Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps

We bound the tail of the series of |a n |. ∞  n=N |a n | = ∞  n=N  |a n | 1/n  n ≤ ∞  n=N r n = r N 1 − r  ∞ n=0 a n is absolutely convergent. Example 12.1.8 Consider the series ∞  n=0 n a b n , where a and b are real constants. We use the root test to check for absolute convergence. lim n→∞ |n a b n | 1/n < 1 |b| lim n→∞ n a/n < 1 |b|exp  lim n→∞ 1 ln n n  < 1 |b| e 0 < 1 |b| < 1 Thus we see that the series converges absolutely for |b| < 1. Note that the value of a does not affect the absolute convergence. Example 12.1.9 Consider the absolutely convergent series, ∞  n=1 1 n 2 . 534 We aply the root test. lim n→∞ |a n | 1/n = lim n→∞     1 n 2     1/n = lim n→∞ n −2/n = lim n→∞ e − 2 n ln n = e 0 = 1 It fails to predict the convergence of the series. Raabe’s Test Result 12.1.5 The series  a n converges absolutely if lim n→∞ n  1 −     a n+1 a n      > 1. If the limit is less than unity, then the series diverges or converges conditionally. If the limit is unity, the test fails. 535 Gauss’ Test Result 12.1.6 Consider the series  a n . If a n+1 a n = 1 − L n + b n n 2 where b n is bounded then the series converges absolutely if L > 1. Otherwise the series diverges or converges conditionally. 12.2 Uniform Convergence Continuous Functions. A function f(z) is continuous in a closed d omain if, given any  > 0, there exists a δ > 0 such that |f(z) − f(ζ)| <  for all |z −ζ| < δ in the domain. An equivalent definition is that f(z) is continuous in a closed domain if lim ζ→z f(ζ) = f(z) for all z in the domain. Convergence. Consider a series in which the terms are functions of z,  ∞ n=0 a n (z). The serie s is convergent in a domain if the series converges for each point z in the domain. We can then define the function f(z) =  ∞ n=0 a n (z). We can state the convergence criterion as: For any given  > 0 there exists a function N(z) such that |f(z) − S N(z) (z)| =       f(z) − N(z)−1  n=0 a n (z)       <  for all z in the domain. Note that the rate of convergence, i.e. the number of terms, N(z) required for for the absolute error to be less than , is a function of z. 536 Uniform Convergence. Consider a series  ∞ n=0 a n (z) that is convergent in some domain. If the rate of convergence is independent of z then the series is said to be uniformly convergent. Stating this a little more mathematically, the series is uniformly convergent in the domain if for any given  > 0 there exists an N, inde pendent of z, such that |f(z) − S N (z)| =      f(z) − N  n=1 a n (z)      <  for all z in the domain. 12.2.1 Tests for Uniform Convergence Weierstrass M-test. The Weierstrass M-test is useful in determining if a series is uniformly convergent. The series  ∞ n=0 a n (z) is uniformly and absolutely convergent in a domain if there exists a convergent series of positive terms  ∞ n=0 M n such that |a n (z)| ≤ M n for all z in the domain. This condition first implies that the series is absolutely convergent for all z in the domain. The condition |a n (z)| ≤ M n also ensures that the rate of convergence is independent of z, which is the criterion for uniform convergence. Note that absolute convergence and uniform convergence are independent. A series of functions may be absolutely convergent without being uniformly convergent or vice versa. The Weierstrass M-test is a sufficient but not a necessary condition for uniform convergence. The Weierstrass M-test can succeed only if the series is uniformly and absolutely convergent. Example 12.2.1 The series f(x) = ∞  n=1 sin x n(n + 1) is uniformly and absolutely convergent for all real x because | sin x n(n+1) | < 1 n 2 and  ∞ n=1 1 n 2 converges. 