Chapter 4 Sequeces III 4. Roots We ca use the results we ve established i the last workbook to fid some iterestig limits for sequeces ivolvig roots. We will eed more techical expertise ad low cuig tha have bee required hitherto. First a simple iequality. Beroulli s Iequality Whe x > ad is atural umber, (+x) +x. Strictly Speakig Beroulli s Iequality is actually strict uless x = 0, = 0 or =. Exercise Sketch a graph of both sides of Beroulli s iequality i the cases = 2 ad = 3. For o-egative values of x Beroulli s iequality ca be easily proved usig the Biomial Theorem, which expads the left-had side: (+x) = +x+ ( ) x 2 + ( )( 2) 2 6 + +x +x +x. What difficulties do we have with this lie of argumet if x < 0? Exercise 2 Fiish off the followig proof of Beroulli s Iequality for x > usig mathematical iductio. Note dow where you use the fact that x >. Proof. We wat to show that (+x) +x where x > ad is a atural umber. This is true for = sice (+x) = +x. Now assume (+x) k +kx. The (+x) k+ = (+x) k (+x) +x 0 (+kx)(+x) = +(k+)x+kx 2 kx2 0 +(k+)x. We are doe. x 3 Biomial Theorem For all real values x ad y ( ) (x+y) = x k y k k k=0 where ( ) k =. k!( k)! The Beroulli Boys Beroulli s Iequality is amed after Jacques Beroulli, a Swiss mathematicia who used it i a paper o ifiite series i689(thoughitcabefoud earlier i a 670 paper by a Eglishma called Isaac Barrow). 35
36 CHAPTER 4. SEQUENCES III 2.8.6.4.2 0.8 0.6 0.4 0.2 0 0 2 4 6 8 0 2 4 6 8 20 Figure 4.: The sequece ( / ). Roots x / = x is the positive th root of x. Exercise 3 Use a calculator to explore the sequeces (2 / ), (0 / ) ad (000 / ). Repeated use of the square root butto will give a subsequece i each case. Propositio If x > 0 the (x / ). Example 00000000000 / ad also 0.00000000000 /. To prove this result you might follow the followig fairly cuig steps (although other proofs are very welcome): Exercise 4. First assume that x ad deduce that x /. 2. Let = x / ad use Beroulli s iequality to show that x +. 3. Use the Sadwich Rule to prove that ( ) is ull sequece. 4. Deduce that (x / ). 5. Show that (x / ) whe 0 < x < by cosiderig (/x / ). Auto Roots / = is the th root of. Natural umbers approach uity by rootig themselves. What happes if we first write = ( / ) = (+( / )) ad apply Beroulli s Iequality? Do we get aywhere? Exercise 5 Use a calulator to explore the limit of (2 +3 ) /. Now fid the limit of the sequece (x +y ) / whe 0 x y. (Try to fid a sadwich for your proof.) Propositio ( / ). See figure 4. for a graph of this sequece.
4.2. POWERS 37 5 4.5 4 3.5 3 2.5 2.5 0.5 0 0 2 4 6 8 0 2 4 6 8 20 Figure 4.2: The sequece (x ) with three slightly differet values of x. Proof. The proof is similar to that of the previous lemma but we have to be cuig ad first show that ( /2 ). Sice we have /2. Therefore, = ( /2 ) = (+( /2 )) +( /2 ) > ( /2 ) usig Beroulli s iequality. Rearragig, we see that /2 < +. So ( /2 ) by the Sadwich Theorem. Hece ( / ) = ( /2 ) 2 by the Product Rule. 4.2 Powers Exercise 6 Explore, with a calculator if ecessary, ad the write dow a cojectured limit for the power sequece (x ). (Warig: you should get four differet possible aswers depedig o the value of x.) To prove your cojectures you ca use Beroulli s iequality agai. Note that if x > the x = (+(x )) +(x ). To prove your cojecture for 0 < x < look at the sequece /x ad the use Exercise of Chapter 2. The treat all other values of x such as x = 0, x =, < x < 0, ad x. May sequeces are ot exactly powers but grow or shrik at least as fast as a sequece of powers so that we ca compare them with (or sadwich them by) a geometric sequece. A useful idea to formalise this is to cosider the ratio of two successive terms: + /. If this is close to a value x the the sequece ( ) might behave like the sequece (x ). We explore this idea. Ratio Lemma, light versio Let ( ) be a sequece of positive umbers. Suppose 0 < l < ad a+ l for all. The 0.
