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Unformatted text preview: SECTION 2.4 Sequences and Summations The ﬁrst half of this exercise set contains a lot of routine practice with the concept of and notation for
sequences. It also discusses telescoping sums; the product notation, corresponding to the summation
notation discussed in the section; the factorial function, which occurs repeatedly in subsequent chapters;
and a few challenging exercises on more complicated sequences. The last part of the exercise set has some
fairly challenging exercises involving inﬁnite sets. Do not be surprised ifyou ﬁnd this last material very strange
and hard to comprehend at ﬁrst going; mathematicians did not understand it at all until the late nineteenth
century, three hundred years after calculus was well understood. To show that an inﬁnite set is countable, you
need to ﬁnd a oneto—one correspondence between the set and the set of positive integers. One way to do this
directly is to provide a listing of the elements of the set. (There is no listing, for instance, of the set of real
numbers.) Various indirect means are also available, such as showing that the set is a subset of a countable
set, or showing that it is the union of a countable collection of countable sets. 64 1. Chapter 2 Basic Structures: Sets, Functions, Sequences, and Sums 3) Un=2(~3)”+5‘3:21+1:3
C) “A? '13li+5“:281+625:787 b) e1:2(—3)1+51=2(—3)+5=~1
d) a, = 2  (—3)5 + 55 z 2 (4243) +3125 : 2639 . In each case we simply evaluate the given function at n = 0, 1, 2, 3. a) a0:20+i:2,a,=21+i:3, a2=22+i:5,a,:23+1=9 b) 0.021121, (11:29:11, (12:33:27, (13:44:256 c) (10 = [0/2] :0, (1.1 = [1/21 : 0, 9.2 : [2/2] :1, (13 :13/2j21 d) «.0 :10/21+[0/21:0+0 : 0, a1 : Ll/2j+i1_/2i = 0+1 : 1, a2 =
or, 2 [3/2] + [3/21 : 1+ 2 : 3. Note that [11/21 + always equals n. L2/2j+[2/21:1+1: 2, . In each case we just follow the instructions. a) 2,5,8,11,14,17,20,23,26,29 b) 1,1,1,2,2,2,3,3,3,4 c) 1,1,3,3,5,5,7,7,9,9 d) This requires a bit of routine calculation. For example, the ﬁfth term is 5! i 2‘3 = 120 i 32 = 88. The ﬁrst
ten terms are —1, ~2, —2, 8,88,656,4912,40064, 362368, 3627776. e) 3,6,12,24,48,.96,192,384,768,1536 f) 1, 1,2,3,5,8, 13, 21, 34, 55 g) For n = 1, the binary expansion is 1, which has one bit, so the ﬁrst term of the sequence is 1. For n : 2,
the binary expansion is 10, which has two bits, so the second term of the sequence is 2. Continuing in this
way we see that the ﬁrst ten terms are 1,2,2,3,3,3,3,4,4,4. Note that the sequence has one 1, two 2‘s,
four 3’s, eight 4‘s, as so on, with 2;“1 copies of Ir. h) The English word for 1 is “one” which has three letters, so the ﬁrst term is 3. This makes a good
brainteaser; give someone the sequence and ask her or him to ﬁnd the pattern. The ﬁrst ten terms are
3,3,5,4,4,3,5.5,4,3. One pattern is that each term is twice the preceding term. A formula for this would he that the 12'“ term
is 2’“1 . Thus to move from the ﬁrst term to the second we add 1; to move from the second to the third we add 2; Another pattern is that we obtain the next term by adding increasing values to the previous term.
then add 3, and so on. So the sequence would start out 1,2, 4, 7,11, 16,22, . we could also have trivial
answers such as the rule that the ﬁrst three terms are 1,2,4 and all the rest are 17 (so the sequence is
1,2,4,17,17,17,...), or that the terms simply repeat 1, 2, 4, 1, 2, 4, 1, 2,4, . . .. Here is anotl'ier pattern: Take
71. points on the unit circle, and connect each of them to all the others by line segments. The inside of the
circle will be divided into a number of regions. lNhat is the largest this number can be? Call that value (in.
If there is one point, then there are no lines and therefore just the one original region inside the circle; thus
(11 2 1. If n 2 2, then the one chord divides the interior into two parts, so 02 = 2. Three points give us a
triangle, and that makes four regions (the inside of the triangle and the three pieces outside the triangle), so
:13 : 4. Careful drawing shows that the sequence starts out 1, 2,4, 8, 16, 31. That’s right: 31, not 32. Creative students may well ﬁnd other rules or patterns with various degrees of appeal. . In some sense there are no right answers here. The solutions stated are the most appealing patterns that the author has found. a) it looks as if we have one 1 and one 0, then two of each, then three of each, and so on, increasing the
number of repetitions by one each time. Thus we need three more 1’s (and then four 0’s) to continue the
sequence. b) A pattern here is that the positive integers are listed in increasing order, with each even number repeated.
