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Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 11 Infinite Sequences and Series
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Slide 11 - 2 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Notice: Extra class 1. 9.30 am, 6 Oct 2007 Sabtu, SK1 2. 8.30 pm, 8 Oct 2007 Isnin, DKQ
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Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11.1 Sequences (2 nd lecture of week 24/09/07- 29/09/07)
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Slide 11 - 4 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What u a sequence A sequence is a list of numbers in a given order. Each a is a term of the sequence. Example of a sequence: 2,4,6,8,10,12,…,2n,… n is called the index of a n
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Slide 11 - 5 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In the previous example, a general term a n of index n in the sequence is described by the formula a n = 2n. We denote the sequence in the previous example by {a n } = {2, 4,6,8,…} In a sequence the order is important: 2,4,6,8,… and …,8,6,4,2 are not the same
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Slide 11 - 6 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Other example of sequences
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Slide 11 - 9 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 15 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 6: Applying theorem 3 to show that the sequence {2 1/n } converges to 0. Taking a n = 1/n, lim n ∞ a n = 0 ≡ L Define f(x)=2 x. Note that f(x) is continuous on x=L, and is defined for all x= a n = 1/n According to Theorem 3, lim n ∞ f(a n ) = f(L) LHS: lim n ∞ f(a n ) = lim n ∞ f(1/n) = lim n ∞ 2 1/n RHS = f(L) = 2 L = 2 0 = 1 Equating LHS = RHS, we have lim n ∞ 2 1/n = 1 the sequence {2 1/n } converges to 1
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Slide 11 - 18 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 7: Applying l’Hopital rule Show that Solution: The function is defined for x ≥ 1 and agrees with the sequence {a n = lnn /n} for n ≥ 1. Applying l’Hopital rule on f(x): By virtue of Theorem 4,
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Slide 11 - 19 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 9 Applying l’Hopital rule to determine convergence
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Slide 11 - 20 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution: Use l’Hopital rule
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Slide 11 - 22 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 10 (a) (ln n 2 )/n = 2 (ln n) / n 2 0 = 0 (b) (c) (d) (e) (f)
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Slide 11 - 23 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 12 Nondecreasing sequence (a) 1,2,3,4,…,n,… (b) ½, 2/3, ¾, 4/5, …,n/(n+1),… (nondecreasing because a n+1 -a n ≥ 0) (c) {3} = {3,3,3,…} Two kinds of nondecreasing sequences: bounded and non-bounded.
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Slide 11 - 24 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 13 Applying the definition for boundedness (a) 1,2,3,…,n,…has no upper bound (b) ½, 2/3, ¾, 4/5, …,n/(n+1),…is bounded from above by M = 1. Since no number less than 1 is an upper bound for the sequence, so 1 is the least upper bound.
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Slide 11 - 25 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 26 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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11.2 Infinite Series (3 rd lecture of week 24/09/07- 29/09/07)
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Slide 11 - 28 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 29 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example of a partial sum formed by a sequence {a n =1/2 n-1 }
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Slide 11 - 30 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 31 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Short hand notation for infinite series The infinite series may converge or diverge
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Slide 11 - 32 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Geometric series Geometric series are the series of the form a + ar + ar 2 + ar 3 + …+ ar n-1 +…= a and r = a n+1 /a n are fixed numbers and a 0. r is called the ratio. Three cases: r 1, r =1.
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Slide 11 - 33 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Proof of for |r|<1
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Slide 11 - 34 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For cases |r|≥1
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Slide 11 - 35 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 36 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 2 Index starts with n=0 The series is a geometric series with a=5, r=-(1/4). It converges to s ∞ = a/(1-r) = 5/(1+1/4) = 4
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Slide 11 - 37 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 5 A nongeometric but telescopic series Find the sum of the series Solution
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Slide 11 - 38 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Divergent series Example 6
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Slide 11 - 39 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The nth-term test for divergence Let S be the convergent limit of the series, i.e. lim n ∞ s n = = S When n is large, s n and s n-1 are close to S This means a n = s n – s n-1 a n = S – S = 0 as n ∞
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Slide 11 - 40 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Question: will the series converge if a n 0?
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Slide 11 - 41 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 7 Applying the nth-term test
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Slide 11 - 42 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 8 a n 0 but the series diverges The terms are grouped into clusters that add up to 1, so the partial sum increases without bound the series diverges Yet a n =2 -n 0
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Slide 11 - 43 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Corollary: Every nonzero constant multiple of a divergent series diverges If a n converges and b n diverges, then a n +b n ) and a n - b n ) both diverges.
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Slide 11 - 44 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Question: If a n and b n both diverges, must a n b n ) diverge?
