“Parameterizations are non-unique”, how can we prove it?
$begingroup$
$$
textbf{"Parameterizations are non-unique"}
$$
I have seen this statement in several books and at Wikipedia.
However, I have never seen a proof of the statement. How can we prove it?
real-analysis calculus analysis proof-writing parametrization
$endgroup$
add a comment |
$begingroup$
$$
textbf{"Parameterizations are non-unique"}
$$
I have seen this statement in several books and at Wikipedia.
However, I have never seen a proof of the statement. How can we prove it?
real-analysis calculus analysis proof-writing parametrization
$endgroup$
add a comment |
$begingroup$
$$
textbf{"Parameterizations are non-unique"}
$$
I have seen this statement in several books and at Wikipedia.
However, I have never seen a proof of the statement. How can we prove it?
real-analysis calculus analysis proof-writing parametrization
$endgroup$
$$
textbf{"Parameterizations are non-unique"}
$$
I have seen this statement in several books and at Wikipedia.
However, I have never seen a proof of the statement. How can we prove it?
real-analysis calculus analysis proof-writing parametrization
real-analysis calculus analysis proof-writing parametrization
edited Jan 15 at 9:12
Bernard
121k740116
121k740116
asked Jan 15 at 8:45
JDoeDoeJDoeDoe
7701614
7701614
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$begingroup$
Actually, what it says is that they are generally nonunique. But if you have a parametrizition of a curve $gammacolon[a,b]longrightarrowmathbb{R}^n$, you can always define, say,$$begin{array}{rccc}gamma^starcolon&[a+1,b+1]&longrightarrow&mathbb{R}^n\&t&mapsto&gamma(t-1)end{array}$$and that's another parametrization.
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Actually, what it says is that they are generally nonunique. But if you have a parametrizition of a curve $gammacolon[a,b]longrightarrowmathbb{R}^n$, you can always define, say,$$begin{array}{rccc}gamma^starcolon&[a+1,b+1]&longrightarrow&mathbb{R}^n\&t&mapsto&gamma(t-1)end{array}$$and that's another parametrization.
$endgroup$
add a comment |
$begingroup$
Actually, what it says is that they are generally nonunique. But if you have a parametrizition of a curve $gammacolon[a,b]longrightarrowmathbb{R}^n$, you can always define, say,$$begin{array}{rccc}gamma^starcolon&[a+1,b+1]&longrightarrow&mathbb{R}^n\&t&mapsto&gamma(t-1)end{array}$$and that's another parametrization.
$endgroup$
add a comment |
$begingroup$
Actually, what it says is that they are generally nonunique. But if you have a parametrizition of a curve $gammacolon[a,b]longrightarrowmathbb{R}^n$, you can always define, say,$$begin{array}{rccc}gamma^starcolon&[a+1,b+1]&longrightarrow&mathbb{R}^n\&t&mapsto&gamma(t-1)end{array}$$and that's another parametrization.
$endgroup$
Actually, what it says is that they are generally nonunique. But if you have a parametrizition of a curve $gammacolon[a,b]longrightarrowmathbb{R}^n$, you can always define, say,$$begin{array}{rccc}gamma^starcolon&[a+1,b+1]&longrightarrow&mathbb{R}^n\&t&mapsto&gamma(t-1)end{array}$$and that's another parametrization.
edited Jan 15 at 9:06
answered Jan 15 at 8:52


José Carlos SantosJosé Carlos Santos
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