18.100C
Fall 2012
Problem Set 5
1. Given a series of real numbers
P1
k=1 xk , dene another series
P
l yl
by
y1 = x 1 ,
1
y2 = x2 ,
2
1
2
n 2
1
x2 +
x3 + +
xn 1 + x n .
yn =
n
n(n 1)
n(n 1)
n(n 1)
P
P
Show that if 1 xk converges, then 1 yk converges to the
18.100C
Fall 2012
Problem Set 1
1. Let m and n be positive integers with no common factor. Prove that
p
if m/n is rational, then m and n are both perfect squares, that is to
say there exist integers p and q such that m = p2 and n = q 2 . (This is
proved i
18.100C
Fall 2012
Practice Midterm 1
No notes, textbooks, calculators, or other materials may be used. Please
switch o all mobile phones or other electronic devices.
Unless the problem specically states otherwise, the rules are as follows: you
can use any
Problem Set 1 Solutions, 18.100C, Fall 2012
September 12, 2012
1
Let m and n be positive integers with no common factor. Prove that if
p
m/n is rational, then m and n are both perfect squares, that is to say
there exist integers p and q such that m = p2 a
Problem Set 8 Solutions, 18.100C, Fall 2012
November 14, 2012
For bounded functions f, g : [a, b] ! R, we use the notation |f | =
supcfw_|f (x)|x 2 [a, b] and d(f, g) = |f g|.
1
We have fn ! f and gn ! g uniformally. We wish to show that fb gn ! f g
unifo
Problem Set 4 Solutions, 18.100C, Fall 2012
October 11, 2012
1
Let X be a complete metric space with metric d, and let f : X ! X be a contraction, meaning that there exists < 1 such that d(f (x), f (y) d(x, y)
for all x, y 2 X. Then there is a unique poin
Problem Set 3 Solutions, 18.100C, Fall 2012
September 26, 2012
1
p
We have a metric space (X, d), and dene the function d0 (x, y) = d(x, y).
We wish to show that (X, d0 ) is also a metric space with the same open sets
as (X, d). We rst check that d0 is a
Problem Set 5 Solutions, 18.100C, Fall 2012
October 18, 2012
1
P
Pn
Let sn = n xi and n = i=1 yi be the partial sums. Then we claim
i=1
that n is the average of the rst n sn , i.e.
n
=
s1 + s2 + sn
n
To see this, we will look at the coe cients of the xk i
18.100C
Fall 2012
Problem Set 4
1. A metric space is called complete if every Cauchy sequence converges.
Let (X, d) be a complete metric space, and f : X ! X a map with the
following property. There is some 0 < 1 such that for all x, y 2 X,
d(f (x), f (y)
September 20, 2012
1
We dene
F = Q = cfw_a +
p
2b|a, b 2 Q R
We wish to show that F is a subeld of R. In order to show this, we need
to show that a) 0, 1 2 F ; b) F is closed under addition and multiplication;
and c) if x 2 F and x = 0, then x 2 F and 1/x
November 30, 2012
1
Write B 1 := B 1 ([a, b]), d(f, g) = supx2[a,b] |f (x) g(x)| the uniform metric on real bounded functions, and d1 (f, g) = d(f, g) + d(f 0 , g 0 ) the given
metric on B 1 . Note that if f, g 2 B 1 , then f, f 0 , g, g 0 are all bounded
18.100C
Fall 2012
Practice final exam
No notes, textbooks, calculators, or other material may be used. Please switch o all mobile
phones and other electronic devices.
Unless the problem specically states otherwise, the rules are as follows: you can use an
Practice Midterm 1 Solution , 18.100C, Fall 2012
1
Suppose F is a eld, and x, y 2 F satisfy x2 = y 2 . We need to show
that either x = y or x = y. For the sake of sparing the proliferation of
parenthesis, we will adopt the standard conventions about multi
Problem Set 7 Solutions, 18.100C, Fall 2012
October 31, 2012
1
We wish to dene !/2 as the smallest zero of cos(x), i.e. as the positive real
number such that cos(!/2) = 0 and for 0 < x < !/2, cos(x) 6= 0. Clearly
such a number, if it exists, is unique; if
Problem Set 6 Solutions, 18.100C, Fall 2012
October 25, 2012
1
Q
Q
Let sk = n xk . Then we say that 1 xk converges to s if the sequence
k=1
k=1
sn converges to s as n ! 1.
Q
6
Now suppose 1 xk = s converges to some s = 0. We wish to show
k=1
that limk xk
Practice final exam solutions
1
We write d = dSN CF for the French Railroad metric on R2 . In this problem
we will often use the easily checked fact that if (xn ) is a Cauchy sequence in
any metric space, then xn ! x if and only if xnk ! x for some subseq
1
The stament is true. Let bk = 2 k ak . By Rudin Theorem 3.23, limk!1 ak =
0. In particular, the ak s are bounded, i.e. for some a > 0, |ak | < a. Then
P
for any k, |bk | < 2 k a. But 1 2 k a = 2a, so |bk | is bounded above by a
P k=0
P
convergent series
Problem Set 10 Solutions, 18.100C, Fall 2012
December 5, 2012
1
We have a continuous K : [0, 1] [0, 1] ! R which is continuous. For each
xed x 2 [0, 1], the function y 7! K(x, y) is thus a continuous function from
[0, 1] ! R, hence is Riemann-integrable.