Mathematical Preliminaries

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1 Matheatical Preliiaries I this chapter we ll review soe atheatical cocepts that will be used throughout this course. We ll also lear soe ew atheatical otatios ad techiques that are iportat for aalysis of algoriths.. Floors ad Ceiligs I doig coputer coputatios, we soeties have to discard the fractioal part of a real uber x, while i other cases we ay have to roud upward to the first iteger larger tha or equal to x. It is coveiet to have siple atheatical otatios that idicate these operatios. The floor ad ceilig otatios were iveted for this purpose. Notatios. We shall use N for the set of cardial or atural ubers, N = {0,,,...}, Z for the set of itegers, Q for the set of ratioal ubers, ad R for the set of real ubers. Defiitio.. The floor of a real uber x, deoted by x, is the largest iteger that is less tha or equal to x. The ceilig of a real uber x, deoted by x, is the sallest iteger that is greater tha or equal to x. We ca express these defiitios i the followig way: x =ax{k : k Z, k x}; x =i{k : k Z, k x}. Exaples =3; 0.04 =0; 0.04 = ; 6 =6;.04 =; 3.7 =4; 0.04 =; 0.04 =0; 6 =6;.04 =3. Spriger Iteratioal Publishig Switzerlad 04 D. Vrajitoru ad W. Kight, Practical Aalysis of Algoriths, Udergraduate Topics i Coputer Sciece, DOI 0.007/ _ 9

2 0 Matheatical Preliiaries I a C or C++ progra, a expressio of the for / idicates iteger divisio whe ad are of a iteger data type. The atheatical expressio deotes the exact value of the C/C++ expressio / whe ad are positive. Thus, whe reasoig about progras i which iteger divisio is perfored, we will ake repeated use of expressios such as. If you cosult the docuetatio for the C/C++ ath.h or cath library header file, you will fid that it cotais the followig two prototypes: double ceil (double x); double floor(double x); Note that the values retured by these two fuctios are of type double, ot log. This ay surprise you util you realize that floatig poit real ubers ca be far larger tha the iteger LONG_MAX, ad so i order to have expressios such as floor(3.8e9) retur a sesible value, the fuctio ust retur the iteger i floatig poit for. The followig lea gives us a property of itegers that we will use ofte i this chapter. Lea..3 Let ad be ay iteger ubers. If <, the + ad. Proof Suppose <. If + were greater tha, the would be a iteger strictly betwee ad +. That is ipossible. Thus + ust be less tha or equal to. It follows fro + that. The followig theore lists soe properties of floor ad ceilig that we will use throughout i this book. It gives us a better idea about how the floor ad the ceilig relate to their arguet ad to each other. Theore..4 Let x ad y be ay real ubers. Let,, ad k be ay itegers. (a) x x < x+; equivaletly, x < x x. (b) x < x x ; equivaletly, x x < x +. (c) If x < + (or x < x), the = x; siilarly, if < x (or x < x + ), the = x. (d) If x y, the x y ad x y. Siilarly for x y. (e) If x, the x. If x the x. (f) If x has a iteger value, the x =x = x. If x has a o-iteger value, the x = x+. (g) ( )x =( ) x ad ( )x =( ) x. Proof (a) Take ay real uber x. The iequality x x is true because the floor of x is a eleet of a set of itegers less tha or equal to x. The iequality x < x+ is true because x is the largest of the itegers less tha or equal

3 . Floors ad Ceiligs to x, ad x+ is a iteger larger tha x,so x+ caot be i the set of itegers less tha or equal to x. Thus x+ is greater tha x. This copletes the proof that x x < x +. To prove that x < x x, we eed oly prove that x < x because we have already proved that x x. Take the iequality x < x+, which we have already proved, ad subtract fro both sides. This gives x < x. (b) Take ay real uber x. The iequality x x is true because the ceilig of x is a eleet of a set of itegers greater tha or equal to x. To prove the iequality x < x, first ote that x is the sallest of the itegers greater tha or equal to x, ad x is saller tha x,so x caot be i the set of itegers greater tha or equal to x. Thus x is saller tha x. This copletes the proof that x < x x. The iequalities x x < x + are proved by takig apart the iequalities above, addig to both sides of x < x, ad the puttig the pieces back together agai i a differet order. (c) Let x be ay real uber, ad let be a iteger satisfyig x < +. The ust be the largest of the itegers that are less tha or equal to x because by Lea..3 every iteger greater tha is at least +. By defiitio of x, it follows that = x. The secod assertio has a siilar proof. (d) Suppose x y. We ll prove that x yad leave the other assertios as exercises. First ote that either x < y or y x. Ifx < y, the we ca cobie this with the iequality x x (see part (a)) to coclude that x < y. If, istead, y x, we ca cobie this with x y ad y < y+ to obtai y x < y +. Sice y is a iteger, it follows fro part (c) that x = y. (e) Suppose x. The by (d), x. Sice has a iteger value, =, which eas that x. The proof of the secod part is siilar. (f) Suppose x has a iteger value. The trivially x belogs to the set {k : k Z, k x} ad is the largest eleet of the set, so x = x. Siilarly, x = x. Now suppose x is ot a iteger. We will prove that x = x+ by showig that x x+ ad x x+. Sice x is ot a iteger, the x < x < x, fro which we get x < x. Sice both x ad x are itegers, we ca use Lea..3 to coclude that x+ x. By part (a): x < x+. Sice x+ is a iteger, we ca apply (e) to deduce that x x+. Fro the two iequalities that we have proved, we ca coclude that x = x+. (g) Let us start with the secod iequality fro (a): x < x x. Multiply through by to reverses the iequalities: ( )( x ) >( )x ( )( x ). This ca be rewritte fro right to left as ( ) x ( )x <( ) x +. We ca see that the iteger ( ) x satisfies the first part of (c) with respect to the real uber ( )x, ad thus it ust be the floor of this uber: ( ) x = ( )x. The secod part is proved i a siilar way.

4 Matheatical Preliiaries The followig theore shows soe properties of the floor ad ceilig fuctios i relatio to arithetic operatios applied to their arguets. Theore..5 Let x ad y be ay real ubers. Let,, ad k be ay itegers. (a) x ± = x ± ad x ± = x ± ; but i geeral, x = x ad x = x ad x + y = x + y ad x + y = x + y. (b) + =. + (c) = ; equivaletly, + =. y y y y (d) If k > 0, the = ad =. I particular, if = 0 we ca k k k k put y = to get these idetities: = ad =. k k k k Proof (a) Take ay particular iteger. We ll prove that x + = x+. Sice x+ is a iteger, by Theore..4 (c) it suffices to prove that ( x+) x + < ( x +) +. These two iequalities are logically equivalet to ( x+) x + <( x+) +, which are i tur equivalet to x x < x +, which are true for all x (see Theore..4 (a)). This copletes the proof for floors. The proof for ceiligs is siilar. The equatio x = x is proved by otig that x = x + ( ) = x +( ) = x because is a iteger. The assertios ivolvig x ad x are deostrated by usig exaples: = ad = (b) Take ay iteger ad real uber x. If is eve, the = for soe iteger, ad thus + = + = + = + = =. If is odd, the = + for soe iteger, ad thus + = = By part (a) of this theore, this last expressio equals = = + =. (c) Take ay iteger. If is eve, the = for soe iteger, so + + = = + = + = + 0 = ad + = = = also, so =.

