Lecture 3 - Tuesday July 5th

Transcription

1 Lecture 3 - Tuesday July 5th jacques@ucsd.edu Key words: Identities, geometric series, arithmetic series, difference of powers, binomial series Key concepts: Induction, proofs of identities 3. Identities Some very useful identities arise in the real numbers. It is very convenient at this point to introduce notation. If f : Z R is a function and m, n 0, then we write f( m + f( m f(n = f(. = m The capital sigma on the right in words means the sum of f( from = m to = n. There are some basic rules for doing sums for functions f, g : Z R. ( For c R, n n = m cf( = c = m f( ( n = m (f( + g( = n = m f( + n = m g(. The first identity is the sum of an arithmetic series: Theorem Arithmetic Series For any n N, n(n + =. This is easily proved by induction. The second identity is the geometric series, with the understanding x 0 =. Theorem Geometric Series For any x R\{} and n N, x = xn+ x. This is proved by induction on n. Clearly P ( is true since for n = both sides are +x. Assuming P (n is true, we have to show P (n is true. Now n x = +x n = xn x + xn = xn+ x as required. This completes the proof. A more general identity is the difference of powers identity, namely

2 Theorem 3 For y, z R and n N, z n y n = (z y z n y. = As an exercise one can prove this by induction on n. formula, by taing z = and y = x. It implies the geometric series Finally we do the binomial series. We define n! to be the product of all natural numbers up to an including n. Thus! =! = = 3! = 3 = 6 4! = 4 3 = 4 and so on. We define the binomial coefficients for n, r Z by n! = if n r 0 r r!(n r! and r = 0 otherwise. It is convenient to define 0! = so that 0 = = n. One of the main identities involving binomial coefficients is Pascal s Triangle Identity: Theorem 4 For, n N with n, ( ( n n = + We can prove this by noting that the left side is n!!(n! = n (n! (!(n! = n (n! (!(n! + (n! (!(n! n =!(n! + (n! (!(n! ( ( n n = + From this we can obtain the Binomial Series. Theorem 5 Binomial Series For x, y R and n N,

3 To prove this, we proceed by induction on n. Clearly P ( is true since both sides are x + y in that case. Suppose P (n is true. Then n (x + y (x + y n = (x + y x y n. By property ( of sums, this is n n x + y n + Replacing with in the first sum we get = n x y n + Since ( ( n n = 0 and n = 0 by definition, we can include = in the first sum and = n in the second sum to get x y n + By property ( of sums this is ( + By Pascal s Triangle Identity, the braceted term is, which completes the proof of the binomial series. The binomial series says ( + x n = n x. This gives another proof that ( + x n + nx when x 0: since x = 0 x 0 + x + + x n n x x = + nx. 3. Sequences Sequences are ordered analogs of sets. Let S N. A sequence whose nth entry for n S is a real number a n is denoted (a n n S. Clearly, sequences are just functions written in a different way: we could define f : S R by f(n = a n for n S, and then f is a function. A sequence is finite if S is finite, and infinite otherwise. For instance, 3, 5, 7, 9 3

4 is a finite sequence whereas the sequence of prime numbers, 3, 5, 7,,... is infinite. In the first case, the underlying function could be f( =, f( = 3, f(3 = 5, f(4 = 7 and f(5 = 9, and in the second case we could define f(n to be the nth prime, so f( =, f( = 3, f(3 = 5, f(4 = 7 and so on. How sequences are represented is very important. The first sequence of odd numbers, for example, could be easily represented by f(n = n + where n N. Using the sequence notation, this is (n + n N. The sequence of primes, however, has no such nice representation that we now of, since there is no explicit formula for the nth prime. It is traditional to let p n be the nth prime, in which case the sequence notation is (p n n N. There is a situation that is inbetween these two: for the odd numbers we have an explicit formula, but for the primes we don t even nown how to get to the next prime p n+ from any given prime p n. Sometimes, we now how to get between consecutive terms in a sequence, and in that case the sequence is defined recursively. 3.3 Recurrence equations A sequence (a n n N is defined recursively if a n is a function of a n, a n,..., a for n N. For instance, the sequence of odd numbers is defined recursively by a = and a n = a n +. The sequence of even numbers is defined recursively by a = and a n = a n +. The primes, however, have no nown recursive definition. One of the main techniques in mathematics is to solve recurrence equations. That is, we are given a sequence (a n n N defined recursively and we want to determine explicitly what the sequence is. From the equations a = and a n = a n +, we see easily that (a n n N is the sequence of odd numbers. Now if we are given a = and a n = a n, we see that a n is the sequence of powers of two. To prove this, and many other solutions of recurrence equations, we could proceed by induction. The problem is to find what the solutions are, and this is beyond the scope of the course. Rather, we are interested in the calculus of sequences. 3.4 Limits A sequence (a n n N has a limit a if successive terms of the sequence get arbitrarily close to a. For example, the sequence 0.0, 0.99, 0.999,... would be expected to have a = as its limit. To mae this mathematically rigorous, we define the limit of a sequence as follows: 4

