Sequences and Summation

Transcription

1 Sequeces ad Summatio

2 Sequeces Iformally, a sequece is a ifiite progressio of objects (usually umbers), cosistig of a first, a secod, a third, ad so o. The members of a sequece are called elemets or terms. Example sequece: 2,4,6,8,0, It is customary to deote sequeces with the letters a, b, c ad to use subscript otatio to refer to idividual terms: a is the th term of the sequece a. The otatio {a } refers to the etire sequece, ot to the set of the terms. A sequece is a ordered list, whereas a set is a uordered collectio of objects. Example: if {a } = 2,4,6,8,0, the a 0 = 2, a = 4. To simplify some of the formulas, the idex will always start with = 0 i this presetatio. So for us, the iitial term of the sequece is a 0 - the zeroth term. This choice comes at a price. Speakig of the first term of the sequece is ow ambiguous. Is it a 0, or is it a? I the rigorous sese defied above - a is the th term of the sequece the first term is a. I a colloquial sese, where first is take as a syoym for iitial the first term is a 0. [Because of these difficulties, may mathematicias prefer the more atural oebased idexig or umberig where the iitial elemet is a ad there is ever ay cofusio what is meat by the first term of a sequece. ]

3 Zero vs Oe-Based Idexig i Programmig The disagreemet amog mathematicias over whether the atural umbers should iclude zero or ot ad whether the first elemet of a sequece is the first or the zeroth - is reflected i a correspodig disagreemet amog programmig laguages cocerig the proper idexig of arrays. I the C family of programmig laguages, as well as i Pytho, Javascript, Ruby ad may others, array idices start at zero. Therefore, if you have a array declared by it umbers[0]; the proper way of loopig through these elemets is for(it i=0; i < 0; i++) umbers[i]. Tryig to access or assig a value to umber[0] is a illegal operatio. Other programmig laguages use oe-based idexig. This icludes prehistoric oes such as COBOL ad Fortra, but also ewer ad almost popular oes like Lua. The Wolfram ad MATLAB laguages are also oe-based.

4 How ot to defie a sequece O a previous slide, we defied a sequece by givig the first 5 terms ad expected that a reasoable reader would uderstad that we mea the sequece of positive eve umbers. This way of defiig a sequece, by givig fiitely may terms of it ad expectig the reader to recogize a patter i them is mathematically idefesible because there is always more tha oe coceivable patter to cotiue a sequece, ad, more importatly, a sequece does ot have to fit ay patter i the first place. Each term is idepedet from all other terms ad ca assume ay value. Give {a } =,2,3, {a } could be the sequece that repeats these 3 umbers i perpetuity: {a } =,2,3,,2,3, or {a } could be costat after the third term: {a } =,2,3,7,7,7,7, If you thik that these examples are far-fetched ad exaggerate the issue of misuderstadig, cosider the followig example: {a } =,2,4, could represet {a } =,2,4, 8, 6, 32, 64, (each term is double the previous) but also {a } =,2,4, 7,,6, where the th term plus produces the ext term, for all.

5 Defiig a sequece properly A proper defiitio of a sequece requires us to defie all terms, ot just fiitely may of them. This ca be doe i two ways, directly ad recursively. (We will discuss recursive defiitio later i this presetatio). A direct defiitio gives each a as a fuctio of. We ofte just give a equatio for a without botherig to quatify the for all N 0. Examples: a = 2 is the sequece of oegative eve umbers. a = 2 is the sequeces of squares. a = 2 is the sequece of powers of 2 that are itegers.

6 Sequeces as Fuctios Techically, a sequece is a special kid of fuctio, amely a fuctio whose domai is N 0. Therefore, we could use stadard fuctio otatio to represet sequeces ad write f() istead a, but we use the latter for reasos of traditio.

