1873_solutions

# Is which occurs when x 1 or x 1 when x 1 the given

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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 n1 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 n1 bn 3n belongs to C and is unequal to x and we have Ý |x bn | |a n y| 3n n1  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 n1 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 n2 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 n1 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 n1 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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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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