By Singer I.

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Example text

5) 0, since gl(xl,x2) equals the 1st entry in AX['] Eqs. 5) yields = x:. Eliminating s2 from s1 = c - s12 and we must invert the integral 7 = J'""c - fs2' In other words, to obtain xl(t) we specialize the indeterminate s,.

Thus, the point m ( t ) = [q(t),u R ( t ) , uc(t)9i R ( r ) , ic(t)] will move in R5,but be constrained to lie only on M . Hence, the dynamics of the circuit take place only on M (see Fig. 5). By virtue of the fact that q ( t ) determines the circuit, the vector field describing the flow m ( t )is determined by the differential equation dq/dt = -f(q/C). C. If we employ the special choice of controls -1, 0, 1, uz(s)= 1:: 0, t I s < 2t, 0 5 s < t , 2t 5 s I 3 t , 3t I s < 44 O I S < t , t I s < 2t, 3t I s < 44 2t 5 s < 3t, 111 SOME IDEAS FROM DIFFERENTIAL GEOMETRY 43 then a slightly involved calculation shows that + o(t3).

Then we have nxi+ = ( A n)xi 0 or, using our earlier notation i = 0,1,2, .... [xi+l]= A[x,], If dim X = n and dim U = p, then it is easy to verify that A is an (n - p x (n - p) matrix, a reduction of dimension that may be crucial in some practical situations. Let us now examine a slightly more concrete application of these ideas involving quotient sets. , (V where F : X + X and G : U + X are linear maps and X and U are n- and rn-dimensional k-vector-spaces, respectively. , y is the subspace generated by the columns of G and define (Fly) = y + F y + F2y + .