Timeline for Convergence of a sequence by iteration
Current License: CC BY-SA 4.0
16 events
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| Aug 4, 2018 at 9:01 | comment | added | Mateusz Kwaśnicki | @MB2009: Oh, I do not quite remember my answer now! I think the construction works for any (small) $c > 0$, and we eventually choose it small enough, so that the bound for second derivative works as desired. | |
| Aug 1, 2018 at 9:48 | comment | added | user111097 | I have one more questions. In Step 6, you claim that we may choose arbitrarily small $c$, but could we ensure that, for any $c>0$, the iterated point $P-\nabla G(P)$ can always be lying on the spiral? My intuition is that, $\nabla G(P)$ denotes the step that is proportional to $c$. So if $c$ is too small, we might have a step which is not long enough to touch the spiral again. Thanks a lot! | |
| Jul 9, 2018 at 22:04 | comment | added | Mateusz Kwaśnicki | @MB2009: If $\hat{r}$ and $\hat{t}$ are the unit vectors in polar coordinates and a point with coordinates $(r, t)$ is translated by $\alpha \hat{r} + \beta \hat{t}$, then the resulting point has coordinates $(r', t')$ such that $r' \cos(t' - t) = r + \alpha$, $r' \sin(t' - t) = \beta$. This is easy to see once you draw a picture. In our case $\alpha$ and $\beta$ are given as the derivatives of $-G$ with respect to $r$ and $t$, respectively (the latter one multiplied by $r$). | |
| Jul 8, 2018 at 21:51 | comment | added | user111097 | Thank you so much for such a detailed proof. It remains one thing that is not clear: Our condition reads $r' \cos(t' - t) = r - \partial_r G(P)$ , $r' \sin(t' - t) = -r \partial_t G(P)$. Why $P'=P-\nabla G(P)$ implies that? I remember that you confirm $\nabla$ corresponds to Cartesian coordinate. Why do you use the partial derivatives w.r.t. $r$ and $t$? Thanks again! | |
| Jul 8, 2018 at 21:00 | comment | added | Mateusz Kwaśnicki | @MB2009: I edited this part. Is it any better now? | |
| Jul 8, 2018 at 20:59 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
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| Jul 6, 2018 at 20:52 | comment | added | user111097 | I'm still confused with the definition of $G$, could you please specify its construction with more details? | |
| Jul 6, 2018 at 14:55 | comment | added | user111097 | Thanks for the quick reply. As for 1), it seems that $G((1+e^{-t})e^{it})-\nabla G((1+e^{-t})e^{it})$ is not a point as $G((1+e^{-t})e^{it})$ is real value and $\nabla G((1+e^{-t})e^{it})$ takes values in $\mathbb R^2$. Is there an error here? Also, could you please explain the meaning of "lying on the same spiral (1+e^{-s})e^{is}"? | |
| Jul 6, 2018 at 14:50 | comment | added | Mateusz Kwaśnicki | @MB2009: 1) Yes, polar coordinates are just to simplify the definition of $G$. I will try to improve the terrible notation in my answer, but I have no time for this now, sorry. 2) Yes, this is the usual $(x, y)$ gradient. 3) Yes (and there is a gap between $r \ge 10$ and $r \le 1 + e^{-t + 2 \pi}$ to be filled in some way). I'll try to include a picture at some point. | |
| Jul 6, 2018 at 14:02 | comment | added | user111097 | 3) The function $G$ is constructed by distinguishing $r\le 1$, $1+e^{-t}\le r< 1+2e^{-t}$, $1+2e^{-t}\le r< 1+e^{-t+2\pi}$ and $r\ge 10$. Is it correct? | |
| Jul 6, 2018 at 13:57 | comment | added | user111097 | 2) Here the polar coordinate is applied, is the gradient $\nabla G$ still corresponding to $(x,y)$ instead of $(r,t)$? | |
| Jul 6, 2018 at 13:53 | comment | added | user111097 | Thanks a lot for the prompt reply. I'm reading your constructed example, and have some questions: 1) What does "the point $G((1+e^{-t})e^{it})-\nabla G((1+e^{-t})e^{it})$ also lies on the same spiral $(1+e^{-s})e^{is}$" mean? | |
| Jul 6, 2018 at 9:30 | comment | added | Mateusz Kwaśnicki | @MB2009: I meant discrete (that is, with no accumulation points). However, since we know in advance that $\nabla F$ is bounded, this does not really make any difference: all critical points of $\nabla G$ are contained in a bounded set, and therefore "discrete" is equivalent to "finite". I added some explanation regarding connectedness of the set of accumulation points. | |
| Jul 6, 2018 at 9:27 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
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| Jul 6, 2018 at 8:42 | comment | added | user111097 | Thank you very kindly for the reply, and I find it a very elegant answer. Could you please explain a bit more where the set of accumulation points of $(x_n)$ is connected? Also, I'd like to confirm in your claim that, the set of critical points is discrete (NOT FINITE). Thanks again! | |
| Jul 5, 2018 at 19:19 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |