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Unformatted text preview: xceeding 1/2 n . Therefore any interval that is included in the Cantor set must have length less than
1/2 n for every positive integer n. In other words, such an interval must be a singleton.
2. Prove that every member of the Cantor set is a limit point of the Cantor set.
Suppose that x belongs to the Cantor set C. We express x in the ternary decimal form
Ý x
n1 an
3n where each number a n is either 0 or 2. To show that x is a limit point of C, suppose that 0.
Choose a positive integer k such that
2 .
3k
Given any positive integer n we define
an if n 2 bn a n if n k k . The number
Ý y
n1 bn
3n belongs to C and is unequal to x and we have
Ý x bn  a n y 3n n1 2k .
3 3. True or false? If a number x in the interval 0, 1 can be expressed as a ternary decimal expansion of the form
Ý x
n1 an
3n and if there exists a value of n for which a n 1 then the number x can’t belong to the Cantor set.
This statement is false. For example, we have seen that
1
3 Ý
n2 2
3n C. 4. At which numbers x is the Cantor function differentiable?
Given any number x
0, 1
C, since the Cantor function is constant in a neighborhood of x it
must be differentiable at x and, in fact, x 0. Now suppose that x C. We express x in the 349 ternary decimal form
Ý x
n1 an
3n where each number a n is either 0 or 2. Now if for a given positive integer k we define
an if n 2 bn a n if n k k and if we define the number y by the equation
Ý y
n1 bn
3n then y belongs to C and is unequal to x and we have
x
x ak bk 2k
ak 2k
bk 3k y
y 3k 1
2 1
2 3
2 k . Therefore, since the difference quotients of at x form an unbounded set, the limit
y x
lim
yx
yx
cannot exist (and be finite) and so fails to be differentiable at x whenever x C. 13 Improper Integrals
Some Exercises on Improper Integrals
1. Evaluate each of the following improper integrals, when possible, and specify those that diverge. If you
can’t see how to evaluate the integral exactly yourself, ask Scientific Notebook to evaluate it for you. (Before
asking Scientific Notebook to evaluate one of these intgrals, remove the arrow sign from the limits of
integration.)
a. Þ Ý 1
1 x 2 0 3/2 dx The substitution u arctan x gives us
Ý
w
1
1
dx w Ý Þ
lim
Þ0
3/2
0 1 x2
1 x2
wÝ
lim
b. Þ Ý
2 1
x x2 1 Þ0 arctan w dx 3/2 1
1 tan 2 u 3/2 sec 2 udu w Ý sin arctan w
lim dx The substitution u arcsec x gives us
Ý
1
dx w Ý
lim
Þ2
2
xx
1
wÝ
lim Þ2 w 1
x Þ arcsec 2 arcsec w 350 dx
1
1
sec u tan udu
sec u tan u
2 x2 arcsec 2. 0 1. c. Þ 2 1 1 x x2 1 dx The substitution u arcsec x gives us Þ1 2 1
x x2 1 Þ
1 w
2 dx lim
w d. Þ Ý 1 1 x x2 1 x dx
1 1du arcsec 2
Þ
1 arcsec w
arcsec 2 lim
w 1
x2 1 dx This integral is defined to be Þ1 2 e. Þ /2
0 1
x x2 1 dx Þ Ý
2 1
dx
2
x x2 1 1. tan xdx This integral is Þ 0 tan xdx wlim
/2
w lim w /2 log sec w Ý which means that the given integral diverges.
f. Þ /2
0 tan x sin x dx This integral is Þ0 w...
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 Fall '08
 STAFF
 Math, Calculus

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