1

If I have a function $y = f(x)$, and I parameterize it so that I get two new functions $x(t)$ and $y(t)$, I'm curious about when I can say that two distinct parameterizations are equal, and in what sense they are equal. I'll give a trivial example:

Say $y = x, x \in [-1,1]$ is my original function. I might parameterize this function as:

$$x(t) = t$$

$$y(t) = t$$

$$t \in [-1,1]$$

This would, when plotted, yield the same points in the same order as the original function, and I think that the parameterization and the original function are equal in that regard. But let's now introduce a new parameterization:

$$x(t) = -cos(t)$$ $$y(t) = -cos(t)$$ $$t \in [0,\pi]$$

This new parameterization still produces the same points (and I think in the same order, but correct me if I'm wrong), and yet is qualitatively different from the original parameterization. In what sense can I say that these two parameterizations are equal, and in what sense are they not equal? How do I quantify that?

Michael Stachowsky
  • 2,943
  • 1
    Actually, there are an infinite number of parameterizations for a function. With y= x, x= f(t), y= f(t) work for any function f. As for saying that they are equal, you simply mean that they are parameterizations for the same function. – user247327 Apr 06 '18 at 13:54
  • Agreed that there are infinitely many, although I'm sure there are restrictions on what the parameterizing functions can be (for instance, $f(t)$ must take on all values in the original function's domain is at least necessary if not sufficient for this. In our case, $f(t) = t^2 + 10$ wouldn't parameterize the original function as given). But that still begs the question - what does "equal" mean? – Michael Stachowsky Apr 06 '18 at 13:57

0 Answers0