Thus the next three terms are 9, 10,10. e) The terms in the odd locations (ﬁrst, third, ﬁfth, etc.) are just the successive terms in the geometric
sequence that starts with 1 and has ratio 2, and the terms in the even locations are all 0. The nth term is
0 if n is even and is 21””11/2 if n is odd. Thus the next three terms are 32,0,64. Section 2.4 11. 13. 15. Sequences and Sunin'iations 65 d) The ﬁrst. term is 3 and each successive term is twice its predecessor. This is a geometric sequence. The
nth term is 3 l 2””. Thus the next three terms are 384, 768, 1536. e) The first term is 15 and each successive term is 7' lees than its predecessor. This is an arithmetic sequence.
The it‘ll term is 22 — Tn. Thus the next three terms are 734,—411, 748. f) The rule is that the first term is 3 and the n'“ term is obtained by adding It to the (n. — 1)”1 term.
One can actually find a quadratic expression for a sequence in which the successive differences form an
arithmetic sequence; here it is (n2 + n + 4)/2. The easiest way to see this is to note that the em term is
3 + 2 + 3 l— 4 + 5 + 0 +    + n. Except. for the initial 3 instead of a l, the nth term is the sum of the ﬁrst 71',
positive integers, which is ri(ri + 1)/2 by a. formula in Table 2. Therefore the at" term is (Mn —1— 1)/2) + 2,
as claimed. We see that the next. three terms are 57, (3'8, 80. g) One should play around with the sequence if nothing is apparent at ﬁrst. Here we note that all the terms
are even, so if we divide by 2 we obtain the sequence 1, 8, 27', 64, 125, 216, 343, . . .. This sequence appears in
Table 1:_ it is the cubes. So the rim term is 2n3. Thus the next three terms are 1024, 1458, 2000. h) These terms look close to the terms of the sequence whose nth term is 11.! (see Table 1). In fact, we see
that the 11‘“ term here is n] + 1. Thus the next three terms are 302881,3628801,:39916801. it is pretty clear that on should be approximately equal to n. —i— since the sequence is just the sequence
of positive integers with perfect squares left out. There are about \/T—1 perfect squares up to 71, so the count
needs to get ahead by about this amount. Proving that this plausibility argument gives the correct formula
involves some careful counting. The sequence begins 2, 3, 5, 6, T, 8,10,11,....
1),, that jumps every time a perfect Square is encountered. Thus {5,} begins 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, . . .; We can write it as the sequence on : it plus a sequence there are two 1's, four 2’s, six 3’s, eight 4’s, and so on. So we must. show that 1),, 2 where means the integer closest to ﬂ (note that there is never an ambiguity here, since this will never be a half—
iuteger). Because of the way the sequence is formed, 1),, g k if and only if 2 + 4 + 6 +    + 2k 2 n. This
is equivalent to His —l— 1.) 2 n. Applying the quadratic formula and recalling that it is an integer, we obtaiu [(71 + \/1 + 4n.)/2l. Simplifying, we have (in : iwé —— “in + . Subtracting % ‘n, + i is a half—integer#see Exercise 43 in 1),, : and then rounding up is the same as rounding to the nearest integer (the smaller one if 1 {\ 1.5+ But it can neVer happen that ME 3 m l— 2 while Section 2.3), so (with. this understanding) b“ :
yin + > in + % for some positive integer ms rsthis would imply that n :4 m? + m. + i and n. > m2 + m, an
   .  . i 1 . impossibility. Thelefoie — { \ n + 3 } , and we are done. An alternative solution is provided in the answer section of the text. a)2+3+4+5+6=20 b)1—2+478+16:11 c)3+3++3~——103=30
d) This series “telescopes”: each term cancels part of the term before it (see also Exercise 19). The sum is
(2— 1)+(4—2)+(874)++(512—256):—1+512=511. n+1 we use the formula for the sum of a geometric progression: Z;‘zoor1 : a(r i 1)/(r i l).
21) Here a = 3, 'r : 2, and ft : 8, so the sum is 3(29 71)/(2 71): 1533.
