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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










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$begingroup$


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?










share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    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?










    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      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?










      share|cite|improve this question











      $endgroup$




      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






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      share|cite|improve this question













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      edited Apr 8 at 3:43







      Z Wang

















      asked Apr 8 at 1:33









      Z WangZ Wang

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          $begingroup$

          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.






          share|cite|improve this answer









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            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 8 at 7:49









                WrzlprmftWrzlprmft

                3,25911335




                3,25911335



























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