A “counterexample” on Takens' embedding theorem for phase space contruction The 2019 Stack Overflow Developer Survey Results Are InConfusion with periodic orbits, classical dynamical systemsAny two Poincare’s maps for a periodic phase space are diffeomorphicEmbedding dimension $m>4$ for high-dimensional chaos.Can the change in phase space volume be periodic?Understanding Takens' Embedding theoremNatural projection of the billiard phase spaceTopological conjugation between autonomous and non-autonomous systemsApproximating the tangent vector in a phase space (or state space) reconstructionWhy Lorenz attractor can be embedded by a 3-step time delay map?Dynamical Systems: Understanding the concept of phase flow
Why does the nucleus not repel itself?
I am an eight letter word. What am I?
Finding the area between two curves with Integrate
Is an up-to-date browser secure on an out-of-date OS?
Mathematics of imaging the black hole
What information about me do stores get via my credit card?
Dropping list elements from nested list after evaluation
What is the meaning of Triage in Cybersec world?
Can a flute soloist sit?
What does もの mean in this sentence?
A word that means fill it to the required quantity
How to charge AirPods to keep battery healthy?
Accepted by European university, rejected by all American ones I applied to? Possible reasons?
Geography at the pixel level
What is the most efficient way to store a numeric range?
How to display lines in a file like ls displays files in a directory?
How to notate time signature switching consistently every measure
Is it safe to harvest rainwater that fell on solar panels?
If I score a critical hit on an 18 or higher, what are my chances of getting a critical hit if I roll 3d20?
Why not take a picture of a closer black hole?
Are spiders unable to hurt humans, especially very small spiders?
Why isn't the circumferential light around the M87 black hole's event horizon symmetric?
Can withdrawing asylum be illegal?
Ubuntu Server install with full GUI
A “counterexample” on Takens' embedding theorem for phase space contruction
The 2019 Stack Overflow Developer Survey Results Are InConfusion with periodic orbits, classical dynamical systemsAny two Poincare’s maps for a periodic phase space are diffeomorphicEmbedding dimension $m>4$ for high-dimensional chaos.Can the change in phase space volume be periodic?Understanding Takens' Embedding theoremNatural projection of the billiard phase spaceTopological conjugation between autonomous and non-autonomous systemsApproximating the tangent vector in a phase space (or state space) reconstructionWhy Lorenz attractor can be embedded by a 3-step time delay map?Dynamical Systems: Understanding the concept of phase flow
$begingroup$
Background: Consider the 1-dimensional submanifold $y=sinx$ for $xinmathbbR$, it is a dynamical system with the flow/dynamics function $f(t,(x,y))=(x+t,sin(x+t))$. Now consider its projection onto the y-axis, $rho(t)=(t,sin(t))$ for $tinmathbbR$.
By the Takens' embedding theorem, I can use some function like $h(t)=(sin(t),sin(t+tau))$ to reconstruct my phase space, where $tau$ is the time delay. However, I see that the image of $h(t)$ is an ellipse for $tau=1$, while the original phase space is the sine curve. The curve and the ellipse are not homeomorphic, and so this is a counter-example to Takens' theorem?
ordinary-differential-equations dynamical-systems time-series chaos-theory
$endgroup$
add a comment |
$begingroup$
Background: Consider the 1-dimensional submanifold $y=sinx$ for $xinmathbbR$, it is a dynamical system with the flow/dynamics function $f(t,(x,y))=(x+t,sin(x+t))$. Now consider its projection onto the y-axis, $rho(t)=(t,sin(t))$ for $tinmathbbR$.