5 . Floors ad Ceiligs 3 + If is odd, the = + for soe iteger, ad thus = + + = + = +, ad = = + = + + = + also, ad thus =. The fact that + = + follows fro part (a) ad siple algebra. y (d) Take ay real uber y ad ay iteger k > 0. By defiitio of the floor, k y y is a iteger. To prove that = it suffices by Theore..4 (c) to k k y y prove that the iteger satisfies the iequalities y y k k k < + k (take x to be y y k ad to be ). k y First, let s prove y y k k. By Theore..4 (a), y y k k,so k y. k y Sice k is the product of two itegers, it is a iteger, ad it is less or equal k y y to y,so k yby Theore..4 (e). That proves y k k k. +, we ote that y k + > y k To prove that y y k < by Theore..4 (a), k ad y k y y k because y yby Theore..4 (a). Thus + > y k k, which is what we wated to prove (i tured-aroud for). The proof of the assertio about ceiligs is quite siilar ad will be oitted here. As a exaple of where soe of the forulas i Theores..4 ad..5 are useful, cosider what happes durig a biary search of a subarray a[first..last]. Here is a fraget of the code for the search (for the coplete fuctio see Fig. 5.3, p. 76): while (first <= last &&!foud) // The value paraeters first ad last will { // be odified durig executio of this code it id = (first + last)/; if (target < a[id])... A siilar calculatio of id = (first + last)/ occurs durig the Merge Sort algorith, which splits a array ito two pieces, sorts the recursively, ad the erges the back together. Whe aalyzig such algoriths, it is useful to have atheatical expressios for the legths of the followig three saller subarrays a[first..id-] a[first..id] a[id+..last] i ters of the legth of the larger subarray a[last].

6 4 Matheatical Preliiaries first id last a: / = ( )/ / / Fig.. Theore..6 0 first- first last Fig.. Theore..6 a Theore..6 Cosider a subarray a[first..last] of soe array a, where 0 first last. Let deote the legth of the subarray, ad let id deote the quatity (first + last) / (this is the average of the subscripts at the eds of the subarray). The (a) is the value of the expressio last - first + (b) the legth of the subarray a[first..id] is ; (c) the legth of the subarray a[first..id-] is = ; (d) the legth of the subarray a[id+..last] is. See Fig.. for a graphical represetatio. Proof Part (a) is proved by cosiderig the full array that starts at locatio 0 ad eds at locatio last. There are last + cells i that subarray. If we the reove the cells i the subarray a[0..first-] we produce the subarray a[first..last]. We are reovig (first-)+ = first cells fro last + cells, leavig (last+)-first = last-first+ cells. See Fig... For the proof of stateet (b), we begi by otig that the subarray a[first.. id] cotais id-first+ cells, ad id = (first+last)/. The we ca write first + last id first + = first + first + last = first + (by Theore..5 (a))

7 . Floors ad Ceiligs 5 first + last first + last first + = = (last first + ) + = + = =. To prove part (c), ote that the subarray a[first..id-] has oe less cell tha a[first..id]. By part (b), the legth of a[first..id] is /, so the legth of a[first..id-] is + = = by Theore..5 (c). To prove part (d), subtract the uber of cells i a[first..id] fro to obtai the uber of cells i a[id+..last]. We get = by Theore..5 (b)... Further Exaples Exaple..7 (a) Solutio (a) =? (b) =? (c) =? (d) =? = 3 + 5/7 =3 (b) 4 (c) 0 (d) 7 Exaple..8 Use part (a) of Theore..4 to prove that 0 x x < for all real ubers x. Solutio By Theore..4 (a) we kow that x x < x + for all real x. We ca pull this expressio apart ito the followig pair of iequalities: x x ad x < x+ for all real x. Fro the first of these it follows that 0 x x (subtract x fro both sides of the iequality); fro the secod it follows siilarly that x x <. Thus we have 0 x x ad x x < for all x. Puttig these back together gives the desired expressio: 0 x x <. Exaple..9 Verify part (a) of Theore..5 i the case where x =.46 ad = 5. Solutio x + =.46+( 5) = 7.46 =7, ad x+ =.46+( 5) = 5 = 7. Thus we have verified i this oe particular case that x + = x+. Siilarly, x + =.46 + ( 5) = 7.46 = 8 ad.46 +( 5) = 3 5 = 8.

8 6 Matheatical Preliiaries Exaple..0 Verify part (c) of Theore..5 for soe odd iteger ad also for soe eve iteger Solutio For = 3, is it true that =, i.e., that 7 = 6.5? Yes: both are equal to 7. For = 4, is it true that =, i.e., that 7.5 = 7? Yes: both are equal to 7. Exaple.. For each of the assertios below, cite oe or ore parts of Theores..4 ad..5 to help you justify the assertio. p + q p + q (a) For all itegers p ad q, + = p + q. (b) For all itegers, + = +. p (c) For all itegers, p, ad q, ifp = ad q =, the q =. More 4 p p p geerally, if p =, p =, p 3 =,...,p =, the p =. + (d) For all itegers, =. Solutio (a) Use part (b), with = p + q. (b) Use part (a), with x = ad =. p (c) Use part (d) with = ad k =. That is, q = = =. The secod setece i (c) is proved by iductio. 4 (d) Use part (c), with =. = Exaple.. I the C++ loop give below, assue that the variable has soe positive iteger value. Whe the loop is executed, how ay ties will the body of the loop be executed? (I this loop, the body of the loop cosists of just the cout stateet.) Express your aswer as a atheatical forula i ters of the variable. it k; for (k = ; (3*k)+ <= ; ++k) cout << k << edl; Solutio First ote that the iitial value coputed for (3*k) + is 4, so if the value of is 3 or less, the body of the loop will be executed 0 ties. So ow let s look at the cases where the value of is at least 4. I these cases the body of the loop will be executed at least oce. Sice the variable k starts with the value ad is icreeted by each tie the body of the loop is executed, k is a loop couter. That is, whe its value is i

