Sentence examples for have a decomposition from inspiring English sources
Let X be a Banach space and have a decomposition: X = Y + Z where Y and Z are two subspaces of X with dim Y < + ∞.
Moreover, let (P = {text {Cone}}(beta [-1]) cong k oplus I[-1]), so that we have a decomposition Open image in new window (11.5 of the form (2.3) of the diagonal map (k rightarrow k oplus k).
Continuing this operation, finally we have a decomposition begin{array}{*{20}l} (mathbb{Z}/Nmathbb{Z})^{n}&=B^{-1}left((mathbb{Z}/Nmathbb{Z} times{0}^{n-1}right)opluscdotsoplus B^{-1}left({0}^{n-1}right. &left.quadtimes (mathbb{Z}/Nmathbb{Z}){vphantom{{0}^{n-1}}}right) end{array}.
Considering (H = H^{2}(mathbb{R}^{N})) and (mathcal{K} = {u in H^{2}(mathbb{R}^{N}); u geq0}), if it is possible to prove that (mathcal{K}^ subset-mathcal{K}), then we have a decomposition of a function (u in H^{2}(mathbb{R}^{N})) in terms of a non-negative and a non-positive function such that many times we can substitute the trivial decomposition not available in our case.
The functionalized polymer has a decomposition temperature of about 400 °C and exhibits good thermal stability.
We prove that any complex Dunkl polyharmonic function has a decomposition of the form, for all, where are complex Dunkl harmonic functions, that is,.
Conversely, assume that f = ( f n ) n ≥ 0 has a decomposition of the form (1). Let M = sup k ∈ Z 2 k F ˜ − 1 ( P ( τ k < ∞ ) ).
For the converse parts of the proof of Theorem 2.2 assume that f = ( f n ) n ≥ 0 has a decomposition of the form (1) and let λ n = ∑ k ∈ Z χ { τ k ≤ n } ∥ S ( a k ) ∥ ∞.
In (11.50), the operator (widetilde{T}^{({mathcal {C}})}) has a decomposition like (11.51) where ({mathcal H}) is replaced by ({mathcal H}^{({mathcal {C}})}), the functions of the differences of the centers of mass of the (C_j).