By the Takens' embedding theorem, I can use some function like $h(t)=(sin(t),sin(t+tau))$ to reconstruct my phase space, where $tau$ is the time delay. However, I see that the image of $h(t)$ is an ellipse for $tau=1$, while the original phase space is the sine curve. The curve and the ellipse are not homeomorphic, and so this is a counter-example to Takens' theorem?
ordinary-differential-equations dynamical-systems time-series chaos-theory
$endgroup$
add a comment |
$begingroup$
Background: Consider the 1-dimensional submanifold $y=sinx$ for $xinmathbbR$, it is a dynamical system with the flow/dynamics function $f(t,(x,y))=(x+t,sin(x+t))$. Now consider its projection onto the y-axis, $rho(t)=(t,sin(t))$ for $tinmathbbR$.
By the Takens' embedding theorem, I can use some function like $h(t)=(sin(t),sin(t+tau))$ to reconstruct my phase space, where $tau$ is the time delay. However, I see that the image of $h(t)$ is an ellipse for $tau=1$, while the original phase space is the sine curve. The curve and the ellipse are not homeomorphic, and so this is a counter-example to Takens' theorem?
ordinary-differential-equations dynamical-systems time-series chaos-theory
$endgroup$
Background: Consider the 1-dimensional submanifold $y=sinx$ for $xinmathbbR$, it is a dynamical system with the flow/dynamics function $f(t,(x,y))=(x+t,sin(x+t))$. Now consider its projection onto the y-axis, $rho(t)=(t,sin(t))$ for $tinmathbbR$.
By the Takens' embedding theorem, I can use some function like $h(t)=(sin(t),sin(t+tau))$ to reconstruct my phase space, where $tau$ is the time delay. However, I see that the image of $h(t)$ is an ellipse for $tau=1$, while the original phase space is the sine curve. The curve and the ellipse are not homeomorphic, and so this is a counter-example to Takens' theorem?
ordinary-differential-equations dynamical-systems time-series chaos-theory
ordinary-differential-equations dynamical-systems time-series chaos-theory
edited Apr 8 at 3:43
Z Wang
asked Apr 8 at 1:33
Z WangZ Wang
786
786
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your second component ($y$) is not fed back into the dynamics:
The value of $f$ does not depend on $y$ at all.
Thus your dynamics can be simplified to $f(t,x) = x+t$, whose topology is a simple (infinite) line.
With other words, your second component is a write-only variable, which can discard information without affecting the dynamics, and indeed it does on account of the sine function not being invertible.
Now, since you choose this write-only variable with discarded observation as an observable, your phase-space reconstruction fails and you do not obtain an embedding of your original dynamics.
However, while your phase-space reconstruction failed, Takens’ theorem does not:
It states only that you obtain an embedding for almost any measurement function $h$, where in your case $h(x,y) = y$ is your projection.
And indeed, for a projection to any other axis, you would have obtained a (proper) embedding.
To better understand this, it may help to extremify your example and consider the dynamics $f(t,(x,y)) = (x+t,0)$ with the same projection.
Your time series would consist only of zeroes and should not be surprising that you cannot reconstruct the dynamics from this.
For any dynamics, you can find a similar extension and projection.
In reality, this corresponds to a broken measurement device or one that measures a completely irrelevant observable.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179020%2fa-counterexample-on-takens-embedding-theorem-for-phase-space-contruction%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your second component ($y$) is not fed back into the dynamics:
The value of $f$ does not depend on $y$ at all.
Thus your dynamics can be simplified to $f(t,x) = x+t$, whose topology is a simple (infinite) line.
With other words, your second component is a write-only variable, which can discard information without affecting the dynamics, and indeed it does on account of the sine function not being invertible.
Now, since you choose this write-only variable with discarded observation as an observable, your phase-space reconstruction fails and you do not obtain an embedding of your original dynamics.
However, while your phase-space reconstruction failed, Takens’ theorem does not:
It states only that you obtain an embedding for almost any measurement function $h$, where in your case $h(x,y) = y$ is your projection.
And indeed, for a projection to any other axis, you would have obtained a (proper) embedding.
To better understand this, it may help to extremify your example and consider the dynamics $f(t,(x,y)) = (x+t,0)$ with the same projection.