7 Arithmetic sequeces A sequece that has a costat differece betwee successive terms is called arithmetic. A arithmetic sequece has the form a, a + d, a + 2d, a + 3d, a + 4d, Where a is the first term ad d is the commo differece betwee successive terms. The geeral formula is a = a + d. (Importat: that s the geeral formula assumig zero-based umberig. Thik about what the correct formula is for oe-based umberig.) A arithmetic sequece is just a liear fuctio with a domai restricted to the atural umbers. Example: a = + 2 is arithmetic with a = ad d = 2.

8 Geometric Sequeces A sequece that has a costat quotiet betwee successive terms is called geometric. A geometric sequece has the form a, aq, aq 2, aq 3, Where a is the first term ad q is the commo quotiet betwee successive terms. The geeral formula is a = aq. Agai, that formula is oly correct for zero-based umberig. Thik about what the formula is for oe-based umberig. A geometric sequece is just a expoetial fuctio with a domai restricted to the atural umbers. Example: a = 3 2 is arithmetic with a = 3 ad q = 2.

9 Recursive Defiitio A recursive defiitio gives each term of a sequece as a fuctio of previous sequece terms: a = f(a, a 2,, a k ) This equatio is called a recurrece relatio, or more precisely, a k-step recurrece relatio. A recursive defiitio ivolvig a k-step recurrece relatio requires the values of the first k terms: a 0, a,, a k. These values are called the iitial coditios. For example, a = a + 2 ad a 0 = 0 defies the sequece of oegative eve umbers recursively. The equatio a = a + 2 is the recurrece relatio. Each arithmetic sequece a = a + d has the recursive defiitio a = a + d, a 0 = a. Each geometric sequece a = aq has the recursive defiitio a = qa, a 0 = a. These three recursive defiitios all ivolve -step relatios.

10 Efficietly Computig Arithmetic Sequeces () Suppose you eed to sample [,3] [2,5], i.e. the rectagle i R 2 cosistig of the poits (x,y) where x goes from to 3 ad y goes from 2 to 5. You decide to create a regular grid by subdividig each iterval ito N smaller parts, ad the you loop over all the grid poits as follows: for(=0;<=n;++) for(m=0;m<=n;m++) { x = + (3-)/N*; y = 2 + (5-2)/N*m; procedure(x,y); } If N is large ad the procedure to be carried out is ot particularly expesive computatioally, the the total ruig time of these ested loops may well be domiated by the repetitive liear trasformatios that compute x from ad y from m. A first speedup is gaied by the realizatio that x does ot chage i the ier loop because it oly depeds o the outer loopig variable. Additioally, there is o reaso to re-compute the slopes of the two liear fuctios x() = + (3-)/N * y() = 2 + (5-2)/N* m agai ad agai. It is more efficiet to compute the slopes oce ad for all at the begiig of the program. Based o this, we optimize our program as follows:

11 Efficietly Computig Arithmetic Optimized program: Sequeces (2) Sx = 2/N; Sy = 3/N; for(=0;<=n;++) { x = + Sx*; for(m=0;m<=n;m++) { y = 2 + Sy*m; procedure(x,y); } } This is better, but still requires oe additio ad oe multiplicatio i each ier loop, ot coutig what the procedure might do with x ad y.

12 Efficietly Computig Arithmetic Sequeces (3) We ca save the multiplicatios by realizig that x() = + Sx* ad y() = 2 + Sy*m are arithmetic sequeces with first terms ad 3, ad costat differeces Sx ad Sy. We ca get each x from the previous by addig Sx, ad each y from the previous by addig Sy, i.e. by usig the recursive defiitios of the sequeces. We just eed to make sure we iitialize x ad y with the correct values x just oce, ad y oce i each outer loop. Sx = 2/N; Sy = 3/N; x = ; for(=0;<=n;++) { y = 2; for(m=0;m<=n;m++) { procedure(x,y); y += Sy; } x += Sx; } The program with this fial optimizatio is prited o the right.

13 A Example of a Multi-Step Recurrece Relatio The Fiboacci Sequece is the sequece {f } defied by the iitial coditios f 0 =, f = ad the recurrece relatio f = f + f 2 for = 2,3,4, : f =,,2,3,5,8,3,2,34,55, f = f + f 2 is a two-step recurrece.

