Mixing
Understanding Arnold's definition of “differentiableâ€
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I'm looking at Arnold's Mathematical Methods of Classical Mechanics at the beginning of chapter 3 p 55 which defines when a functional is differentiable. Slightly paraphrasing and skipping a couple of details:
A functional $Phi$ is said to be differentiable if
$Phi(gamma+h)-Phi(gamma)=F+R$
where $F(gamma,h)$ is linear in $h$ and $Rsim O(h^2)$ in the sense that
for $|h|<epsilon$ and $left|fracdhdtright|<epsilon$ then $|R|<Cepsilon^2$.
I've been telling myself to think of $gamma$ as a curve, $h$ as a slight variation of the curve, and $F$ as the differential or "principal variation" of the functional, and $R$ the "error."
This is my first time seeing this definition of "differentiable" and also of $O(h^2)$, and I have a pair of questions about the latter.
In the 50 pages preceding, I can't seem to find out what $|cdot |$ means here. Is $|h|$ total variation of the curve $h$, or something like that?
How can I make sense of the constraints $|h|<epsilon$ and $left|fracdhdtright|<epsilon$ controlling $R$? The best I've come up with is "if $h$ does not go up and down too much and the speed doesn't go up and down too much, then $R$ will be under control." It might be helpful to have a prototypical example of a situation where $h$ slow but too wavy, and a situation where $h$ is not wavy but the speed varies wildly. Good heuristics are welcome too.
calculus-of-variations variational-analysis
asked Jul 26 at 19:34
rschwieb
99.7k1191225
 |Â
show 13 more comments
up vote
2
down vote
favorite
I'm looking at Arnold's Mathematical Methods of Classical Mechanics at the beginning of chapter 3 p 55 which defines when a functional is differentiable. Slightly paraphrasing and skipping a couple of details:
A functional $Phi$ is said to be differentiable if
$Phi(gamma+h)-Phi(gamma)=F+R$
where $F(gamma,h)$ is linear in $h$ and $Rsim O(h^2)$ in the sense that
for $|h|<epsilon$ and $left|fracdhdtright|<epsilon$ then $|R|<Cepsilon^2$.
I've been telling myself to think of $gamma$ as a curve, $h$ as a slight variation of the curve, and $F$ as the differential or "principal variation" of the functional, and $R$ the "error."
This is my first time seeing this definition of "differentiable" and also of $O(h^2)$, and I have a pair of questions about the latter.
In the 50 pages preceding, I can't seem to find out what $|cdot |$ means here. Is $|h|$ total variation of the curve $h$, or something like that?
How can I make sense of the constraints $|h|<epsilon$ and $left|fracdhdtright|<epsilon$ controlling $R$? The best I've come up with is "if $h$ does not go up and down too much and the speed doesn't go up and down too much, then $R$ will be under control." It might be helpful to have a prototypical example of a situation where $h$ slow but too wavy, and a situation where $h$ is not wavy but the speed varies wildly. Good heuristics are welcome too.
calculus-of-variations variational-analysis
asked Jul 26 at 19:34
rschwieb
99.7k1191225
$|h|$ is $|h|_infty$ on $[t_0,t_1]$ for the next theorem. The definition is left ambiguous to adjust it to each case. see, for example, footnote 26. Think of it as a template of many definitions, parametrized by the spaces of curves chosen, the norms chosen, etc.
– user577471
Jul 26 at 20:35
@HGLandcaster ....each case? well... I would think it is quite unusual if the definition of $O(h^2)$ is variable based on the problem presented. Are you sure it's not just the sup norm everywhere?
– rschwieb
Jul 26 at 20:40
It might be enough for all what follows, I didn't continue reading much more.
– user577471
Jul 26 at 20:43
It is not unusual for $O(h^2)$ and differentials to depend on the norms. It just happens that in finite dimensions they are equivalent.
– user577471
Jul 26 at 20:47
@HGLandcaster ok, a bit of a surprise, but good to know. Thanks!
