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A function $f$ with the argument $x$ is commonly written $f_x : A\to B, x\mapsto f(x)$, or $f_x : \mathbb{R} \to \mathbb{R}, x\mapsto x^2$, but say I don't want to specify the argument, how would I write this ? $f : \cdot\to\cdot, \mapsto ^2$ seems unnatural.

Git Gud
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Frank Vel
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    There's no 'argument specification' in the notation $f\colon A\to B, x\mapsto f(x)$, the variable is bound. – Git Gud Jan 26 '15 at 22:08
  • @GitGud I'm afraid I'm no good with terminology as English is not my first language. What I'm trying to have is a function without the $x$ specified, instead replaced by an unspecified variable/argument/parameter (whichever of these is the correct term). – Frank Vel Jan 26 '15 at 22:16
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    I don't think that's possible. Why do you need this? Perhaps if you answer this question, people can try to convince you that your undertaking is unnecessary. – Git Gud Jan 26 '15 at 22:19
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    What's wrong with $x$ as a representation of whatever the input is? Anyway, are you looking for $f(\phantom{x})=(\phantom{x})^2$, or $f(\cdot)=\cdot^2$? – 2'5 9'2 Jan 26 '15 at 22:22
  • If your user wants to use $y$ for the input, and you have declared $f(x)=x^2$, then they are welcome to let $x$ take the value $y$. – 2'5 9'2 Jan 26 '15 at 22:23
  • @GitGud I want a notation to distinguish the function $f(x) = x^2$ from the expression $f(x) = x^2$. If there was a symbol to replace the $x$ in the function, lets say $#$, then it would be easier to distinguish the function $#^2$ from the expression $x^2$. I'd like this to better explain why it's wrong to differentiate both sides of an equation, even though it may appear to be correct. – Frank Vel Jan 26 '15 at 22:24
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    "The function $f$ is given by squaring the argument". – Arthur Jan 26 '15 at 22:25
  • @alex.jordan Possibly, although I find $\cdot$ to be hard to read when writing. Maybe $( )$ would suit fine. – Frank Vel Jan 26 '15 at 22:25
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    Use this LaTeX: f(\phantom{x})=(\phantom{x})^2 for $f(\phantom{x})=(\phantom{x})^2$. – 2'5 9'2 Jan 26 '15 at 22:26
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    " I'd like this to better explain why it's wrong to differentiate both sides of an equation, even though it may appear to be correct" See this. You can't actually differentiate $x^2$or $f(x)$ because these objects are not functions, they are numbers. – Git Gud Jan 26 '15 at 22:27
  • @Arthur That would do fine too, but I'd like to write it in a similar manner you'd write $f_x : A\to B, x\mapsto f(x)$, instead of using words. – Frank Vel Jan 26 '15 at 22:27
  • @GitGud Yes, but I find the notation confusing, and often misused/abused. Textbooks may often define $(x^2)' = 2x$, so it is a source of confusion that would be avoided if it was rather defined $(#^2)' = 2#$. (Or $((\phantom{x})^2)' = 2(\phantom{x})$ as alex.jordan suggested.) – Frank Vel Jan 26 '15 at 22:31
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    @fvel It's an abuse. I too find it confusing. Once you understand it you just have to mentally translate it for yourself every time you see that abuse. If you want to explain it to someone, then you have two choices. You avoid the abuse or explain the abuse. If you want to avoid it, one possibility is to do it as I explained in my answer to the linked question. Edit: In any case, $\lambda$-calculus deals with this sort of matter. But people don't do mathematics in $\lambda$-calculus. – Git Gud Jan 26 '15 at 22:33
  • @GitGud My attempt is to explain it by using other notation to distinguish them when teaching the different concepts. When the difference is taught I can switch back to common notation. But I think functions without parameters should have a common notation anyways, as it's useful to talk about the function itself sometimes. – Frank Vel Jan 26 '15 at 22:48

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To address the explicit question, you can define the function $f: x \mapsto x^2$ without reference to variables by defining it as $f = (id)^2 = id \cdot id$, where $id$ denotes the identity function and $\cdot$ denotes multiplication of functions.

If you respond "But how do you define multiplication of functions without reference to variables?", you can always go back to the fact that a function is, fundamentally, nothing more than a set of ordered pairs (meeting certain conditions of course), and with that in mind one can define $$f \cdot g = \{(a,bc): (a,b)\in f,(a,c) \in g\}$$ If you respond to this with "But $a$ is just the variable in disguise!" then I think you are going to have to eventually acknowledge that there is no way to define an operation on an object without at some level dealing with what the object is.

mweiss
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