Math 425 - Solution Set for Homework #10
7) Let f be bounded on [a, b] and P = cfw_x0 , x1 , . . . , xn be a partition of [a, b]. Then
s(P ) =
n
X
mj (xj xj1 ),
j=1
where
mj =
inf
xj1 xxj
f (x).
An arbitrary Riemann sum of f over P is of the form
=
n
X
f
Math 425 - Solution Set for Homework #6
8) Let f be a function which assumes only finitely many values.
Suppose f is constant on some interval (x0 , x0 +). Then f is continuous at x0 as constant functions
are continuous.
Now suppose f is continuous at a p
Math 425 - Solution Set for Homework #12
1) Suppose f is integrable on [a, b] and c is a constant. We can assume that c 6= 0 because if c = 0, then
Rb
cf = 0 and is clearly integrable and its integral is 0. Let L = a f (x)dx. Then for all > 0 there
exists
Math 425 - Solution Set for Homework #9
Module 2, pp. 74-75
In each of the following problems, LHospitals rule is used to find to specified limit.
3)
lim
x0
sin x
1 cos x
= lim
log(1 + x2 ) x0 2x
1 + x2
sin(x)(1 + x2 )
= lim
x0
2x
=
(Using LHospitals rul
Math 425 - Solution Set for Homework #4
1.4)
a) = 1/4
b) = 1/6
c) = 6
d) = 1
1.5)
a) The set S is neither open nor closed.
S 0 = (1, 2) (3, )
(S c )o = (, 1) (2, 3)
(S o )c = (, 1] [2, 3]
b) The set S is open.
S o = (, 1) (2, )
(S c )o = (1, 2)
(S o )c =
Math 425 - Solution Set for Homework #11
1) This proof follows as a natural extension of the arguments from the text.
Suppose that
|f (x)| M,
a x b,
0
and let P be a partition of [a, b] obtained by adding r points to a partition P = cfw_x0 , x1 , . . . ,
Math 425 - Solution Set for Homework #5
12) Suppose lim f (x) = L. Then for all > 0, there exists > 0, such that for all x Df \ cfw_x0 ,
xx0
|f (x) f (x0 )| < whenever 0 < |x x0 | < .
Thus, for all x Df \ cfw_x0 ,
|f (x) L| < whenever x0 < x < x0 .
So the
Math 425 - Solution Set for Homework #7
24) Suppose there is no x2 [a, b] such that f (x2 ) = . Then f (x) < for all x [a, b]. Suppose t [a, b].
Then
f (t) +
f (t) <
< .
(1)
2
By the continuity of f there is an open interval of It containing t such that
Math 425 - Solution Set for Homework #3
In each of these problems, induction is used. To simplify notation B denotes the base case and I denotes
the induction step.
n
X
1) We want to show
m2 =
m=1
B.
1
X
m2 = 1 =
m=1
I. Suppose that
n(n + 1)(2n + 1)
.
6
1
Math 425 - Solution Set for Homework #2
3) Let c > 1 and S = cfw_x R, x 1, x3 < c. The set S is non-empty since 1 S.
Given x S, it follows that x3 < c < c3 since c > 1. Then 0 < c3 x3 = (c x)(c2 + cx + x2 ). Since
c2 + cx + x > 0, it follows from the posi
Math 425 - Solution Set for Homework # 1
1.1) Write the following expressions in equivalent forms not involving absolute values.
a)
(
a+b+ab
ab>0
a + b + |a b| =
a + b (a b) a b < 0
(
2a a > b
=
2b b > a
= 2 max(a, b)
b)
(
a+ba+b
ab>0
a + b |a b| =
a + b
FUNCTIONS DEFINED BY
IMPROPER INTEGRALS
William F. Trench
Andrew G. Cowles Distinguished Professor Emeritus
Department of Mathematics
Trinity University
San Antonio, Texas, USA
wtrench@trinity.edu
This is a supplement to the authors
Introduction to Real A
Math 425/525 - Fall 2016 - Review for Test II
For each of the problems below:
1. Prepare a written solution in a form that can be shared with the rest of the class.
2. Include a brief summary of what the problem is about.
3. Give complete proofs and state
Midterm 2 Review
5. Assume that f : R R is differentiable and that its derived is bounded. Show that f is uniformly
continuous on R.
Call the bound of f 0 M . We know |f 0 (x)| M .
Since the f 0 is bounded, we know f satisfies the Lipschitz condition.
Tha
Math 425 - Solution Set for Homework #8
8) Suppose f and g are differentiable at x0 , f (x0 ) = g(x0 ) = 0, and that g 0 (x0 ) 6= 0. From Lemma 3.2,
f (x) = f (x0 ) + [f 0 (x0 ) + Ef (x)](x x0 )
and
g(x) = g(x0 ) + [g 0 (x0 ) + Eg (x)](x x0 )
where Ef and