13) Here. a. : 1, r' 2 2, and n : 8. The sum taken over all the values of j from 0 to n. is, by the formula,
(29 * 1)/(2 * 1) : 511. However, our sum starts at j : 1, so we must subtract out the term that isn’t there,
namely 20. Hence the answer is 511 — 1 2 510. c) Again we have to subtract the missing terms, so the sum is ((—3)9 — 1)/((—3) — 1) — (—3)“ — [—3)1 :
4921. i 1.7 (*3) : 4923, d) 2((—3)9 — 1)/((43) A 1) : 98.42 66 17. 19. 21. 23. 25. 27. 29. 31. Chapter 2 Basic Structures: Sets, Functions, Sequences, and Sums The easiest way to do these sums. since the number of terms is reasonably small, is just to write out the
summands explicitly. Note that the inside index runs through all of its values for each value of the outside
index a) (1+1)+(1+2)+(1+3)+(2+1)+(2+2)+(2+3)=21 b) (0+3+6+9)+(2+5+8+11)+(4+7+10+13):78 c) (1+1+1)+(2+2+2)+(3+3+3)= 18 d) (0+0+0)+(1 +2+3)+(2+4+6) = 18 If we just write out what the sum means, we see that parts of successive terms cancel, leaving only two terms:
72
E (61;; “(Edi—i) = (21 “ac +02 r 01+ 0‘3 r 612 + ' ' ' +ﬂa—1 ‘ 0:14 +Gn _ (In—1 = on man
j:1 a) \Ve use the hint. where at = #22: H. :1 2(23; *1) 2 Z“; _ (k _1)2) : n2 _ 02 2 n2 kil k=1
b) We can use the distributive law to rewrite TLLQk — 1) (which we know from part (:1) equals 112) in
terms of the sum we want, 8 : 22:] 1:: Ti 112:2(2k:~1)=2ik—i1=2S—n.
19:1 k:l L‘zl Now we solve for S, obtaining S = (n12 + n)/27 which is usually expressed as + 1) /2. This exercise is like Example 15. From Table 2 we know that 2:031 k = 200 I 201/2 = 20100, and 23:1 1: :
99  100/2 : 4950. Therefore the desired sum is 20100 — 4950 2 15150. If we write down the ﬁrst few terms of this sum we notice a pattern. It starts (1 + 1 + 1) + (2 + 2 + 2 +
2+2)+(3+3+3+3+3+3+3)+.
in general there are + 1)2 — i2 = 21i+ 1 copies of So we need to sum i(2i + l) for an appropriate There are three l’s, then live 255, then seven 3'ss, and so on; range of values for We must ﬁnd this range. It gets a little messy at the end if m. is such that the — 1. Then there are
n + 1 blocks, and (71+ 1)2 — l is where the next—to—last block ends. The sum of those complete blocks is
21:1 i(2:i + l) : 2:; 27? + = n(n + 1)(2'n. + 1)/3 + n.(n. + l)/2. The remaining terms in our summation
all have the value n + 1 and the number of them present is m — + 1)2 ~ 1). Our ﬁnal answer is therefore n(n +1)(2n +1)/3+n(n+1)/2 +(n+1)(m ~ (21+ 1)2 +1). sequence stops before a complete range of the last value is present. Let n. = b) 5678:1680 c) Each factor is either 1 or ~1, so the product is either 1 or —1. To see which it is, we need to determine a) 0 (anything times 0 is 0)
how man}r of the factors are —1. Clearly there are 50 such factors, namely when 1'. : 1, 3, 5, . . ., 99. Since
(—1)50 : 11 the product is 1. d) 222=210 =1024 01+1l+2i+35+4i=1+1+l2+123+1234=1+1+2+6+24234 a) The negative integers are countable. Each negative integer can be paired with its absolute value to give
the desired one—tooue correspondence: 1 H —1, 2 H —2, 3 H —3, etc. Section 2. 4 33. 35. 37. 39. 41. Sequences and Summations 67 b) The even integers are countable. \Ve can list the set of even integers in the order 0, 2, —2,4, 74,6, —6,. . .,
and pair them with the positive integers listed in their natural order. Thus 1 H 0, 2 <—> 2, 3 <—> i2, 4 H 4,
etc. There is no need to give a. formula for this correspondeneeithe discussion given is quite sufﬁcient; but it.