537 Dirichlet Test. Consider a sequence of monotone decreasing, positive constants c n with limit zero. If all the partial sums of a n (z) are bounded in some closed domain, that is      N  n=1 a n (z)      < constant for all N, then  ∞ n=1 c n a n (z) is uniformly convergent in that closed domain. Note that the Dirichlet test does not imply that the series is absolutely convergent. Example 12.2.2 Consider the series, ∞  n=1 sin(nx) n . We cannot use the Weierstrass M-test to determine if the series is uniformly convergent on an interval. While it is easy to bound the terms with |sin(nx)/n| ≤ 1/n, the sum ∞  n=1 1 n does not converge. Thus we will try the Dirichlet test. Consider the sum  N−1 n=1 sin(nx). This sum can be evaluated in closed form. (See Exercise 12.9.) N−1  n=1 sin(nx) =  0 for x = 2πk cos(x/2)−cos((N−1/2)x) 2 sin(x/2) for x = 2πk The partial sums have infinite discontinuities at x = 2πk, k ∈ Z. The partial sums are bound ed on any closed interval that does not contain an integer multiple of 2π. By the D irich let test, the sum  ∞ n=1 sin(nx) n is uniformly convergent on any such closed interval. The series may not be uniformly convergent in neighborhoods of x = 2kπ. 538 12.2.2 Uniform Convergence and Continuous Functions. Consider a series f(z) =  ∞ n=1 a n (z) that is uniformly convergent in s ome domain and whose terms a n (z) are continuous functions. Since the series is uniformly convergent, for any given  > 0 there exists an N such that |R N | <  for all z in the domain. Since the finite sum S N is continuous, for that  there exists a δ > 0 such that |S N (z) −S N (ζ)| <  for all ζ in the domain satisfying |z −ζ| < δ. We combine these two results to show that f(z) is continuous. |f(z) − f(ζ)| = |S N (z) + R N (z) − S N (ζ) − R N (ζ)| ≤ |S N (z) − S N (ζ)| + |R N (z)| + |R N (ζ)| < 3 for |z −ζ| < δ Result 12.2.1 A uniformly convergent series of continuous terms represents a continuous function. Example 12.2.3 Again consider  ∞ n=1 sin(nx) n . In Example 12.2.2 we showed that the convergence is uniform in any closed interval that does not contain an integer multiple of 2π. In Figure 12.2 is a plot of the first 10 and then 50 terms in the series and finally the function to which the series converges. We see that the function has jump discontinuities at x = 2kπ and is continuous on any closed interval not containing one of those points. 12.3 Uniformly Convergent Power Series Power Series. Power series are series of the form ∞  n=0 a n (z −z 0 ) n . 