Mixing
Why $v_amapsto D_{v}|_a$ (the directional derivative) is a linear map $mathbb R^n_ato T_amathbb R^n$?
$begingroup$
Let $mathbb R^n_a={(a,v)=v_amid ainmathbb R^n, vin mathbb R^n}$ the set of vector of $mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=left.frac{d}{dt}right|_{t=0}f(a+tv)$$
where $D_v|_a:mathcal C^infty (mathbb R^n)to mathbb R$. Let $T_amathbb R^n$ the set of all derivatives of $mathcal C^infty (mathbb R^n)$ at $a$.
Why $v_amapsto D_v|_a$ is a linear map ?
Attempts
Let $fin mathcal C^infty (mathbb R^n)$.
$$D_{v+w}|_af=lim_{tto 0}frac{f(a+t(v+w))-f(a)}{t}$$
$$=lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}+underbrace{lim_{tto 0}frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$
Now, how can I prove that $$lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af ?$$
derivatives manifolds
asked Jan 12 at 16:28
user623855user623855
1507
$endgroup$
add a comment |
$begingroup$
Let $mathbb R^n_a={(a,v)=v_amid ainmathbb R^n, vin mathbb R^n}$ the set of vector of $mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=left.frac{d}{dt}right|_{t=0}f(a+tv)$$
where $D_v|_a:mathcal C^infty (mathbb R^n)to mathbb R$. Let $T_amathbb R^n$ the set of all derivatives of $mathcal C^infty (mathbb R^n)$ at $a$.
Why $v_amapsto D_v|_a$ is a linear map ?
Attempts
Let $fin mathcal C^infty (mathbb R^n)$.
$$D_{v+w}|_af=lim_{tto 0}frac{f(a+t(v+w))-f(a)}{t}$$
$$=lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}+underbrace{lim_{tto 0}frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$
Now, how can I prove that $$lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af ?$$
derivatives manifolds
asked Jan 12 at 16:28
user623855user623855
1507
$endgroup$
$begingroup$
It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
$endgroup$
– Ian
Jan 12 at 16:32
$begingroup$
@Ian: Thank you. It's not possible like I tried to ?
$endgroup$
– user623855
Jan 12 at 16:34
$begingroup$
Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
$endgroup$
– Ian
Jan 12 at 16:34
add a comment |
$begingroup$
Let $mathbb R^n_a={(a,v)=v_amid ainmathbb R^n, vin mathbb R^n}$ the set of vector of $mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=left.frac{d}{dt}right|_{t=0}f(a+tv)$$
where $D_v|_a:mathcal C^infty (mathbb R^n)to mathbb R$. Let $T_amathbb R^n$ the set of all derivatives of $mathcal C^infty (mathbb R^n)$ at $a$.
Why $v_amapsto D_v|_a$ is a linear map ?
Attempts
Let $fin mathcal C^infty (mathbb R^n)$.
$$D_{v+w}|_af=lim_{tto 0}frac{f(a+t(v+w))-f(a)}{t}$$
$$=lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}+underbrace{lim_{tto 0}frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$
Now, how can I prove that $$lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af ?$$
derivatives manifolds
asked Jan 12 at 16:28
user623855user623855
1507
$endgroup$
Let $mathbb R^n_a={(a,v)=v_amid ainmathbb R^n, vin mathbb R^n}$ the set of vector of $mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=left.frac{d}{dt}right|_{t=0}f(a+tv)$$
where $D_v|_a:mathcal C^infty (mathbb R^n)to mathbb R$. Let $T_amathbb R^n$ the set of all derivatives of $mathcal C^infty (mathbb R^n)$ at $a$.
Why $v_amapsto D_v|_a$ is a linear map ?
Attempts
Let $fin mathcal C^infty (mathbb R^n)$.
$$D_{v+w}|_af=lim_{tto 0}frac{f(a+t(v+w))-f(a)}{t}$$
$$=lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}+underbrace{lim_{tto 0}frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$
Now, how can I prove that $$lim_{tto 0}frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af ?$$
derivatives manifolds
derivatives manifolds
asked Jan 12 at 16:28
user623855user623855
1507
asked Jan 12 at 16:28
user623855user623855
1507
asked Jan 12 at 16:28
user623855user623855
1507
asked Jan 12 at 16:28
user623855user623855
1507
asked Jan 12 at 16:28
user623855user623855
1507
1507
$begingroup$
It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
$endgroup$
– Ian
Jan 12 at 16:32
$begingroup$
@Ian: Thank you. It's not possible like I tried to ?
$endgroup$
– user623855
Jan 12 at 16:34
$begingroup$
Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
$endgroup$
– Ian
Jan 12 at 16:34
add a comment |
$begingroup$
It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
$endgroup$
– Ian
Jan 12 at 16:32
$begingroup$
@Ian: Thank you. It's not possible like I tried to ?
$endgroup$
– user623855
Jan 12 at 16:34
$begingroup$
Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
$endgroup$
– Ian
Jan 12 at 16:34
$begingroup$
It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
$endgroup$
– Ian
Jan 12 at 16:32
$begingroup$
It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
$endgroup$
– Ian
Jan 12 at 16:32
$begingroup$
@Ian: Thank you. It's not possible like I tried to ?
$endgroup$
– user623855
Jan 12 at 16:34
$begingroup$
@Ian: Thank you. It's not possible like I tried to ?
$endgroup$
– user623855
Jan 12 at 16:34
$begingroup$
Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
$endgroup$
– Ian
Jan 12 at 16:34
$begingroup$
Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
$endgroup$
– Ian
Jan 12 at 16:34
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint
If $vin mathbb R^n$,
$$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$
answered Jan 12 at 16:39
SurbSurb
38k94375
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint
If $vin mathbb R^n$,
$$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$
answered Jan 12 at 16:39
SurbSurb
38k94375
$endgroup$
add a comment |
$begingroup$
Hint
If $vin mathbb R^n$,
$$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$
answered Jan 12 at 16:39
SurbSurb
38k94375
$endgroup$
add a comment |
$begingroup$
Hint
If $vin mathbb R^n$,
$$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$
answered Jan 12 at 16:39
SurbSurb
38k94375
$endgroup$
Hint
If $vin mathbb R^n$,
$$f(a+tv)=f(a)+tnabla f(a)cdot v+o(t).$$
answered Jan 12 at 16:39
SurbSurb
38k94375
answered Jan 12 at 16:39
SurbSurb
38k94375
answered Jan 12 at 16:39
SurbSurb
38k94375
answered Jan 12 at 16:39
SurbSurb
38k94375
38k94375
add a comment |
add a comment |
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$begingroup$
It is easier to first show that the directional derivative is given by the dot product of the gradient with $v$. Then it follows from the bilinearity of the dot product. In the end your approach is consistent but finishing the proof from where you're at now is not really any easier than it is to do it from scratch in a different way.
$endgroup$
– Ian
Jan 12 at 16:32
$begingroup$
@Ian: Thank you. It's not possible like I tried to ?
$endgroup$
– user623855
Jan 12 at 16:34
$begingroup$
Your attempt will work but the last identity is basically the same difficulty to prove as the original one, so you haven't really made the problem any easier.
$endgroup$
– Ian
Jan 12 at 16:34