can some one tell me the steps to find a cricles center x:y from 2 points and its radius... i calculated the straight line distance of point1 and point2 but don't know what i have to calculate next?
So, I was playing around with a question in game theory, and ended up proving a statement about automorphisms of compact metric spaces which I can't think of an elementary way of proving.
Let $M$ be a compact metric space. $M$ can not admit both an order-$2$ automorphism with a unique fixed point, and an order-$2$ automorphism with no fixed points.
assuming I haven't made any mistakes with my game theory proof anyway
click "launch" when the two objects are apart a bit and it should draw a curved line joining them but im calculating the red circle's center in the wrong place
ah ok well basically ive got distance of two points that are on the circle's circumfrence and the circles radius and in order to draw the circle on the screen i need to know the circle's center x:y
i dont know what step i do once i calculate the distance between the two points
Hm, let's see if I can find something non trivial and non easy to solve...
LOL at this one:
Let $S$ be the surface given the graph of $f(x,y)=(1+x^2+y^2)^{-1}$ over $\lVert (x,y)\rVert \leqslant 1$, and let $${\bf F}(x,y,z)=\left(\frac{zx}{x^2+y^2},\frac{zy}{x^2+y^2},0\right)$$ Find $\displaystyle \int_S {\bf F}d{\bf S}$. Think before acting.
Find a sequence of closed, connected subsets $C_1, C_2, \dots$ of $\mathbb{R}^2$ such that $C_1$ contains $C_2$ which contains $C_3$ (and so on) and $\bigcap_{n = 1}^{\infty} C_n$ is not connected. Isn't this also proving a point isn't connected?
(I've read a good chunk of Gunning's first volume, but considering a main goal is to prove Riemann-Roch, I'd think it more an algebraic geometry text.)
@Don: you're working with all closed sets, so I don't see the relevance. @Pedro is hinting tgat you want to keep pushing the intersections further and further out.
Pick A,B disjoint and closed so that the intersection is their union. Pick A', B' disjoint and open that contain the respective closed sets (we agree you can do this?)
@Sanchez now that I'm on a computer and not artificially shortening what I write: I meant to say "Given a descending sequence of closed, connected subsets of a compact space, i.e., $C_1 \supset C_2 \supset ... \supset C_i \supset ...$, do we have that $\cap_{i=1}^\infty C_j$ is connected?"
Oh, @TedShifrin: All compact Hausdorff spaces are normal
hey guys, just going over some notes and my prof used an inequality that I can't seem to prove. The inequality is as follows: $$ |x|^{2N} \leq n^N \sum_{\gamma = N}|x^\gamma|^2$$ where $\gamma$ is a multiindex, and $x \in \mathbb{R}^n$.
Sanchez, I proved the statement above by picking some disconnected open sets that contained $A$ and $B$ where $A \cup B$ is the intersection of that sequence (and $A \cap B = \emptyset$)
