MATH 132, FALL 2009
HOMEWORK 9 SOLUTIONS
1. (Sec. 4.1, ex. 1) The matrix of L is
[L] =
Since det
1 15
.
2 0 1
11
= 2 = 0, L has rank 2. The kernel of L can be computed by solving the
2 0
system
x1 + x2 + 5x3 = 0
2x1 + x3 = 0.
All solutions are scalar mult
MATH 132, FALL 2009 HOMEWORK 1 SOLUTIONS
1. (Sec. 1.2, ex. 5) Starting with the right-hand side, we obtain: |x + y |2 + |x y |2 = (x + y ) (x + y ) + (x y ) (x y ) = |x|2 + 2x y + |y |2 + |x|2 2x y + |y |2 = 2|x|2 + 2|y |2 . This means: the sum of the squ
MATH 132, FALL 2009
HOMEWORK 4 SOLUTIONS
1. (Sec. 2.5, ex. 6) () Suppose f is continuous at x0 and let (xm ) be a sequence in S converging to
x0 . Let us show that f (xm ) f (x0 ), as m . Let > 0 be arbitrary. Since f is continuous
at x0 , there exists >
San Jos State University
e
Math 132, Fall 2009
Final Exam Solutions
Assigned on December 7, 2009
Due on December 16, 2009 by 2 PM
Happy holidays and enjoy your winter break!
Name:
Granwyth Hulatberi
Score
1
25
2
25
3
25
4
25
5
25
6
25
Total 150
You are al
MATH 132, FALL 2009 HOMEWORK 3 SOLUTIONS
1. (Sec. 2.2., ex. 7(a) Let > 0 be arbitrary. Since f is continuous at x0 and f (x0 ) = 0, there exists > 0 such that |f (x) f (x0 )| = |f (x)| < 1/a , for all |x x0 | < . This implies that |g (x) g (x0 )| = |f (x)
MATH 132, FALL 2009
HOMEWORK 12 SOLUTIONS
Convention: We will say that f satises some property P almost everywhere if the set of points on
which P does not hold has measure zero.
(1) Let Q be a rectangle in En and assume that f : Q E1 is integrable.
(a) S
MATH 132, FALL 2009
HOMEWORK 5 SOLUTIONS
1. (Sec. 3.1, ex. 1) (a) If f (x, y ) = x log(xy ) (xy > 0), then
1 f (x, y ) = log(xy ) + 1,
2 f (x, y ) =
x
.
y
(b) If f (x, y, z ) = (x2 + 2y 2 + x)3 , then
1 f = 6x(x2 + 2y 2 + x)2 ,
(c) For f (x) = x x =
2
i x
MATH 132, FALL 2009
HOMEWORK 7 SOLUTIONS
1: (Sec. 3.4, ex. 7) (a) Showing 12 f (x, y ) = 21 f (x, y ), for (x, y ) = (0, 0) amounts to a calculation,
which is left as an exercise to the reader.
(b) For any (x, y ) = (0, 0), we have:
1 f (x, y ) =
x4 y + 4
MATH 132, FALL 2009
HOMEWORK 8 SOLUTIONS
1. (Sec. 3.3, ex. 4) (a) Let f (x) = x0 x. Assuming that x0 = (a1 , . . . , an and x = (x1 , . . . , xn ), we
obtain f (x) = i ai xi . Therefore, i f = ai , so the gradient of f is
f (x) = a.
x2 + + x2 . Then for x
San Jose State University
Math 132, Fall 2009
Midterm 1 Solutions
Assigned on October 1, 2009, due October 8, 2009.
Name:
Granwyth Hulatberi
Score
1
25
2
25
3
25
4
25
Total 100
Explain your work
1. (25 points) Find the limit of f at the origin if it exis
MATH 132, FALL 2009
HOMEWORK 12, DUE DECEMBER 3, 2009
(1) Let Q be a rectangle in En and assume that f : Q E1 is integrable.
(a) Show that if f (x) 0, for all x Q, then Q f 0.
(b) Show that if f (x) > 0 for all x Q, then Q f > 0.
(2) Suppose that f : [a,
MATH 132, FALL 2009
HOMEWORK 10 SOLUTIONS
1. (Sec. 4.4, ex. 2) If F (x, y ) = f (x, y, g (x, y ) and f and g are C 2 , then (thinking of f as a function
of x, y and z ):
Fx = fx + fz gx ,
Fy = fy + fz gy .
Dierentiating again, we obtain:
Fxx = fxx + fxz g
MATH 132, FALL 2009
HOMEWORK 6 SOLUTIONS
1. (Sec. 3.3, ex. 8) () Suppose f dierentiable away from the origin and homogeneous of degree
p, i.e., f (tx) = tp f (x), for all x = 0 and t > 0. Then dierentiating at t = 1 and using the
(current version of) the
BANACHS FIXED POINT THEOREM
Solving equations is, arguably, the most important problem in mathematics. Equations can often be
written in the form
F (x) = 0,
(1)
where F is some map. How does one solve (1)? While there is no general answer that works for a