38 CHAPTER 4. SEQUENCES III Proof. We have that l l 2 2... l a. The 0 l a. Sice l a 0 as, ad usig the Sadwich Theorem (Chapter 3), we obtai that 0. Exercise 7 Cosider the sequece = +. The + + = + + = +2 + + = (+2) (+) 2 = 2 + 2 +2+ <, for all. The 0 by the Ratio Lemma. Is that really so? If ot, what is wrog? A much more powerful of the above lemma is as follows: Ratio Lemma, full versio Let ( ) be a sequece of ozero umbers. Suppose 0 < l < ad a+ l evetually. The 0. The precise defiitio of a+ l evetually is that there exists N such that a+ l for all > N. Proof. This proof follows the same lies as those for the light versio of the lemma, but it looks more difficult. We kow that + l for all > N. The if > N, l l 2 2... l N a N+ = l a N+ l N+. By choosig C large eough, we have l C for all N. Suppose also that C a N+ l N+. The we have for all, 0 l C. Sice l C 0 as, we have 0 by the Sadwich Theorem (Chapter 3). Examples Powerful Powers All icreasig power sequeces grow faster tha ay polyomial sequece. Powerless Powers All power sequeces are powerless agaist the factorial sequece ().. Show that 2 2 0 2. Show that 000 Proof. The ratio of successive terms is a+ = (+)2 /2 + 2 /2 = 2 ( ) + 2 2. So, takig ǫ = 4 i the defiitio of covergece, we have 4 a+ 3 4 for large. The Ratio Lemma the implies that 2 2 0.
4.2. POWERS 39 2. Let = 000. The a+ = 000+ /(+)! 000 / = 000 + 2 The Ratio Lemma says that 000 0 so that 000. for all 999. ( a+ ) I both of the examples above we showed that a for 0 a <, ad the used this to show that a+ l evetually. The followig corollary to versio 2 of the Ratio Lemma allows us to cut out some of this work. Corollary ( ) Let a 0,a,a 2,... be a sequece of positive umbers. If a+ a with 0 a < the ( ) 0. Proof. If a+ a, the for all ε > 0 there exists N such that a+ for all > N. We choose ε = a 2. The a + a < a+ε = 2 + 2 < for all > N. The 0 by the Ratio Lemma (full versio). a < ε Exercise 8 State whether the followig sequeces ted to zero or ifiity. Prove your aswers:. 000 2 2..000 3. 000 4. () 2 (2)! Exercise 9 Try usig the Ratio Lemma to prove that the sequece 0. Why does the lemma tell you othig? The sequeces ( k ) for k =,2,3,... ad (x ) for x > ad () all ted to ifiity. Which is quickest? The above examples suggest some geeral rules which we prove below. Exercise 0 Prove that ( x ) 0 for all values of x. Notice that this result implies that () /, sice for ay value of x, evetually we have that ( ) x < givig that x < () /. Exercise Prove that ( ) 0 as. Proof. We use the Ratio Lemma. Let =. We have + = (+)! (+) + = (+)! (+) + = (+) = (+ ) 2. The last iequality follows from ( + ) 2 (Beroulli s iequality). The 0 by the Ratio Lemma. Alteratively, we could have remarked that = = ( ) ( 2) 2.
40 CHAPTER 4. SEQUENCES III The 0 by the Sadwich Theorem. Exercise 2 Fid the limit of the sequece ( ) x as for all values of k x > 0 ad k =,2,... Exercise 3 Fid the limits of the followig sequeces. Give reasos. ( ) (. 4 + 9 9 (4 7 2 + 2. 0 +2 ) ) / ( ) ( (3 3. 4. 2 + ) ) / 3 3 +cos 2 2 +si 2 Stirlig & de Moivre The Stirlig formula is due to Abraham de Moivre (667 754), a Frech Protestat who lived i Eglad because of persecutios from the Catholic kig Louis XIV. De Moivre could improve the formula thaks to commets by the Scottish James Stirlig (692 770), who was from a Jacobite family ad therefore had difficulties with the Eglish (Jacobites were supporters of the kig James II of Britai, who was deposed because of his Catholic faith ad wet ito exile i Frace). 4.3 Applicatio - Factorials Factorials occur throughout mathematics ad especially where coutig argumets are used. The last sectio showed that the factorial sequece () is more powerful tha ay power sequece (x ), but earlier you showed that 0. Oe gais much iformatio about the speed of divergece of with Stirlig s formula: e. Here is a amazigly precise versio of Stirlig s formula: Theorem For all, 2π e ( + 2 ) 2π e ( + 2 + 288 2 ). I this sectio we will get a partial proof of the above theorem by usig two clever but very useful tricks. The first trick is to chage the product = 2 3 4 ( ) ito a sum by takig logarithms: log = log2+log3+ +log( )+log. The secod trick, which we shall use repeatedly i future sectios, is to approximate the sum by a itegral, see figure 4.3. Here is a graph of the fuctio log(x) with a series of blocks, each of width oe, lyig udereath the graph. The sum of the areas of all the blocks is log2+log3+ +log( ). But the area of the blocks is less tha the area uder the curve betwee x = ad x =. So we have: log = log2+log3+ +log( )+log logxdx+log = [xlogx x] +log = (+)log +. Takig expoetials of both sides we get the woderful upper boud + e +. This should be compared to claim of the theorem, amely that + 2e.