14 Computig The Fiboacci Sequece: A Cautioary Tale about Recursio (I) Not every recursio leds itself directly to a computatioally efficiet recursive implemetatio. Based o the recursive defiitio of the Fiboacci sequece, some studets will implemet the sequece as give i the box o the right. it F(it ) { if (==0) or (==) retur ; else retur F(-)+F(-2); } If you ru this code, you will fid that it produces the first few terms of the sequece just fie. But oce the values get eve moderately large, the fuctio takes a log time to complete. That is because this implemetatio is t just a little iefficiet. It is disastrously, mostrously, iefficiet. The problem is that the recursive computatio of F(-) requires the full recursive computatio of F(-2), which is the repeated agai to compute F(-2). This iefficiecy compouds itself ad causes this fuctio to require a umber of additios that expoetially icreases with.

15 Computig The Fiboacci Sequece: A Cautioary Tale about Recursio (II) Cosider how our fuctio computes F(5). I the recursio tree, the etire F(3) tree gets idepedetly computed twice, which subsequetly causes the F(2) tree to get computed three times. The total umber of additios is 7, whe really oly 4 additios were eeded: F(4) F(5) F(3) F(3) F(2) F(2) F() F(2) = F() + F(0), F(3) = F(2) + F(), F(4) = F(3) + F(2), F(5) = F(4) + F(3). F() F(2) F() F() F(0) F() F(0) F(0)

16 Computig The Fiboacci Sequece: A Cautioary Tale about Recursio (III) It is ot difficult to come up with a recursive formula for computig the umber of additios required by our iefficiet recursive algorithm. The computatio of each F() requires the full ad idepedet computatio of F(-) ad F(-2), followed by a additio. Therefore, if we let N() be the umber of additios required to compute F(), the N() = N(-) + N(-2) +, with N() = N(0) = 0. F(-) Usig this formula, we ca determie, for example, that computig F(00) usig our iefficiet recursive algorithm requires 573,47,844,03,87,084,00 additios. A Itel Core i7 6950X ruig at 3 Ghz would take about 57 years to carry out this may additios. A efficiet, iterative algorithm that loops the recursio formula by repeatedly addig the last two terms together to compute the ext oe oly requires 99 additios to compute F(00). F() F(-2)

17 Summatio The sigma otatio is a coveiet way to express legthy sums that follow a patter: N f(k) = f + f f(n) k= The idex variable k always rus from the iteger to the iteger N i steps of. Examples: 5 8 k 2 = k = k=4 5 k 2 = Sice the terms i a sum ca be arbitrarily rearraged, ad commo multiplicative costats ca be factored out, we have the geeral laws N f k + g(k) = f(k) + g(k) k= k= k= N N N cf k = c f k. k= k= N

18 Sums of Cosecutive Itegers There is a coveiet summatio formula available for the sum of the first positive itegers: + k = ( ) + = 2 Such a formula for a sigma sum is kow as a closed form formula. For eve, this formula has a simple explaatio. The first ad the last term have a sum of +. The secod ad the secod-to-last term also have a sum of +, ad so o. Sice there are + such pairs, the sum is. The formula is also valid for odd 2 2. Thik about how this explaatio eeds to be adjusted for that case. Examples: = 00 k= k = k k = = 50 0 =

19 The idex variable i a sigma sum is ivisible to the outside! Later, whe we study proofs of summatio formulas by iductio, we will eed to cosider sequeces defied by sigma sums, such as a = k. Observe that a is a fuctio of aloe. It is ot a fuctio of k. k is like a local variable used i a fuctio i a computer program that is used to compute the output, but ivisible to the outside. Evaluatig this a for differet values meas chagig the upper limit i the summatio oly, while leavig the lower idex ad the expressio of what is beig summed aloe. For example, + a + = k. The followig equatios are icorrect: + a + = (k + ), + a + = (k + ) k=2

