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What is the norm of a complex number? [duplicate] We can define the norm of a complex number in other ways, provided they satisfy the following properties Positive homogeneity Triangle inequality Zero norm iff zero vector We could define a $3$-norm where you sum up all the components cubed and take the cubic root The infinite norm simply takes the maximum component's absolute value as the
What is the difference between the Frobenius norm and the 2-norm of a . . . Frobenius norm = Element-wise 2-norm = Schatten 2-norm Induced 2-norm = Schatten $\infty$-norm This is also called Spectral norm So if by "2-norm" you mean element-wise or Schatten norm, then they are identical to Frobenius norm If you mean induced 2-norm, you get spectral 2-norm, which is $\le$ Frobenius norm (It should be less than or
How are $C^0,C^1$ norms defined - Mathematics Stack Exchange Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
Understanding L1 and L2 norms - Mathematics Stack Exchange The L1 norm is the sum of the absolute value of the entries in the vector The L2 norm is the square root of the sum of the squares of entries of the vector In general, the Lp norm is the pth root of the sum of the entries of the vector raised to the pth power
normed spaces - How are norms different from absolute values . . . The norm you describe in your post, $||\epsilon||=\max|\epsilon_i|$ is a particular norm that can be placed on $\mathbb{R}^n$; there are many norms that can be defined on $\mathbb{R}^n$ The notion of norm on a vector space can be done with any field that is contained in $\mathbb{C}$, by restricting the modulus to that field Added
linear algebra - 2-norm vs operator norm - Mathematics Stack Exchange So every vector norm has an associated operator norm, for which sometimes simplified expressions as exist The Frobenius norm (i e the sum of singular values) is a matrix norm (it fulfills the norm axioms), but not an operator norm, since no vector norm exists so that the above definition for the operator norm matches the Frobenius norm
Definition of $L_\infty$ norm - Mathematics Stack Exchange Hence $\cup_{ p \ge 1 } B_p(0,1)$ is bounded, balanced, convex and contains $0$ in its interior, and so is the unit ball of some norm, in this case, $\|\cdot \|_\infty$ The point here is that there is a natural motivation in terms of the unit balls which 'converge' (in the above sense) to the $\|\cdot \|_\infty$ unit ball
Intuitive explanation of $L^2$-norm - Mathematics Stack Exchange Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
Why is the 2 norm special? - Mathematics Stack Exchange Clearly, then, if we use the 1-norm and the 2-norm more than other vector norms, it's not some arbitrary whim -- these really are God's favorite norms! (He argues that the 2-norm is even more special than the 1-norm ) These lectures eventually became Quantum Computing Since Democritus
1 and 2 norm inequality - Mathematics Stack Exchange I know the definitions of the $1$ and $2$ norm, and, numerically the inequality seems obvious, although I don't know where to start rigorously Thank you analysis