9 . Floors ad Ceiligs 7 the cout stateet, the body of the loop is beig executed for the first tie. Whe its value is i the cout stateet, the body of the loop is beig executed for the secod tie. Etc. Let L deote the value of k i the cout stateet the last tie the body of the loop is executed. The that uber L is the total uber of ties the body of the loop is executed. That is, L is the uber we wat to copute. Let deote the value of the progra variable. The 3L + because whe k has the value L, the loop cotrol coditio (3 k)+ <= ust be true. (If that were ot so, the body of the loop would ot be executed whe k gets the value L.) Also ote that whe k is the icreeted to L + (i the expressio k++), the loop cotrol coditio ust the be false i order to ake the loop halt. Thus 3(L + ) + >. We ow have the followig two iequalities ivolvig L ad : 3L + < 3(L + ) +. Subtractig fro each the part of this expressio, ad the dividig each part by 3 gives the iequalities L < L + 3 It follows that the iteger L ust bethe floor of the fractio (-)/3 (see Theore..4 (c)). That is, L =. This is the uber of ties the body of 3 the loop is executed whe 4. I fact, this forula gives the correct uber of ties eve i the case where is,, or 3. Exaple..3 Prove that for all real ubers x ad y we have x + y x + y +. Solutio Take ay real ubers x ad y. By Theore..4 (b), x < x + ad y < y +. It follows that x + y < x + y +. By Theore..4 (b), x + y x + y, sox + y + x + y +. Puttig together the iequalities i the two precedig seteces we get x + y < x + y +. Note that the left ad right sides of this iequality are itegers. If i < j for itegers i ad j, the we kow that i j, so x + y x + y +... Exercises..4 Let a be the ae of a array of legth > 0 i a C or C++ progra. Explai why the expressio a[/] will be accepted by the copiler (eve though ay be a odd iteger) but a[floor(/.0)] will ot copile...5 Use part (b) of Theore..4 to prove that 0 x x < for all real ubers x...6 Is it always true that x = + x? If so, prove this fact. If ot, give a couterexaple, i.e., a exaple of a real uber for which the equatio is ot true.

10 8 Matheatical Preliiaries..7 Theore..4 (d) says that if x y the x y for all real ubers x ad y. Ca we coclude that if x < y the x < y?..8 (a) Give a exaple of a iteger ad a real uber x for which x = x. (See part (a) of Theore..5.) (b) Give a exaple of two real ubers x ad y for which x + y = x+ y. (See part (a) of Theore..5.)..9 Verify part (b) of Theore..5 for the odd iteger = ad also for the eve iteger =...0 (a) Verify part (d) of Theore..5 i the case where = 37, = 8, ad k = 3. (b) Verify part (d) of Theore..5 i the case where = 6, = 7, ad k =... Is it true that x = x for all itegers ad real ubers x? Aswer the sae questio for the equatio x = x... For each of the assertios below, cite oe or ore parts of Theores..4 ad..5 to help you justify the assertio. + (a) For all itegers, + =. (b) For all o-egative real ubers x, x = x. + (c) For all itegers, + = +. (d) For all real ubers x ad o-egative itegers p, x + p = x + p. (e) For all real ubers x, x + x = x...3 I the C++ loop give below, assue that the iteger variable has soe positive value. How ay ties is the body of the loop executed? Express your aswer as a atheatical expressio i ters of the value of. Justify your aswer by a arguet siilar to the oe i Exaple.. o p. 7. Also check your aswer carefully by deteriig what itegers are set to output whe = 50. it k; for (k = ; k*k <= ; ++k) cout << k << edl;..4 I the C++ loop give below, assue that the iteger variable has soe positive value. How ay ties is the body of the loop executed? Express your aswer as a atheatical expressio i ters of the value of. Justify your aswer by a arguet siilar to the oe i Exaple.. o p. 7. Also check your aswer carefully by deteriig what itegers are set to output whe =0ad also whe =.

11 . Floors ad Ceiligs 9 it k; for (k = ; *k < ; ++k) cout << k << edl;..5 (a) Prove that for all real ubers x ad y we have x+ y x + y x+ y+. (b) Prove that for all real ubers x ad y we have x + y x + y x + y. (c) Prove that for all o-egative real ubers x ad y, x y xy x y+ ( x+ y)...6 Let ad be two positive iteger ubers. Prove that + ad Let be a iteger uber. If is odd, the = ad = +.. Logariths Defiitio.. For a fixed real uber b >, the logarith base b fuctio is defied o positive real ubers x as follows: log b (x) = the power o b that produces x. I other words, log b (x) = y if ad oly if b y = x. Exaples: log 3 (9) = because 3 = 9. log 5 (5) = 3 because 5 3 = 5. log 4 (4) = because 4 = 4. log 6 () = 0 because 6 0 =. Two iportat equatios follow istatly fro this defiitio. Theore.. For ay fixed real uber b >, (a) b log b (x) = x for all real x > 0; (b) log b (b y ) = y for all real y. Logariths were iveted early 400 years ago because they have several iportat properties that could be used to siplify certai difficult arithetic calculatios: they covert products to sus, they covert quotiets to differeces, ad they covert powers to products. Here is a precise stateet of these properties. Theore..3 For ay fixed real uber b >,

12 0 Matheatical Preliiaries (a) log b (s t) = log b (s) + log b (t) for all positive real ubers s ad t; (b) log b (s/t) = log b (s) log b (t) for all positive real ubers s ad t; (c) log b (t p ) = p log b (t) for all real ubers p ad all positive real ubers t. Proof The proofs are give i algebra or pre-calculus textbooks ad will ot be repeated here. Exaple..4 Most of us et logariths base 0 i our high school algebra class. log 0 (000) = 3; log 0 (00) = ; log 0 (0) = ; log 0 () = 0 because 0 0 = ; log 0 (0.0) = because 0.0 = 0 ; log 0 ( 00) is udefied because 0 x > 0 for all real x; < log 0 (5) < because 0 < 5 < 0 ; by calculator, log 0 (5).4; log 0 (/t) = log 0 (t). Exaple..5 Here is the logarith you et i calculus; it s called the atural logarith: log e (x) = l(x) = x t dt for all real x > 0, where e.788. Of course, e l(x) = x if x > 0, ad l(e y ) = y for all real ubers y. I coputer sciece, the logarith that we ecouter ore ofte tha ay other is the logarith base fuctio, deoted by lg(x) or log (x). Here are soe easily coputed values of this fuctio: lg(3) = 5; lg(8) = 3; lg(04) = 0; lg() = 0; lg() = ; lg(/6) = 4. What is the uerical value of lg(00)? We ca see that it ust be betwee 6 ad 7 because 6 < 00 < 7. If we wat a ore exact aswer we will eed to cosult a table of logariths or use a electroic calculator. Most pocket calculators, however, do t have a lg butto. This is ot a proble because, as we will ow prove, all logarith fuctios are just costat ultiples of each other. Theore..6 For all real ubers a ad b greater tha ad all positive real ubers x, log b (x) = log a (x) log a (b). Proof Take the logarith, base a, of both sides of the idetity b log b (x) = x.