20 Applicatio to Arithmetic Sums Usig the summatio formula we just leared, we ca evaluate all arithmetic sums, i.e. all sums of the form ( + ) (a + kd) = a + d k = a + d 2 Example: 00 (2 + 3k) =

21 Sums of Cosecutive Squares Summatio formulas are also available for the sum of the squares ad cubes of the first positive itegers: k 2 = = + (2 + ) k 3 = = Such formulas for σ k p exist, i fact, for all positive itegers p. Example: =

22 The Laws of Arithmetic Have Not Bee Repealed Just Because We Are Usig The Sigma Symbol For Sums Some studets are tempted to simplify sigma sums of products as the product of the sums, like this: k(k + ) = k (k + ) This is bad algebra which igores the distributive law. (a+b)(x+y) is ot ax + by. Whe you distribute o the right side, you multiply every term i k sum by every term i the k+ sum (a total of 2 terms), ot just correspodig terms with each other. The correct simplificatio here is to distribute k(k + ) : k(k + ) = k 2 + k = k 2 + k.

23 Let us cosider aother summatio example: Idex Shiftig 50 (k + ) 2 = Sice we have a summatio formula for the k 2, we could evaluate this sum by expadig (k + ) 2 ito k 2 + 2k + : 50 (k + ) 2 = k k + = = There is a better way though, which is to perform a idex shift. Idex shiftig meas to icrease the limits of the idex variable k by some iteger costat c ad simultaeously substitute k c for k i the expressio beig summed: N f(k) = k= N+c k=+c f(k c) If we apply a idex shift with c = to our example sum, we get (k + ) 2 = k 2 = k 2 = k= = 45525

24 Geometric Sums We shall determie a summatio formula that helps us evaluate geometric sums, i.e. sums of the form aq k Sice the costat multiplier a ca be factored out, we oly require a formula for k=0 q k = + q + q q k=0 Let us multiply that expressio by q ad distribute: + q + q q q = q + q q + q q Every positive term here is caceled by a egative term, except for two terms that remai: If q, we ca divide by q ad obtai For q =, + q + q q =. + q + q q q = q + k=0 q k = q+ q

25 Geometric Sums II To evaluate a geometric sum where the expoet does ot start at zero, we could use the differece approach we have already ecoutered earlier: N k= N q k = k=0 q k k=0 There is a better way, however. We factor out the commo highest factor of q, which is q ad the perform a idex shift: q k N k= N q k = q k= N q k = q k=0 q k = q qn + q = qn+ q q

26 Geometric Sums III Let us work a more complex example of a summatio ivolvig the geometric sum k+ k= k 3 5 3k = 5 3k 5 = 5 25 k = 5 k=5 k=5 k= k k We ow use the formula we just discovered to evaluate: 20 5 k= k =

27 The geometric series () If we let i the geometric sum σ k=0 series: k=0 q k q k, we obtai the geometric Techically, this quatity is the limit of the geometric sums as. It is a calculus fact that we shall ot explai that this limit oly exists whe q <. I that case, the limit of q + as is zero. Therefore, k=0 q k q + = lim q = q for q <. Example: = the form.. = 0.) 2 = 2. (I biary, this equatio takes

28 The geometric series (2) Just like i the case of geometric sums, we ca hadle geometric series that do t start with a power of 0, by factorig out the first power ad the idex shiftig the remaiig series: k=k q k = q K k=k q k K = q K The we use the kow summatio formula to obtai k=k q k = qk q k=0 for q <. Observe that the umerator i this formula is just the first term of the series. Example: = =. (I biary, this equatio takes the form = 0..) q k

29 Geometric Series ad Ifiite Repeatig Decimal Expasios () You might already have realized by ow that ifiite, repeatig decimal expasios are all actually geometric series. Here is a famous decimal example: k = 9 = =. 0 Geerally, whe you have a base-b umber that starts with 0., followed by ifiitely may digits (b-), that umber equals. For example, i biary, 0. =. I octal, =. I hexadecimal, 0.FFFFFFFF =.

