How to make a simul • page 1/2 • Off-Topic Discussion • lichess.org

agg144

#1

How to make a simul? Does someone know? Or is it only for titled players cause i only have seen them hosting simuls

Feuer1

#2

@agg144 On the Simul page

agg144

#3

@Feuer1 where can i find this page?

Feuer1 edited

#4

@agg144 https://lichess.org/simul Or Play und then Simulous exhibition

krasnaya edited

#5

@agg144: under "PLAY" in the top menu is a menu entry "Simultaneous exhibitions" how much clearer labelled should that be, hm?

Furthermore, this is the "off-Topic" forum, if you have a questiona about Lichess, go to the respective forum and look there.

Also, there is a search function and how about - instead of asking a question answered for the umpteenth time - looking at these answers instead of pestering us with the result of your lazyness?

At last: mostly titled players host simuls because they - unlike you - are strong enough to take on several others at the same time. With the playing level according to your rating you should play in a simul someone else hosts, not host one yourself.

krasnaya

@agg144: under "PLAY" in the top menu is a menu entry "Simultaneous exhibitions" how much clearer labelled should that be, hm? Furthermore, this is the "off-Topic" forum, if you have a questiona about Lichess, go to the respective forum and look there. Also, there is a search function and how about - instead of asking a question answered for the umpteenth time - looking at these answers instead of pestering us with the result of your lazyness? At last: mostly titled players host simuls because they - unlike you - are strong enough to take on several others at the same time. With the playing level according to your rating you should play in a simul someone else hosts, not host one yourself. krasnaya

RiKo2020_ChEsS edited

#6

Click https://lichess.org/simul. Then, click the blue button that says, "Host a new Simul." By the way, https://lichess.org/@/krasnaya you were pretty mean and could have been nicer.

. . . .

first click on this link https://lichess.org/simul. or hover on play and click Simultaneous exhibitions and click host a simul.

Alexander_501_Q300

#9

It's actually quite a hard thing to do, here's a break down:

For every closed piecewise C1 curve in D must be holomorphic on D.

The assumption of Morera's theorem is equivalent to f having an antiderivative on D.

The converse of the theorem is not true in general. A holomorphic function need not possess an antiderivative on its domain, unless one imposes additional assumptions. The converse does hold e.g. if the domain is simply connected; this is Cauchy's integral theorem, stating that the line integral of a holomorphic function along a closed curve is zero.

The standard counterexample is the function f(z) = 1/z, which is holomorphic on ℂ − {0}. On any simply connected neighborhood U in ℂ − {0}, 1/z has an antiderivative defined by L(z) = ln(r) + iθ, where z = reiθ. Because of the ambiguity of θ up to the addition of any integer multiple of 2π, any continuous choice of θ on U will suffice to define an antiderivative of 1/z on U. (It is the fact that θ cannot be defined continuously on a simple closed curve containing the origin in its interior that is the root of why 1/z has no antiderivative on its entire domain ℂ − {0}.) And because the derivative of an additive constant is 0, any constant may be added to the antiderivative and it's still an antiderivative of 1/z.

In a certain sense, the 1/z counterexample is universal: For every analytic function that has no antiderivative on its domain, the reason for this is that 1/z itself does not have an antiderivative on ℂ − {0}.

It's actually quite a hard thing to do, here's a break down: For every closed piecewise C1 curve in D must be holomorphic on D. The assumption of Morera's theorem is equivalent to f having an antiderivative on D. The converse of the theorem is not true in general. A holomorphic function need not possess an antiderivative on its domain, unless one imposes additional assumptions. The converse does hold e.g. if the domain is simply connected; this is Cauchy's integral theorem, stating that the line integral of a holomorphic function along a closed curve is zero. The standard counterexample is the function f(z) = 1/z, which is holomorphic on ℂ − {0}. On any simply connected neighborhood U in ℂ − {0}, 1/z has an antiderivative defined by L(z) = ln(r) + iθ, where z = reiθ. Because of the ambiguity of θ up to the addition of any integer multiple of 2π, any continuous choice of θ on U will suffice to define an antiderivative of 1/z on U. (It is the fact that θ cannot be defined continuously on a simple closed curve containing the origin in its interior that is the root of why 1/z has no antiderivative on its entire domain ℂ − {0}.) And because the derivative of an additive constant is 0, any constant may be added to the antiderivative and it's still an antiderivative of 1/z. In a certain sense, the 1/z counterexample is universal: For every analytic function that has no antiderivative on its domain, the reason for this is that 1/z itself does not have an antiderivative on ℂ − {0}.

RiKo2020_ChEsS

#10

What do you mean @Alexander_501_Q300