Problem Set 6 Solutions

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1 6.04/18.06J Mathmatics for Computr Scinc March 15, 005 Srini Dvadas and Eric Lhman Problm St 6 Solutions Du: Monday, March 8 at 9 PM in Room Problm 1. Sammy th Shark is a financial srvic providr who offrs loans on th following trms. Sammy loans a clint m dollars in th morning. This puts th clint m dollars in dbt to Sammy. Each vning, Sammy first chargs a srvic f, which incrass th clint s dbt by f dollars, and thn Sammy chargs intrst, which multiplis th dbt by a factor of p. For xampl, if Sammy s intrst rat wr a modst 5% pr day, thn p would b (a) What is th clint s dbt at th nd of th first day? Solution. At th nd of th first day, th clint ows Sammy (m + f)p = mp + fp dollars. (b) What is th clint s dbt at th nd of th scond day? Solution. ((m + f)p + f)p = mp + fp + fp (c) Writ a formula for th clint s dbt aftr d days and find an quivalnt closd form. Solution. Th clint s dbt aftr thr days is (((m + f)p + f)p + f)p = mp 3 + fp 3 + fp + fp. Gnralizing from this pattrn, th clint ows mp d + d fp k dollars aftr d days. Applying th formula for a gomtric sum givs: ( ) p mp d d f p 1 1 Problm. Find closd-form xprssions qual to th following sums. Show your work.

2 Problm St 6 (a) 9 i 7 i i Solution. Split th xprssion into two gomtric sris and thn apply th formula for th sum of a gomtric sris. 9 i 7 i i = ( ) i 9 = 1 ( = ) n+1 ( 9 ( ) i 7 1 ( ) n+1 ) n ( 7 ) n (b) n i=1 3 4i+5 Solution. Taking th logarithm rducs this product to an asy sum. n i=1 3 4i+5 = 3 log 3( Q n i=1 34i+5 ) = 3 P n i=1 4i+5 = 3 n(n+1)+5n (c) j=1 j 5/3 ( 1 1 ) i j 1/3 Solution. This farsom-looking sum is a papr tigr; w just apply th formula for th sum of a gomtric sris followd by th formula for th sum of an arithmtic sris. j=1 j 5/3 ( 1 1 ) i = j 1/3 = j 5/3 1 j j=1 j=1 = n(n + 1 )(n + 1) 3 1 ( ) 1 1 j 1/3

3 Problm St 6 3 Problm 3. Thr is a bug on th dg of a 1-mtr rug. Th bug wants to cross to th othr sid of th rug. It crawls at 1 cm pr scond. Howvr, at th nd of ach scond, a malicious first-gradr namd Mildrd Andrson strtchs th rug by 1 mtr. Assum that hr action is instantanous and th rug strtchs uniformly. Thus, hr s what happns in th first fw sconds: Th bug walks 1 cm in th first scond, so 99 cm rmain ahad. Mildrd strtchs th rug by 1 mtr, which doubls its lngth. So now thr ar cm bhind th bug and 198 cm ahad. Th bug walks anothr 1 cm in th nxt scond, laving 3 cm bhind and 197 cm ahad. Thn Mildrd striks, strtching th rug from mtrs to 3 mtrs. So thr ar now 3 (3/) = 4.5 cm bhind th bug and 197 (3/) = 95.5 cm ahad. Th bug walks anothr 1 cm in th third scond, and so on. Your job is to dtrmin this poor bug s fat. (a) During scond i, what fraction of th rug dos th bug cross? Solution. During scond i, th lngth of th rug is 100i cm and th bug crosss 1 cm. Thrfor, th fraction that th bug crosss is 1/100i. (b) Ovr th first n sconds, what fraction of th rug dos th bug cross altogthr? Solution. Th bug crosss 1/100 of th rug in th first scond, 1/00 in th scond, 1/300 in th third, and so forth. Thus, ovr th first n sconds, th fraction crossd by th bug is: 1 100k = H n/100 (This formula is valid only until th bug rachs th far sid of th rug.) (c) Approximatly how many sconds dos th bug nd to cross th ntir rug? Solution. Th bug arrivs at th far sid whn th fraction it has crossd rachs 1. This occurs whn n, th numbr of sconds lapsd, is sufficintly larg that H n / Now H n is approximatly ln n, so th bug arrivs about whn: ln n ln n 100 n sconds

4 4 Problm St 6 Problm 4. Us intgration to find lowr and uppr bounds on th following infinit sum that diffr by at most 0.1. Show your work. S = To achiv this accuracy, add up th first fw trms xplicitly and thn us intgration to bound all rmaining trms. Solution. Th sum of th first thr trms is: An uppr bound on th rmaining trms is: And a lowr bound is: s = = x dx = (x + 1) dx = 1 4 Ovrall, w hav: S = Ths bounds diffr by 1/1 < 0.1. Th actual valu of th sum is π /6, though th proof is not asy. Problm 5. A sasond MIT undrgraduat can: Complt a problm st in days. Writ a papr in days. Tak a -day road trip. Study for an xam in 1 day. Play foosball for an ntir day. An n-day schdul is a squnc of activitis that rquir a total of n days. For xampl, hr ar thr possibl 7-day schduls: pst, papr, pst, foosball papr, study, foosball, pst, study road trip, road trip, road trip, study

