Assignment 4. u n+1 n(n + 1) i(i + 1) = n n (n + 1)(n + 2) n(n + 2) + 1 = (n + 1)(n + 2) 2 n + 1. u n (n + 1)(n + 2) n(n + 1) = n

Transcription

1 Assignment 4 Arfken 5..2 We have the sum Note that the first 4 partial sums are n n(n + ) s 2, s 2 2 3, s 3 3 4, s so we guess that s n n/(n + ). Proving this by induction, we see it is true for n, we assume it is true for n and verify it for n + s n+ n+ i i(i + ) n n + + (n + )(n + 2) n(n + 2) + (n + )(n + 2) n + n + 2 which completes the proof. Now, as n, s n and we conclude that the sum of our infinite series is. Arfken (a) (n + )(n + 2) + n(n + ) which, by Gauss test converges (h > ). (b) Use the integral test: which diverges. Therefore n ln n (c) By the ratio test: dx x ln x also diverges. + 2 n ln ln x x2 + n2 n (n + )2 n+ n 2 n + which goes to /2 as n and thus the series converges. (d) Use the integral test: ln + x dx ln( + x) ln x dx (x + ) ln x + (x + ) x ln x + x ln(x + ) + x ln( + x )

2 where the last two terms 0 at the upper limit, but notice that the first term in the last line diverges. Thus the integral and the sum diverge. (e) Use comparison with the sum in (b). If n n /n > n ln n then the sum will diverge. Simplifying the inequality, we only need to show (consider n > ) As n, ln n but ln n > n /n n /n e n ln n and the exponent goes to 0 (use, e.g. l Hospital s rule), the exponential goes to and the sum diverges. Arfken The hypergeometric series has as its n th term xn n! (α + n )!(β + n )!(γ )! (α )!(β )!(γ + n )! To find the range of convergence, use the ratio test initially + xn+ (α + n)(β + n) x n n + γ + n x n2 + n(α + β) + αβ n 2 + n(γ + ) + γ Note, that for x >, this diverges by the ratio test while for x <, by the same test, it converges as n. For x, we must use a more sensitive test such as the Gauss test. For it we need the inverse of this + x n 2 + n(γ + ) + γ n 2 + n(α + β) + αβ which is convergent for x provided γ + > α + β +. Arfken Anticipating using Gauss test, we construct (2s )!! (2s + 2)!!(2s + 3) + (2s)!!(2s + ) (2s + )!! (2s + 2)(2s + 3) (2s + )(2s + ) s s s 2 + s + 4 Since 5/2 > +, Gauss test is satisfied for convergence. Arfken 5.3. (a) The series is ( ) s (2s )!! (4s + 3) (2s + 2)!! s0 2

3 Testing for absolute convergence, construct + 4s + 3 (2s )!! (2s + 4)!! 4s + 7 (2s + )!! (2s + 2)!! 4s + 3 2s + 4 4s + 7 2s + s2 + s/4 + 3/2 s 2 + 9s/4 + 7/8 Since /4 < 9/4 +, Gauss test tells us that this series diverges. However, note that as an alternating series it may converge conditionally if we can show that the terms in the series are monotonically decreasing. To this end, consider the following infinite product definition of π (Eq. 5.24) π 2 (2n) 2 (2n + )(2n ) which we can interpret in the following way Using this, we have lim n n lim n (2n)!! (2n )!! 2 2n + (2n)!! π (2n )!! lim (2n + ) n 2 )!! lim (4s + 3)(2s s (2s + 2)!! lim (4s + 3) π s (2s)(2s + 2) 2 2s 0 Thus terms are monotonically decreasing and the series satisfies the Leibniz criteria for a conditionally convergent series. (b) The series is ( ) s (2s )!! (4s + 3) (2s)!! s0 Testing for absolute convergence, construct + 4s + 3 (2s )!! (2s + 2)!! 4s + 7 (2s + )!! (2s)!! 4s + 3 2s + 2 4s + 7 2s + s2 + 7s/4 + 3/4 s 2 + 9s/4 + 7/8 Since 7/4 < 9/4 +, Gauss test tells us that this series diverges. However, note that as an alternating series it may converge conditionally if we can show that the terms in the series are monotonically decreasing. Using the argument from part (a), we have )!! lim (4s + 3)(2s lim s (2s)!! s (4s + 3) π 2s 2 2s Thus it does not satisfy the Leibniz criteria for a conditionally convergent series and hence diverges. 3

4 Arfken ζ(n) k n n2 n2 k n2 k2 k2 n2 k2 k2 k2 l k n k n ( k 2 + k 3 + k 4 + k 2 k k(k ) (l + )l where the last line uses the result of problem Part (b) is virtually the same: ( ) n ζ(n) ( ) n k n n2 n2 n2 k2 k2 n2 k2 k2 ( k k ) n ( ) n k k 2 + k k(k + ) k k(k + ) ) 4

5 Arfken For the series, n0 /(+xn ), decide for what range of positive x values, it converges and is uniformly convergent. Use the ratio test, + + xn + x n+ If x >, this goes to /x < as n. On the other hand, if x, this ratio will go to as n. So we conclude that the series is absolutely convergent for x >. For uniform convergence, we must compare this to a series of numbers, M n, which is convergent and for which M n. Try the following p x p n < + x n n + x n < p n + x n < p n Since the series of numbers converges for p > (it s just a geometric series), our series of functions, converges uniformly for all < p x <. Arfken 5.6. ( + x) m/2 + ( m ) ( m)( m x ) x 2 2! + ( m )( m 2 2 )( m 2 2) x 3 + ( m )( m 2 2 )( m 2 2)( m 2 3) x 4 4! + + ( m )( m 2 2 ) ( m 2 (n 2))( m 2 (n )) x n n! + ( ) n x n m(m + 2)(m + 4) (m + 2n 4)(m + 2n 2) n! 2n n ( ) n x n (m + 2n 2)!! n! 2 n (m 2)!! n n Arfken The Doppler shift formulas are ν (a) ν ± v c + v2 c 2 ± ν (b) ν ± v c ν (c) ν ± v c + v2 2c 2 ± 5

6 Arfken Two binomial expansions: are added together to give x x n x x 0 n0 n which is obviously false. The problem comes in the ranges over which the expansions are convergent. The first expansion converges for x < while the second converges for x >. Thus, to add the resulting expansions together, defined as they are over different ranges, makes no sense; indeed it is an undefined operation. Arfken The linear combination is y(x 2h) 8y(x h) +8y(x + h) y(x + 2h) y(x) + 2hy (x) + (2h)2 2! + y(x) 2hy (x) + (2h)2 2! + 8 y(x) + hy (x) + h2 8 y(x) hy (x) + h2 x n x n x n y (x) + (2h)3 y (x) (2h)3 y (x) + (2h)4 4! y (x) + (2h)4 4! y (4) (x) + (2h)5 y (5) (x) + y (4) (x) (2h)5 y (5) (x) + 2! y (x) + h3 y (x) + h4 4! y(4) (x) + h5 y(5) (x) + 2! y (x) h3 y (x) + h4 4! y(4) (x) h5 2 2hy (x) + (2h)3 y (x) + (2h)5 y (5) (x) hy (x) + (h)3 y (x) + (h)5 y (5) (x) + 2h y 48 h5 (x) y (5) (x) + so that dividing by 2h gives the answer in the text. y(5) (x) + 6