Timeline for What is the attracted locus in this recursion?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2017 at 7:05 | comment | added | Bobby Shen | That definitely describes the recurrence S, thank you. Unfortunately, T has four eigenvalues of 1 and is likely to be much harder | |
| Jan 11, 2017 at 3:46 | comment | added | Vaughn Climenhaga | I didn't check the equations in your question, but the curve you're describing is usually called a "local stable manifold". Assuming the fixed point at the origin is hyperbolic (the Jacobian matrix has one eigenvalue inside the unit circle and the other outside) then it will indeed be a curve; this is the Hadamard-Perron theorem. With those keywords you should be able to find some literature on how to compute it. | |
| Jan 10, 2017 at 23:17 | history | edited | Bobby Shen | CC BY-SA 3.0 |
added 220 characters in body
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| Jan 10, 2017 at 22:59 | history | edited | Bobby Shen | CC BY-SA 3.0 |
edited body
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| Jan 10, 2017 at 22:58 | comment | added | YCor | Julia sets might suggest that your set can be far from the set of zeros of a reasonable equation. Since convergence is fast, you could at least draw a picture of your set: maybe it will clearly show off fractal behavior. | |
| Jan 10, 2017 at 22:40 | review | Close votes | |||
| Jan 10, 2017 at 23:48 | |||||
| Jan 10, 2017 at 22:15 | history | asked | Bobby Shen | CC BY-SA 3.0 |