Timeline for Is there a systematic procedure to Solve Abel's, Böttcher's, or Schröder's Equation
Current License: CC BY-SA 4.0
13 events
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| Aug 30, 2023 at 22:15 | comment | added | Sidharth Ghoshal | Ahhh typo, $\phi, 1-\phi$ should be $\frac{1\pm i\sqrt{3}}{2}$ respectively…. Which are the negatives of the two nontrivial cubic roots of unity. Again estimating $|L’|$ gives us that both points are attractors and so a candidate $H=(x^2-x+1)e^{-x^2}$ still works! | |
| Aug 30, 2023 at 22:01 | comment | added | Sidharth Ghoshal | In general you want to take the set of fixed points of the Riemann sphere under the maps $g$ or $g^{-1}$ and let $H$ go to 0 rapidly for those points. | |
| Aug 30, 2023 at 21:57 | comment | added | Sidharth Ghoshal | ONE of them must be a attractor, and so letting H go to 0 for that fixed point along with infinity gives you a function which will approach converge at least for a decent chunk of the complex plane (though maybe not the whole thing). $H = (x^2 - x + 1)^2e^{-x^2}$ then seems like a good place to start. | |
| Aug 30, 2023 at 21:54 | comment | added | Sidharth Ghoshal | So it’s easy to see that $g = x^2+1$ tends to infinity when applied so $H$ should go to 0 at infinity. $L = g^{-1} = \sqrt{x-1}$ is quite a bit trickier. This has two potentially two fixed points $x_0 = \phi$ and $x_1 = 1-\phi$ where $\phi$ is the golden ratio (which can be seen by solving $\sqrt{x-1} = x$. Using a result from here: math.stackexchange.com/questions/421269/… we can check if $|L’(x_i)| < 1$ for either point. Wolfram Alpha confirms both points satisfy that so up to convention with how you define the square root | |
| Aug 30, 2023 at 18:48 | comment | added | Anthony Corsi | One last thing, I was playing around with this method to solve various Schröder equations, one I ran into was f(x^2+1)=sf(x). Trying to construct H that met the criteria was exceedingly hard. After reviewing the problem I noticed that the infinite iteration of the inverse function sqrt(x^2-1) (we'll call this function G) was undefined over R. I then considered G as a holomorphic function, but even then I could not figure out where it mapped the reals or if it even converged to a complex number. In a case like this how does one construct H or figure out G? -(ps thanks for all the help) | |
| Aug 30, 2023 at 10:47 | comment | added | Sidharth Ghoshal | See the example I just added, we derive that any $H$ that goes to 0 quickly at $0,1,\infty$ gives us a solution and so we sort of manufacture an $H$ but its hardly "solving" its basically just making an $H$ that meets those criteria. | |
| Aug 30, 2023 at 10:46 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
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| Aug 30, 2023 at 10:17 | comment | added | Sidharth Ghoshal | You don't really solve for $H$, the space of $H$'s is very large, you just need to find ANY $H$ that decays to 0 fast enough | |
| Aug 29, 2023 at 22:40 | comment | added | Anthony Corsi | I understand, my main question is given an arbitrary g how does one solve for H | |
| Aug 29, 2023 at 17:02 | comment | added | Sidharth Ghoshal | The $H$ is selected to make the series converge. Formally speaking (without worrying about convergence) you can do whatever you want including foregoing the use of $H$ but you usually want to be able to graph / talk about your function over R or C. I will include some examples later to explain further | |
| Aug 29, 2023 at 16:31 | comment | added | Anthony Corsi | This was very helpful, but there a few things I don't understand. In Schröder's Equation how does one find the H function used in the series for f(x)? Also why is the H function needed considering that without it the series representation of f(x) would still satisfy the Schröder Equation? | |
| Aug 29, 2023 at 8:09 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
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| Aug 29, 2023 at 8:02 | history | answered | Sidharth Ghoshal | CC BY-SA 4.0 |