– rschwieb
Jul 26 at 22:59
 |Â
show 13 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm looking at Arnold's Mathematical Methods of Classical Mechanics at the beginning of chapter 3 p 55 which defines when a functional is differentiable. Slightly paraphrasing and skipping a couple of details:
A functional $Phi$ is said to be differentiable if
$Phi(gamma+h)-Phi(gamma)=F+R$
where $F(gamma,h)$ is linear in $h$ and $Rsim O(h^2)$ in the sense that
for $|h|<epsilon$ and $left|fracdhdtright|<epsilon$ then $|R|<Cepsilon^2$.
I've been telling myself to think of $gamma$ as a curve, $h$ as a slight variation of the curve, and $F$ as the differential or "principal variation" of the functional, and $R$ the "error."
This is my first time seeing this definition of "differentiable" and also of $O(h^2)$, and I have a pair of questions about the latter.
In the 50 pages preceding, I can't seem to find out what $|cdot |$ means here. Is $|h|$ total variation of the curve $h$, or something like that?
How can I make sense of the constraints $|h|<epsilon$ and $left|fracdhdtright|<epsilon$ controlling $R$? The best I've come up with is "if $h$ does not go up and down too much and the speed doesn't go up and down too much, then $R$ will be under control." It might be helpful to have a prototypical example of a situation where $h$ slow but too wavy, and a situation where $h$ is not wavy but the speed varies wildly. Good heuristics are welcome too.
calculus-of-variations variational-analysis
asked Jul 26 at 19:34
rschwieb
99.7k1191225
I'm looking at Arnold's Mathematical Methods of Classical Mechanics at the beginning of chapter 3 p 55 which defines when a functional is differentiable. Slightly paraphrasing and skipping a couple of details:
A functional $Phi$ is said to be differentiable if
$Phi(gamma+h)-Phi(gamma)=F+R$
where $F(gamma,h)$ is linear in $h$ and $Rsim O(h^2)$ in the sense that
for $|h|<epsilon$ and $left|fracdhdtright|<epsilon$ then $|R|<Cepsilon^2$.
I've been telling myself to think of $gamma$ as a curve, $h$ as a slight variation of the curve, and $F$ as the differential or "principal variation" of the functional, and $R$ the "error."
This is my first time seeing this definition of "differentiable" and also of $O(h^2)$, and I have a pair of questions about the latter.
In the 50 pages preceding, I can't seem to find out what $|cdot |$ means here. Is $|h|$ total variation of the curve $h$, or something like that?
How can I make sense of the constraints $|h|<epsilon$ and $left|fracdhdtright|<epsilon$ controlling $R$? The best I've come up with is "if $h$ does not go up and down too much and the speed doesn't go up and down too much, then $R$ will be under control." It might be helpful to have a prototypical example of a situation where $h$ slow but too wavy, and a situation where $h$ is not wavy but the speed varies wildly. Good heuristics are welcome too.
calculus-of-variations variational-analysis
asked Jul 26 at 19:34
rschwieb
99.7k1191225
edited Jul 26 at 19:41
asked Jul 26 at 19:34
rschwieb
99.7k1191225
asked Jul 26 at 19:34
rschwieb
99.7k1191225
asked Jul 26 at 19:34
rschwieb
99.7k1191225
99.7k1191225
$|h|$ is $|h|_infty$ on $[t_0,t_1]$ for the next theorem. The definition is left ambiguous to adjust it to each case. see, for example, footnote 26. Think of it as a template of many definitions, parametrized by the spaces of curves chosen, the norms chosen, etc.
– user577471
Jul 26 at 20:35
@HGLandcaster ....each case? well... I would think it is quite unusual if the definition of $O(h^2)$ is variable based on the problem presented. Are you sure it's not just the sup norm everywhere?
– rschwieb
Jul 26 at 20:40
It might be enough for all what follows, I didn't continue reading much more.
– user577471
Jul 26 at 20:43
It is not unusual for $O(h^2)$ and differentials to depend on the norms. It just happens that in finite dimensions they are equivalent.