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Slide 11 - 45 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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11.3 The Integral Test (3 rd lecture of week 24/09/07- 29/09/07)
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Slide 11 - 47 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Nondecreasing partial sums Suppose {a n } is a sequence with a n > 0 for all n Then, the partial sum s n+1 = s n +a n ≥ s n The partial sum form a nondecreasing sequence Theorem 6, the Nondecreasing Sequence Theorem tells us that the series converges if and only if the partial sums are bounded from above.
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Slide 11 - 49 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 1 The harmonic series The series diverges. Consider the sequence of partial sum The partial sum of the first 2 k term in the series, s n > k/2, where k=0,1,2,3… This means the partial sum, s n, is not bounded from above. Hence, by the virtue of Corollary 6, the harmonic series diverges
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Slide 11 - 52 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 4 A convergent series
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Slide 11 - 53 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Caution The integral test only tells us whether a given series converges or otherwise The test DOES NOT tell us what the convergent limit of the series is (in the case where the series converges), as the series and the integral need not have the same value in the convergent case.
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Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11.4 Comparison Tests (1 st lecture of week 01/10/07- 06/10/07)
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Slide 11 - 55 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 56 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 57 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Caution The comparison test only tell us whether a given series converges or otherwise The test DOES NOT tell us what the convergent limit of the series is (in the case where the series converges), as the two series need not have the same value in the convergent case
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Slide 11 - 59 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 60 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 2 continued
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Slide 11 - 61 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Caution The limit comparison test only tell us whether a given series converges or otherwise The test DOES NOT tell us what the convergent limit of the series is (in the case where the series converges)
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Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11.5 The Ratio and Root Tests (1 st lecture of week 01/10/07- 06/10/07)
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Slide 11 - 65 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Caution The ratio test only tell us whether a given series converges or otherwise The test DOES NOT tell us what the convergent limit of the series is (in the case where the series converges)
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Slide 11 - 66 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To skip
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Slide 11 - 67 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To skip
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Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11.6 Alternating Series, Absolute and Conditional Convergence (2 nd lecture of week 01/10/07-06/10/07)
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Slide 11 - 69 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Alternating series A series in which the terms are alternately positive and negative
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Slide 11 - 70 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The alternating harmonic series converges because it satisfies the three requirements of Leibniz’s theorem.
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Slide 11 - 73 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 75 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In other words, if a series converges absolutely, it converges.
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Slide 11 - 76 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Caution All series that are absolutely convergent converges. But the converse is not true, namely, not all convergent series are absolutely convergent. Think of series that is conditionally convergent. These are convergent series that are not absolutely convergent.
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Slide 11 - 77 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley p series with p=2
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Slide 11 - 78 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To skip
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11.7 Power Series (2 nd lecture of week 01/10/07- 06/10/07)
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Slide 11 - 82 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Mathematica simulation
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Slide 11 - 83 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Continued on next slide Note: To test the convergence of an alternating series, check the convergence of the absolute version of the series using ratio test.
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Slide 11 - 84 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 11 - 85 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The radius of convergence of a power series
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Slide 11 - 86 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley a a+R x a-R R R | x – a | < R
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Slide 11 - 87 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley R is called the radius of convergence of the power series The interval of radius R centered at x = a is called the interval of convergence The interval of convergence may be open, closed, or half-open: [a-R, a+R], (a-R, a+R), [a-R, a+R) or (a-R, a+R] A power series converges for all x that lies within the interval of convergence.
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Slide 11 - 88 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley See example 3 (previous slides and determine their interval of convergence
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Slide 11 - 91 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Caution Power series is term-by-term differentiable However, in general, not all series is term- by-term differentiable, e.g. the trigonometric series is not (it’s not a power series)
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Slide 11 - 92 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A power series can be integrated term by term throughout its interval of convergence
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Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11.8 Taylor and Maclaurin Series (3 rd lecture of week 01/10/07- 06/10/07)
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Slide 11 - 98 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Series Representation In the previous topic we see that an infinite series represents a function. The converse is also true, namely: A function that is infinitely differentiable f(x) can be expressed as a power series We say “the function f(x) generates the power series” The power series generated by the infinitely differentiable function is called Taylor series. The Taylor series provide useful polynomial approximations of the generating functions
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Slide 11 - 99 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding the Taylor series representation In short, given an infinitely differentiable function f(x), we would like to find out what is the Taylor series representation of f(x), i.e. what is the coefficients of b n in In addition, we would also need to work out the interval of x in which the Taylor series representation of f(x) converges. In generating the Taylor series representation of a generating function, we need to specify the point x=a at which the Taylor series is to be generated.
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Slide 11 - 100 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Note: Maclaurin series is effectively a special case of Taylor series with a = 0.
