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Apr 17, 2019 at 17:09 comment added Phillip Do you know any reference for this formula in the language of tensors?
Apr 17, 2019 at 14:16 comment added Dan Fox @Phillip: indeed, a $\det h$ was missing from the right-hand side. Corrected now.
Apr 17, 2019 at 14:16 history edited Dan Fox CC BY-SA 4.0
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Apr 17, 2019 at 8:52 comment added Phillip Do you know any reference where the formula for $\nabla \det(h)$ is proven?
Apr 17, 2019 at 8:41 comment added Phillip Is this formula $\nabla \det (h) = h^{-1}\nabla h$ correct? Shouldn't it be $\nabla \det (h) = \det(h) \text{tr}(h^{-1}\nabla h)$ ?
Apr 15, 2019 at 14:38 history edited Dan Fox CC BY-SA 4.0
added 20 characters in body
Apr 15, 2019 at 10:44 history answered Dan Fox CC BY-SA 4.0