Proof of equivalence well defined function
$begingroup$
This is the definition:
Definition 1:
A function $f:Dsubseteq Rto R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.
Definition 2:
A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] to R^n$, the points f(a) and f(b) are then called the end points of the arc$
Now comes the theorem:
Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]subseteq Rto R^n $ Then any smooth parametrisation $h:[c,d]subseteq Rto R^n$ of C is 1-1 and is equivalent to f.
Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:
It starts as follows:
Proof
The function $Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=fcircPhi$.
Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $Phi$ is well defined.
Question
1) Why is $Phi$ well defined?
2) Why does it satisfy $h=fcircPhi$
proof-explanation curves parametrization
edited Nov 29 '18 at 8:38
José Carlos Santos
159k22126231
asked Nov 29 '18 at 8:15
ALEXANDERALEXANDER
8951921
$endgroup$
add a comment |
$begingroup$
This is the definition:
Definition 1:
A function $f:Dsubseteq Rto R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.
Definition 2:
A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] to R^n$, the points f(a) and f(b) are then called the end points of the arc$
Now comes the theorem:
Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]subseteq Rto R^n $ Then any smooth parametrisation $h:[c,d]subseteq Rto R^n$ of C is 1-1 and is equivalent to f.
Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:
It starts as follows:
Proof
The function $Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=fcircPhi$.
Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $Phi$ is well defined.
Question
1) Why is $Phi$ well defined?
2) Why does it satisfy $h=fcircPhi$
proof-explanation curves parametrization
edited Nov 29 '18 at 8:38
José Carlos Santos
159k22126231
asked Nov 29 '18 at 8:15
ALEXANDERALEXANDER
8951921
$endgroup$
$begingroup$
It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
$endgroup$
– AnyAD
Nov 29 '18 at 8:31
$begingroup$
So if f is one to one it follows that $f^{-1}$ is well defined?
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:33
add a comment |
$begingroup$
This is the definition:
Definition 1:
A function $f:Dsubseteq Rto R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.
Definition 2:
A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] to R^n$, the points f(a) and f(b) are then called the end points of the arc$
Now comes the theorem:
Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]subseteq Rto R^n $ Then any smooth parametrisation $h:[c,d]subseteq Rto R^n$ of C is 1-1 and is equivalent to f.
Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:
It starts as follows:
Proof
The function $Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=fcircPhi$.
Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $Phi$ is well defined.
Question
1) Why is $Phi$ well defined?
2) Why does it satisfy $h=fcircPhi$
proof-explanation curves parametrization
edited Nov 29 '18 at 8:38
José Carlos Santos
159k22126231
asked Nov 29 '18 at 8:15
ALEXANDERALEXANDER
8951921
$endgroup$
This is the definition:
Definition 1:
A function $f:Dsubseteq Rto R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.
Definition 2:
A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] to R^n$, the points f(a) and f(b) are then called the end points of the arc$
Now comes the theorem:
Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]subseteq Rto R^n $ Then any smooth parametrisation $h:[c,d]subseteq Rto R^n$ of C is 1-1 and is equivalent to f.
Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:
It starts as follows:
Proof
The function $Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=fcircPhi$.
Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $Phi$ is well defined.
Question
1) Why is $Phi$ well defined?
2) Why does it satisfy $h=fcircPhi$
proof-explanation curves parametrization
proof-explanation curves parametrization
edited Nov 29 '18 at 8:38
José Carlos Santos
159k22126231
asked Nov 29 '18 at 8:15
ALEXANDERALEXANDER
8951921
edited Nov 29 '18 at 8:38
José Carlos Santos
159k22126231
asked Nov 29 '18 at 8:15
ALEXANDERALEXANDER
8951921
edited Nov 29 '18 at 8:38
José Carlos Santos
159k22126231
edited Nov 29 '18 at 8:38
José Carlos Santos
159k22126231
edited Nov 29 '18 at 8:38
José Carlos Santos
159k22126231
159k22126231
asked Nov 29 '18 at 8:15
ALEXANDERALEXANDER
8951921
asked Nov 29 '18 at 8:15
ALEXANDERALEXANDER
8951921
asked Nov 29 '18 at 8:15
ALEXANDERALEXANDER
8951921
8951921
$begingroup$
It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
$endgroup$
– AnyAD
Nov 29 '18 at 8:31
$begingroup$
So if f is one to one it follows that $f^{-1}$ is well defined?
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:33
add a comment |
$begingroup$
It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
$endgroup$
– AnyAD
Nov 29 '18 at 8:31
$begingroup$
So if f is one to one it follows that $f^{-1}$ is well defined?
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:33
$begingroup$
It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
$endgroup$
– AnyAD
Nov 29 '18 at 8:31
$begingroup$
It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
$endgroup$
– AnyAD
Nov 29 '18 at 8:31
$begingroup$
So if f is one to one it follows that $f^{-1}$ is well defined?
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:33
$begingroup$
So if f is one to one it follows that $f^{-1}$ is well defined?
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
- It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.
- Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.
answered Nov 29 '18 at 8:33
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35
$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37
$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41
1
$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59
1
$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25
|
show 6 more comments
Your Answer
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1 Answer
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$begingroup$
- It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.
- Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.
answered Nov 29 '18 at 8:33
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35
$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37
$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41
1
$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59
1
$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25
|
show 6 more comments
$begingroup$
- It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.
- Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.
answered Nov 29 '18 at 8:33
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35
$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37
$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41
1
$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59
1
$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25
|
show 6 more comments
$begingroup$
- It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.
- Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.
answered Nov 29 '18 at 8:33
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
- It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.
- Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.
answered Nov 29 '18 at 8:33
José Carlos SantosJosé Carlos Santos
159k22126231
answered Nov 29 '18 at 8:33
José Carlos SantosJosé Carlos Santos
159k22126231
answered Nov 29 '18 at 8:33
José Carlos SantosJosé Carlos Santos
159k22126231
answered Nov 29 '18 at 8:33
José Carlos SantosJosé Carlos Santos
159k22126231
159k22126231
$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35
$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37
$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41
1
$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59
1
$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25
|
show 6 more comments
$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35
$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37
$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41
1
$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59
1
$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25
$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35
$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35
$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37
$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37
$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41
$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41
1
1
$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59
$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59
1
1
$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25
$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25
|
show 6 more comments
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$begingroup$
It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
$endgroup$
– AnyAD
Nov 29 '18 at 8:31
$begingroup$
So if f is one to one it follows that $f^{-1}$ is well defined?
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:33