# Sequences and Series

Size: px
Start display at page:

Transcription

1 CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their properties; for example, trascedetal fuctios. This is the techique of Ifiite Series. Computer algorithms for determiig the value of a fuctio deped upo the usual arithmetic operatios; thus a exact determiatio ca oly be achieved for ratioal fuctios (quotiets of polyomials). If a fuctio is trascedetal, its values ca oly be approximated. For example, we kow that (9.) e x lim This expressio tells us that if for ay we do the calculatio described by the expressio o the right, that these umbers will, for large eough, be close to the true value of e x. Now, it turs out that this is a very iefficiet way to calculate e x, ad the expressio as a ifiite series (which we will discover later i this chapter) (9.2) e x x x x 2 2! x 3 3! x! is far better. Equatio 9.2 is take to mea: add up the umbers of the form x!, startig with 0. If we add up eough terms, we have a good approximatio to e x. Of course, it is importat to have estimates o how good this approximatio is, ad more geerally, to have ways of fidig these approximatig sums. That is what we study i this chapter, startig with the idea of covergece i the sese of good approximatio. Defiitio 9. A sequece is a list of umbers, deoted a, where a is the th term of the sequece. A sequece may be defied by a specific formula or a algorithm for determiig the members of the sequece successively, or recursively. 28

4 9. Covergece: Defiitio ad Examples 3 Propositio 9.4 a) (Squeeze theorem) Give three sequeces a b c, if evetually (9.9) a b c for all ad lim a the also (9.0) lim b b) Suppose that a bouded ad L lim c L b c evetually, that is, for all larger tha some N. If the sequece b is (9.) lim c the also (9.2) lim a p Example 9.5 For ay positive iteger p, lim 0! The idea here is that the umerator is a product of p terms, whereas the deomiator is a product of terms, so grows faster tha the umerator. To make this precise, write (9.3) p! 0 0 p p! Now, if is so large that p 2 2p will do), the the first factor is bouded by 2 p. Thus, for 2p, that is, evetually, (9.4) p! 2p p! Sice p! 0 ad, the result follows from the squeeze theorem. Fially, we ote that the limit of a sum is the sum of the limits: Propositio 9.5 If a b c, ad the sequeces b ad c coverge, the so does the sequece a, ad (9.5) lim a lim b lim c Series For may sequeces, i fact, the most importat oes, the geeral term is formed by addig somethig to its predecessor; that is, the sequece is formed by the recursio s s a, where a is from aother sequece. Such a sequece is called a series. Explicitly, the terms of the series are (9.6) a a a 2 a a 2 a 3 a a 2 a 3 a

5 Chapter 9 Sequeces ad Series 32 It is useful to use the summatio symbol: (9.7) a a 2 a 3 a k a k Defiitio 9.3 The series (9.8) a k is to be cosidered as the limit of the sequece (9.9) s a k If the limit L of the sequece s exists, the series is said to coverge, ad L is called its sum. If the limit does ot exist, the series diverges. The terms of the sequece s are called the partial sums of the series. Example 9.6 k 2 k Let s look at a few partial sums: (9.20) We see that each term adds half the distace of its predecessor from, from which we guess that the partial sum: s 2. Let s ow show that to be true. As we have see it is true for the first four terms. If it is true for the th term, it is also true for the th term: (9.2) s s Thus, our guess holds for the fifth, ad the the sixth, ad, by cotiued applicatio of equatio 9.2, ultimately, for all terms. So the result is easy to coclude: (9.22) k 2 k lim s lim 2 Now, remember that the idex is a way of relatig the partial sums of the series to the geeral term from which it is defied, so if we chage that relatio cosistetly, we do t chage the series. For example, (9.23) k a k a a k m 9a m 8 ad so forth. Each represetatio comes about by replacig the idex with a ew idex. For example, if we substitute for k, we get the first equality; if we substitute k for we get the secod equality, ad