13 . Logariths Exaple..7 l(x) = log 0 (x) log 0 (e).303 log 0 (x) by usig a calculator. Siilarly, lg(x) = l(x) l() l(x) l(x); also, lg(x) = log 0 (x) log 0 () 3.3 log 0 (x). I aalysis of algoriths, we seldo have occasio to copute lg(x) whe x is a o-iteger real uber. Mostly we will ecouter lg() where is a positive iteger. Moreover, the lg usually appears i coectio with the floor or ceilig fuctio. The followig theore gives a uber of useful idetities ivolvig the lg fuctio cobied with floors ad ceiligs. Theore..8 Let be a positive iteger. (a) lg() < lg()+ ; equivaletly, < lg(). (b) lg() = ad lg()+ = 0. Moreover, the oly iteger k for which k = is k = lg(). We ca iterpret this as sayig that lg() is the uber of iteger divisios by that are required to reduce dow to. (c) lg( + ) = + lg() for all. (d) lg( /) = + lg() for all. (e) lg( / ) = + lg() for all. (f) lg( / ) = + lg( + ) for all. (g) lg( /) = + lg( ) for all 3. Proof To prove part (a), we ca start fro the fact that lg() lg() < lg()+, which is a istace of Theore..4 (a), with x = lg(). Fro this it follows that lg() lg() < lg()+. Sice lg() =, this proves that lg() < lg()+. To prove the first of the two equatios i part (b), divide the first double iequality i part (a) through by lg() to obtai <. lg() By Theore..4 (c), this iediately iplies that the iteger is the floor of lg() ; that is lg() =. To prove the secod part we start fro the double iequality above ad divide it through by to obtai lg()+ <. Sice 0 <, we ca deduce that 0 < lg()+ <, which iplies that lg()+ = 0.

14 Matheatical Preliiaries To prove the last assertio i part (b), let s start with ad divide it repeatedly by, discardig reaiders. We ll give aes to the itegers we produce alog the way: 0 = = 0 (here we are usig the fact that = ad 0 = ) 0 = = = = (here we are usig Theore..5 (d); ote we ve divided by twice) 3 = = 3 (agai we re usig Theore..5 (d); ote we ve divided by three ties) etc. For ay positive iteger k, after k divisios by we obtai k k = = k. I particular, whe k = lg(), we see by the first equatio i part (b) that we will arrive at the value. That is, by dividig by (discardig reaiders) lg() ties, we reduce to. To prove the equatio i part (c), it suffices by Theore..4 (c) to prove that the iteger + lg() satisfies the iequalities ( + lg()) < lg( + ) + lg(), which are equivalet to lg() < lg( + ) ad lg( + ) + lg(). The first oe is easy because lg() lg() <lg( + ). For the secod, start with the iequality < lg()+ that we proved i part (a). Both sides of the iequality are itegers, so it ust be true that + lg()+. Takig logariths o both sides gives the desired iequality above. To prove part (d) we begi with the first iequalities i part (a) ad divide by to obtai lg() / < lg() for all positive itegers. If, the lg() ad lg() are both itegers, ad thus lg() / < lg() for all itegers. Takig the logarith of every ter gives lg() lg( /) < lg(), which is exactly what is eeded to show that lg() = lg( /) (see Theore..4 (c)). To prove part (e), ote that ( lg ( / ) = lg ( ) = + lg ) + (by Theore..5 (c)) (by part (c) above) = + ( + lg( )) (by part (d) above)

15 . Logariths 3 Fig..3 Plot for the fuctio y = lg(x) = lg( ) = + lg(( )+) (by part (c) above, with replaced by ) = + lg(). The proofs of parts (f) ad (g) are siilar, ad left as a exercise. Exaple..9 To illustrate part (b) of Theore..8, take = 00. The 00/ =50; 50/ =5; 5/ =; / =6; 6/ =3; 3/ =. There are 6 iteger divisios by i that sequece, ad ideed, lg(00) =6 because 6 < 00 < 7,so6< lg(00) <7. Sice we ecouter ay divide ad coquer algoriths that ivolve cotiually splittig a list of legth ito two roughly equal pieces, Theore..8 (b) gives us soe isight ito why these algoriths ofte have ruig ties that ivolve lg() i oe way or aother. The ost iportat thig to uderstad about lg() is how very slowly it grows as icreases. Eve whe gets extreely large, lg() teds to reai relatively sall. The reaso is this: doublig the arguet i lg() just adds to the value of the logarith lg() = + lg(). This is illustrated i the graph of y = lg(x) show i Fig..3. You ight woder whether there is soe horizotal asyptote for this graph, i.e., a horizotal lie that the graph approaches as. I fact, there is o such asyptote. If we draw a horizotal lie at height h, the graph of y = lg(x) will itersect that lie whe x reaches the value h, ad will cotiue to rise slowly above the lie thereafter. I atheatical otatio, li lg(x) =+ (.) x Later i this book we will aalyze ethods for sortig data ito key order. Soe sortig algoriths require a aout of tie proportioal to i order to sort eleets, while certai other algoriths require a aout of tie proportioal to

16 4 Matheatical Preliiaries lg() i order to sort eleets. Which of these algoriths would require the least tie whe sortig eleets? The aswer is that the algorith requirig tie proportioal to lg() would require less tie whe is large. The reaso is this: for ay fixed positive costats of proportioality A ad B, A lg() B= B for all large values of (the sybol is read is uch saller tha ). The followig theore states a curious idetity that will prove to be useful. Theore..0 For all positive real ubers x ad y ad b > we have x log b (y) = y log b (x). Proof Sice the logarith fuctio is iductive (oe-to-oe) o its doai (the set of all positive real ubers), we ca prove that two positive ubers are equal by provig that their logariths (i ay base) are equal. Thus the forula i the theore is true if ad oly if log b (x log b (y) ) = log b (y log b (x) ) which is equivalet to log b (y) log b (x) = log b (x) log b (y). The latter is surely true for ay positive x ad y... Further Exaples Exaple.. For all positive itegers, 3 lg() = lg(3).58. Exaple.. Do the followig ultiple choice exercises without usig a calculator. (a) l(e 7x ) = Choose oe: (i) 7l(x) (ii) 7x (iii) 7e x (iv) l(7) + l(x) (b) l(x 7 ) = Choose oe: (i) 7l(x) (ii)7x (iii)7e x (iv) l(7) + l(x) (c) l(7x) = Choose oe: (i) 7l(x) (ii) 7x (iii) 7e x (iv) l(7) + l(x) (d) e l(7x) = Choose oe: (i) 7l(x) (ii) 7x (iii) 7e x (iv) 7 + x (e) e x+l(7) = Choose oe: (i) 7l(x) (ii) 7x (iii) 7e x (iv) 7 + x (f) lg( ) = Choose oe: (i) (ii) / (iii) (iv) (/) lg() (g) lg(8) = Choose oe: (i) 3 (ii) 8 (iii) 3 + (iv) 3 + lg() (h) lg(8 + ) = Choose oe: (i) 3 (ii) 8 (iii) 3 + (iv) 3 + lg() (i) lg(/3) = Choose oe: (i) /5 (ii) 5 (iii) 5 (iv) lg() 5 (j) lg(8) = Choose oe: (i) 9 (ii) 8 (iii) (iv) 36 (k) 3lg(3) = Choose oe: (i) 7 (ii) 64 (iii) 8 (iv) 8 (l) lg() = Choose oe: (i) (ii) / (iii) (iv) () lg( 8 ) = Choose oe: (i) 8 (ii) 36 (iii) lg(8) (iv) 8 () lg( ) = Choose oe: (i) (ii) 0.5 (iii) lg() (iv) 0.5 (o) l(e x ) = Choose oe: (i) l(x) (ii) e l(x) (iii) x (iv) x (p) e l(+) = Choose oe: (i) + e (ii) e (iii) (iv) + (q) e l(x) = Choose oe: (i) x (ii) x (iii) x (iv) l(x)