4.4. * APPLICATION - SEQUENCES AND BEYOND * 4 l- l3 l4 l2 2 3 4 5 3 2 Figure 4.3: Approximatig logxdx from below. Exercise 4 Use figure 4.4 to obtai a lower boud o. I this case the area of the blocks is greater tha the area uder the graph. Exercise 5 Use your upper ad lower bouds o to fid the followig limits: (i) ( ) 2 (ii) ( ) 4 4.4 * Applicatio - Sequeces ad Beyod * I Chapter 2 we defied a sequece as a ifiite list of umbers. However, the cocept of a ifiite list of other objects is also useful i mathematics. With that i mid, we defie a sequece of objects to be a ifiite list of those objects. For example, let P be the regular -sided polygo of area. The (P ) is a sequece of shapes. The mai questio we asked about sequeces was whether they coverged or ot. To examie covergece i geeral, we eed to be able to say whe two objects are close to each other. This is ot always a easy thig to do. However, give a sequece of objects, we may be able to derive sequeces of umbers ad examie those sequeces to lear somethig about the origial sequece. For example, give a sequece of tables, we could look at the umber of legs of each table. This gives us ew sequece of umbers which is related to the origial sequece of tables. However, differet related sequeces ca behave i wildly differet ways.
42 CHAPTER 4. SEQUENCES III l l3 l4 l5 l2 2 3 4 5 3 2 Figure 4.4: Approximatig logxdx from above. We shall demostrate this by cosiderig the set of solid shapes i R 2. Give a sequece of shapes, (S ), two obvious related sequeces are the sequece of perimeters, (p(s )), ad the sequece of areas, (a(s )). Example Let P be the regular -sided polygo cetred at the origi which fits exactly iside the uit circle. This sequece starts with P 3, which is the equilateral triagle. Let = a(p ) be the sequece of areas of the polygos. I Chapter we saw that = 2 si( ) 2π ad i Chapter 2 we saw that this coverges to π as which is the area of the uit circle. Exercise 6 Let p = p(p ) be the perimeter of P. Show that p = 2si ( ) ( π = 2si 2π 2) = 2a2. Exercise 7 Show that limp = 2π. We see that i both cases we get what we would expect, amely that as the shape looks more ad more like the circle, so also the area ad perimeter ted to those of the circle. Exercise 8 Cosider the sequece of shapes i figure 4.5. Each is produced from the former by replacig each large step by two half-sized oes. Draw the limitig shape. Give that the origial shape is a square of area, what is the perimeter ad area of the th shape? Compare the limits of these sequeces with the perimeter ad area of the limitig shape.
4.4. * APPLICATION - SEQUENCES AND BEYOND * 43 Figure 4.5: A famous example of this type of behaviour is the Koch curve, see figure 4.6. The iitial figure is a equilateral triagle of area A ad perimeter p. To each side of the triagle is attached aother equilateral triagle at the trisectio poits of the triagle. This process is the applied to each side of the resultig figure ad so o. Figure 4.6: Let p be the perimeter of the shape at the th stage ad A the area. Exercise 9. What is the umber of sides of the shape at the th stage (for = the aswer is 3). 2. Show that p + = 4 3 p. 3. Prove that (p ). 4. Show that, i makig the (+) th shape, each little triagle beig added has area 9 A. 5. Show that A + = A + 3 9 ( 4 ) A 9. 6. Prove that (A ) 8 5 A. (Recall the sum of a geometric series.) Check Your Progress By the ed of this chapter you should be able to: Uderstad, memorise, prove, ad use a selectio of stadard limits ivolvig roots, powers ad factorials.
44 CHAPTER 4. SEQUENCES III