6 9. Covergece: Defiitio ad Examples 33 if we replace k examples show. Example 9.7 For (9.24) Example 9.8 (9.25) by m 8, we get the last oe. It is ofte useful to make a chage of idex as the ext 2 k 2 2 k k 2 k 2 k 2 k 2 First, chage the idex by k m, ad the factor out 2 : k 2 k m 0 2 m 2 m 0 2 m Propositio 9.6 (Geometric Series) : (9.26) x k x for x (9.27) x k diverges for x To show this, we obtai (by a clever little observatio) a formula for the partial sums (9.28) s Note that (9.29) s (9.30) s Equatig these expressios for s x k x x 2 x x x 2 x x x x 2 x, we obtai s x s x xs ad xs. Solvig this for s : (9.3) s so (9.32) x k lim s x k x x x lim x

7 Chapter 9 Sequeces ad Series 34 which equals x if x ad diverges if x. We look at the cases x separately. For x s, so the series diverges. For x, the sequece s is the sequece 0 0 0, so caot coverge to ay particular umber. Example 9.9 (9.33) k k Thus the partial sum s ca be calculated: (9.34) s 2 We first use the fact that k k 3 2 k k 4 3 (9.35) (9.36) which coverges to as goes to ifiity. This is a example of a telescopig series. Fially, we observe that if a series coverges, its geeral term must go to zero. Be careful: there are may series whose geeral term goes to zero which do ot coverge. Propositio 9.7 If a k coverges, the lim a k To see this, let s a k t k 0 a k. The, sice these are both sequeces of the partial sums of the series, but idexed differetly, lim s lim t. Thus lim s t 0. But s t a. Fially, Propositio 9.5 gives us: Propositio 9.8 If a b c, ad the series b ad c coverge, the so does the series a, ad a b c Tests for covergece Throughout this sectio, uless otherwise specified, we will be cosiderig series, all of whose terms are positive. For such a series, the sequece of partial sums is icreasig. If they remai bouded, the - just as i the assertio of theorem 8. for fuctios - the sequece of partial sums will coverge. Propositio 9.9 If a k 0 for all k, ad there is a M k 0 such that a M for all, the a k coverges.

8 9.2 Tests for covergece 35 Because of this propositio, for a series with positive terms, the statemets a k coverges, a k diverges, are usually writte simply as (9.37) a k coverges Here is a importat applicatio of this propositio: a k diverges 0 Propositio 9.0 (Compariso Test). Give two sequeces a k b k with a a) If b k the a k b) If a k the b k k b k. The As for (a), the sequece of partial sums s a k is bouded by 0 b k, so coverges by Propositio 9.9. I the secod case, sice the sequece of partial sums a k has o boud, either does the sequece of partial sums of b k. It is importat to observe that it is ot ecessary that the iequalities i the hypothesis of propositio 9.0 hold for all k, oly that they evetually hold. That is because the issue of covergece series is determied by the ed of the series, ad ot affected by ay fiite umber of terms. Example 9.0 r k r if 0 r Sice r k r k r, (9.38) so the compariso test applies. Example 9. k r k if r r k r r k Now, here the trouble is that the umerator grows without boud - but it does t grow as fast as a power. So, what we do is borrow somethig from the deomiator to compesate for the umerator. We ote that evetually k k r 2 ; i fact, this is true as soo as k 2lk lr (which evetually happes, sice k lk ). The for all k larger tha this umber (9.39) k r k Sice r, we also have r, ad so the series k r k r k r k (9.40) r k coverges, ad thus, by compariso, our origial series coverges. Example 9.2 (9.4) Now, 2 2

9 Chapter 9 Sequeces ad Series 36 so our series is domiated by a telescopig series which coverges (see example 9.9. above). A very useful applicatio of the compariso test is the followig. Propositio 9. (The Itegral Test). Suppose that f is a oegative, oicreasig fuctio defied o a iterval M. Suppose the a is a sequece such that for M, a a) If f x dx the a M b) If M f x dx the a Let (9.42) b f x dx f. The The, sice the fuctio is oicreasig, f b f a ; that is a b a compariso theorem. For example, if f x dx, the b coverges, so by compariso a coverges. Example 9.3 (The harmoic series). We apply the itegral test usig the fuctio f x (9.43) as we saw i chapter 8, the result follows. dx x x. Sice Now, use the also If we apply example 7 of chapter 8 to series via the itegral test we have a result which is very useful for comparisos: Propositio 9.2 Let p be a positive umber. a) p if p p b) p if This follows from the same result for the itegral of x p. Example The fuctio f x l p x lx p is decreasig. We itegrate usig the substitutio u lx: (9.44) A 2 dx la du x lx p l2 u p We kow (agai from example 7, chapter 8) that this coverges if p, ad otherwise diverges. Thus, by the itegral test, (9.45) 2 l if p p

