Solutions to Homework for Weeks 13-14 - Math 1540 - Spring, 2004 - Rubin
Wade, pg. 454, # 2c. The set y = |x| makes 2 cuts through the cylinder, one along y = x
for x > 0 and another along y = x for x < 0. On both pieces, x2 + 3z 2 = 1, and
z changes sign
Solutions to Homework for Week 1 - Math 1540 - Spring, 2004 - Rubin
pg. 317, #8. Pointwise convergence of continuous functions on a compact set to a
continuous limit DOES NOT imply uniform convergence on the set. In fact, we saw
a counterexample in Sectio
Solutions to Homework for Week 2 - Math 1540 - Spring, 2004 - Rubin
pg. 283, #8. a) Let : M M such that d(x), (y ) < d(x, y ) for all x = y M ,
where M is compact. Given > 0, if d(x, y ) < /2, then d(x), (y ) < /2. Now,
note that the function d(x), x) is
Solutions to Homework for Week 3 - Math 1540 - Spring, 2004 - Rubin
Section 5.8, #1. Since f is a continuous function, Theorem 5.8.1 tells us that there
exists a sequence qn of polynomials converging uniformly to f on [0, 1]. Let pn (x) =
x
> 0. Then ther
Solutions to Homework for Week 4 - Math 1540 - Spring, 2004 - Rubin
Section 6.3, #1 = pg. 338, #2. The local Lipschitz condition does not imply differentiability. For example, f (x) = |x| is Lipschitz (with Lipschitz constant 1)
but not dierentiable at 0.
Solutions to Homework for Week 5 - Math 1540 - Spring, 2004 - Rubin
Section 6.5, 1 - pg. 386, # 13. b) G(x, y, z ) = sin(x2 + yz )(y 3 + xy ) = sin(x3 y + x2 y 3 +
xy 2 z + y 4 z ). The chain rule gives
DG(x, y, z ) = (3x2 y + 2xy 3 + y 2 z ) cos(x3 y + x
Solutions to Homework for Week 7 - Math 1540 - Spring, 2004 - Rubin
Section 7.1 - pg. 439, #3. The Jacobian determinant for the transformation takes the
form
f (x0 )
0
= f (x0 ) = 0
x0 f (x0 ) + f (x0 ) 1
Hence, invertibility follows from the Inverse Func
Solutions to Homework for Week 10 - Math 1540 - Spring, 2004 - Rubin
pg. 493, # 34. Note that for any partition P of [0, 1], U (f, P ) = 1. Suppose that we consider a uniform partition P with n subintervals [(i 1)/n, i/n] for i = 1, . . . , n.
Then the in
Solutions to Homework for Week 11 - Math 1540 - Spring, 2004 - Rubin
pg. 469, #1. Let gn : [0, 1] I be a sequence of nonnegative functions such that each
R
1
improper integral 0 gn (x) dx exists and is nite. Suppose that 0 gn+1 gn and
that gn (x) 0 for ea
Solutions to Homework for Week 12 - Math 1540 - Spring, 2004 - Rubin
Wade, pg. 447, # 2. For x0 , a I m , with a = 0, set (t) = ta + x0 and C = (, I
R
R).
This curve is smooth since (t) = a = 0 for all t I It is simple, since t1 = t2
R.
(t1 ) (t2 ) = (t1
Solutions to Homework for Weeks 8-9 - Math 1540 - Spring, 2004 - Rubin
pg. 491, # 17a. If f is integrable, then we can apply Darbouxs theorem. This says that
for any > 0, there exists a corresponding > 0 such that as long as our partition has
subrectangle
Two Exercises relating to Marsden and Homan Section 6.6 Math 1540 - February 6, 2004
1. Given that Df (x0 ) is the derivative of f : I n I m at x0 I n , show that there exists
R
R
R
K > 0 such that |Df (x0 ) (x x0 )| K |x x0 |.
2. Given that f is dierenti