1
(Math 360) Free Response Final:
April 28, 2009
2
Rules
Preparation:
In preparing for this part of the nal exam you are allowed to use any books
as well as any source you nd on the internet. You are also allowed to talk
with anyone involved with this cla
MATH 360
University of Pennsylvania
Fall 2015
Midterm 1 Study Guide
by Elaine So
1. General tips.
(1) You can use any fact or theorem provided that it does not make the problem trivial. For
example, if you are proving that [2, 3) is not closed, you can as
MATH 360 Homework 6
Due 1 March 2013
Denition. Given two metrics d and on the same set M , we say d is equivalent to if there are positive
real numbers C1 and C2 so that for any x, y M , we have
C1 (x, y ) d(x, y ) C2 (x, y )
1. Show that metric equivalen
MATH 360 Homework 6.5
Not due, but do!
Dene
R = (x1 , x2 , . . . , xk , . . .) xi R, all but nitely many xi = 0
Note that under pointwise operations, R is a vector space.
1. For x, y R , show that x, y
2
xi y i denes an inner product on R . Write d2 for t
MATH 360 Homework 5
Due 22 February 2013
Denition. A sequence (xn )n2N in a metric space is called eventually
constant if there is some N so that
! N guarantees x = x .
1. Let A & B be subsets of a metric space (M; d). Show that A is dense in B if and onl
MATH 360 Homework 4
Due 8 February 2013
Recall that our denition of a closed subset diers from Rudins:
Denition. A subset of a metric space A (M, d) is closed if its complement is open.
1.
i. Show that the union of a nite number of closed sets is closed.
MATH 360 Homework 3
Due 1 February 2013
1. Suppose (an )nNis a sequence of real numbers with an A and A > 0. Consider the sequence
bn = n + an n.
i. Let 0 < c < 1 . Show that there is some N N, depending on c and the sequence (an )nN , so that
2
n N guara
MATH 360 Homework 2
Due 25 January 2013
1. (Versions of the Archimedean property) Prove that the following properties are equivalent in an ordered
eld, i.e. any ordered eld F which possesses one of the properties must possess the other two as well:
i. If
MATH 360 Homework 1
Due 18 January 2012
Properties of the Real Numbers
1. Show that the following properties hold in any eld.
i. (Unique identities) If a + x = a for some a, then x = 0. If ax = a for some a, then x = 1.
ii. (Unique inverses) If a + x = 0,
MATH 360 Homework 11
Due 23 April 2013
1. For a xed n N, consider the sequence of functions n : [0, 1] R, given by: if x = p is a rational
q
number in reduced form, with q n, then n (x) = q12 ; for other x (that is, the irrationals and any
rational with r
MATH 360 Homework 10
Due 12 April 2013
1. (Marsden-Homans problem 4.42) For x > 0, dene L(x) =
this denition of L and theorems about the denite integral.
x1
dt.
1t
Show the following using only
i. L(x) is a strictly increasing function.
ii. L(xy ) = L(x)
MATH 360 Homework 9
Due 29 March 2013
1. Show that f is dierentiable at x0 i lim
h0
f (x0 + h) f (x0 )
exists. Moreover if this limit exists, it
h
must be f (x0 ).
2. Let f : R R.
i. Suppose f is dierentiable at x0 (a, b). Show that
f (x0 ) = lim
h0
f (x0
MATH 360 Homework 7
Due 15 March 2013
1. Suppose (V; kk) is a normed space. Show that kk : V 3 R is continuous with respect to the metric
induced by kk and the standard metric on R. (Hint. Use the reverse triangle inequality.)
2. Suppose (M1 ; d1 ) and (M
MATH 360 Homework 8
Due 22 March 2013
1.
i. Give a precise meaning to the statement, Disconnections pull back. Is the statement true?
ii. Show that the image of a connected set under a continuous map is connected.
2. Dene S n = x Rn+1 x = 1 .
i. Show that
1. A - It is possible to place a linear order on C so that we do NOT get
an ordered eld, and impossible to place a linear order of C so that we DO
get an ordered eld.
2. D - |x + y | = |x| + |y | doesnt have to happen. For example, let
y = x = 0.
3. Yes,
MATH 360
University of Pennsylvania
Fall 2015
Midterm Solutions, 1st Draft
by Elaine So
1.a. (2 pts): Let A, B be any (finite) sets. For set powering, we know that |AB | = |A|B| . For disjoint
unions, we know that |A B| = |A| + |B|. For products, we know
MATH508. ADVANCED CALCULUS
LECTURE 5. DIFFERENTIATION AND INTEGRATION
A. A. KIRILLOV
The goal of this lecture is to recall the basic notions and rules from Calculus and to give them rigorous definitions and proofs.
1. Differentiation
We shall consider rea
MATH 360 (ADVANCED CALCULUS I)
FALL 2015. TIME: TU-TH 10:30-12:00
A. A. KIRILLOV
1. Introduction
1.1. Notations and symbols. .
S1 S2 , S2 S1
the statement S1 implies the statement S2
S1 S2
the statement S1 is equivalent to the statement S2
A := B, B =: A
Math 360: Homework 9
Due Friday, November 15
Midterm 2 Redux Question:
1. Prove that for any real constants a0 , a1 , a2 ,
x3 + a2 x2 + a1 x + a0
= .
x
x2 + 1
lim
In class on November 5th, we dened the functions sin x and cos x on the real line and establ
Math 360: Homework 10
Due Friday, November 22
1. Fix k 1, and suppose g (t) is real-valued and k -times dierentiable on some neighborhood of the
origin in R (we specically do not require that g (k) is continuous).
(a) Suppose that g (0) = g (0) = = g (k1)
MATH 360
University of Pennsylvania
Fall 2015
HW #4 Solutions, 1st Draft
by Elaine So
P
1.a. (4 pts): Recall that a power series n=1 an xn is differentiable term-by-term in its radius of conver
1
1
1
. However, in our case
=
= n n! >
gence (R, R), where
MATH 360
University of Pennsylvania
Fall 2015
HW #5 Solutions, 1st Draft
by Elaine So
1.a. (2 pts): Let X be the set of all sequences in N of finite length. Let pn denote the nth prime number.
Then define the map f : X N such that:
n
f (x1 , x2 , . . . ,
MATH 360
University of Pennsylvania
Fall 2015
HW #3 Solutions, 1st Draft
by Elaine So
1.a. (3 pts): Let xn = n define a sequence cfw_xn Z. Suppose that there is a subsequence cfw_xnk cfw_xn
such that xnk L Z. Let = 1, so that there is some N N, k N |xn
MATH 360
University of Pennsylvania
Fall 2015
HW #1 Solutions, Draft 1
by Elaine So
1. (3 pts): We use the following identities (note that this is quite overkill):
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
X \Y =X YC
(X Y ) Z = (X Z) (X Y )
(X Y ) Z = (X Z) (Y
MATH 360
University of Pennsylvania
Fall 2015
HW #2 Solutions, Draft 1
by Elaine So
1 (5 pts): Let = 1. Let N N be any positive integer. Then certainly 2N + 1 N , but
|x2N +1 1| = |(1)2N +1 1| = | 1 1| = 2 > 1 = . Thus, xn 6 1.
2.a. (4 pts): We first esta