10 9.2 Tests for covergece 37 ad otherwise diverges. Fially, we eed a tool to test for covergece whe we caot realize the geeral term of the series i the form f for some fuctio f. For example, if the expressio for a ivolves the factorial, we proceed to the followig. Propositio 9.3 (Ratio Test). Give the series a, cosider (9.46) lim a a if the limit exists. If L, the series coverges; if L, the series diverges. For the case L draw o coclusio. Suppose that L. The there is a umber r with L r such that evetually a there is a iteger N such that a a r for all N. We coclude (9.47) a N a N r a N 2 a N r a N r2 a N 3 a N 2 r a N r3 L, we ca a r. That is, ad so forth. Thus, we have, for all k, a N k a N r k, so by compariso with the geometric series, our series coverges. If o the other had, L, there is a umber r, L r, such that evetually a a r. Followig the same argumet but with the iequalities reversed, we coclude that for all k, a N k a N r k, so we have divergece by compariso with the geometric series. We ca coclude othig if L. This is the case for the all the series of the type p, ad as we have see, for some p we get covergece, ad divergece for other p. Example 9.5 a! We try the ratio test. (9.48) a a a!! a a 0 as, so the ratio test gives us covergece. Example 9.6 (9.49) Try the ratio test: a a so we have covergece. Example 9.7 (9.50) r Here the ratio test gives a a r

11 Chapter 9 Sequeces ad Series 38 so we coclude that the series coverges if r, ad diverges if r. This may seem to be a simplificatio of propositio 9.6, but i fact it is a fraud. The argumet is circular, for we have used the covergece of the geometric series to derive the ratio test. We observe that we did t really eed to kow that the limit of a a exists, oly that evetually these ratios are either less tha some umber less tha to coclude covergece, or greater tha some umber greater tha, for divergece Absolute covergece There are ew difficulties whe we have to cosider series of egative as well as positive terms. For example, although the harmoic series diverges, if we alterately chage sigs, the series ow coverges. Example 9.8 The series coverges. To see this, we look at the sequeces of eve partial sums ad odd partial sums separately. Sice (9.5) s 2 s the sequece of eve partial sums is icreasig. Similarly, (9.52) s 2 s 2 2 s s 2 tells us that the sequece of odd partial sums is decreasig. Now (9.53) s 2 s 2 2 s 2 that is, the odd partial sums are all greater tha all the eve partial sums. So both sequeces are bouded, ad thus coverge. But, they coverge to the same limit, as we see by takig the limit i equatio 9.53: (9.54) lim s 2 lim s 2 lim 2 lim s 2 sice 2 0. Sice they both coverge to the same limit, the full sequece also coverges, ad to the same limit. This argumet actually geeralizes to ay alteratig series, a series whose terms alterate i sig. Propositio 9.4 If a is a decreasig sequece, ad lim a 0 the the series a coverges. Defiitio 9.4 Give a sequece a, we say the series a coverges absolutely if, for the series formed of the absolute values a, we have covergece: a. Propositio 9.5 If a series coverges absolutely, it coverges. That is, (9.55) If a the a coverges

