New PDF release: [Article] Why does neurorehabilitation fail?

By M. D. van den Broek

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Extra info for [Article] Why does neurorehabilitation fail?

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Matrix inside always lies in B (see appendix). C a s e 3 : w — wu For notational convenience, we set S = 7r(A^)^(wn).

W I 1 mod Vh -TW I I \ since m a x X > m a x W and the constraint Z € Vh h x lie in V ~ . , h Multiplying by Y then puts them in V . Thus, we get \$(u ) = §(uxuzuYuw) (( ! = \$ 1+ IZ tiytijy (1 + XZ)-1 = \$ \l + XZ (1 + XZ)" Uy^W Chris Jantzen 26 since XZ G Vh~\ (1 + XZ)~XX = I m o d Vh. Thus, (i + xz)- 1 \$( w -) = \$ X 1 + XZ Y W V x-^i + x^^Cu-) The rest of the argument is similar to case A. As Z runs over Vh h x 1 runs over V ~ . Then, \$ ( « -) = x " ^ ^ ^ ^ ) ^ ! ^ " ) f o r a11 s u c n l *, XZ ^ implies \$(u") = 0, 1 1 since x is not constant on 1 + P' " .

Degenerate Principal construct a basis for VB*. Series 15 The problem breaks itself into three cases. 4. The reader is advised that the third case is substantially longer than the other two. We start with the following lemma, which is sort of a Bruhat decomposition. We let B be the parahoric subgroup corresponding to P (obtained by adjoining 5 2 , . . , sn to the Iwahori). 1 Let w00 = identity, I ( 1. u>io '1 •1 wn '1 \ 0 \ (wio = si). Note that for n = 1, u>10 does not exist. K = B U (Bw10B) U (BwuB) Then, for n > 2 and K = B U (BwnB) Proof for n = 1.