17 . Logariths 5 (r) e l() l() = Choose oe: (i) (ii) / (iii) l( )(iv) e l( ) (s) 4lg(3) = Choose oe: (i) (ii) 4 (iii) 48 (iv) 8 (t) lg(3 0 ) = Choose oe: (i) 5 (ii) 5 (iii) 5lg(0) (iv) 00 Solutio (a) (ii) because l(e r ) = r for ay real uber r (b) (i) because l(x p ) = p l(x) for ay real uber p (c) (iv) because l(st) = l(s) + l(t) (d) (ii) because e l(t) = t (e) (iii) because e a+b = e a e b by the properties of expoets (f) (i) because lg( p ) = p (g) (iv) because lg(8) = lg(8) + lg() = 3 + lg() (h) There is o correct aswer. It was a trick questio. (i) (iv) because lg(/3) = lg()g() 5 (j) (ii) because lg(x) = x (k) (i) because 3lg(3) = lg(33) = lg(7) = 7 (l) (i) because lg() = = lg( ) = () (i) () (ii) because = 0.5 (o) (iii) (p) (iv) (q) (i) because l(x) = l(x ) (r) (ii) because l() l() = l(/) (s) (iv) because 4 lg(3) = lg(3 4 ) = lg(8) (t) (iv) because lg(3 0 ) = 0 lg(3) Exaple..3 Show how to copute the value of lg(500) without usig a calculator. Solutio 0 = 04 < 500 < 048 =,so0 < lg(500) <, so lg(500) =0. Exaple..4 I the block of C++ code show below, assue that the variable has a positive iteger value. Whe the code is executed, how ay ties is the body of the loop executed? Express the aswer i ters of the value of. it k = ; it k, p; for (k =, p = ; p <= ; p *= ) // Double p after body of loop is executed. { cout << k << edl; ++k; } Solutio First cosider the special case where has the value. I that case, the loop cotrol coditio <= is iediately false, so the body of the loop is executed 0 ties. Now let s look at the cases where the value of is at least, so that the body of the loop will be executed at least oce. As i Exaple.. o p. 7, the variable k is a loop couter. Let L deote the value of k i the cout stateet the last tie the body of the loop is executed. The L is the uber we wat to copute.

18 6 Matheatical Preliiaries Whe the body of the loop is executed, the iitial value of k is set to output ad the k is icreeted to. Next, the value of p is doubled to 4 (= ) ad the the loop cotrol coditio p <= is tested. If it is foud to be true, the the value ofkwill be set to output, k will be icreeted to 3, ad p will be doubled to 8(= 3 ). Agai the loop cotrol coditio will be tested. Whe k gets the value L (which we wat to copute), the variable p will the get the value L, ad the loop cotrol coditio will be true for the last tie. The uber L will be set to output, the variable k will be icreeted to L +, ad p will be doubled to L+. The loop coditio will ow be foud to be false. If we let deote the (fixed) value of the progra variable, the the followig iequalities ust be true: L < L+. Takig the logarith base of each part gives us L lg() <L +. As we kow fro Theore..4 (c), this eas that L = lg(). This is the uber of ties the body of the loop is executed. Note that this forula is valid eve whe has the value, for i that case lg() = 0. I the solutio of the precedig exaple, the key observatio was that there exists a siple atheatical relatioship betwee the values k ad p of the progra variables k ad p each tie the loop cotrol coditio is tested. That relatioship is p = k, ad it is true eve whe the loop cotrol is tested for the last tie ad foud to be false (at that poit k = L + ad p = L+ ). Ay coditio that s true for the progra variables every tie a loop cotrol coditio is tested is called a loop ivariat for the loop. I the block of code i Exaple A..4, the loop ivariat p = k is easy to spot, ad seeig it was the key to solvig the give proble. There are other loop ivariats for the block of code i Exaple..4. Exaples are p > ad k < p, but these ivariats are ot useful i solvig the proble that we were give... Exercises..5 Do the followig ultiple choice exercises without usig a electroic calculator. (a) 5+lg(t) = Choose oe: (i) 3 + t (ii) 3t (iii) 5 + t (iv) 5t (b) lg( 5 ) = Choose oe: (i) 5 + (ii) 3 + (iii) 5 (iv) 3 (c) lg(3p) = Choose oe: (i) 5lg(p) (ii) 3 lg(p) (iii) 5p (iv) 5 + lg(p) (d) lg( 8 ) = Choose oe: (i) 8lg() (ii) 3lg() (iii) 3 (iv) 3 + lg() (e) 3lg(k) = Choose oe: (i) 8lg(k) (ii) 8k (iii) k 3 (iv) 3k (f) lg(8 k ) = Choose oe: (i) 8/k (ii) 3k (iii) /(3k) (iv) 3lg(k)

19 . Logariths 7 (g) lg(3x) = Choose oe: (i) 5 + lg(x) (ii) 5x (iii) 3 + x (iv) 5 lg(x) (h) lg(a bc) = Choose oe: (i) a + b + c (ii) abc (iii) a+b+c (iv) lg(a) + lg(b) + lg(c) (i) lg(8/ ) = Choose oe: (i) 6 lg() (ii) 3/ lg( ) (iii) 3/ (iv) 3 lg() (j) lg(5) = Choose oe: (i) 5 (ii) 3 (iii) 5 (iv) lg(5) (k) 3lg(5) = Choose oe: (i) 3 (ii) 5 (iii) 40 (iv) 5 (l) lg(8) = Choose oe: (i) 8 (ii) /(8) (iii) 8/ (iv) 64 () lg( 8 ) = Choose oe: (i) 56 (ii) 3 (iii) 8 (iv) 6 () lg( 38) = Choose oe: (i) 9 (ii) lg(38) (iii) lg(38) (iv) 38 (t) lg(8 5 ) = Choose oe: (i) 5 (ii) 40 (iii) 8lg(5) (iv) 3 + lg(5)..6 For each of the followig, give a sigle uerical value without usig a calculator. (a) lg(36) lg(9) (b) lg(56) lg(7) (c) lg( 8) (d) lg( 3 3)..7 (a) Show how to copute the value of lg(500) without usig a calculator. (b) Show how to copute the value of lg(900) without usig a calculator...8 (a) Express log 3 (x) as soe costat C ties log 0 (x). Use a pocket calculator to deterie the uerical value of the costat C. (b) Express log 5 (x) as soe costat C ties l(x). Use a pocket calculator to deterie the uerical value of the costat C...9 (a) Use Theore..0 to express l() as a power of. Use a calculator to fid the uerical value of the power. (b) Use Theore..0 to express 5 l() as soe costat ties a power of. Use a calculator to fid the uerical value of the power...0 We oted that the first of the two equatios i Theore..8 (b) ca be iterpreted as sayig that lg() is the uber of iteger divisios by that are required to reduce dow to. Give a siilar iterpretatio for the secod of the two equatios i Theore..8 (b)... Cosider the followig block of C++ code i which is a positive iteger. it k; for (k = ; k > ; k = k/) cout << "Hello" << edl; (a) How ay ties will the cout stateet be executed if the value of is whe the loop begis? (b) How ay ties will the cout stateet be executed if the value of is whe the loop begis?