5 Definition of the limit. Let (a n n N be a sequence of real numbers. Then (a n n N has the limit a, namely, lim n a n = a if and only if for every real number ε > 0 there exists N R such that a n a < ε for all natural numbers n > N. This relies on the archimedean property that for any natural number N there are natural numbers n N. From this definition, one can prove the existence of limits of certain sequences. 3.5 Examples of limits The sequence,,,... surely has zero as its limit. To prove this, first write the sequence 3 explicitly as a n = for n N. Then for any ε > 0, we must find N R such that n a n 0 < ε for n > N. Now a n 0 < ε is the same as < ε which is the same as n >. n ε Taing N =, we are done. ε Suppose a sequence is given by a n = n. Determine lim n+ n a n. A good guess for the limit would be since writing out the first few terms we get n We have to show for every ε > 0, that there is N such that for n N, < ε. n+ Now this is the same as < ε which is n >. So this is a suitable value of N. 4n+ 4ϵ We show that the limit of the sequence 0.9, 0.99, 0.999,... is a =. We may write the sequence explicitly as a n = 0 n for n N. According to the definition, we have to show that for any real number ε > 0, there is N N such that 0 n a < ε for n > N. In other words, we have to find an N as a function of ε for which 0 n < ε whenever n > N. Now 0 n < ε is the same as 0 n > ε. Now 0 n = ( + 9 n + 9n using the inequality ( + y n + yn for y 0. So if n >, then 9ε 0n > ε are done with N =. 9ε and so we 3.6 Unbounded sequences A sequence (a n n N is unbounded if the set S = {a n n N} has no upper bound. There are two ways that the limit of (a n n N can fail to exist: first when the sequence is unbounded 5

6 and second when the sequence oscillates. For instance, if a n = n for n N, then clearly the sequence is unbounded since N, the set of natural numbers, is unbounded as proved before using the completeness axiom. Other examples include a n = n, a n = n and even a n = ( n. We now that a n = ( n n oscillates wildly, the first few terms being and this sequence is certainly unbounded. We can chec as an exercise that all the sequences just mentioned are unbounded. Tae a n = n for instance. If S = {n n N} has an upper bound c, then n c for all n N. However, we now that there is a natural number n > c since N is unbounded. Now (n n n > c, and this is a contradiction. Therefore the given sequence a n = n is unbounded. Similarly one can show a n = ( n n and a n = n are unbounded. Another more complicated example of an unbounded sequence is a n = ω(n, which is the number of different prime factors of n N. The first few terms of this sequence are a = 0 a = a 3 = a 4 = a 5 = a 6 = and the sequence is unbounded since a n = when n = p p... p, the product of the first primes. Theorem 6 Unbounded sequences cannot have limits. To see this, suppose a n has no upper bound, and yet has a limit a. Then for every ϵ > 0, there exists N such that a n a < ε when n > N. Taing ε =, we get a n a < for all n > N, which means a n < a + for all n > N. Then set {a n n N} is finite, and therefore it is bounded above by its maximum element, b. Let c be the maximum of a + and b. Then we have a n c for all n N, contradicting that a n has no upper bound. There are bounded sequences which also have no limits: for example a n = ( n has no limit, and we can prove it. The contrapositive to the definition of the limit is that there exists ε > 0 such that for every a R and N R, there exists n > N such that a n a > ε. In the current situation, we have to show that there exists ε > 0 such that for every a R and N R, we can choose n > N such that ( n a > ε. Pic ε =. For any N and n > N odd, we have ( n a = a = a + and ( n+ a = a. Now it is the case that a+ + a > (exercise and therefore either a n a > or a n+ a >. 6