MATH 425b SAMPLE MIDTERM 2 SOLUTIONS SPRING 2008 Prof. Alexander (1) Let (t) = ty + (1 - t)x = x + t(y - x) for 0 t 1, and g(t) = f (t). By the chain rule, g (t) = f (t) (t) = f (t)(y - x). By the usual MVT for function on R, f (y) - f (x) = g(1) -
MATH 425b SAMPLE MIDTERM 1 SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) The problem does not actually say if we are dealing with real or complex-valued functions (though C[0, 1] implicitly means real-valued), so I will give solutions for both cases.
MATH 425b IN-CLASS FINAL EXAM SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) We calculate d = 2z dx dy dz, all other terms being 0. By Stokes Theorem, =
A A
d.
We can parametrize A by the identity so
a a -a a a a -a a
d =
A -a -a
z dx dy dz =
-a
MATH 425b ASSIGNMENT 9 SOLUTIONS SPRING 2007 Prof. Alexander Chapter 10: (16) Let ij = [p0 , ., pi-1 , pi , ., pj-1 , pj+1 , ., pk ] (the simplex with pi , pj missing.) Then
k
=
i=0 k
(-1)i [p0 , ., pi-1 , pi+1 , ., pk ],
=
i=0
2
(-1)i
j<i
(-
MATH 425b ASSIGNMENT 7 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 9: (23) f (x, y1 , y2 ) = x2 y1 +ex +y2 . Clearly f (0, 1, -1) = 0. We have (D1 f )(x, y1 , y2 ) = 2xy1 +ex so (D1 f )(0, 1, -1) = 1 = 0. In the notation of the Implicit Function Th
MATH 425b ASSIGNMENT 6 SOLUTIONS SPRING 2007 Prof. Alexander Chapter 9: (7) Suppose E R is open, f : E R, and there exists M such that |(Dj f )(x)| M for all x E and all j M . Let x E and let B be a neighborhood of x with B E. Suppose x + h B
MATH 425b ASSIGNMENT 5 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 9: (3) If x = y then Ax - Ay = A(x - y) = 0, that is, Ax = Ay. This shows A is one-to-one. (4) Let T L(X, Y ), and let z Z = range(T ) = {y Y : y = T x for some x X}. We show Z
MATH 425b ASSIGNMENT 4 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 8: (9)(a) Define f on [0, ) by f (x) = 1/n for x [n - 1, n), for all n 1. Then f (x) Also sN =
1 N +1
1 f (x - 1) for all x 1. x
N N
(1)
f (x - 1) dx =
0
f (x) dx = 1 +
1
MATH 425b ASSIGNMENT 3 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7 (20) Let > 0. There exists a polynomial with |P - f | < (sup norm), say P (x) = N n n=0 cn x . f is bounded since it is continuous on the compact set [0, 1], so there exists M suc
MATH 425b ASSIGNMENT 2 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7 (8) Since |cn I(x - xn )| |cn | and |cn | < , the series defining f converges uniformly by the Weierstrass M -test (Theorem 7.10.) Letting fn (t) = n cn I(x - xn ), this means th
MATH 425b ASSIGNMENT 1 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7 (1) Suppose fn f uniformly. Let > 0. There exists an N such that |fN - f | < . Since fN is bounded, there exists an MN such that |fN (x)| MN for all x. Then for all x, |f (x)|
MATH 425b TAKE-HOME FINAL EXAM SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) Let (t) = pk + t(pk - pk ), t [0, 1], which traces a line from pk to pk . Let t = [p0 pk-1 (t)] be the corresponding simplex. Since these points are on opposite sides on