12 9.4 Power Series 39 To see that, let s be the th partial sum of the sequece, p the sum of all the positive terms makig up s, ad q the sum of the absolute values of all the egative terms. The (9.56) s p q Both sequeces p ad q are icreasig, ad bouded by a, so coverge, to, say p, q respectively. The (9.57) a lim s lim p lim q p q Because of this peculiarity of sequeces of terms with alteratig sigs, we shall be most iterested i absolute covergece. We ca use the tests of sectio 9.3 (applied to the series of absolute values), to test for absolute covergece. Example 9.9 x coverges for x This is because the sum of the absolute values is just the geometric series. Example x coverges for x Here we use the ratio test for the absolute values; a (9.58) a 2 x 2 x Thus, we get covergece for x of absolute value less tha. 2 x x 9.4. Power Series Defiitio 9.5 the power series. A power series is a series of the form a x c The poit c is called the ceter of A power series defies a fuctio o the set of poits for which it coverges by (9.59) f x a x c The series provides a effective way of approximately evaluatig the fuctio f ; our goal i these last sectios is to show that most fuctios do have a power series represetatio. We ca use the ratio test to determie the questio of covergece. We take the ratio of successive terms of (4): a (9.60) x c a a x c a x c L x c

15 Chapter 9 Sequeces ad Series 42 where the last equatio is obtaied by replacig the idex by. Thus f x differetial equatio, y y, defiig the expoetial fuctio. Sice f 0 fuctio. f x, so satisfies the also, it is the expoetial Example 9.29 e x 2 x2! for all x Just replace x i example 9.28 by x Taylor Series Fially we tackle the questio: how do we fid the power series represetatio of a give fuctio? Recallig that the purpose of the power series is to have a effective way to approximate the values of a fuctio by polyomials, we tur to that questio: what is the best way to so approximate a fuctio? We start with a fuctio f that has derivatives of all orders defied i a iterval about the origi. To begi with, we recall the defiitio of the derivative i this cotext: (9.7) lim x 0 f x f 0 x If we rewrite this as f 0 (9.72) lim x 0 f x f 0 f 0 x x we see that the liear fuctio y f 0 f 0 x approximates f x to first order: f 0 f 0 x is closer to f x tha x is to zero, ad by a order of magitude. We ow ask, ca we fid a quadratic polyomial which approximates f to secod order? Let y a bx cx 2 be such a polyomial. The we wat f x a bx cx 2 (9.73) lim x 0 x 2 0 We calculate this limit usig l Hôpital s rule. First of all, for l Hôpital s rule to apply, we have to have a f 0. The f x f 0 bx cx 2 (9.74) lim l H f x b 2cx x 0 x 2 lim x 0 2x We ca apply l Hôpital s rule agai, if we have b f 0 : f x f 0 2cx (9.75) lim l H f x 2c lim x 0 2x x 0 2 if c f 0 2. We coclude that the polyomial (9.76) f 0 f 0 x approximates f to secod order: this is closer to f x tha x is to 0 by two orders of magitude. Furthermore, it is the uique quadratic polyomial to do so. f 0 2 x 2 0 0

16 9.5 Taylor Series 43 We ca repeat this procedure as may times as we care to, cocludig Propositio 9.8 The polyomial which approximates f ear 0 to th order is (9.77) f 0 f 0 x f 0 2 x 2 f 0! Of course we ca make the same argumet at ay poit, ot just the origi. To summarize: Defiitio 9.6 Suppose that f is a fuctio with derivatives at all orders defied i a iterval about the poit c. The Taylor polyomial of degree of f, cetered at c is (9.78) T c f x f k c k! x c k Propositio 9.9 The Taylor polyomial T c f is the polyomial of degree which approximates f ear c to th order. So, we ca compute effective approximatios to the values of f x ear c by these Taylor polyomials; but the questio is, how effective is this? More precisely, what is the error? We use this estimate: Propositio 9.20 Suppose that f is differetiable to order i the iterval c a c a cetered at the poit c. The the error i approximatig f i this iterval by its Taylor polyomial of degree, T c f is bouded by (9.79) M! x c where M is a boud of the values of f over the iterval c a c a. To be precise, we have the iequality (9.80) f x Tc M f x! x c I the ext chapter we will show how the error estimate is obtaied, ad see how to work with it. What we wat ow is to cocetrate o the represetatio by series. Defiitio 9.7 Let f be a fuctio which is differetiable to all orders i a eighborhood of the poit c. The Taylor series for f cetered at c is (9.8) T c f x f c x c! If c is the origi, this series is called the Maclauri series for f. Propositio 9.2 Suppose that f is a fuctio which has derivatives of all orders i the iterval c a c a, Let M be a boud for the th derivative of f i the iterval. If the sequece (9.82) M! x c 0

