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I know that $x$ is called the fixed point of a function $f$ if it satisfies the constraint $f(x) = x$.

However, for a function $f$ if there exists some value $x$ such that $f(x) = f$ then what is the term for the value $x$ with respect to $f$.


Consider the following function in JavaScript:

var bind = Function.prototype.bind;
var bindable = bind.bind(bind);

Now bindable is the function $f$ that I'm talking about. It satisfies the constraint $f(x) = f$ for the value bind:

bindable(bind) = bindable;

I know that I shouldn't express mathematics in terms of programming but I didn't know any other way to put it.


Consider that bind has the following type definition:

(a b -> c) a -> (b -> c)

It takes a function of type a b -> c and zero or more arguments which grouped together have the type a and returns another function of type b -> c where b has the type of the rest of the arguments grouped together.

Hence bind(bind) has the following type definition:

(a b -> c) -> (a -> (b -> c))

Let bindable = bind(bind), thenbindable(bind) also has the same type definition:

(a b -> c) -> (a -> (b -> c))

i.e. bindable(bind) = bindable.

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Why do you write mathematical expressions like this? Enclose them in dollar signs for them to show properly. Also, what do you mean by $f(x)=f$? It is not possible in standard set theory (it's a violation of regularity), and I can't think of a case when this would be useful. –  tomasz Jun 4 '13 at 12:14
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If $\,f\,$ is a function, then $\,f:A\to B\,$ , with $\,A\,,\,B\,$ some sets. If $\,f\notin B\,$ , then the equation $\;f(x)=f\;$ makes absolutely no sense at all. –  DonAntonio Jun 4 '13 at 12:21
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Given that he is talking computer science, why is everybody telling him that $f(x)=f$ is not possible? This is quite common in lambda calculus, for example. –  Thomas Andrews Jun 4 '13 at 12:29
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Unfortunately, your example with bind is not particularly useful, since it requires knowledge of JavaScript. –  Thomas Andrews Jun 4 '13 at 12:32
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@Aadit: You may have better luck dropping the idea that you're talking about the evaluation of a generalized notion of function, and instead asking something even more general. e.g. "if you have a binary operation with the property that $fx=f$ for some values $f$ and $x$, how might you describe the situation?" If you really need terminology specialized to your particular context of interest, then you're probably better off asking someplace more specialized. (Or, at least, explicitly stating the field of interest you are asking about) –  Hurkyl Jun 4 '13 at 12:58

2 Answers 2

up vote 3 down vote accepted

I doubt that there is a name for this. If there is, it would be something from lambda calculus.

Unlike fixed points, where $f(x)=x$, I'm not sure what the value would be in such a concept.

There is a sense in which $f(x)=f$ is also a fixed point. Namely, if $D=\lambda y.\lambda g. g(y)$, then $f$ is a fixed point of $Dx$. But I'm not sure if that really gives you anything.

You might consider asking this question at the StackExchange site dedicate to computer science questions.

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This situation never occurs in ordinary mathematics, because in order to specify a function you need to specify the codomain $C$ (set of possible values the function could take) first, and that set cannot contain any functions that have $C$ as codomain. Or in terms of set theory (though that point of view does not seem the most important to me) the set of functions $X\to Y$ is a subset of $X\times Y$; if some $x \in X$ and $f\in Y$ satisfied $f(x)=f$ then one would have $(x,f)\in f$ and since also $f \in^+ (x,f)$ where $\in^+$ is the "transitive closure" of the relation $\in$ (just how depends on the definition of ordered pairs) one would have $f\in^+ f$ violating the axiom of regularity.

This being said one can imagine "function application" is just some externally defined operator that takes $f$ and $x$ to produce a value written $f(x)$, and in this case $f(x)=f$ is possible. To give it a name I would say $f$ is a fixed point for $x$ under reverse function application (being-applied-to).

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What do you call "ordinary mathematics?" Lambda calculus is something I learned as an undergraduate. It doesn't occur in the standard definition of function in set theory, though. –  Thomas Andrews Jun 4 '13 at 13:02
    
@ThomasAndrews: Lambda calculus is certainly mathematics, but lambda expression are not functions in the ordinary mathematical sense of the term. Indeed this is exactly an example of an external operation defined on the set of pairs of (equivalence classes of?) lambda terms that is somewhat arbitrarily called "function application" (one could have called it multiplication or something else). –  Marc van Leeuwen Jun 4 '13 at 13:07
    
And yet, given the poster is talking about computer functions, which are "not functions" in this set theory sense, then that would seem to be his intent. –  Thomas Andrews Jun 4 '13 at 13:11
    
@ThomasAndrews: Which is why I wrote the second paragraph. But this being a math forum, I felt I needed to address it for mathematicians as well. –  Marc van Leeuwen Jun 4 '13 at 13:38

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