30 Geometric Series ad Ifiite Repeatig Decimal Expasios (2) If a patter of more tha oe digit repeats ifiitely, we ca also evaluate the umber usig the geometric series, thereby turig it ito a fractio. Example (i decimal): = = = = k = = = = We could use the geometric series to prove the geeral formula: if s = a a is a fiite sequece of decimal digits, the 0. sssssssss = a a digits

31 A Applicatio of the Geometric Series to Percetage Problems () The price of a product was raised by 2% ad is ow $00. What was the origial price? May studets who first lear percetages will thik that the aswer is 2% less of $00, or $98. This is wrog because of the shiftig baselie: the icrease was 2% of the origial price, ot 2% of the icreased price. Reducig the icreased price by 2% reduces it by too much. The correct origial price must therefore have bee more tha $98. The correct solutio you would lear i a lower-level class is to set up a equatio for the origial price p, takig ito accout that icreasig a quatity by 2% meas multiplyig it by.02: p.02 = $00 From this, it follows that p = $00.02 $98.04 rouded to the earest cet. It turs out that our wrog aswer $98 was actually a fairly good approximatio. This should ot be too much of a surprise sice the percet icrease was small, so the differece betwee 2% of the origial price ad 2% of the icreased price was small as well. The geometric series ca help us uderstad this pheomeo quatitatively.

32 A Applicatio of the Geometric Series to Percetage Problems (2) O the previous page, we saw that the origial price ca be recovered as p = $ Raisig the price by 2% meat multiplyig by.02; u-doig that icrease meat dividig by.02. Dividig by.02 however is ot the same as decreasig the icreased amout by 2%. The geometric series sheds some light o this. Dividig by.02 meas multiplyig by /.02 = /(+0.02) = /(-(-0.02)). The reaso we would write the multiplier that way is because that is the format of the right side of the geometric series: q k = q k=0 It follows that ( 0.02) = Sice the umber 0.02 is small i absolute value, its powers quickly become egligible. Retaiig just the first three terms ad simplifyig, we get the approximatio ( 0.02)

33 A Applicatio of the Geometric Series We foud ( 0.02) to Percetage Problems (3) = 00% 2% % = 00%.96% Remember that this is the multiplier that takes us back to the old price after a 2% icrease was applied. What this formula is sayig is that to recover the old price approximately, we just eed to subtract 2% from the ew price (i.e. apply the popular wrog solutio) ad correct it by addig 0.04% of the ew price. The 0.04% does ot come out of owhere, it s simply the 2% squared. Applied to the ew price of $00, this produces the old price correctly to the earest cet: $00-$2+$0.04=$ The percetage decrease of the old price relative to the ew price as a baselie is.96%. I other words, kowledge of the geometric series empowers you to solve a iverse percetage problem, which ordiarily requires divisio by a decimal fractio ad therefore the use of a calculator, solely with simple arithmetic i your head with sufficiet accuracy, based o the approximatio + x x + x2 = (x x 2 ) for small x. Let us look at oe more problem like this ad apply our shortcut.

34 A Applicatio of the Geometric Series to Percetage Problems (4) What percetage reductio is eeded to udo a 3% icrease? Our approximatio approach produces the aswer 3%-0.09% = 2.9%. (0.09% is , which is 3%=0.03 squared.) The exact aswer is more complicated ad requires a calculator: the multiplier eeded to udo a 3% icrease is = = %. Therefore, the percetage.03 reductio eeded is %. It turs out that our back-of-the-apki aswer 2.9% is a good approximatio.

35 Telescopig Sums Let us cosider k(k + ) = The partial fractio decompositio of k(k+) is By substitutig this idetity ito the sum we get k(k + ) = k k +. k(k + ) = We ca see that all terms i this sum cacel except ad + telescope ad is therefore amed a telescopig sum. Therefore, k(k + ) = +.. The sum collapses like a old-style