5 Problm St 6 5 (a) Exprss th numbr of possibl n-day schduls using a rcurrnc quation and sufficint bas cass. Solution. S(0) = 1, S(1) =. Any schdul for n > 1 days nds with on of 3 possibl -day activitis or on of possibl 1-day activitis. So S(n) = S(n 1) + 3S(n ) for n > 1. (b) Find a closd-form xprssion for th numbr of possibl n-day schduls by solving th rcurrnc. Solution. Th charactristic polynomial for this linar homognous rcurrnc is x x 3 = (x + 1)(x 3). Hnc th solution is of th form S(n) = a( 1) n + b3 n. Ltting n = 0, w conclud that a+b = 1, and ltting n = 1, w conclud a+3b =, so b = 3/4, a = 1/4, and th solution is: S(n) = 3n+1 + ( 1) n. 4 Problm 6. Find a closd-form xprssion for T (n), which is dfind by th following rcurrnc: T (0) = 0 T (1) = 1 T (n) = 5T (n 1) 6T (n ) + 6 for all n Solution. Th charactristic quation is x 5x + 6 = 0, which has roots x = and x = 3. Thus, th homognous solution is: T (n) = A n + B 3 n For a particular solution, lt s first guss T (n) = c: c = 5c 6c + 6 c = 3 Our guss was corrct; T (n) = 3 is a particular solution. Adding this to th homognous solution givs th gnral solution: T (n) = A n + B 3 n + 3

6 6 Problm St 6 Substituting n = 0 and n = 1 givs: 0 = A + B = A + 3B + 3 Solving this systm givs A = 7 and B = 4. Thrfor: T (n) = 7 n n + 3 Problm 7. Dtrmin which of ths choics Θ(n), Θ(n log n), Θ(n ), Θ(1), Θ( n ), Θ( n ln n ), non of ths dscribs ach function s asymptotic bhavior. Proofs ar not rquird, but brifly xplain your answrs. (a) n + ln n + (ln n) Solution. Both n > ln n and n > (ln n) hold for all sufficintly larg n. Thus, for all sufficintly larg n: n < n + ln n + (ln n) < n + n + n So n + ln n + (ln n) = Θ(n). (b) n + n 3 n 7 Solution. Obsrv that: n + n 3 lim = 1 n n 7 This mans, that for all sufficintly larg n, th fraction lis, for xampl, btwn, 0.99 and 1.01 and is thrfor Θ(1). (c) i+1 Solution. Gomtric sums ar dominatd by thir largst trm, which is n+1 = 4 n. This is Θ(4 n ), which dos not appar in th list providd. (d) ln(n!) Solution. By Stirling s formula: n! πn ( n ) n

7 Problm St 6 7 Taking logarithms givs: ln(n!) ln( πn ( n ) n = ln( ( n πn ) + ln Th first trm is tiny compard to th scond, which w can rwrit as: ( n ln ) n ) ) n ( ) n = n ln = Θ(n ln n) () k (1 1 ) k Solution. Th xprssion in parnthss is always at last 1/ and at most 1. Thus, w hav th bounds: 1 k k (1 1 ) k k Sinc th first xprssion and th last ar both Θ(n ), so is th on in th middl. Problm 8. A triangular numbr is an intgr n of th form whr k is a positiv intgr. n = k = k(k + 1) (a) Dscrib a solution to th four-pg Towrs of Hanoi puzzl with k(k + 1)/ disks that rquirs T k movs, whr: Solution. T 1 = 1 T k = T k 1 + k 1 Mov all but th k largst disks to anothr pg rcursivly. This rquirs T (k 1) movs. Mov th k largst disks to anothr pg using th thr-pg stratgy. This rquirs k 1 movs. Now mov all th othr disks on top of th k largst disks rcursivly. This rquirs T (k 1) movs.

8 8 Problm St 6 Thus, with this stratgy, th total numbr of movs rquird to mov a stack of k(k + 1)/ disks is T (k) = T (k 1) + k 1. (b) Find a closd form xprssion qual to T k. Solution. This is an inhomognous linar quation. Lt s bgin by trying to find a particular solution. Thr is both an xponntial trm ( k ) and a constant trm, so w might guss somthing of th form a k + c: a k + c = (a k 1 + c) + k 1 = (a + 1) k + c 1 0 = k + (c 1) Evidntly, th constant trm is c = 1, but th xponntial part is mor complicatd. Our rcip says w should nxt try a particular solution of th form a k + bk k + 1: a k + bk k + 1 = (a k 1 + b(k 1) k 1 + 1) + k 1 = (a b + 1) k + bk k 1 Equating th cofficints of th k trms givs a = a b + 1, which implis b = 1. Thus, a k +k k +1 is a particular solution for all a. As long as w hav this dgr of frdom, w might as wll choos a so this solution is consistnt with th boundary condition T 1 = 1 and b don: a = 1 a = 1 Thrfor, th solution to th rcurrnc is T k = (k 1) k + 1. (c) Approximatly how many movs ar rquird to solv th four-pg, n-disk Towrs of Hanoi puzzl as a function of n? Assum n is a triangular numbr. (For styl points, mak corrct us of asymptotic notation.) Solution. W hav k = 1 ( 8n + 1 1) = n + O(1). So th numbr of movs rquird is Θ( n n ).

Homework #3. 1 x. dx. It therefore follows that a sum of the

Homework #3. 1 x. dx. It therefore follows that a sum of the Danil Cannon CS 62 / Luan March 5, 2009 Homwork # 1. Th natural logarithm is dfind by ln n = n 1 dx. It thrfor follows that a sum of th 1 x sam addnd ovr th sam intrval should b both asymptotically uppr-