1
$\begingroup$

I am wondering if there is a common notation for a function that does not depend on a particular parameter. I am wondering about notation both for applying the function ($f(x, y)$) as well as defining it ($f(x, y) = 2y$). In programming, the underscore is often used for this purpose (for example, function(_, y) = 2*y).

For example, suppose I am considering $f: \mathbb{R}^2 \to \mathbb{R}$ and have proven that $f(x_1, y) = f(x_2, y)$ for all $x_1, x_2$. Then later, I would like to use the expression $f(x, y)$ while making it clear that the result does not depend on $x$. I was thinking, maybe something like $f(\cdot\ , y)$.

I am also curious if there is an established notation for defining a function that is constant with respect to a parameter. Maybe, defining $f(\cdot\ , y) = 2y$ rather than $f(x, y) = 2y$ to make the lack of dependence on $x$ more explicit.

Would the use of $\cdot$ in these cases be clear, or is there another syntax that would be better for this use?

Improve this question
$\endgroup$
1
  • $\begingroup$ I think words serve better than notation here, but one way to emphasise the dependence would be to use fewer variables. If you are willing to regard $f$ as a function-valued function of one scalar variable rather than as a scalar-valued function of two scalar variables, you could (and a lambda-theoretician, to coin an ungainly word, might) write something like $\operatorname{flip}(f)(y) = \operatorname{const}(2y)$ for all $y \in \mathbb R$. But I think it is better just to say something like "$f(x, y) = 2y$ for all $x, y \in \mathbb R$. In particular, $f$ depends only on its second argument." $\endgroup$
    – LSpice
    Commented Oct 2, 2022 at 3:05

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .