Math 3110 Homework 6 Solutions
Exercise 7.2.1
1
Let an = 2 . Dene e(n) by e(n) = 1 if n is even and e(n) = 0 if n is odd. Because an 0,
n
we get 0 e(n)an an and 0 (1 e(n)an an for all n. Then by Theorem 7.2D, because
an converges, so do
n=1
(1 e(n)an . By
Math 3110 Homework 2 Solutions
Exercise 3.1.2
Claim. If an is a nonnegative sequence, lim an = 0 = lim an = 0.
Proof. Let an be a nonnegative sequence. Assume lim an = 0 (this means that for any
> 0, there is a natural number N such that for all n N ,
Math 3110 Homework 3 Solutions
Exercise 4.1.1
(i)
Claim. If an L and bn M , then an + bn L + M .
Proof. Let > 0 and let dn = an L and en = bn M denote the error terms for the given
sequences.
Since an L and bn M , then by the errorform principle, dn 0 an
Math 3110
Spring 09
Homework 7: Selected Solutions [10.11]: Claim: If f (x) = cos x then (a) f (, ) = [1, 1], (b) f (0, 1) = (1, 1), (c) 13 f ( , ) = [1, 0) 22
Proof Omitted; selfevident. x2 1 is not bounded on its domain D(f ) = (, 2) x2 4 (2, 2) (2, )
Math 3110 Homework 11 Solutions
Exercise 12.1.1
Let f be a continuous function whose values are always rational numbers. Assume for
contradiction that f is not a constant function. Then there exist x1 , x2 with x1 < x2 such
that f (x1 ) = f (x2 ). Without
Math 3110: Introduction to Analysis, Fall 2010
Class Meetings
MWF 11:15 AM  12:05 PM Malott 207
Course Sta
Instructor: Al Schatz
Teaching Assistant: Fatima Mahmood (fm88@cornell.edu)
Oce Hours
Professor Schatz: Wednesdays 12:30 PM  2:30 PM Malott 557
Fa
Math 3110 Homework 10 Solutions
Exercise 11.1.1
We can compute
f (x) f (1) =
=
=
x
1
1+x 1+1
1+x
2x
2(1 + x) 2(1 + x)
x1
.
2(x + 1)
Given > 0, let = mincfw_1, 2 . If x 1 < 1, then we have x 1 < 1, from which
0 < x < 2, and so 1 < x + 1 < 3. Hence,

Math 3110 Homework 9 Solutions
Exercise 10.1.4
We can use the triangle inequality and that  cos(x) 1 for all x to get
k
k
k
an cos(nx)
an cos(nx)
n=1
n=1
an ,
n=1
k
an .
so the function is bounded by
n=1
Exercise 10.1.9
(a)
True. If f is bounded
Math 3110 Homework 8 Solutions
Exercise 9.2.3
1
, we must have (f (x)2 = 1, and so f (x) = 1. Clearly any function that
Since f (x) =
f (x)
has either f (x) = 1 or f (x) = 1 for each real x satises the equation. This must hold for
each real x, but dierent
Math 311
HW 7 Solutions
April 11, 2008
7.3.2 Let bn  < B for all n. By hypothesis, such a B exists. Then,
an bn  < an B
for all n,
and so an bn  converges by comparison with an B = B
Then by absolute convergence theorem,
an bn converges.
7.3.3 a
Math 3110 Homework 7 Solutions
Exercise 8.1.1
In each case, let an denote term n of the series.
(a)
We compute
Since
an+1
xn+1
2n n
n
x
= n+1
.
=
n
an
2
2
n + 1 x
n+1
n
x
1 as n , we want
< 1, so the radius of convergence is 2.
2
n+1
(b)
We com
Math 3110 Homework 5 Solutions
Exercise 6.2.2
The terms of a sequence cfw_xn take on only nitely many values a1 , a2 , . . . , ak . That is, for
every n, xn = ai for some i (the index i will depend on n).
Claim. The sequence cfw_xn has a cluster point.
Math 3110 Homework 12 Solutions
Exercise 13.1.1
(a) At least one of the intervals must contain innitely many terms of the sequence, for
otherwise, the sequence would only have nitely many terms in each of nitely many intervals,
and hence only nitely many
Math 3110 Homework 13 Solutions
Exercise 14.1.5
Let > 0 be given, and let = . If x < , then
f (x) f (0) = x sin
1
x
x < = .
This shows that f is continuous at 0.
In order for f (x) to be dierentiable at 0, we would need for
x sin
f (x) f (0)
= lim
Math 3110 Preliminary Exam Solutions
1.
Prove that there exists
Let
xR
f (x) = x, g(x) = cos x.
such that
x = cos x.
Observe that
f (0) = 0 < 1 = g(0)
and
f (/2) = /2 > 0 = g(/2).
Hence, by the intersection principle, there exists
x [0, /2]
such that
x =
Math 3110: Introduction to Analysis, Fall 2010
Prelim 1
This takehome exam is due in class on Friday, October 15, 2010.
You must work independently, but you may use anything from your textbook and notes.
1.
2.
3.
4.
5.
6.
7.
8.
Problem: 14 (pg. 14)
Exer
Math 3110 Homework 14 Solutions
Exercise 15.1.3
If f (x) changes sign from negative to positive on [a, b], then there are some h, k [a, b] with
h < k , f (h) < 0, and f (k ) > 0. Since f (x) is dierentiable on [a, b] and [h, k ] [a, b],
f (x) is dierentia
Math 3110 Homework 4 Solutions
Exercise 5.1.4
an
Claim. If
L, bn = 0 for any n, and bn 0, then an 0.
bn
Proof. Since lim
n
an
= 0 and lim bn = 0, then by the product theorem (5.1(2),
n
bn
lim
n
But since an =
an
bn
bn
an
bn , then lim an = lim
n
n
bn
=