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- Note that $||f(t) - L||$ is a number (see [[Norm]]). &\leq \sqrt{2} {\rm max}\{ |f_1(x) - L_1|, |f_2(x) - L_2| \} {\rightarrow} 0 {\rm (orange)}.34 KB (5,665 words) - 15:12, 13 November 2012
- This number is also called the ''magnitude'' or the ''norm'' of the vector. '''Definition.''' The ''norm'' of a vector113 KB (19,680 words) - 00:08, 23 February 2019
- $$f(c) = M = \max _{x\in [a,b]} f(x),$$ In turn, it implies that the sup-norm is well-defined.19 KB (3,207 words) - 13:06, 29 November 2015
- and, further, in multidimensional calculus the norm<!--\index{norm}--> replaced the absolute value: $$M(x):=\max \{f(x),g(x)\},\ \ m(x):=\min \{f(x),g(x)\}.$$42 KB (7,138 words) - 19:08, 28 November 2015
- $$winner:=\arg\max _{i\in K^{(0)}} a^0(i).$$ ...,y_i$ are the coordinates of $x,y$ with respect to the basis, as well as a norm.47 KB (8,030 words) - 18:48, 30 November 2015
- However, the [[norm]] might vary. Yes on the first one, but only with respect to the [[max norm]]. The derivatives don't converge (and remember the derivative does appear10 KB (1,593 words) - 13:20, 8 April 2013
- '''Exercise.''' What does this construction have to do with the norm of a linear operator? $$winner:=\arg\max _{i\in K^{(0)}} a^0(i).$$45 KB (6,860 words) - 16:46, 30 November 2015
- $$\min a_i \le b\le \max a_i$$ ...combinations of segments do converge as functions (with respect to the sup-norm) but their derivatives don't! Considering that the derivative appears in th21 KB (3,664 words) - 02:02, 18 July 2018
- $$winner:=\arg\displaystyle \max _{i\in K^{(0)}} a^0(i).$$ To find the function to be minimized, we replace the norm with its square and then expand:41 KB (6,942 words) - 05:04, 22 June 2016