Math 1540 Midterm Exam Solutions
February 22, 2012
Instructions: No books or notes may be used during the exam. Attempt all problems, giving complete and clear explanations of your answers.
1. (15 points) Recall that two norms on a vector space are equiva
Exercises on Dierential Forms
April 14, 2012
1. Stokes Theorem asserts that for any k -chain and any continuously
dierentiable k form on the range of we have
In the special case of a curve : [a, b] Rn , this reduces to
df = f (b) f (a)
for any contin
Solutions to Homework 7
March 31, 2012
Chapter 10 Number 8
Map the unit square I 2 = [0, 1] [0, 1] to R by
(x, y ) = (u, v ) = (1, 1) + u(2, 1) + v (1, 3) = (1 + 2u + v, 1 + u + 3v ).
J = det
By the Change of Variables Theorem,
More Exercises on the Lebesgue Integral
March 16, 2012
1. (a) Let (Ek ) be a sequence of measurable sets satisfying Ek+1 Ek
for all k and (E1 ) < . Show that ( Ek ) = limk (Ek ).
Let E =
Ek . Let Fk = E1 \ Ek and
Fk = E1 \ E.
Supplementary Exercises on the Lebesgue
March 8, 2012
You should solve these problems working directly from the denitions.
Do not appeal to more advanced properties of Lebesgue measure that we
have not proved.
1. In this exercise, you m
Solutions to Homework 3
February 17, 2012
Chapter 9 Number 16
f (0) = lim
and for t = 0
f (t) = 1 + 4t sin(1/t) 2 cos(1/t).
For t (1, 1) we have |f (t)| 7. Let tn = 1/(n ). Then f (tn ) = 3 when n
is odd, and f (tn ) = 1
Solutions to Homework 2
February 3, 2012
Chapter 9 Number 13
Write f = (f 1 , . . . , f n ) so that
|f |2 =
(f i )2 .
Since each f i is dierentiable, so is (f i )2 , and its derivative is 2f i (f i ) . Therefore, |f |2 is dierentiable, with deri
Solutions to Homework 1
January 15, 2012
Supplement Number 1
We will check the three dening properties of a norm.
Positivity: Since is a norm,
|v| = (v) 0,
with equality if an only if (v) = 0. Moreover, since is a linear bijection, we
have (v) =