rngantner
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 Jan24 comment Minimum number of function evaluations to numerically estimate gradient The gradient is defined as the vector of the $N$ partial derivatives, so you have to estimate all $N$ values to get the complete gradient. As far as I know, there does not exist an estimate that only requires one point per partial derivative, so combining forward (or backward) differences yields the least number of evaluations, $N+1$. However, if you have additional information on the structure of the gradient you might get away with less, but not in general. Jan23 awarded Teacher Jan23 answered Minimum number of function evaluations to numerically estimate gradient Nov20 comment Is the set of points of equal distance to the surface of an ellipsoid again an ellipsoid? Thanks for the references.. I'll have a look at them, maybe there's something there I can use. Nov19 comment Is the set of points of equal distance to the surface of an ellipsoid again an ellipsoid? after computing the distance in the 2d case by just inserting into the formula of an ellipse this is apparently not true. But how could an expression for the error (distance of point on $A'$ to point exactly $\varepsilon$ from A) be derived in the general case ($d$ dimensions)? Nov19 asked Is the set of points of equal distance to the surface of an ellipsoid again an ellipsoid? Nov12 comment Tight bounds for harmonic measure Thanks for the comment- I managed to solve my problem using the book and some references. There's a chapter on estimating Harmonic Measure in there. Nov7 awarded Student Nov7 asked Tight bounds for harmonic measure