20 8 Matheatical Preliiaries (c) How ay ties will the cout stateet be executed if the value of is 3 whe the loop begis? (d) Give a expressio ivolvig the variable for the uber of ties that the code above will sed Hello to output. Cite oe of the theores i this sectio to justify your aswer... (a) Use (aog other thigs) Theore..4 o floors ad ceiligs to prove that for all positive itegers, lg() <. Warig: i geeral, lg() = lg(). Hit: look at the proof for Theore..8 (a). (b) Use (aog other thigs) Theore..4 o floors ad ceiligs to prove that for all positive itegers, 0 < 0 log 0 (). Warig: i geeral, 0 log 0 () = 0 log 0 (). Hit: look at the proof for Theore..8 (a)...3 (a) Use Theore..5 (c) ad part (d) of Theore..8 to prove part (f) of Theore..8. (b) Use Theore..5 (c) ad part (e) of Theore..8 to prove part (g) of Theore (a) How ay ay ties is the body of the loop below executed? (Note: the block differs fro the code i Exaple..4 oly i the iitial value of p.) Justify your aswer by usig a loop ivariat. Assue that has a positive iteger value. it k, p; for (, p = ; p <= ; p *= ) // Double p after body is executed. { cout << k << edl; ++k; } (b) How ay ay ties is the body of the loop below executed? Assue that has a positive iteger value. You ay base your aswer o your aswer for part (a). it p; for (p = ; p <= ; p *= ) // Double p after body is executed. cout << p << edl; (c) How ay ties is the body of the loop below executed? Assue that has a positive iteger value. it p; for (p = ; p < ; p *= ) // Double p each tie body of loop is executed. cout << p << edl;

21 . Logariths 9 Hit. Iagie that a loop-coutig variable k with iitial value is itroduced ito the code. Give a loop ivariat ivolvig the values k ad p of the progra variables k ad p. Let L deote the value of k the last tie the body of the loop is executed. Use your loop ivariat to calculate L. You will eed to cite Theore..4 (c) at soe poit..3 Sus ad Suatio Notatio What is the value of the su ? To aswer this questio, you first have to fill i the issig ters that are idicated by the 3 dots i the su. I this case that s ot hard to do. The 3 dots ust surely stad for What about the su ? Here we ay or ay ot see that the ters ca be writte as , i which case the issig ters are = It is useful to have a otatio that lets us express sus i a ore precise way that eliiates the guesswork associated with the 3 dots otatio. The siga suatio otatio was iveted for this purpose. The two sus give above ca be writte this way: 0 (k ) ad 9 k(k + ) More geerally, suppose a real fuctio f (k) is defied o itegers k i the rage,...,. The expressio f (k) k= where is the upper case Greek letter siga, is a precise ad cocise way of writig the su f () + f ( + ) + + f (). The uerical value represeted by these expressios is exactly the value that would be produced i the variable su by executig the followig fraget of C code:

22 30 Matheatical Preliiaries su = 0; for (k = ; k <= ; ++k) su += f(k); I particular, by covetio, above). k= f (k) has the value 0 if < (look at the loop 5 k Exaples.3. (a) k + = = k= 7 (b) k k = = 769 (c) (d) k=0 4 5/k = 5/4 =3 k=4 5 3 = = 48 (because there are 5 + ters here, ot 0 just 5). More geerally, c = c( + ) for ay fixed real uber c ad k= itegers. ( 3 j ) ( ) ( 3 ) (e) ( j k) = ( k) + ( k) + (3 k) j= = (0 ) + ( + 0 ) + ( ) = 6 The variable k i the expressio k= f (k) is called the suatio idex. Itis iportat to uderstad that it is a duy idex that ca be replaced by ay other letter that has o pre-assiged eaig. Thus, for exaple, all four of the followig expressios have the sae value: k(k + ) i(i + ) t(t + ) a(a + ). k= k= k= i= Aother iportat fact is that suatio is a liear operator ; that is, ( ) c f (k) = c f (k) ad [ f (k)±g(k)] = f (k)± g(k). t= k= (.) Occasioally we have to iterchage the order of suatio i a double su. That is, we have to use the fact that ( q q ) f ( j, k) = f ( j, k). k= j=p j=p k= a= k= k=

23 .3 Sus ad Suatio Notatio 3 This equatio siply says that if we su over each row separately i the followig array, ad the add up the sus we get, we obtai the sae aswer as if we su over each colu separately ad the add up those sus. We ca do this as log as the boudaries p ad q for the iterior su are idepedet of the idex variable of the exterior su k. f (p, ) f (p +, ) f (p +, )... f (q, ) f (p, + ) f (p +, + ) f (p +, + )... f (q, + ) f (p, + ) f (p +, + ) f (p +, + )... f (q, + ) f (p, ) f (p +, ) f (p +, )... f (q, ) We are ow goig to look at several types of sus that occur frequetly i aalysis of algoriths. Our goal will be to fid soe shortcuts for coputig the values of these sus. Our first exaple is the siple su , where is a positive iteger. (Whe is or or 3, the by we ea siply or + or+ + 3 respectively.) We ca write this su i the siga suatio otatio this way: k. This is called a triagular su because if we draw a bar graph with bars of heights,, 3,...,, they for a rough triagle. It turs out that there is a siple forula for the value of the triagular su. It ca be obtaied by the followig algebraic trick. Let S() deote the su k. The S() = ( ) +. I reverse order we have S() = + ( ) + ( ) Addig the gives S() = ( + ) + ( + ) + ( + ) + +( + ) + ( + ) = ( + ) so ( + ) S() =. Theore.3. For ay positive iteger we have ( + ) =. More geerally, for ay itegers ad satisfyig we have + ( + ) + ( + ) + + = ( + ) +. Proof We have already proved the first of these suatio forulas. The secod ca be proved i a siilar way. These forulas should be eorized. It will help you to ote that i both forulas above the first factor o the right side is the uber of ters o the left side, while

