Eclipse" is as good a candidate for a name as any....
$endgroup$
– fleablood
9 hours ago
$begingroup$
This is loosely related to the concept of en.wikipedia.org/wiki/Coequalizer, so maybe one could call $f$ "co-equalizing $g$ and the identity"
$endgroup$
– Hagen von Eitzen
9 hours ago
add a comment |
$begingroup$
I've been saying function $f$ "eclipses" $g$ if $f(g(x)) = f(x)$ for all $x$. For example, if $f(x) = lvert x rvert$ and $g(x) = -x$, then $f$ eclipses $g$.
Is there an established word for this property?
functions terminology
asked 9 hours ago
DoradusDoradus
1836
$endgroup$
I've been saying function $f$ "eclipses" $g$ if $f(g(x)) = f(x)$ for all $x$. For example, if $f(x) = lvert x rvert$ and $g(x) = -x$, then $f$ eclipses $g$.
Is there an established word for this property?
functions terminology
functions terminology
asked 9 hours ago
DoradusDoradus
1836
asked 9 hours ago
DoradusDoradus
1836
asked 9 hours ago
DoradusDoradus
1836
asked 9 hours ago
DoradusDoradus
1836
asked 9 hours ago
DoradusDoradus
1836
1836
2
$begingroup$
Don't know that there is one. "Eclipse" is as good a candidate for a name as any....
$endgroup$
– fleablood
9 hours ago
$begingroup$
This is loosely related to the concept of en.wikipedia.org/wiki/Coequalizer, so maybe one could call $f$ "co-equalizing $g$ and the identity"
$endgroup$
– Hagen von Eitzen
9 hours ago
add a comment |
2
$begingroup$
Don't know that there is one. "Eclipse" is as good a candidate for a name as any....
$endgroup$
– fleablood
9 hours ago
$begingroup$
This is loosely related to the concept of en.wikipedia.org/wiki/Coequalizer, so maybe one could call $f$ "co-equalizing $g$ and the identity"
$endgroup$
– Hagen von Eitzen
9 hours ago
2
2
$begingroup$
Don't know that there is one. "Eclipse" is as good a candidate for a name as any....
$endgroup$
– fleablood
9 hours ago
$begingroup$
Don't know that there is one. "Eclipse" is as good a candidate for a name as any....
$endgroup$
– fleablood
9 hours ago
$begingroup$
This is loosely related to the concept of en.wikipedia.org/wiki/Coequalizer, so maybe one could call $f$ "co-equalizing $g$ and the identity"
$endgroup$
– Hagen von Eitzen
9 hours ago
$begingroup$
This is loosely related to the concept of en.wikipedia.org/wiki/Coequalizer, so maybe one could call $f$ "co-equalizing $g$ and the identity"
$endgroup$
– Hagen von Eitzen
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:
Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.
Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.
answered 9 hours ago
punctured duskpunctured dusk
14.8k32887
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:
Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.
Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.
answered 9 hours ago
punctured duskpunctured dusk
14.8k32887
$endgroup$
add a comment |
$begingroup$
I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:
Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.
Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.
answered 9 hours ago
punctured duskpunctured dusk
14.8k32887
$endgroup$
add a comment |
$begingroup$
I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:
Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.
Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.
answered 9 hours ago
punctured duskpunctured dusk
14.8k32887
$endgroup$
I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:
Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.
Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.
answered 9 hours ago
punctured duskpunctured dusk
14.8k32887
edited 9 hours ago
answered 9 hours ago
punctured duskpunctured dusk
14.8k32887
answered 9 hours ago
punctured duskpunctured dusk
14.8k32887
answered 9 hours ago
punctured duskpunctured dusk
14.8k32887
14.8k32887
add a comment |
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$begingroup$
Don't know that there is one. "Eclipse" is as good a candidate for a name as any....
$endgroup$
– fleablood
9 hours ago
$begingroup$
This is loosely related to the concept of en.wikipedia.org/wiki/Coequalizer, so maybe one could call $f$ "co-equalizing $g$ and the identity"
$endgroup$
– Hagen von Eitzen
9 hours ago