Is parametric form of a given function unique?

Question

Can we say that for any given function in single/multivariable, it is always possible to have a parametric form? (Elementary functions, complicated functions?)
Given any function, is parametric form uniquely determined?

@AhaanS.Rungta I see you created the tag (concept) in your edit. I don't think that such tag is needed. But if you do think it could be useful, feel free to start a discussion about the new tag on meta. (This is irrelevant to the original question, so if some more discussion is needed, let us continue in chat.) — Martin Sleziak, Dec 19 '14 at 08:25

Ahaan S. Rungta · Answer 1 · 2014-12-17T13:02:54.303

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In general, no. For example, many piecewise functions have no parametric forums. You will have to be more specific about this and if you clarify, I will edit this answer. For example, consider the Dirichlet function. I don't know if you consider that in the set of functions you are asking about.
No. Quick counterexample. Take $ y = x $. The parametric equations $$ \begin {eqnarray*} x &=& t, \\ y &=& t \end {eqnarray*} $$and $$ \begin {eqnarray*} x &=& 2t, \\ y &=& 2t \end {eqnarray*} $$There are many other counterexamples.

edited Dec 17 '14 at 13:02

answered Dec 17 '14 at 12:42

Ahaan S. Rungta

For part 2. You get the same curve only if the domain is contained in $[0, \infty]$. – bubba Dec 17 '14 at 12:52
@bubba No? In both cases, $y=x$ for all $t$. – Ahaan S. Rungta Dec 17 '14 at 12:54
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Yes, but the first parameterization gives you the entire line $y=x$, but the second one only gives you half of it. – bubba Dec 17 '14 at 12:57
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Or, in fancy mathematician jargon: ${(t,t) : t \in \mathbb{R}} \ne {(t^2,t^2) : t \in \mathbb{R}}$. – bubba Dec 17 '14 at 13:00
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Thanks; good point. Edited. Also, +1 on your answer. – Ahaan S. Rungta Dec 17 '14 at 13:03

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

No. Any parametric curve is a connected set (since it is the image of a connected set under a continuous mapping). But the set $\{(x,y) \in \mathbb{R}^2 : x^2 = 1\}$ consists of two disconnected lines, so it can't be represented using parametric equations (using continuous functions, anyway).
No. The parameterizations $t \mapsto(0,t)$ and $t \mapsto(0,1-t)$ both represent the same curve, for example. Even if you restrict $t$ to $[0,1]$, it's still true that both parameterizations give you the same curve segment.

See also this question, which provides an example of two different parameterizations of a quadrant of a circle.

2 Answers2