– user577471
Jul 26 at 20:47
@HGLandcaster ok, a bit of a surprise, but good to know. Thanks!
– rschwieb
Jul 26 at 22:59
 |Â
show 13 more comments
$|h|$ is $|h|_infty$ on $[t_0,t_1]$ for the next theorem. The definition is left ambiguous to adjust it to each case. see, for example, footnote 26. Think of it as a template of many definitions, parametrized by the spaces of curves chosen, the norms chosen, etc.
– user577471
Jul 26 at 20:35
@HGLandcaster ....each case? well... I would think it is quite unusual if the definition of $O(h^2)$ is variable based on the problem presented. Are you sure it's not just the sup norm everywhere?
– rschwieb
Jul 26 at 20:40
It might be enough for all what follows, I didn't continue reading much more.
– user577471
Jul 26 at 20:43
It is not unusual for $O(h^2)$ and differentials to depend on the norms. It just happens that in finite dimensions they are equivalent.
– user577471
Jul 26 at 20:47
@HGLandcaster ok, a bit of a surprise, but good to know. Thanks!
– rschwieb
Jul 26 at 22:59
$|h|$ is $|h|_infty$ on $[t_0,t_1]$ for the next theorem. The definition is left ambiguous to adjust it to each case. see, for example, footnote 26. Think of it as a template of many definitions, parametrized by the spaces of curves chosen, the norms chosen, etc.
– user577471
Jul 26 at 20:35
$|h|$ is $|h|_infty$ on $[t_0,t_1]$ for the next theorem. The definition is left ambiguous to adjust it to each case. see, for example, footnote 26. Think of it as a template of many definitions, parametrized by the spaces of curves chosen, the norms chosen, etc.
– user577471
Jul 26 at 20:35
@HGLandcaster ....each case? well... I would think it is quite unusual if the definition of $O(h^2)$ is variable based on the problem presented. Are you sure it's not just the sup norm everywhere?
– rschwieb
Jul 26 at 20:40
@HGLandcaster ....each case? well... I would think it is quite unusual if the definition of $O(h^2)$ is variable based on the problem presented. Are you sure it's not just the sup norm everywhere?
– rschwieb
Jul 26 at 20:40
It might be enough for all what follows, I didn't continue reading much more.
– user577471
Jul 26 at 20:43
It might be enough for all what follows, I didn't continue reading much more.
– user577471
Jul 26 at 20:43
It is not unusual for $O(h^2)$ and differentials to depend on the norms. It just happens that in finite dimensions they are equivalent.
– user577471
Jul 26 at 20:47
It is not unusual for $O(h^2)$ and differentials to depend on the norms. It just happens that in finite dimensions they are equivalent.
– user577471
Jul 26 at 20:47
@HGLandcaster ok, a bit of a surprise, but good to know. Thanks!
– rschwieb
Jul 26 at 22:59
@HGLandcaster ok, a bit of a surprise, but good to know. Thanks!
– rschwieb
Jul 26 at 22:59
 |Â
show 13 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863733%2funderstanding-arnolds-definition-of-differentiable%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
$|h|$ is $|h|_infty$ on $[t_0,t_1]$ for the next theorem. The definition is left ambiguous to adjust it to each case. see, for example, footnote 26. Think of it as a template of many definitions, parametrized by the spaces of curves chosen, the norms chosen, etc.
– user577471
Jul 26 at 20:35
@HGLandcaster ....each case? well... I would think it is quite unusual if the definition of $O(h^2)$ is variable based on the problem presented. Are you sure it's not just the sup norm everywhere?
– rschwieb
Jul 26 at 20:40
It might be enough for all what follows, I didn't continue reading much more.
– user577471
Jul 26 at 20:43
It is not unusual for $O(h^2)$ and differentials to depend on the norms. It just happens that in finite dimensions they are equivalent.
– user577471
Jul 26 at 20:47
@HGLandcaster ok, a bit of a surprise, but good to know. Thanks!
– rschwieb
Jul 26 at 22:59