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(Continuation of Example 4.8) Let k = 2.
Denote by Sym D the set of all symmetries of a distribution D. (Continuation of Example 3.4) Let k = 2.
(Continuation of Example 4.9) In this case, we have ∂ x ≡ − p ∂ u − r ∂ p − F ∂ q, ∂ r ≡ 0, ∂ y ≡ − q ∂ u − F ∂ p − t ∂ q, ∂ t ≡ 0, Open image in new window.
(Continuation of Example 2.3) If N is an integral curve of the distribution, then x can be chosen as a coordinate on N, and therefore, N = { ( x, h 0 ( x ), h 1 ( x ), ⋯, h k − 1 ( x ) ) | x ∈ R }. Open image in new window.
(Continuation of Example 2.4) This distribution in not a CID because there is no 4-dimensional integral manifold, and dim D = 4. For, if N be a 4-dimensinal integral manifold of the distribution, then (x,y,u,p) can be chosen as coordinates on N, and therefore, N : q = h ( x, y, u, p ), r = l ( x, y, u, p ), t = m ( x, y, u, p ), s = F ( x, y, u, p, h ).
A continuation of Example 1 gives an illustration.
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Example 4 As a continuation of the previous example, we consider the same model with the same parameters and initial condition.
The process is repeated and stops when a final diagnosis is reached or when new clinical data is not available (see the continuation of the numerical example in section Methods-How the system works-).
As a continuation of Theorem 1.1 and example (2) above, we prove the following result.
Example 4 This example is a continuation of the previous one.
Let me give you one concrete example: the continuation of the Bush tax cuts.
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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