17 Chapter 9 Sequeces ad Series 44 coverges to zero for all x i the iterval, the f is give by its Taylor series: (9.83) f x i c c a. As a example, e x has the Maclauri series f c! x c (9.84) e x! x We have already show by other meas. We ca verify this usig propositio 9.2 as well, sice the th derivative of e x is still e x, ad the value at x 0 is. By a parallel calculatio we obtai the power series represetatio of e x cetered at ay poit: Example 9.30 For c ay poit, the fuctio e x has the Taylor series represetatio cetered at c: (9.85) e x e c! x c We do have to verify that the remaiders coverge to zero; that is the terms 9.82 coverge to zero. Sice e x is a icreasig fuctio, its maximum i the iterval a c a c is at x a c, so we ca take M e a c. The, for the expoetial fuctio we have M (9.86) lim! x c by example It is useful to make the followig observatio e a c x c lim! Propositio 9.22 Suppose that f has a power series represetatio: (9.87) f x The, this is its Taylor series. More precisely: a x c (9.88) a f c! 0 This is easy to see; if we differetiate 9.87 k times we obtai: (9.89) f k x Now, let x c ad obtai f k c k k a x c k k!a k, for all terms but the first have the factor x c. So, if we have foud a power series represetative of a fuctio, the that is automatically the Taylors series for the fuctio.

18 9.5 Taylor Series 45 Example 9.3 Fid the Maclauri series for the fuctio f x x 5x 2 x 3. Sice a polyomial is already expressed as a sum of powers of x, that expressio is a power series, ad thus the Maclauri series for the polyomial. Example 9.32 Fid the Taylor series cetered at c for the fuctio f x have to fid the values of the derivatives of f at c : (9.90) f 4 x 5x 2 x 3. We (9.9) f x 0x 3x 2 so f 6 (9.92) f x 0 6x so f (9.93) f x 6 so f 6 4 ad all higher derivatives are zero. Thus the Taylor series is (9.94) f x 4 6 x 4 2! x 2 6 3! x x 2 x 2 x 3 Now, we ca fid the Maclauri series for may fuctios, so log as we kow how to differetiate them. Followig is a list of the most importat Maclauri series. Propositio 9.23 a) x b) e x! x x x c) cosx d) six 2! x2 2! x2 e) arctax 2 x2 We have already see how to get (a), (b) ad (d). For the trigoometric fuctios, we proceed as follows. First, the cosie: (9.95) f 0 (9.96) f x six so f (9.97) f x cosx so f 0

19 Chapter 9 Sequeces ad Series 46 (9.98) f x (9.99) f iv x Thus, up to four terms we have six so f cosx so f iv 0 (9.00) cosx x 2 2! x 4 4! But, ow, sice we have retured to cosx, the cycle 0 0 coclude that (9.0) cosx x 2 2! x 4 x 6 4! 6! x 8 8! repeats itself agai ad agai. We which ca be rewritte as iii of propositio 9.23 above. Oe fial Taylor series is worth otig: sice the itegral of x is l x, we ca fid the Taylor series cetered at for lx as follows: (9.02) t t (9.03) l t Now, make the substitutio x t, so t x: x (9.04) lx t x

### In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008

I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces

INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal

### SAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx

SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval

### Our aim is to show that under reasonable assumptions a given 2π-periodic function f can be represented as convergent series

8 Fourier Series Our aim is to show that uder reasoable assumptios a give -periodic fuctio f ca be represeted as coverget series f(x) = a + (a cos x + b si x). (8.) By defiitio, the covergece of the series

### Infinite Sequences and Series

CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...

