21-356 Lecture 23

# Fix h r then the segment joining the point x hei with

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Unformatted text preview: and x ! U , there exists B (x, r) ) U . Fix |h| &lt; r, then the segment joining the point x + hei with x is contained in B (x, r). Deﬁne the curve . : [a, b + 1] * RN as follows &amp; - (t) if t ! [a, b] , . (t) := x + (t ' b) hei if t ! [b, b + 1] . Using (ii), we have that A A f (x + hei ) = g = f (x) + ' = f (x) + b A b+1 A j =1 gj (x + (t ' b) hei ) h)ij dt gi (x + (t ' b) hei ) h dt = b = f (x) + N b+1 + h gi (x + sei ) ds, 0 where in the last equality we have used the change of variable s = (t ' b) h. It follows by the mean value theorem that A 1h f (x + hei ) ' f (x) = gi (x + sei ) ds = gi (x + sh ei ) , h h0 where sh is between 0 and h. As h * 0, we have that sh * 0 and so x+sh ei * x. Using the continuity of gi , we have that there exists lim h\$0 f (x + hei ) ' f (x) = lim gi (x + sh ei ) = gi (x) , h\$0 h which proves the claim. The equivalence between (ii) and (iii) is left as an exercise. Next we give a simple necessary condition for a ﬁeld g to be conservative. 61...
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