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 Apr 6 revised Determine if a matrix can be transformed to a nonnegative matrix edited body Apr 5 asked Determine if a matrix can be transformed to a nonnegative matrix Feb 12 awarded Yearling Feb 11 answered Extrinsic Curvature of Surface of Codimension > 1 Sep 12 awarded Nice Answer Jul 6 comment How to embed $U(1)$ (or other groups) into a bigger group, using Dynkin diagrams @Oscar Maybe the way to think about it is that when you are erasing a node, you are reducing the dimension of the root vector space. So you lose all the root vectors that had a component parallel to the root you erased. However, as you said, the node you erased had a corresponding generator in the maximal torus, which is sitting at the origin in the root vector space (i.e. its weight is the zero vector). So it can be included in the subspace, but will commute with everything else in the subalgebra that is left. Jun 19 answered How to embed $U(1)$ (or other groups) into a bigger group, using Dynkin diagrams Dec 8 awarded Caucus Dec 1 revised Function such that $f(x) f(\pi/2 - x) = 1$ edited tags Dec 1 awarded Scholar Dec 1 accepted Function such that $f(x) f(\pi/2 - x) = 1$ Dec 1 revised Function such that $f(x) f(\pi/2 - x) = 1$ added 454 characters in body Dec 1 comment Function such that $f(x) f(\pi/2 - x) = 1$ @StevenStadnicki okay, I see. However, I have to choose the $C^\infty$ function such that its power series works at the midpoint. Setting all the derivatives to zero is like making it locally the constant function, and I could also locally make it match a $\tan$ function. Maybe I should have asked for how many analytic functions satisfy the requirement, since each analytic function would characterize the type of power series the $C^\infty$ function can have at the midpoint. Dec 1 revised Function such that $f(x) f(\pi/2 - x) = 1$ added 13 characters in body Dec 1 comment Function such that $f(x) f(\pi/2 - x) = 1$ by smooth I mean $C^\infty$. Looking at the second derivative at $x=pi/4$ (or I guess $1$ in your example) I think gives a nontrivial restriction on the function at that point, $f''-(f')^2=0$. Higher derivatives give still more restrictions, and I'm wondering if $tan(x)$ and the constant function are the unique solutions satisfying all these smoothness conditions. Nov 30 comment Function such that $f(x) f(\pi/2 - x) = 1$ I changed the interval to \$0