Your time series would consist only of zeroes and should not be surprising that you cannot reconstruct the dynamics from this.
For any dynamics, you can find a similar extension and projection.
In reality, this corresponds to a broken measurement device or one that measures a completely irrelevant observable.
$endgroup$
add a comment |
$begingroup$
Your second component ($y$) is not fed back into the dynamics:
The value of $f$ does not depend on $y$ at all.
Thus your dynamics can be simplified to $f(t,x) = x+t$, whose topology is a simple (infinite) line.
With other words, your second component is a write-only variable, which can discard information without affecting the dynamics, and indeed it does on account of the sine function not being invertible.
Now, since you choose this write-only variable with discarded observation as an observable, your phase-space reconstruction fails and you do not obtain an embedding of your original dynamics.
However, while your phase-space reconstruction failed, Takens’ theorem does not:
It states only that you obtain an embedding for almost any measurement function $h$, where in your case $h(x,y) = y$ is your projection.
And indeed, for a projection to any other axis, you would have obtained a (proper) embedding.
To better understand this, it may help to extremify your example and consider the dynamics $f(t,(x,y)) = (x+t,0)$ with the same projection.
Your time series would consist only of zeroes and should not be surprising that you cannot reconstruct the dynamics from this.
For any dynamics, you can find a similar extension and projection.
In reality, this corresponds to a broken measurement device or one that measures a completely irrelevant observable.
$endgroup$
add a comment |
$begingroup$
Your second component ($y$) is not fed back into the dynamics:
The value of $f$ does not depend on $y$ at all.
Thus your dynamics can be simplified to $f(t,x) = x+t$, whose topology is a simple (infinite) line.
With other words, your second component is a write-only variable, which can discard information without affecting the dynamics, and indeed it does on account of the sine function not being invertible.
Now, since you choose this write-only variable with discarded observation as an observable, your phase-space reconstruction fails and you do not obtain an embedding of your original dynamics.
However, while your phase-space reconstruction failed, Takens’ theorem does not:
It states only that you obtain an embedding for almost any measurement function $h$, where in your case $h(x,y) = y$ is your projection.
And indeed, for a projection to any other axis, you would have obtained a (proper) embedding.
To better understand this, it may help to extremify your example and consider the dynamics $f(t,(x,y)) = (x+t,0)$ with the same projection.
Your time series would consist only of zeroes and should not be surprising that you cannot reconstruct the dynamics from this.
For any dynamics, you can find a similar extension and projection.
In reality, this corresponds to a broken measurement device or one that measures a completely irrelevant observable.
$endgroup$
Your second component ($y$) is not fed back into the dynamics:
The value of $f$ does not depend on $y$ at all.
Thus your dynamics can be simplified to $f(t,x) = x+t$, whose topology is a simple (infinite) line.
With other words, your second component is a write-only variable, which can discard information without affecting the dynamics, and indeed it does on account of the sine function not being invertible.
Now, since you choose this write-only variable with discarded observation as an observable, your phase-space reconstruction fails and you do not obtain an embedding of your original dynamics.
However, while your phase-space reconstruction failed, Takens’ theorem does not:
It states only that you obtain an embedding for almost any measurement function $h$, where in your case $h(x,y) = y$ is your projection.
And indeed, for a projection to any other axis, you would have obtained a (proper) embedding.
To better understand this, it may help to extremify your example and consider the dynamics $f(t,(x,y)) = (x+t,0)$ with the same projection.
Your time series would consist only of zeroes and should not be surprising that you cannot reconstruct the dynamics from this.
For any dynamics, you can find a similar extension and projection.
In reality, this corresponds to a broken measurement device or one that measures a completely irrelevant observable.
answered Apr 8 at 7:49
WrzlprmftWrzlprmft
3,25911335
3,25911335
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179020%2fa-counterexample-on-takens-embedding-theorem-for-phase-space-contruction%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown