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(vi) Finally, let x and y be arbitrary elements of (C I,mathbb {R})) and (z:=max{x,y}).
Proof Let v and z be arbitrary elements of D ( y ε θ, θ ) ⊂ H 0 1.
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Assume that (vec{f}(x)) and (vec{r}(x)) are arbitrary elements of (L_{2} (Omega )).
end{aligned} Since u was arbitrary element of X in (3.13), we obtain omega^{T}bigl(F X),varepsilonbigr) leq omega^{T}(p,varepsilon)+frac{1}{(n-1-k)!} bigl[varepsilon T^{n}U_{r_{0}}^{T}+phi^{T}( varepsilon TU_{r_{0}}^{T} bigr].
Let and be arbitrary elements in and a constant.
Let x, y ∈ E 1 be arbitrary elements.
Let (x=(x_{n})) and (y= y_{n})) be two arbitrary elements of the space (c^{I}(hat{F})), and let α, β are scalars.
And let be any continuous linear functional, that is, is an arbitrary element of the dual space of.
Therefore, K commutes with every subgroup of H. Let x be
This is a simple fact because we may take to be the identity operator: Then, we may take to be an arbitrary element of with and define by induction (3.1).
In this section, we will define two transformations for the points and lines of P K 2 ( Q and also we will show that these transformations are collineations. Similar transformations can be found in [6]. Let a = a 1 + a 2 ε be an arbitrary element of Q.
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.
Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com