PHYSICS 116A Homework 1 Solutions. The distance traveled in 10 bounces (i.e. to reach the maximum height after the 10th bounce) is

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1 . The total distance is + ( 4 + PHYSICS 6A Homework Solutions ( ) + ) = + (/4) = +6 = 7m. 4 /4 The distance traveled in bounces (i.e. to reach the maximum height after the th bounce) is [ ( ) ( ) ] 9 ( ) ( ) = + (/4)[ (/4)9 ] + = = 6.6m /4 4. Boas, problem.. Find the limit of (n +5n )/(n + 4+n 6 ) as n. For n we have lim n n +5n n + 4+n 6 = lim n 5n (+/(5n)) n +n (+/n 6 ) = lim 5n n 5n =. (). Boas, problem.4. Write formulas for the sequences a n, S n, and R n and find the limit of the sequence as n (if the limit exists). This is a geometric series. The general term in this series is a n = ( ) n. Note that a n as n. Using eq. (.4) of Boas with a = and r = /, one obtains: The sum of the infinite series is given by and the remainder is given by R n = S S n = ) n S n = ( ( ) = S = lim n S n =, [ ( [ ( ) n ]. ) n ] = ( ) n 4. Boas, problem.5 7. Use the preliminary test to decide whether n= ( ) n n n + as n. n

2 is divergent or requires further testing. As n, n n + n n /. n/ Thus, this series passes the preliminary test, and requires further testing. In fact, as this is an alternating series, it passes the alternating series test(see section.7 of Boas), and we can conclude that the series is convergent. 5. Boas, problem.6 5. Test the following series for convergence using the comparison test: (a) n= n, (b) n= lnn. For any positive integer n, we have n < n so that n > n. Hence, using the comparison test, n= n > n= n =, since the harmonic series diverges. We conclude that series (a) also diverges. Likewise, for any positive integer n, we have lnn < n so that Hence, using the comparison test, lnn > n. n= and we conclude that series (b) also diverges. lnn > n =, 6. Boas, problem.6 5. Use the integral test to prove the following so-called p-series test. The series { n p is convergent, if p >, divergent, if p. n= n= First, consider p >. Then, n= (/np ) is convergent, since the integral, dx x p = ( p)x p = p,

3 is finite. Next, consider p <. In this case, n= (/np ) is divergent, since the integral, dx x p = ( p)x p =, is divergent (since lim x x p = if p < ). Finally, the case of p = corresponds to the harmonicseries. Thiswastreatedonp.ofBoas. Theintegraltest also exhibits the divergence, as dx x = lnx =. 7. Boas, problem.6. Use the ratio test to find whether converges or diverges. n! (n)! Using the ratio test, we evaluate: ρ n = (n+)!(n)! n!(n+)! = (n+), as n, (n+)(n+) where we used the definition of the factorial to obtain (n+)! = (n+)n! and similarly (n+)! = (n+)(n+)(n)!. Since ρ n ρ < as n, we conclude that the series converges. 8. Boas, problem.7. The following alternating series are divergent (but you are not asked to prove this). Show that a n. Why doesn t the alternating series test prove (incorrectly) that these series converge? (a) (b) In series (a), the general term is given by n if n is an odd positive integer, a n = if n is an even positive integer. n As n, we see that a n for both even and odd n. However, the a n are not monotonically decreasing as n increases. For example, if n is a positive even number, then a n = /n and a n+ = /(n+), but we have n+ > n, and so the condition a n+ a n is violated. Hence the alternating series test does not tell us whether the series converges or not. (In fact it diverges.)

4 In series (b), the general term is given by n+ if n is an odd positive integer, a n = n+ if n is an even positive integer. As n, we see that a n for both even and odd n. However, once again, the a n are not monotonically decreasing as n increases. For example, if n is a positive even number, then a n = /(n+) and a n+ = /(n+4), but we have n+4 > n+, so the condition a n+ a n is violated. Hence, again, the alternating series test does not tell us whether the series converges or not. 9. Boas, problem. 8. Find he interval of convergence the power series and check the endpoints of the interval for convergence or divergence. n= ( ) n x n n, () Using the ratio test, we compute ρ n = x n+ / n+ x n / n n = x x, as n. n+ The interval of convergence is determined by the condition ρ lim n ρ n <. Hence, the power series above converges for x < and diverges for x >. We now examine the endpoints x = and x = separately. At x =, eq. () is (conditionally) convergent, since it is an alternating series, n= ( )n / n, whose terms monotonically approach zero as n. At x =, eq. () reduces to n= n, which is divergent due to the p-series test [cf. problem 6 above]. Hence, it follows that the interval of convergence for the power series given in eq. () is: < x.. Boas, problem. parts (a) and (b). Find the first few terms of the Maclaurin series for the function e t dt. Find the general term of the Maclaurin series and write the series in summation form. Using the Maclaurin series for e y, where y t, it follows that e t = ( t ) n = n! Integrating from t = to t = x, and using t n dt = xn+ n+, 4 ( ) n tn n!.

