Sorry for my stupidity, but I cannot figure out what you are talking about. First, for electrical circuits, whatever you wanted to do with a planar dual is useless if the graph is nonplanar (which for a complicated circuit, it always will be). Second, for trusses, if we have a V-vertex planar graph it has E edges where E<=3*V-6. Now seems to me, if you demand zero vector force at each vertex (2D universe) that is 2 equations per vertex, but you've got E unknown forces. So if E>2*V, such as E=3*V-6, then how are you going to solve for the forces? No comprende. Maybe you have in mind an "outerplanar" graph (all vertices on one face). Then looks like can do. Seems like the solution for general planar graph is going to be highly non-unique but by minimizing elastic energy over all solutions, you'd uniquify. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Sorry for my stupidity, but I cannot figure out what *you* are talking about, either, Warren. Could you please quote what it is you're responding to, or at least paraphrase it in your own words? Thanks, Dan On 2013-05-20, at 1:46 PM, Warren D Smith wrote:
Sorry for my stupidity, but I cannot figure out what you are talking about.
First, for electrical circuits, whatever you wanted to do with a planar dual is useless if the graph is nonplanar (which for a complicated circuit, it always will be).
Second, for trusses, if we have a V-vertex planar graph it has E edges where E<=3*V-6. Now seems to me, if you demand zero vector force at each vertex (2D universe) that is 2 equations per vertex, but you've got E unknown forces. So if E>2*V, such as E=3*V-6, then how are you going to solve for the forces? No comprende. Maybe you have in mind an "outerplanar" graph (all vertices on one face). Then looks like can do. Seems like the solution for general planar graph is going to be highly non-unique but by minimizing elastic energy over all solutions, you'd uniquify.
The following © 2013 Daniel Asimov ---------------------------------- In the complex plane C, consider the differential equation dz/dt = i(z^3-z) for z in C. It's not hard to find an explicit formula for the solution of this for any initial condition z = z(0). ≠≠≠ Putting them all together in one function F, let F(z,t) : = the solution z(t) of [dz/dt = i(z^3-z)] for which z(0) = z. for all t for which F(z,t) is defined. It follows from the form of the solutions that there is a nonempty open set U in C such that (*) for all z in U, we have F(z,pi) = z . But it also follows from the form of the solution that there is another nonempty open set V in C such that (**) for all z in V, we have F(z,pi) ≠ z . But wait, there's more. The Identity Theorem (sometimes known as the Permanence Theorem) in complex variables implies that if two analytic functions (like F(z,pi) and z) are equal on a nonempty open set in C, then they are equal everywhere. So, how can both (*) and (**). Is mathematics inconsistent? --Dan
Dan Asimov wrote:
In the complex plane C, consider the differential equation
dz/dt = i(z^3-z) ... F(z,t) : = the solution z(t) of [dz/dt = i(z^3-z)] for which z(0) = z. ... (*) for all z in U, we have F(z,pi) = z ... (**) for all z in V, we have F(z,pi) ≠ z ... But wait, there's more. The Identity Theorem (sometimes known as the Permanence Theorem) in complex variables implies that if two analytic functions (like F(z,pi) and z) are equal on a nonempty open set in C, then they are equal everywhere.
... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... ... spoiler space ... Why should we assume F(--,pi) is analytic? -- g
(Partly in response to Gareth): Given a differential equation dz/dt = f(z) where f is complex analytic in some open set of C, we know from the existence theorem for such differential equations that for any (z,t) where the function F(z,t) is defined, there is an open set N of CxC = C^2 such that (z,t) lies in N and such that F(z,t) is analytic jointly in z and t (hence in each one separately, holding the other one fixed) in N. --Dan On 2013-05-21, at 1:30 AM, Gareth McCaughan wrote:
Dan Asimov wrote:
In the complex plane C, consider the differential equation
dz/dt = i(z^3-z) ... F(z,t) : = the solution z(t) of [dz/dt = i(z^3-z)] for which z(0) = z. ... (*) for all z in U, we have F(z,pi) = z ... (**) for all z in V, we have F(z,pi) ≠ z ... But wait, there's more. The Identity Theorem (sometimes known as the Permanence Theorem) in complex variables implies that if two analytic functions (like F(z,pi) and z) are equal on a nonempty open set in C, then they are equal everywhere.
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Why should we assume F(--,pi) is analytic?
-- g
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On 21/05/2013 15:16, Dan Asimov wrote:
(Partly in response to Gareth):
Given a differential equation dz/dt = f(z) where f is complex analytic in some open set of C, we know from the existence theorem for such differential equations that for any (z,t) where the function F(z,t) is defined, there is an open set N of CxC = C^2 such that (z,t) lies in N and such that F(z,t) is analytic jointly in z and t (hence in each one separately, holding the other one fixed) in N.
Of course this doesn't actually answer the question I asked :-). -- g
participants (3)
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Dan Asimov -
Gareth McCaughan -
Warren D Smith