MA 1A (SECTION 1) HW5 SOLUTIONS
Problem 1. (Apostol 4.15.4)
2
Let f (x) = 1 x 3 . Show that f (1) = f (1) = 0, but that f (x) is never zero in the
interval [1, 1]. Explain how this is possible, in view of Rolles theorem.
2
1
Solution. f (1) = 1 1 3 = 0, a
1
The Real Number System
The rational numbers are beautiful, but are not big enough for various purposes, and the set R of real numbers was constructed in the late nineteenth century, as a kind of an envelope of Q. (More later on this.) For a nonconstruct
Lecture 2: The real numbers
The purpose of this lecture is for us to develop the real number system. This might
seem like a very strange thing for us to be doing. It must seem to you that you have been
studying real numbers most of your life. However, som
Lecture 6 Power series
A very important class of series to study are the power series. They are interesting
in part because they represent functions and in part because they encode their coecients
which are a sequence. At the end of this lecture, we will
Lecture 5 Innite Series
In todays lecture, we will restrict our attention to innite series, which we will view as
special kinds of sequences. We will bring what we learned about convergence of sequence
to bear on innite series.
An innite series is a forma
Lecture 4: Cauchy sequences, Bolzano-Weierstrass, and the Squeeze theorem
The purpose of this lecture is more modest than the previous ones. It is to state
certain conditions under which we are guaranteed that limits of sequences converge.
Denition We say
Lecture 1: Induction and the Natural numbers
Math 1a is a somewhat unusual course. It is a proof-based treatment of Calculus,
for all of you who have already demonstrated a strong grounding in Calculus at the high
school level. You may have heard complain
Lecture 15: Integrability and uniform continuity
Sorry for this abbreviated lecture. We didnt complete the proof of properties of the
Riemann integral from last time.
We could write the denition of continuity as follows: A function f is continuous at
x if
Lecture 16: The fundamental theorem
Dierentiation and integration are not unrelated, which is why calculus is a subject.
As their names suggest, in a certain sense, they are opposites. We will be precise about
what sense in what follows as we state and pr
1
Problem 1.1.6
Prove the identity
(
) (
) ( )
n+1
n
n
=
+
.
k
k1
k
Use this to prove by induction that
n ( )
n
j=0
j
= 2n .
Solution. We compute:
( ) (
)
n
n
n!
n!
+
=
+
k
k1
(n k)!k! (n (k 1)!(k 1)!
n!
n!
=
+
(n k)!k! (n k + 1)!(k 1)!
(n k + 1)n!
k n!
=
Lecture 3 Limits
In the previous lecture, we used the least upper bound property of the real numbers to
dene the basic arithmetic operations of addition and multiplication. In eect, this involved
nding sequences which converged to the sum and product. In
Lecture 9: The mean value theorem
Today, well state and prove the mean value theorem and describe other ways in which
derivatives of functions give us global information about their behavior.
Let f be a real valued function on an interval [a, b]. Let c be
Lecture 7 Limits and Continuity
Today, for the rst time, well be discussing limits of functions on the real line and
for this reason we have to modify our denition of limit. For the record:
Denition A function f from the reals to the reals is a set G of o
Lecture 13: Inverse functions
Today, well begin with a classical application of the calculus: obtaining numerical
solutions to equations.
Situation: We would like to solve an equation
f (x) = 0.
Here f is a function and we should imagine that its rst and
Lecture 14 The Riemann integral dened
Our goal for today is to begin work on integration. In particular, we would like to
b
dene a f (x)dx, the denite Riemann integral of a function f on the interval [a, b]. Here
f should be, at least, dened and bounded o
Lecture 12: Formal Taylor Series
For the last several lectures, we have been building up the notion of Taylor approximation. We proved
Theorem Let f, f , f , . . . , f (n2) be dened and continuous everywhere on a closed
interval I having c in the interior
Lecture 11: Exponentiation
Todays lecture is going to focus on exponentiation, something you may consider one
of the basic operations of arithmetic. However there is subtle limiting process that takes
place when dening exponentiation which we need to full
Lecture 8: Dierentiation: local theory.
Today well dene the derivative of a function and describe its familiar local theory.
Before doing this well introduce a bit of notation, common in some applied elds like
the analysis of algorithms, but not often use
Lecture 19: Arclength and trigonometric functions.
We begin this lecture by dening the arclength of the graph of a dierentiable function
y = f (x) between x = a and x = b. We motivate our denition by calculating the length
of the part of the line y = mx b
Lecture 24 Roots of Polynomials
Last time, we described the complex numbers. Classically, one of the reasons complex
numbers were studied is that in the complex numbers all polynomials have roots. We
discuss this a bit today.
Last time, we derived Eulers
1
4.3.1
Define f (x) = (x log x log log x)1 and F (x) = log log log x. Observe that
(
)(
)( )
1
1
1
F (x) =
= f (x).
log log x
log x
x
Also, since 2015 > e, log log x > 0 when x 2015. Restricting ourselves to the interval [2015, ), f ,
being the product o
Midterm Exam 2015 solutions: Do not read if you have not taken midterm
exam.
1a. We will first show by induction that any polynomial p of degree n 1 is equal to
its Taylor Series about c. (This was essentially done in Cranks section 1.2, although the
indu
1
Problem 4.1.2
1 2
n1
, ,.,
, 1 be the partition of [0, 1] into n equally spaced pieces. Write Ln for the lower
n n
n
Riemann sum corresponding to Pn . Since x2 is decreasing we have
Let Pn = cfw_0,
Ln (x ) =
2
n1
(
j=0
j
n
)2
1
n
n1
1 2
= 3
j
n j=0
(
)
Solution to HW 1, Ma 1a Fall 2016
Section 1.2 Exercise 4: Prove the generating rule for Pascals triangle. That
is, show for any natural numbers 0 < k < n that the formula
n
n+1
n
+
=
.
k1
k
k
Hint: Put both terms of the left hand side under a common d
Midterm 2016 solutions
1(a). Observe that the base case is true since
13 =
1 2 2
(1 )(2 ).
4
The inductive step is true since
1 2
(n (n + 1)2 n2 (n 1)2 ) = n3 .
4
1(b). Observe that
n
X
n
n+1
X
X
1
3
j + 3j + 3j =
(j + 1) 1 = (
j 3 ) n 1 = (n + 1)2 (n + 2
Solutions for Problem set 2
1.3.6 Either there exists n so that tn (x) < tn (y) or x = y. If it is the latter, then
since y z and x is just the same as y, we know that x z. Similarly, either there exists
m with tm (y) < tm (z) or y = z. If it is the latte
Solution to HW 3, Ma 1a Fall 2016
Section 2.1 Exercise 2: Let C be a subset of the real numbers consisting
of those real numbers x having the property that every digit in the decimal
expansion of x is 1, 3, 5, or 7. Let cfw_cn be a sequence of elements o
Lecture 26 Cauchys Theorem
Last time, we introduced the notion of a dierential of a function of two variables
f (x, y), namely
f
f
df =
dx +
dy.
x
y
We said that a complex-valued function of a complex variable is analytic if
df = f (z)dz,
i.e. the dierent
Lecture 25: Analytic functions
Today we will explore the calculus of complex valued functions of a complex number.
On your last problem set, you have frequently met complex-valued functions f (t) of a real
variable t. Complex numbers have real and imagina