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answered What's happening at $a=-1$ in $\int x^a dx$?
Jun
26
comment Solving $2^x - 3^x + 6^x =0$.
there may also be something to be said about the distribution of the real parts
Jun
26
comment Solving $2^x - 3^x + 6^x =0$.
At the moment I'm really not sure how to argue properly for the top and bottom integrals. If nobody chimes in I'll probably get back to it in a few days.
Jun
26
revised Solving $2^x - 3^x + 6^x =0$.
added 14 characters in body
Jun
26
answered Solving $2^x - 3^x + 6^x =0$.
Jun
26
awarded  Nice Answer
Jun
25
awarded  Pundit
Jun
25
comment Is the complement of countably many disjoint disks path connected?
But um.. the $d1$-thickening is an upper bound for the next steps, so it doesn't matter I think if they get close ? Though I have to say, after chasing disks for so long i don't even know anymore what i wanted to do in any of this. I want to add "and disks of radius >= ?? are either outside (K2,d2) or inside K2" in my mid goals, but I've actually no idea what it is i'm doing. Time to curl up and see this tomorrow. And thank you for your diligent reviewing.
Jun
25
revised Is the complement of countably many disjoint disks path connected?
added 573 characters in body
Jun
25
revised Is the complement of countably many disjoint disks path connected?
added 1462 characters in body
Jun
25
comment Is the complement of countably many disjoint disks path connected?
I think $d_0/3$ was meant to be $d_0/4$ everywhere o_o
Jun
25
revised Is the complement of countably many disjoint disks path connected?
added 227 characters in body
Jun
25
comment Is the complement of countably many disjoint disks path connected?
To fix the problem I need to introduce a third step after the constuction of $L_1$ where we look for small disks intersecting two consecutive arc neighbourhoods and then push out a whole corner between the two arcs up to the border of that small disk. I need to control the maximal pushing distance to make sure that it's still summable. We only go up to the border and don't try to thicken it yet. Then we treat it like any other intersecting disk.
Jun
25
comment Is the complement of countably many disjoint disks path connected?
@san : for your 1st comment, we can choose a smaller $e_1$ to avoid being tangent to any disk, because there are countably many disks and uncountably many small reals. for your 2nd comment yes I think your $M$ corresponds to the condition I state right after the choice of $d_1$. Also thanks for noticing the typo. As for your $4$th comment we can instead pick a reunion of $2$ segments that avoid every point in $C$ again because $C$ is countable. But all of this doesn't fix the flaw in the proof.
Jun
25
comment Is the complement of countably many disjoint disks path connected?
oh damn lol, okay so replace the 3rd ball with $B((0,0.5),0.25)$ which should intersect the $0.05$-neighbourhood of the first two balls. And this would make $R_3$ disconnected (and $R_2$ if you scale everything by $2$)
Jun
24
comment Is the complement of countably many disjoint disks path connected?
I don't think $R_2$ is always path-connected. If $\gamma_0$ is a straight segment from $(-2,0)$ to $(+2,0)$ and there are $3$ disks $B((-0.6,0),1), B((+0,6,0),1),B(0,1.4),0.5)$ I can pick $\epsilon_1$ as large as I want, so I can pick $\delta_1 = 1/10$. Then, $J_2$ contains all the disks and if I pick $\epsilon_2$ too small, $R_2$ will not be path connected.
Jun
22
comment Is the complement of countably many disjoint disks path connected?
pfff I spend half an hour writing things very carefully, reread 2 times, post, and poof, five minutes later I think it's flawed.
Jun
22
answered Is the complement of countably many disjoint disks path connected?
Jun
21
comment Periodic orbits of “even” perturbations of the differential system $x'=-y$, $y'=x$
hmm if it's the case shouldn't there be some potential energy function that's invariant ? i tried to write down the differential equation it should have but i'm bad at it and don't see anything i can do
Jun
21
comment The Passare-Tsikh solution to the principal quintic
@TitoPiezasIII : since (unlike the case in your paper) the coefficients all have the same sign, the series converge in a product of disks $D(r_a)*D(r_b)$ iff it converges for $a= +r_a,b = +r_b$. Now it should be easy to show that the convergence domain is delimited by the determinant curve for positive $a,b$ (and not positive $a$ and negative $b$) (also what picture do you want ?)