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Best rational approximants for π (green circle), e (blue diamond), ϕ (pink oblong), (√3)/2 (grey hexagon), 1/√2 (red octagon) and 1/√3 (orange triangle) calculated from their continued fraction expansions, plotted as slopes y/x with errors from their true values (black dashes)
This page was last edited on 1 February 2025, at 08:48 (UTC).; Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply.
The lemma was proved by Bramble and Hilbert [1] under the assumption that satisfies the strong cone property; that is, there exists a finite open covering {} of and corresponding cones {} with vertices at the origin such that + is contained in for any .
In mathematics, approximation theory is concerned with how functions can best be approximated with simpler functions, and with quantitatively characterizing the errors introduced thereby. What is meant by best and simpler will depend on the application.
Approximation is a key word generally employed within the title of a directive, for example the Trade Marks Directive of 16 December 2015 serves "to approximate the laws of the Member States relating to trade marks". [11] The European Commission describes approximation of law as "a unique obligation of membership in the European Union". [10]
This page was last edited on 27 September 2021, at 13:10 (UTC).; Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply.
Any non-linear differentiable function, (,), of two variables, and , can be expanded as + +. If we take the variance on both sides and use the formula [11] for the variance of a linear combination of variables (+) = + + (,), then we obtain | | + | | +, where is the standard deviation of the function , is the standard deviation of , is the standard deviation of and = is the ...
Let (V, ||·||) be a normed vector space, U a subspace of V, and P a linear projector on U.Then for each v in V: ‖ ‖ (+ ‖ ‖) ‖ ‖. The proof is a one-line application of the triangle inequality: for any u in U, by writing v − Pv as (v − u) + (u − Pu) + P(u − v), it follows that