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Slide 11 - 101 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 1 Finding a Taylor series Find the Taylor series generated by f(x)=1/x at a= 2. Where, if anywhere, does the series converge to 1/x? f(x) = x -1 ; f '(x) = -x -2 ; f (n) (x) = (-1) n n! x (n+1) The Taylor series is
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Slide 11 - 102 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Mathematica simulation
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Slide 11 - 103 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Taylor polynomials Given an infinitely differentiable function f, we can approximate f(x) at values of x near a by the Taylor polynomial of f, i.e. f(x) can be approximated by f(x) ≈ P n (x), where P n (x) = Taylor polynomial of degree n of f generated at x=a. P n (x) is simply the first n terms in the Taylor series of f. The remainder, |R n (x)| = | f(x) - P n (x)| becomes smaller if higher order approximation is used In other words, the higher the order n, the better is the approximation of f(x) by P n (x) In addition, the Taylor polynomial gives a close fit to f near the point x = a, but the error in the approximation can be large at points that are far away.
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Slide 11 - 105 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 2 Finding Taylor polynomial for e x at x = 0 (To be proven later)
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Slide 11 - 106 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Mathematica simulation
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Slide 11 - 108 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Mathematica simulation
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Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11.9 Convergence of Taylor Series; Error Estimates (3 rd lecture of week 01/10/07-06/10/07)
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Slide 11 - 110 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When does a Taylor series converge to its generating function? ANS: The Taylor series converge to its generating function if the |remainder| = |R n (x)| = |f(x)-P n (x)| 0 as n ∞
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Slide 11 - 111 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley R n (x) is called the remainder of order n x x a c f(x)f(x) y 0 f(a)f(a)
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Slide 11 - 112 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley f(x) = P n (x) + R n (x) for each x in I. If R n (x) 0 as n ∞, P n (x) converges to f(x), then we can write
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Slide 11 - 113 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example 1 The Taylor series for e x revisited Show that the Taylor series generated by f(x)=e x at x=0 converges to f(x) for every value of x. Note: This can be proven by showing that |R n | 0 when n ∞
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Slide 11 - 114 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley x 0 c 0 x c y=e x exex ecec e0e0 e0e0 ecec exex
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Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11.10 Applications of Power Series (1 st lecture of week 08/10/07- 10/10/07)
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Slide 11 - 121 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The binomial series for powers and roots Consider the Taylor series generated by f(x) = (1+x) m, where m is a constant:
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Slide 11 - 122 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The binomial series for powers and roots This series is called the binomial series, converges absolutely for |x| < 1. (The convergence can be determined by using Ratio test, In short, the binomial series is the Taylor series for f(x) = (1+x) m, where m a constant
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Slide 11 - 125 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Taylor series representation of ln x at x = 1 f(x)=ln x; f '(x) = x -1 ; f '' (x) = (-1) (1)x -2 ; f ''' (x) = (-1) 2 (2)(1) x -3 … f (n) (x) = (-1) n-1 (n-1)!x -n ; Mathematica simulation
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11.11 Fourier Series (1 st lecture of week 08/10/07- 10/10/07)
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Slide 11 - 130 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley ‘Weakness’ of power series approximation In the previous lesson, we have learnt to approximate a given function using power series approximation, which give good fit if the approximated power series representation is evaluated near the point it is generated For point far away from the point the power series being generated, the approximation becomes poor In addition, the series approximation works only within the interval of convergence. Outside the interval of convergence, the series representation fails to represent the generating function Fourier series, our next topic, provide an alternative to overcome such shortage
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Slide 11 - 133 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A function f(x) defined on [0, 2 ] can be represented by a Fourier series x y 0 22 y = f(x) Fourier series representation of f(x)
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Slide 11 - 134 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley x y 0 22 … 44 88 -2
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Slide 11 - 135 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Orthogonality of sinusoidal functions
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Slide 11 - 136 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Derivation of a 0
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Slide 11 - 137 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Derivation of a k, k ≥ 1
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Slide 11 - 138 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Derivation of b k, k ≥ 1
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Slide 11 - 139 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Fourier series can represent some functions that cannot be represented by Taylor series, e.g. step function such as
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Slide 11 - 145 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Fourier series representation of a function defined on the general interval [a,b] For a function defined on the interval [0,2 ], the Fourier series representation of f(x) is defined as How about a function defined on an general interval of [a,b] where the period is L=b-a instead of 2 Can we still use to represent f(x) on [a,b]?
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Slide 11 - 146 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Fourier series representation of a function defined on the general interval [a,b] For a function defined on the interval of [a,b] the Fourier series representation on [a,b] is actually L=b - a
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Slide 11 - 147 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Derivation of a 0
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Slide 11 - 148 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Derivation of a k
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Slide 11 - 149 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example: 0 x y L 2L2L -L y=mL a=0, b=L
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Slide 11 - 151 Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley n=1 n=4 n=10 n=30 n=50 Click for mathematica
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