### Section 11.3: The Integral Test

Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult

Physics 6A Witer 20 Theorems About Power Series Cosider a power series, f(x) = a x, () where the a are real coefficiets ad x is a real variable. There exists a real o-egative umber R, called the radius

### 4.3. The Integral and Comparison Tests

4.3. THE INTEGRAL AND COMPARISON TESTS 9 4.3. The Itegral ad Compariso Tests 4.3.. The Itegral Test. Suppose f is a cotiuous, positive, decreasig fuctio o [, ), ad let a = f(). The the covergece or divergece

### Soving Recurrence Relations

Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree

### Trigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is

0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values

### Example 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).

BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook - Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly

### Properties of MLE: consistency, asymptotic normality. Fisher information.

Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout

### Lecture 4: Cauchy sequences, Bolzano-Weierstrass, and the Squeeze theorem

Lecture 4: Cauchy sequeces, Bolzao-Weierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits

### SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES

SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,

### Convexity, Inequalities, and Norms

Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for

### FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. 1. Powers of a matrix

FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. Powers of a matrix We begi with a propositio which illustrates the usefuless of the diagoalizatio. Recall that a square matrix A is diogaalizable if

### Basic Elements of Arithmetic Sequences and Series

MA40S PRE-CALCULUS UNIT G GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED NO NEED TO COPY NOTES FROM OVERHEAD) Basic Elemets of Arithmetic Sequeces ad Series Objective: To establish basic elemets of arithmetic

### THE ARITHMETIC OF INTEGERS. - multiplication, exponentiation, division, addition, and subtraction

THE ARITHMETIC OF INTEGERS - multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,

### Chapter 5: Inner Product Spaces

Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples

### a 4 = 4 2 4 = 12. 2. Which of the following sequences converge to zero? n 2 (a) n 2 (b) 2 n x 2 x 2 + 1 = lim x n 2 + 1 = lim x

0 INFINITE SERIES 0. Sequeces Preiary Questios. What is a 4 for the sequece a? solutio Substitutig 4 i the expressio for a gives a 4 4 4.. Which of the followig sequeces coverge to zero? a b + solutio

### 2-3 The Remainder and Factor Theorems

- The Remaider ad Factor Theorems Factor each polyomial completely usig the give factor ad log divisio 1 x + x x 60; x + So, x + x x 60 = (x + )(x x 15) Factorig the quadratic expressio yields x + x x

### CS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations

CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad

### Asymptotic Growth of Functions

CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll

### BINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients

652 (12-26) Chapter 12 Sequeces ad Series 12.5 BINOMIAL EXPANSIONS I this sectio Some Examples Otaiig the Coefficiets The Biomial Theorem I Chapter 5 you leared how to square a iomial. I this sectio you

### .04. This means \$1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth

Questio 1: What is a ordiary auity? Let s look at a ordiary auity that is certai ad simple. By this, we mea a auity over a fixed term whose paymet period matches the iterest coversio period. Additioally,

### S. Tanny MAT 344 Spring 1999. be the minimum number of moves required.

S. Tay MAT 344 Sprig 999 Recurrece Relatios Tower of Haoi Let T be the miimum umber of moves required. T 0 = 0, T = 7 Iitial Coditios * T = T + \$ T is a sequece (f. o itegers). Solve for T? * is a recurrece,

### Lecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)

18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the Bru-Mikowski iequality for boxes. Today we ll go over the

### WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER?

WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I tought Topological Groups at the Göttige Georg August Uiversity. This

### Approximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find

1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1.

### Building Blocks Problem Related to Harmonic Series

TMME, vol3, o, p.76 Buildig Blocks Problem Related to Harmoic Series Yutaka Nishiyama Osaka Uiversity of Ecoomics, Japa Abstract: I this discussio I give a eplaatio of the divergece ad covergece of ifiite

### Class Meeting # 16: The Fourier Transform on R n

MATH 18.152 COUSE NOTES - CLASS MEETING # 16 18.152 Itroductio to PDEs, Fall 2011 Professor: Jared Speck Class Meetig # 16: The Fourier Trasform o 1. Itroductio to the Fourier Trasform Earlier i the course,

### Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at http://tutorial.math.lamar.edu/terms.asp. The olie versio

### 7.1 Finding Rational Solutions of Polynomial Equations

4 Locker LESSON 7. Fidig Ratioal Solutios of Polyomial Equatios Name Class Date 7. Fidig Ratioal Solutios of Polyomial Equatios Essetial Questio: How do you fid the ratioal roots of a polyomial equatio?

