Proof check: Show that $x_n + 1 = f(x_n) = 15 - x_n^2$ is a (Smale) horseshoe

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However, I believe it to be incorrect because $f(K_1) = f(K_2) neq (-sqrt15, sqrt15)$.



Attempt at a solution:



Referring to the diagram from the question...



By the IVT there is some $xi_3 in (sqrt15, 5)$ such that $f(s) = x_2$. Further, by the IVT there is some $xi_1 in (-sqrt15, 0)$ and some $xi_2 in (0, sqrt15)$ such that $f(xi_1) = f(xi_2) = xi_3$.



Let $K_1 = (x_2, xi_1)$ so $f(K_1) = (x_2, xi_1)$ and let $K_2 = (xi_2, xi_3)$ so $f(K_2) = (xi_2, xi_3)$. Letting $K = (x_2, xi_3) approx (-4.405, 4.405)$, note that $K_1 subset (x_2, 0) subset K$ and $K_2 subset (0, xi_3) subset K$ and that $K_1 cap K_2 = emptyset$. Hence, since $f(K_1) = f(K_2) = K$, the map is indeed a horseshoe.



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nonlinear-system chaos-theory nonlinear-analysis non-linear-dynamics




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    up vote
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    down vote

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    A solution is given here:



    Solution



    However, I believe it to be incorrect because $f(K_1) = f(K_2) neq (-sqrt15, sqrt15)$.



    Attempt at a solution:



    Referring to the diagram from the question...



    By the IVT there is some $xi_3 in (sqrt15, 5)$ such that $f(s) = x_2$. Further, by the IVT there is some $xi_1 in (-sqrt15, 0)$ and some $xi_2 in (0, sqrt15)$ such that $f(xi_1) = f(xi_2) = xi_3$.



    Let $K_1 = (x_2, xi_1)$ so $f(K_1) = (x_2, xi_1)$ and let $K_2 = (xi_2, xi_3)$ so $f(K_2) = (xi_2, xi_3)$. Letting $K = (x_2, xi_3) approx (-4.405, 4.405)$, note that $K_1 subset (x_2, 0) subset K$ and $K_2 subset (0, xi_3) subset K$ and that $K_1 cap K_2 = emptyset$. Hence, since $f(K_1) = f(K_2) = K$, the map is indeed a horseshoe.



    Question



    Is my attempt at the solution correct?




    nonlinear-system chaos-theory nonlinear-analysis non-linear-dynamics




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      A solution is given here:



      Solution



      However, I believe it to be incorrect because $f(K_1) = f(K_2) neq (-sqrt15, sqrt15)$.



      Attempt at a solution:



      Referring to the diagram from the question...



      By the IVT there is some $xi_3 in (sqrt15, 5)$ such that $f(s) = x_2$. Further, by the IVT there is some $xi_1 in (-sqrt15, 0)$ and some $xi_2 in (0, sqrt15)$ such that $f(xi_1) = f(xi_2) = xi_3$.



      Let $K_1 = (x_2, xi_1)$ so $f(K_1) = (x_2, xi_1)$ and let $K_2 = (xi_2, xi_3)$ so $f(K_2) = (xi_2, xi_3)$. Letting $K = (x_2, xi_3) approx (-4.405, 4.405)$, note that $K_1 subset (x_2, 0) subset K$ and $K_2 subset (0, xi_3) subset K$ and that $K_1 cap K_2 = emptyset$. Hence, since $f(K_1) = f(K_2) = K$, the map is indeed a horseshoe.



      Question



      Is my attempt at the solution correct?




      nonlinear-system chaos-theory nonlinear-analysis non-linear-dynamics




      share|cite|improve this question













      A solution is given here:



      Solution



      However, I believe it to be incorrect because $f(K_1) = f(K_2) neq (-sqrt15, sqrt15)$.



      Attempt at a solution:



      Referring to the diagram from the question...



      By the IVT there is some $xi_3 in (sqrt15, 5)$ such that $f(s) = x_2$. Further, by the IVT there is some $xi_1 in (-sqrt15, 0)$ and some $xi_2 in (0, sqrt15)$ such that $f(xi_1) = f(xi_2) = xi_3$.



      Let $K_1 = (x_2, xi_1)$ so $f(K_1) = (x_2, xi_1)$ and let $K_2 = (xi_2, xi_3)$ so $f(K_2) = (xi_2, xi_3)$. Letting $K = (x_2, xi_3) approx (-4.405, 4.405)$, note that $K_1 subset (x_2, 0) subset K$ and $K_2 subset (0, xi_3) subset K$ and that $K_1 cap K_2 = emptyset$. Hence, since $f(K_1) = f(K_2) = K$, the map is indeed a horseshoe.



      Question



      Is my attempt at the solution correct?





      nonlinear-system chaos-theory nonlinear-analysis non-linear-dynamics





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 17 at 17:52
























      asked Jul 17 at 15:43









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