539 Figure 12.2: Ten, Fifty and all the Terms of  ∞ n=1 sin(nx) n . Domain of Convergence of a Power Ser ies Consider the series  ∞ n=0 a n z n . Let the series converge at some point z 0 . Then |a n z n 0 | is bounded by some constant A for all n, so |a n z n | = |a n z n 0 |     z z 0     n < A     z z 0     n This comparison test shows that the series converges absolutely for all z satisfying |z| < |z 0 |. Suppose that the series diverges at some point z 1 . Then the series could not converge for any |z| > |z 1 | since this would imply convergence at z 1 . Thus there exists some circle in the z plane such that the power series converges absolutely inside the circle and diverges outside the circle. Result 12.3.1 The domain of convergence of a power series is a circle in the compl ex plane. Radius of Convergence of Power Series. Consider a power series f(z) = ∞  n=0 a n z n 540 Applying the ratio test, we see that the series converges if lim n→∞ |a n+1 z n+1 | |a n z n | < l lim n→∞ |a n+1 | |a n | |z| < 1 |z| < lim n→∞ |a n | |a n+1 | Result 12.3.2 Ratio formula. The radius of convergence of the power series ∞  n=0 a n z n is R = lim n→∞ |a n | |a n+1 | when the limit exists. Result 12.3.3 Cauchy-Hadamard formula. The radius of convergence of the power series: ∞  n=0 a n z n is R = 1 lim sup n  |a n | . 541 Absolute Convergence of Power Series. Consider a power series f(z) = ∞  n=0 a n z n that converges for z = z 0 . Let M be the value of the greatest term, a n z n 0 . Consider any point z such that |z| < |z 0 |. We can bound the residual of  ∞ n=0 |a n z n |, R N (z) = ∞  n=N |a n z n | = ∞  n=N     a n z n a n z n 0     |a n z n 0 | ≤ M ∞  n=N     z z 0     n Since |z/z 0 | < 1, this is a convergent geometric series. = M     z z 0     N 1 1 − |z/z 0 | → 0 as N → ∞ Thus the power series is absolutely convergent for |z| < |z 0 |. Result 12.3.4 If the power series  ∞ n=0 a n z n converges for z = z 0 , then the series c onverges absolutely for |z| < |z 0 |. Example 12.3.1 Find the radii of convergence of the following series. 542 [...]... formula √ 1 1 1 √ =√ z+3 z +2 z 2 + 5z + 6 1 1 √ =√ 3 1 + z/3 2 1 + z /2 1 −1 /2 z 2 −1 /2 z = √ 1+ + + ··· 3 2 3 1 6 1 z 3z 2 z z2 = √ 1− + + ··· 1− + + ··· 6 24 4 32 6 1 5 17 = √ 1 − z + z2 + · · · 12 96 6 12. 6 1+ −1 /2 z −1 /2 + 1 2 2 z 2 2 + ··· Laurent Series Result 12. 6.1 Let f (z) be single-valued and analytic in the annulus R1 < |z − z0 | < R2 For points in the annulus, the function has the convergent... ln(n + 20 ) 4n + 1 3n − 2 ∞ (Logπ 2) n 8 n=0 ∞ 9 n =2 ∞ 10 n =2 n2 − 1 n4 − 1 n2 (ln n)n ∞ (−1)n ln 11 n =2 ∞ 12 n =2 ∞ 13 n =2 1 n (n! )2 (2n)! 3n + 4n + 5 5n − 4n − 3 5 62 ∞ 14 n =2 ∞ 15 n =2 ∞ 16 n=1 ∞ 17 n=1 ∞ 18 n=1 ∞ 19 n=1 ∞ 20 n =2 n! (ln n)n en ln(n!) (n! )2 (n2 )! n8 + 4n4 + 8 3n9 − n5 + 9n 1 1 − n n+1 cos(nπ) n ln n n11/10 Hint, Solution Exercise 12. 4 (mathematica/fcv/series/constants.nb) Show that the... n=1 1 2n−1 1 1 3n 5n+1 Hint, Solution Exercise 12. 12 Evaluate the following sum ∞ ∞ ∞ ··· k1 =0 k2 =k1 kn =kn−1 Hint, Solution 56 5 1 2kn 12. 7 .2 Uniform Convergence 12. 7.3 Uniformly Convergent Power Series Exercise 12. 13 Determine the domain of convergence of the following series ∞ 1 n=0 ∞ 2 n =2 ∞ 3 n=1 ∞ 4 n=1 ∞ 5 n=1 ∞ 6 n=1 ∞ 7 n=0 ∞ 8 n=0 zn (z + 3)n Log z ln n z n (z + 2) 2 n2 (z − e)n nn z 2n 2nz... 0 and z = 3 but diverges for z = 2 20 There exists a power series an (z − z0 )n which converges for z = 0 and z = 2 but diverges for z = 2 Hint, Solution Exercise 12. 