### 3. Greatest Common Divisor - Least Common Multiple

3 Greatest Commo Divisor - Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd

### Chapter 7 Methods of Finding Estimators

Chapter 7 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 011 Chapter 7 Methods of Fidig Estimators Sectio 7.1 Itroductio Defiitio 7.1.1 A poit estimator is ay fuctio W( X) W( X1, X,, X ) of

### 1 Computing the Standard Deviation of Sample Means

Computig the Stadard Deviatio of Sample Meas Quality cotrol charts are based o sample meas ot o idividual values withi a sample. A sample is a group of items, which are cosidered all together for our aalysis.

### 1. MATHEMATICAL INDUCTION

1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1

### SEQUENCES AND SERIES

Chapter 9 SEQUENCES AND SERIES Natural umbers are the product of huma spirit. DEDEKIND 9.1 Itroductio I mathematics, the word, sequece is used i much the same way as it is i ordiary Eglish. Whe we say

### Incremental calculation of weighted mean and variance

Icremetal calculatio of weighted mea ad variace Toy Fich faf@cam.ac.uk dot@dotat.at Uiversity of Cambridge Computig Service February 009 Abstract I these otes I eplai how to derive formulae for umerically

### CHAPTER 3 THE TIME VALUE OF MONEY

CHAPTER 3 THE TIME VALUE OF MONEY OVERVIEW A dollar i the had today is worth more tha a dollar to be received i the future because, if you had it ow, you could ivest that dollar ad ear iterest. Of all

### Chapter 7 - Sampling Distributions. 1 Introduction. What is statistics? It consist of three major areas:

Chapter 7 - Samplig Distributios 1 Itroductio What is statistics? It cosist of three major areas: Data Collectio: samplig plas ad experimetal desigs Descriptive Statistics: umerical ad graphical summaries

### Cooley-Tukey. Tukey FFT Algorithms. FFT Algorithms. Cooley

Cooley Cooley-Tuey Tuey FFT Algorithms FFT Algorithms Cosider a legth- sequece x[ with a -poit DFT X[ where Represet the idices ad as +, +, Cooley Cooley-Tuey Tuey FFT Algorithms FFT Algorithms Usig these

### AP Calculus BC 2003 Scoring Guidelines Form B

AP Calculus BC Scorig Guidelies Form B The materials icluded i these files are iteded for use by AP teachers for course ad exam preparatio; permissio for ay other use must be sought from the Advaced Placemet

### Discrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13

EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may

### A Recursive Formula for Moments of a Binomial Distribution

A Recursive Formula for Momets of a Biomial Distributio Árpád Béyi beyi@mathumassedu, Uiversity of Massachusetts, Amherst, MA 01003 ad Saverio M Maago smmaago@psavymil Naval Postgraduate School, Moterey,

### Department of Computer Science, University of Otago

Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS-2006-09 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly

### GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook. Version: 1.4

GCE Further Mathematics (660) Further Pure Uit (MFP) Tetbook Versio: 4 MFP Tetbook A-level Further Mathematics 660 Further Pure : Cotets Chapter : Comple umbers 4 Itroductio 5 The geeral comple umber 5

### Hypothesis testing. Null and alternative hypotheses

Hypothesis testig Aother importat use of samplig distributios is to test hypotheses about populatio parameters, e.g. mea, proportio, regressio coefficiets, etc. For example, it is possible to stipulate

### Remarques sur un beau rapport entre les series des puissances tant directes que reciproques

Aug 006 Traslatio with otes of Euler s paper Remarques sur u beau rapport etre les series des puissaces tat directes que reciproques Origially published i Memoires de l'academie des scieces de Berli 7

### 1. C. The formula for the confidence interval for a population mean is: x t, which was

s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : p-value

### Repeating Decimals are decimal numbers that have number(s) after the decimal point that repeat in a pattern.

5.5 Fractios ad Decimals Steps for Chagig a Fractio to a Decimal. Simplify the fractio, if possible. 2. Divide the umerator by the deomiator. d d Repeatig Decimals Repeatig Decimals are decimal umbers