is not hard to see that we are pairing the positive integer n with the even integer f(n), where f(n) : n if n
is even and f(n) : 1 k n. if n is odd. (3) The proof that the set of real numbers between 0 and 1 is not countable (Example 21) can easily be
modified to show that the set of real numbers between 0 and 1 / 2 is not countable. We need to let the digit d,
be something like 2 if (if; 7% 2 and 3 otherwise. The number thus constructed will be a real number between
0 and 1/2 that is not in the list. (1) This set is countable, exactly as in part (b); the only diﬁ'erence is that there we are looking at the multiples
of 2 and here we are looking at the multiples of 7. The correspondence is given by pairing the positive integer
n with 711/2 if n. is even and —7(n i 1)/2 if n, is odd. 3) The. bit strings not containing 0 are just the bit strings consisting of all 1’s, so this set. is {A, 1,11,111,
1111, . . where A denotes the empty string (the string of length 0). Thus this set is countable, where the
correSpondence matches the natural number n. with the string of n, 1‘s. b) This is a subset of the set ofrational numbers, so it is countable (see Exercise 36). To find a correspondence,
we can just follow the path in Example 20, but omit fractions in the top three rows (as well as continuing to
omit these fractions not in lowest terms). c) This set is uncountable, as can be shown by applying the diagonal argument. of Example 21. d) This set is uncountable, as can be shown by applying the diagonal argument of Example 21. Yes. We need to look at this from the other direction, by noting that A 2 B U (A 7 B). We are given that
B is countable. If A — B were also countable, then, since the union of two countable sets is countable (which
we are asked to prove as Exercise 40), we would conclude that A is countable. But we are given that A is
not countable. Therefore our assumption that A — B is countable is wrong, and we conclude that A — B is
uncountable. (This is an example of a proof by contraposition.) This is just the contrapositive of Exercise 36 and so follows directly from it. In more detail, suppose that B
were countable, say with elements ()1, b2, . . .. Then since A Q B, we can list the elements of A using the order
in which they appear in this listing of B. Therefore A is countable, contradicting the hypothesis. Thus B is not countable. By what we are given, we know that there are bijections f from A to B and g from C to D. Then we can
deﬁne a bijection from A x C to B X D by sending (a, c) to (f(a),g(c)). This is clearly onetoone and onto,
so we have shown that A X C and B X D have the same cardinality. Since empty sets do not contribute any elements to unions, we can assume that none of the sets in our
given countable collection of countable sets is the emptyr set. If there are no sets in the collection, then
the union is empty and therefore countable. Otherwise let the countable sets be A], A2, (If there
are only a finite number k of them, then we can still assume that they form an inﬁnite sequence by taking
AH] : A5,.” :   ~ : A] Since each set A, is countable and nonempty, we can list its elements in a sequence
as on, (1,2, again, if the set is finite we can list its elements and then list ail repeatedly to assure an
inﬁnite sequence. Now we just need a systematic way to put all the elements aid, into a sequence. We do
this by listing ﬁrst all the elements rig. in which i+j = 2 (there is only one such pair, (1,1)), then all the
elements in which i +j = 3 (there are only two such pairs, (1, 2) and (2, 1)), and so on; except that we do
not list any element that we have already listed. 80, assuming that these elements are distinct, our list starts on, (1.12, (131, (1,13, 0.22, 0.31, 0.14, (If any of these terms duplicates a previous term, then it is simply 68 43. 45. 47. Chapter 2 Basic Structures: Sets, Functions, Sequences, and Sums omitted.) The result of this process will be either an inﬁnite sequence or a ﬁnite sequence containing all the
elements of the union of the sets A,. Thus that union is countable. There are only a ﬁnite number of bit strings of each ﬁnite length, so we can list all the bit strings by listing ﬁrst
those of length 0, then those of length 1, etc. The listing might be A, 0, 1, 00, 01, 10, 11,000, 001, . . .. (Recall
that A denotes the empty string.) Actually this is a special case of Exercise 41: the set of all bit strings is the
union of a countable number of countable (actually finite) sets, namely the sets of bit strings of length n. for
n20,l,2,.... “Te argued in the solution to Exercise 43 that the set of all strings of symbols from the alphabet {0,1} is
countable, since there are only a ﬁnite number of bit strings of each length. There was nothing special about
the alphabet {0,1} in that argument. For any ﬁnite alphabet {for example, the alphabet consisting of all
upper and lower case letters, numerals, and punctuation and other mathematical marks typically used in a
programming language), there are only a ﬁnite number of strings of length 1 (namely the number of symbols
in the alphabet), only a ﬁnite number of strings of length 2 (namely, the square of this number), and so on.
Therefore, using the result of Exercise 41 again, we conclude that there are only countably many strings from
any given ﬁnite alphabet. Now the set of all computer programs in a particular language is just a subset of
the set of all strings over that alphabet (some strings are meaningless jumbles of symbols that are not valid
programs), so by Exercise 36, this set, too, is countable. In Exercise 45 we saw that there are only a countable number of computer programs, so there are only a
countable number of computable functions. In Exercise 46 we saw that there are an uncountable number
of functions. Hence not all functions are computable. Indeed, in some sense, since uncormtable sets are so
much bigger than countable sets, almost all functions are not computable! This is not really so surprising; in
real life we deal with only a small handful of useful functions, and these are computable. Note that this is
a nonconstructive proof—we have not exhibited even one noncomputable function, merely argued that they
have to exist. Actually ﬁnding one is much harder, but it can be done. For example, the following function
is not computable. Let T be the function from the set of positive integers to {0,1} deﬁned by letting T(n)
be 0 if the number 0 is in the range of the function computed by the 71”" computer program (where we list
them in alphabetical order by length) and letting T07.) = 1 otherwise. ...
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This note was uploaded on 02/18/2012 for the course MATH 640:244 taught by Professor Ming during the Spring '09 term at Rutgers.
 Spring '09
 Ming

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