10.1 The Geometric Series
CHAPTER 10
10.1
(page 373)
INFINITE SERIES
The Geometric Series
The advice in the text is: Learn the geometric series. This is the most important series and also the
simplest. The pure geometric series starts with the constant te
\:;ii/>Integrabilitx g: bounded Eiecewisemonotonic functions.
The definition of "piecewise-monotonic" is given on p.
77 of the text.
Lemma. If f is bounded on [a,b] and monotonic on
(a,b), then f is integrable on [a,b].
(Note that we need to assume f
18 014 Problem Set 2 Solutions
Total: 24 points
P
Problem : Let f (x) = n=0 ck xk be a polynomial of degree n.
k
(d) If f (x) = 0 for n + 1 distinct real values of x, then every coe cient ck of f is
zero and f (x) = 0 for all real x.
P
(e) Let g(x) = m bk
Exam
Problem 1 Evaluate limx!0
1
x
1
log(x+1)
Solutions
.
Solution We begin by writing the problem as a single fraction, limx!0 log(x+1) x .
x log(x+1)
Observe that both numerator and denominator have limit zero, and thus we can apply
LHopitals rule to se
Exam
Solutions
October 29, 2010
Total: 60 points
Problem : Find the derivative of each of the following functions
g(x) = log(cos(x2 )
p
h(x) = e
x sin x
Solution For the rst problem, we invoke the chain rule twice. Thus,
d
1
( sin(x2 ) 2x =
(log(cos(x2
Continuity oi the sguare root function.
The following theorem shows that the square-root
function is continuous for x Z 0. We will give a different
proof,based on the intermediate-value theorem, shortly.
Theorem. (i) lim /§ = O.
x+0+
(ii) If a > 0, lim
The Riemann condition.
The most useful criterion for determining whether f is
integrable on [a,b] is given in the following theorem. It is
called the Riemann condition for existence of the integral.
Theorem 1. Suppose f is defined on [a,b]. Then f
is
G.l
Rational exponents - an application of the intermediate-value
theorem.
It is a consequence of the intersediate~va1ue theorem, that;
given a positive integer n and a real nulber a > 0,
there is
exactly one real number b 2 0 such that
We denote b by I3,
18 014 Problem Set 3 Solutions
Total: 12 points
Problem : Find all values of c for which
Rc
(a) R0 x(1 x)dx = 0.
c
(b) 0 |x(1 x)|dx = 0.
Solution (4 points)
(a) Computing, we get
Z
0
c
x(1
x) =
Z
c
(x
0
1
x2 ) = c2
2
1 3
c
3
Setting the right hand side eq
Drl
Progertiesgg integrals
In this section, we prove the four basic properties of the integral
that we shall need.
Theorem. (Progerties g: the integral)
(l) (Linearitx Erogertx.) g; f and g are integrable
99 [a,b], then g2 ig cf + dg (here c and d are con
18 014 Problem Set 1 Solutions
Total: 24 points
Problem : If ab = 0, then a = 0 or b = 0.
Solution (4 points)
Suppose ab = 0 and b = 0. By axiom 6, there exists a real number y such that
6
by = 1. Hence, we have
a = 1 a = a 1 = a(by) = (ab)y = 0 y = 0
usi
18 014 Problem Set 4 Solutions
Total: 24 points
Problem : Establish the following limit formulas. You may assume the formula
limx!0 sinx x) = 1.
(a)
sin(5x)
lim
= 5.
x!0 sin(x)
(b)
lim
sin(5x)
x
x!0
(c)
lim
sin(3x)
1
x!0
p
1
x2
x2
= 2.
1
= .
2
Solution (4
11.1 Vectors and Dot Products
(page 405)
CHAPTER 11
VECTORS AND MATRICES
Vectors and Dot Products
11.1
(page 405)
One side of the brain thinks of a vector as a pair of numbers (x, y) or a triple of numbers (x, y, z ) or a string
n numbers ( x l , x 2 , .