24 3 Matheatical Preliiaries the secod factor o the right side is the average of the first ad last ubers o the left side. I geeral, whe we are cofroted with a expressio X ivolvig a sigasuatio sybol or a 3-dots ellipsis we like to have a closed for for the expressio, by which we ea a atheatical expressio Y whose value is the sae as that of X, but Y is copact ad easily coputed with a fixed uber of operatios icludig expoetiatio ad foratio of logariths. (We regard as a closed for because it ca be evaluated i two steps: let x = l(); the = exp(x).) The calculatio i Theore.3. shows that is a closed for for the expressios ( + ) ad k. Our ext exaple of a frequetly occurrig su looks like this: + t + t + t 3 + +t, where t is ay real uber ad is ay o-egative iteger. We ca write the su i siga otatio this way: t k. Note that the lower liit of the suatio k=0 is k = 0. By covetio, i siga sus of this for, t 0 is uderstood to stad for, eve whe t = 0. (Norally i atheatics the expressio 0 0 is cosidered to be udefied.) Sus of this for are called (fiite) geoetric sus. Wecafida closed for for the su by a trick siilar to the oe used for triagular sus. Let G(t, ) deote t k. The k=0 G(t, ) = + t + t + t 3 + +t. Multiplyig by t gives tg(t, ) = t + t + t 3 + t 4 + +t +. Subtractig the gives G(t, ) tg(t, ) = t + (other ters cacel). Factorig gives G(t, )( t) = t +.Ift =, the we obtai G(t, ) = t+. t That is, + t + t + t 3 + +t = t+ t = t+. whe t =. t Theore.3.3 For all real ubers t ad o-egative itegers, t + t k = if t = ; t + if t =. k=0 Proof See paragraphs precedig this theore. Our third exaple of a frequetly occurrig su looks like this: , which ca be writte i siga suatio otatio this way:. We call a su of k

25 .3 Sus ad Suatio Notatio 33 this for a haroic su ad deote it by H(). (I usic, if strigs of like aterial, diaeter, ad torsio have legths, /, /3, etc., the whe set i vibratio they produce haroic toes.) Sadly, there is o closed for for haroic sus. Nevertheless, we ca approxiate their values very closely by usig a techique that is applicable both here ad i other siilar situatios. The techique is give by the followig theore. Theore.3.4 Let f (x) be a strictly decreasig fuctio defied over a iterval [, ], where ad are itegers such that <. The f () + f (x)dx < k= f (k) < f () + Siilarly, if f (x) is strictly icreasig o [, ], the f () + f (x)dx < k= f (k) < f () + f (x)dx. f (x)dx. Proof The proof is easier if we begi by assuig that f (x) 0 for all x i [, ]. At the ed of the proof we will see how to take care of the cases where f (x) has soe egative values. For the proof whe f (x) 0 ad decreasig we use a graph of the equatio y = f (x) o [, ], draw soe rectagles, ad copare their areas with the area of the regio uder the curve y = f (x) show i Fig..4. Note that the area of the left-ost rectagle i Fig..4 is f () = f (), the area of the ext rectagle is f ( + ) = f ( + ), ad so o to the last oe, which has area f ( ) = f ( ). Now cosider the regio R bouded by the x-axis, the curve y = f (x), ad the vertical lies at x = ad x =. This regio R lies iside the collectio of rectagles show, ad thus the su of areas of Fig..4 Theore.3.4

26 34 Matheatical Preliiaries Fig..5 Theore.3.4 these rectagles is greater tha the area of R. This ca be expressed as follows: area of regio R = f (x)dx < f ()+ f ( +)+ + f ( ) = k= f (k). Addig f () to both sides of the iequality above gives the first of the iequalities about decreasig fuctios. For the secod of the iequalities about decreasig fuctios, cosider the graph show i Fig..5. All the rectagles i Fig..5 lie iside the regio S bouded by the x-axis, the curve y = f (x), ad the vertical lies at x = ad x =, so the su of the areas of the rectagles is less tha the area of S. This ca be expressed as follows: f ( + ) + f ( + ) + + f () < area of regio S = f (x) dx. Addig f () to both sides of the iequality above gives the secod of the iequalities about decreasig fuctios. Now let s take care of the case where soe or all of the values of f (x) are egative. Let C deote the value of f (). Sice f (x) is decreasig o the iterval [, ], it follows that f (x) C for all x i that iterval. This iplies that f (x) C 0 o that iterval. Now defie a ew fuctio g(x) by the forula g(x) = f (x) C. The g(x) 0 everywhere o [, ]. Moreover, g(x) is strictly decreasig o that iterval. By the first part of this proof, g() + g(x)dx < k= Replacig g(x) by f (x) C above gives us f () C + [ f (x) C] dx < g(k) < g() + g(x)dx. [ f (k) C] < f () C + k= [ f (x) C] dx.

27 .3 Sus ad Suatio Notatio 35 Now we use properties of suatio etioed i Exaples.3. (d) ad i forula (.3) o p. 9, together with stadard itegratio forulas to produce the followig iequalities: f () C + f (x)dx C( ) < f (k) C( + ) < f () C + f (x)dx C( ). k= Addig C( + ) to each part of the iequalities above gives the iequalities for f (x) stated i the theore. The proofs for the iequalities ivolvig icreasig fuctios are siilar ad will be left as a exercise. Theore.3.5 (Approxiatio for Haroic Sus). For all itegers > we have + l() <H() < + l(). Proof We wat to approxiate H() = for >. k To do this we apply Theore.3.4 to the strictly decreasig fuctio f (x) = x, with = ad >. We get + x dx < k < + dx for all >, x which iediately yields the iequalities stated i the theore. The iportat thig to reeber fro Theore.3.5 is that for all large we have H() l(). I rare cases where we wat a ore accurate approxiatio for H() whe is large, we ca use the followig theore discovered by Leohard Euler. Theore.3.6 (Euler) li [H() l()] =0.577 approxiately ad therefore H() l() for all large. Proof See alost ay advaced calculus textbook. The uber give i this theore is a approxiatio to the actual liit, which is kow to be a irratioal uber.

