Timeline for Solving a non linear equation
Current License: CC BY-SA 3.0
28 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2014 at 22:36 | comment | added | Pietro Majer | @guest: the conclusion holds true in the generality you said, by the same computation. Write the equation in the form $g(x^a)=g(x^{a+1})$. | |
| Sep 14, 2014 at 23:08 | answer | added | Pietro Majer | timeline score: 19 | |
| Sep 14, 2014 at 18:10 | comment | added | Eckhard | Is there an intuition as to why the fixed point behaves like $K^{-2/3}$ asymptotically? | |
| Sep 14, 2014 at 16:28 | comment | added | freedome321 | @MoritzFirsching In the problem that I face, K is an integer. Whichever makes the proof of uniqueness easier (integer or real) is fine... | |
| Sep 14, 2014 at 16:19 | comment | added | Lucia | If $K$ is reasonably large then the fixed point seems very close to $K^{-2/3}$ -- even for $K=6$ this is a pretty close approximation. Some asymptotic analysis around here (which might be unpleasant but straightforward) should settle the issue -- first show that the fixed points have to be reasonably close to $K^{-2/3}$ and then use the derivative to check uniqueness. | |
| Sep 14, 2014 at 12:44 | comment | added | Moritz Firsching | @MohammadAkbarpour In the case where $K$ is not an integer I don't see how this can be a polynomial (except if $2K$ is an integer of course). | |
| Sep 14, 2014 at 6:28 | comment | added | Manfred Weis | @Noah please accept my apologies; I didn't see the unedited version of the question. | |
| Sep 14, 2014 at 6:24 | comment | added | Noah Schweber | @ManfredWeis, the original version of the question specifically took $K$ to be an integer. | |
| Sep 14, 2014 at 6:12 | comment | added | Suvrit | @NoahS: Ok, I take back my comment; I implicitly assumed a closed interval, while writing an open one. However, it seems that the map $F$ is strictly monotonic... | |
| Sep 14, 2014 at 6:03 | comment | added | Manfred Weis | @Noah as I read the question, there is no restriction of $K$ to integers; it is "a number greater than $1$", which only implies that the numbers must be real (because of the ordering relation); th restriction to natural numbers is an option, if that would make things easier. Using $K$ to denote real numbers is however a bit misleading; maybe replacing it by $t$ would help. | |
| Sep 14, 2014 at 3:12 | comment | added | guest | What about $$ x = \frac{(1 - (1-x^{a+1})^b)^c}{(1 - (1-x^a)^b)^c} $$ with $0<a, 1<b, 1<c$? | |
| Sep 14, 2014 at 2:33 | comment | added | guest | Is this the cdf of a known unimodal distribution on the unit interval? It looks kind of like en.wikipedia.org/wiki/…. | |
| Sep 14, 2014 at 2:13 | history | edited | freedome321 | CC BY-SA 3.0 |
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| Sep 14, 2014 at 2:07 | comment | added | freedome321 | The restriction to integer K's is not important. We can simply relax it if it helps. | |
| Sep 14, 2014 at 1:05 | comment | added | guest | Can you show that it is sigmoidal with f'(0)=0, f'(1)=0, f'(x) increasing between x=0 and x=c for some inflection point c, and f'(x) decreasing between x=c and x=1? Even if c is not at the fixed point this would still convince me. | |
| Sep 13, 2014 at 22:56 | comment | added | Noah Schweber | I'm curious, why the restriction to integral $K$? It'd be interesting to see what happens to the fixed point(s) as $K$ varies continuously. | |
| Sep 13, 2014 at 22:21 | comment | added | Todd Trimble | @MichaelRenardy I agree; despite the fact that the language needed to ask the question is elementary, the problem doesn't look trivial. | |
| Sep 13, 2014 at 22:19 | comment | added | Michael Renardy | Why the votes to close? Is there a simple argument for uniqueness that some of us are missing? | |
| Sep 13, 2014 at 21:13 | comment | added | freedome321 | Yes. I think existence should be doable that way, uniqueness is the trickier part. | |
| Sep 13, 2014 at 21:08 | comment | added | Christian Remling | Existence is easy since $\textrm{RHS}<x$ for small $x>0$ and $\textrm{RHS}>x$ near $x=1$. | |
| Sep 13, 2014 at 20:40 | comment | added | freedome321 | Survit, I think Noah is right. Moritz, do you have any idea of how to rewrite it as a polynomial for general K? Thanks. | |
| Sep 13, 2014 at 20:35 | history | edited | freedome321 | CC BY-SA 3.0 |
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| Sep 13, 2014 at 20:19 | comment | added | Noah Schweber | Suvrit, $x\mapsto x^2$ sends (0, 1) to (0, 1) yet has no fixed point in that interval, so I don't understand your comment. | |
| Sep 13, 2014 at 20:16 | review | Close votes | |||
| Sep 13, 2014 at 21:45 | |||||
| Sep 13, 2014 at 19:53 | comment | added | Moritz Firsching | Maybe rewrite this as polynomial equation and then use Sturm's theorem. It should be easy to get exact symbolic solutions (instead of numerical solutions) this way. | |
| Sep 13, 2014 at 19:38 | comment | added | Suvrit | If $F$ denotes the rhs of your map, then first notice that $F(I) \subset I$, where $I=(0,1)$ --- so existence is easy. You may wish to consider the map $F'$ also... | |
| Sep 13, 2014 at 19:23 | review | First posts | |||
| Sep 13, 2014 at 19:26 | |||||
| Sep 13, 2014 at 19:19 | history | asked | freedome321 | CC BY-SA 3.0 |