3 Determine if the following series converge ∞ 1 n =2 ∞ 2 n =2 1 n ln(n) 1 ln (nn ) ∞ ln 3 √ n ln n n =2 56 1 ∞ 4 1 n(ln n)(ln(ln n)) n=10 ∞ 5 n=1 ∞ 6 n=0 ∞ 7 n=0 ln (2n ) ln (3n ) + 1 1 ln(n + 20 ) 4n + 1 3n − 2 ∞ (Logπ 2) n 8 n=0 ∞ 9 n =2. .. Example 12. 5. 3 Find the Taylor series expansion of 1/(1 + z) about z = 0 For |z| < 1, 1 −1 −1 2 −1 3 =1+ z+ z + z + ··· 1+z 1 2 3 = 1 + (−1)1 z + (−1 )2 z 2 + (−1)3 z 3 + · · · = 1 − z + z2 − z3 + · · · Example 12. 5. 4 Find the first few terms in the Taylor series expansion of √ z2 1 + 5z + 6 about the origin 55 4 We factor the denominator and then apply Newton’s binomial formula √ 1 1 1 √ =√ z+3 z +2 z 2 + 5z... 1 = 1−z ∞ zn, n=0 5 52 for |z| < 1  12. 5. 1 Newton’s Binomial Formula Result 12. 5 .2 For all |z| < 1, a complex: (1 + z)a = 1 + where a r = a a 2 a 3 z+ z + z + ··· 1 2 3 a(a − 1)(a − 2) · · · (a − r + 1) r! If a is complex, then the expansion is of the principle branch of (1 + z)a We define r 0 0 r = 1, = 0, for r = 0, 0 0 = 1 Example 12. 5 .2 Evaluate limn→∞ (1 + 1/n)n First we expand (1 + 1/n)n using... 2 n=−∞ C f (ζ) dζ z n ζ n+1 For the case of arbitrary z0 , simply make the transformation z → z − z0 55 7 Im(z) Im(z) r2 R2 r1 R1 C1 C2 R2 Re(z) R1 z Re(z) C Cz Figure 12. 5: Contours for a Laurent Expansion in an Annulus Example 12. 6.1 Find the Laurent series expansions of 1/(1 + z) For |z| < 1, 1 −1 −1 2 −1 3 =1+ z+ z + z + ··· 1+z 1 2 3 = 1 + (−1)1 z + (−1 )2 z 2 + (−1)3 z 3 + · · · = 1 − z + z2... Exercise 12. 8 Show that the harmonic series, ∞ n=1 1 1 1 = 1 + α + α + ··· , nα 2 3 converges for α > 1 and diverges for α ≤ 1 Hint, Solution 56 4 Exercise 12. 9 Evaluate N −1 sin(nx) n=1 Hint, Solution Exercise 12. 10 Evaluate n n kz k k2z k and k=1 k=1 for z = 1 Hint, Solution Exercise 12. 11 Which of the following series converge? Find the sum of those that do 1 1 1 1 1 + + + + ··· 2 6 12 20 2 1 + (−1)... (−1)3 z 3 + · · · = 1 − z + z2 − z3 + · · · 55 8 For |z| > 1, 1 1/z = 1+z 1 + 1/z 1 −1 −1 −1 2 = 1+ z + z + ··· z 1 2 = z −1 − z 2 + z −3 − · · · 55 9 12. 7 Exercises 12. 7.1 Series of Constants Exercise 12. 1 Show that if an converges then limn→∞ an = 0 That is, limn→∞ an = 0 is a necessary condition for the convergence of the series Hint, Solution Exercise 12. 2 Answer the following questions true or false... 2nz z n! (n! )2 z ln(n!) n! 56 6 ∞ 9 n=0 ∞ 10 n=0 (z − π)2n+1 nπ n! ln n zn Hint, Solution Exercise 12. 14 Find the circle of convergence of the following series 1 z + (α − β) ∞ 2 n=1 z2 z3 z4 + (α − β)(α − 2 ) + (α − β)(α − 2 )(α − 3β) + · · · 2! 3! 4! n (z − ı)n 2n ∞ nn z n 3 n=1 ∞ 4 n=1 n! n z nn ∞ (3 + (−1)n )n z n 5 n=1 ∞ (n + αn ) z n 6 (|α| > 1) n=1 Hint, Solution 56 7 Exercise 12. 15 Find the circle . closed form. (See Exercise 12. 9.) N−1  n=1 sin(nx) =  0 for x = 2 k cos(x /2) −cos((N−1 /2) x) 2 sin(x /2) for x = 2 k The partial sums have infinite discontinuities at x = 2 k, k ∈ Z. The partial. |R N (ζ)| < 3 for |z −ζ| < δ Result 12. 2.1 A uniformly convergent series of continuous terms represents a continuous function. Example 12. 2.3 Again consider  ∞ n=1 sin(nx) n . In Example 12. 2 .2 we. of 2 . By the D irich let test, the sum  ∞ n=1 sin(nx) n is uniformly convergent on any such closed interval. The series may not be uniformly convergent in neighborhoods of x = 2kπ. 53 8 12. 2.2