pl answer EACH part Let f: U rightarrow R^m be differentiable, [p, q] subset U s
ID: 3109167 • Letter: P
Question
pl answer EACH part
Let f: U rightarrow R^m be differentiable, [p, q] subset U subset R^n, and ask whether the direct generalization of the one-dimensional Mean Value Theorem is true: Does there exist a point theta elementof [p, q] such that f(q) - f(p) = (Df) theta (q - p)? (a) Take n = 1, m = 2, and examine the function f(t) = (cos t, sin t) for pi lessthanorequalto t lessthanorequalto 2 pi. Take p = pi and q = 2 pi. Show that there is no theta elementof [p, q] which satisfies. (b) Assume that the set of derivatives {(Df)_x elementof L(R^n, R^m): x elementof [p, q]} is convex. Prove there exists theta elementof [p, q] which satisfies. (c) How does (b) imply the one-dimensional Mean Value Theorem?Explanation / Answer
(a) f(x)=(cosx,sinx) on [,2].
f'(x) = (-sinx,cosx)
take q=2 , p=
f(q) = f(2) = (1,0) and f(p)=f()=(-1,0)
q-p = , f'(q-p)=f'()=(0,-1)
By Equation (28),
(2,0)=f(q)f(p)=f()=(sin,cos) for some (,2).
This is impossible, since the magnitude of the right-hand side is , while the magnitude of the left-hand side is 2.
(b) Let f:UR be a differentiable function (U is an open subset of Rn). Let a and b be points in U such that the entire line segment between them is contained in U. Define h:[0,1]U in the following way:
h(t)=(a1+(b1a1)t,…,an+(bnan)t). ///So that set of derivatives is convex
This function is differentiable on (0,1)and continuous on [0,1], so is fh. If we apply Mean Value Theorem to fh we get
(fh)(c)=(fh)(1)(fh)(0)
where c(0,1) and
f(h(c))(ba)=f(b)f(a).
If we set =h(c) we get
f(b)f(a)=f(c)(ba)
for some c[a,b]
Hence,Proved
(c) If we compare the result in part (b) to one dimensional mean value theorem:
f(b)f(a)=f(c)(b-a) for c(a,b)
we can see it implies MVT for higher dimensions.
DO THUMBS UP ^_^
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