28 36 Matheatical Preliiaries Whe 5, Euler s approxiatio is correct to withi % of the actual value. Of course, for sall values of, we ca calculate H() very accurately with a pocket calculator. Closed fors for several other types of sus are occasioally eeded i aalysis of algoriths. They do ot, however, occur with sufficiet frequecy to ake the worth eorizig. We give four such sus below. Theore.3.7 For all positive itegers ad real ubers t =, + t + 3t + +t = + t+ ( + )t (t ). Proof The left side is just the derivative with respect to t of the fiite geoetric su + t + t + t 3 + +t. Thus we ca obtai the forula above siply by differetiatig both sides of the forula give i Theore.3.3 with respect to t. Theore.3.8 For all positive itegers, lg(k) = lg()+( + ) lg()+ ad k= lg(k) =( + ) lg( + )+ lg(+)+. k= Proof Whe k = wehave lg(k) =. Whe k = 3ork = 4wehave lg(k) =. Whe k = 5or6or7or8wehave lg(k) =3. Etc. Thus, the ters i lg(k) ca be grouped as follows k= k=3,4 k=5,6,7,8= 3 k=9,0,...,6= 4 k= lg() +,..., lg() + {}}{{}}{{}}{{}}{ ( + ) }{{} + ( ) }{{} + ( ) }{{} + + ( lg() + lg()) }{{} + ore 0 ter ters ters 3 ters lg() ters The ore (above) gives the reaiig ters, correspodig to k = lg() + up to k = (this set of ters is epty if is a power of ). Each of these ters is equal to lg()+. Thus we have lg(k) = k= lg() p= k= ( p p + lg()) ( lg()+). (.3)

29 .3 Sus ad Suatio Notatio 37 The siga su o the right side is just the su give i Theore.3.7, but with k replaced by p, replaced by lg(), ad t replaced by. Thus we get lg() p= p p = + lg() lg()+ ( lg()+) lg() ( ). Isertig this ito Eq. (.3) gives us lg(k) =+ lg() lg + ( lg()+) lg + lg()+ lg() lg lg, k= which reduces to the first of the two sus i the stateet of the theore. To prove the secod idetity i this theore, begi by replacig the sybol by the sybol + everywhere i the first equatio stated i the theore. This gives + lg(k) =(+) lg(+)+(+) lg(+)+ for all positive itegers. k= By Theore..8 (c), with replaced by k, we get lg(k) = + lg(k ), so the su above is + ( + lg(k )) = ( + ) lg( + )+( + ) lg(+)+. (.4) k= The left side ca be expressed as + k= + + k= lg(k ) = +0+ lg()+ lg(3)+ + lg() = + Puttig this ito Eq. (.4) gives + lg(k) =( + ) lg( + )+( + ) lg(+)+. k= Subtractig fro both sides of this equatio gives the desired equatio. lg(k). k= Exaple.3.9 May coputer sciece textbooks discuss the represetatio of positive itegers usig various bases. For exaple, the iteger (base 0 otatio) ca also be writte i biary otatio (base otatio) as or i octal otatio (base 8) as or i hexadecial otatio (base 6) as 78D 6. Of course, it is also possible to write this iteger i these bases with oe or ore leadig zeroes: , , , 78D 6. Let s agree to use the phrase stadard represetatio of base b to describe the represetatio of a positive iteger without leadig zeroes i base b otatio. Note that the uber of digits required by the stadard represetatio of depeds o the base used as well as o the size of. For exaple, the stadard represetatios of the uber discussed i the secod setece of this paragraph require 4, 3, 5, ad 4 digits, respectively, i base 0, base, base 8, ad base 6.

30 38 Matheatical Preliiaries How ay digits are there i the stadard decial (i.e., base 0) represetatio of the iteger 34 95? How ay bits (biary digits) does its stadard biary represetatio cotai? More geerally, give ay positive itegers ad b, ca we fid a forula for the uber of digits that will be required for the stadard represetatio of base b? Solutio Let s solve the geeral case first. Assue we are give positive itegers ad b. Let k deote the (ukow) uber of digits i the stadard represetatio of base b. Deote those digits by d 0, d, d,...,d k, where d 0 is the least sigificat digit ad d k is the ost sigificat (ad is therefore ot zero). Each of these digits lies i the rage fro 0 to b. (For exaple, if b is 0, the each digit lies i the rage fro 0 to 9.) Moreover, d k is at least. The = d k b k + +d b + d b + d 0 b 0. Sice d k ad all other d i 0, it follows that b k That is, b k. Siilarly, sice every d i b, it follows that (b )b k + +(b )b + (b )b + (b )b 0 (b )[b k + +b + b + ] (b ) bk b b k. Fro this it follows that < b k. We have ow established the iequalities b k < b k, where k is the uber of digits i the stadard represetatio of base b. Takig logariths base b of each of the quatities above gives us these iequalities: k log b () <k. Sice k is a iteger, it follows fro Theore..4 (c) that k = log b (), or equivaletly, k = + log b (). By Theore..8 (c), k = log b ( + ). Thus we have proved the followig useful fact: Theore.3.0 Let ad b be positive itegers with b. The uber of digits i the stadard represetatio of base b is + log b () = log b ( + ). Proof This forula is derived i Exaple.3.9.

31 .3 Sus ad Suatio Notatio 39 Now we ca aswer the questio we asked i Exaple.3.9 about the uber of digits i the stadard decial represetatio of the iteger The aswer, obtaied with the aid of a calculator, is + log 0 (34 95 )= + 95 log 0 (34) = = =45. The uber of bits i the stadard biary represetatio of is + log (34 95 )= + 95 log (34) = = =50. The su i the followig theore appears here ad there i aalysis of algoriths. Ufortuately it has o siple closed for, so it is helpful to have good lower ad upper bouds for it. Theore.3. The su the iequalities lg() lg() < k is asyptotic to. I particular, it satisfies lg() k. (.5) Proof Usig the fact that x < x x for all real ubers x, we ca write that lg() ( ) k < These iequalities are equivalet to The su lg() lg() k lg() < lg() lg() k k lg() lg() k. k. (.6) that appears i these iequalities ca be coputed by factorig k out the that occurs i every ter ad the usig the geoetric su forula. Note that the ter for k = 0 is issig fro the su ad ust therefore be subtracted fro it: lg() = lg() k = ( lg() ( ) ( k ( ) lg()+ = ) = lg(). Isertig this last forula ito lie.6 we get lg() < lg() lg() ) = k ( ) lg()+ lg(). (.7) lg()

32 40 Matheatical Preliiaries To get fro iequatio (.7) to the desired iequatio (.5) we ust estiate the ter. Theore..8 (a) gives us these useful iequalities: lg() lg() < lg()+. Dividig all ters by lg() ad the ultiplyig throughout by gives >. lg() It follows that lg() < lg() ad. lg() lg() Cobiig this pair of iequalities with iequality (.7) gives the desired result (.5)..3. Further Exaples Exaple.3. Evaluate the followig sus. I each case, give a sigle uerical aswer. 5 3 r + 4 p (a) (b) k(k + ) r (c) p 50 lg(k) (d) q k r=0 p= q= 00 Solutio (a) = 5 (b) = ( 4 (c) = + j j j j + ) ( ) 3 j= j= j= j= ( ) = (d) The value of this su is 0 because the lower liit of suatio (00) is larger tha the upper liit (50). Exaple.3.3 Write each of the followig sus i siga suatio otatio: (a) () + 3 () + 4 (3) + 5 (4) (b) Solutio (a) 4 k + k or 3 j=0 j + 5 ( j + ) or r (r ) (b) r= 0 k=0 3 k Exaple.3.4 For each of the followig sus, evaluate it uerically or express it i a algebraic closed for. (a) (b)