5 one obtains e t dt = The first few terms of the series are given by: ( ) n x n+ (n+) n!. e t dt = x x! + x5 5! x7 7! +O(x9 ).. Boas, problem.4 6. Using Theorem 4.4 on p. 5 of Boas, show that ln( x) = x with an error less than.56 for x <.. Theorem 4.4 on p. 5 of Boas states that if S a nx n converges for x < and if a n+ < a n for all n > N, then N S a n x n < a N+, x <. x The Maclaurin series for ln( x) is given by: ln( x) = This series satisfies the conditions of Theorem 4.4 of Boas. Thus, if we take N =, we conclude that: x ln( x) x <, x <. ( x ) For x =., this inequality reads: n= x n n. ln( x) x < 5 9 <.56.. Boas, problem.5. Use a power series to evaluate: +x 4 cos(x ), at x =.. Compare your results with a calculation performed either with a computer (e.g. Mathematica) or a calculator. The series expansions are and +x 4 = x4! + 4 x x +. cos(x ) = x4! + x8 4! x! +. A proof of this theorem is given in the hints for Boas, 4, problem (c). 5

6 If < x, then it follows that +x 4 cos(x ) = [ 4 ] [ 5 x 8 4! 4 6 ] x +O(x 6 ) 6! = x x +O(x 6 ). Inserting x =. then yields +x 4 cos(x ) x=. I attempted the same computation by evaluating the left hand side above using using Mathematica 8 to evaluate the same quantity on a Mac (64 bit operating system). The result is. 6, which not correct. The answer is wrong because of rounding errors. By default, Mathematica uses 6-digit precision for numerical calculations. However, when two numbers which are almost equal are subtracted the result is much less accurate than this. Here each of the two numbers is close to, but the difference is only around 6 which is the limit of precision for each of these numbers separately. Hence virtually all precision is lost when they are subtracted.. Boas, problem.5 9. Use the Maclaurin series to evaluate sinx x lim x x. Using the Maclaurin series expansion for sin x, sinx = ( ) n xn+ (n+)! = x x! +O(x5 ), it follows that: sinx x x = x ] [x x! +O(x5 ) x = x ] [ x! +O(x5 ) = 6 +O(x ). In the limit of x, any term of O(x ) vanishes. Hence, sinx x lim x x = Boas.6-. Show that n= /n/ is convergent. What is wrong with the following proof that it diverges? > which is = 4 ( ). Since the harmonic series diverges, the original series diverges. Hint: compare n and n n. 6

7 Use the integral test. which is finite, and so the series converges. [ dn n / = n / (/) ] = Thefalse proof compares/n / with/(n),andclaimsthat/n / > /(n). Whilethisistrue forthefirstfewterms(includingthosethatareshown), the inequality is the other way round n > 9. In proofs of convergence the behavior of the first few terms is irrelevant; what matters is the behavior for n. Hence the proof is false. All one can show from this comparison is that the series is smaller than a divergent series, from which no conclusion about the convergence of the original series can be drawn. 5. Boas.6-4. Test for convergence n= n n!. Use the ratio test. using (n+)! = (n+)n!. Hence Since ρ <, the series converges. ρ n = a n+ a n = n+ n! n (n+)! = n+, ρ = lim ρ n = lim n n n+ =. 6. Boas, problem.6. Use series you know to show that: π! π4 5! + π6 =. 7! This result is a consequence of the Maclaurin expansions for sin x, sinx = Using this series, it follows that: ( ) n xn+ (n+)! = x x! + x5 5! x7 7! + sinx x = x! x4 5! + x6 7! Substituting x = π, and noting that sin(π) =, it follows that as requested. = π! π4 5! + π6 7!, 7