### Chapter 6: Variance, the law of large numbers and the Monte-Carlo method

Chapter 6: Variace, the law of large umbers ad the Mote-Carlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value

### A probabilistic proof of a binomial identity

A probabilistic proof of a biomial idetity Joatho Peterso Abstract We give a elemetary probabilistic proof of a biomial idetity. The proof is obtaied by computig the probability of a certai evet i two

### CS103X: Discrete Structures Homework 4 Solutions

CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible six-figure salaries i whole dollar amouts are there that cotai at least

### SEQUENCES AND SERIES CHAPTER

CHAPTER SEQUENCES AND SERIES Whe the Grat family purchased a computer for \$,200 o a istallmet pla, they agreed to pay \$00 each moth util the cost of the computer plus iterest had bee paid The iterest each

### University of California, Los Angeles Department of Statistics. Distributions related to the normal distribution

Uiversity of Califoria, Los Ageles Departmet of Statistics Statistics 100B Istructor: Nicolas Christou Three importat distributios: Distributios related to the ormal distributio Chi-square (χ ) distributio.

### Simple Annuities Present Value.

Simple Auities Preset Value. OBJECTIVES (i) To uderstad the uderlyig priciple of a preset value auity. (ii) To use a CASIO CFX-9850GB PLUS to efficietly compute values associated with preset value auities.

### where: T = number of years of cash flow in investment's life n = the year in which the cash flow X n i = IRR = the internal rate of return

EVALUATING ALTERNATIVE CAPITAL INVESTMENT PROGRAMS By Ke D. Duft, Extesio Ecoomist I the March 98 issue of this publicatio we reviewed the procedure by which a capital ivestmet project was assessed. The

### The Stable Marriage Problem

The Stable Marriage Problem William Hut Lae Departmet of Computer Sciece ad Electrical Egieerig, West Virgiia Uiversity, Morgatow, WV William.Hut@mail.wvu.edu 1 Itroductio Imagie you are a matchmaker,

### Fast Fourier Transform

18.310 lecture otes November 18, 2013 Fast Fourier Trasform Lecturer: Michel Goemas I these otes we defie the Discrete Fourier Trasform, ad give a method for computig it fast: the Fast Fourier Trasform.

### Overview of some probability distributions.

Lecture Overview of some probability distributios. I this lecture we will review several commo distributios that will be used ofte throughtout the class. Each distributio is usually described by its probability

### I. Chi-squared Distributions

1 M 358K Supplemet to Chapter 23: CHI-SQUARED DISTRIBUTIONS, T-DISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad t-distributios, we first eed to look at aother family of distributios, the chi-squared distributios.

### AP Calculus AB 2006 Scoring Guidelines Form B

AP Calculus AB 6 Scorig Guidelies Form B The College Board: Coectig Studets to College Success The College Board is a ot-for-profit membership associatio whose missio is to coect studets to college success

### THE REGRESSION MODEL IN MATRIX FORM. For simple linear regression, meaning one predictor, the model is. for i = 1, 2, 3,, n

We will cosider the liear regressio model i matrix form. For simple liear regressio, meaig oe predictor, the model is i = + x i + ε i for i =,,,, This model icludes the assumptio that the ε i s are a sample

### Lecture 3. denote the orthogonal complement of S k. Then. 1 x S k. n. 2 x T Ax = ( ) λ x. with x = 1, we have. i = λ k x 2 = λ k.

18.409 A Algorithmist s Toolkit September 17, 009 Lecture 3 Lecturer: Joatha Keler Scribe: Adre Wibisoo 1 Outlie Today s lecture covers three mai parts: Courat-Fischer formula ad Rayleigh quotiets The

### Escola Federal de Engenharia de Itajubá

Escola Federal de Egeharia de Itajubá Departameto de Egeharia Mecâica Pós-Graduação em Egeharia Mecâica MPF04 ANÁLISE DE SINAIS E AQUISÇÃO DE DADOS SINAIS E SISTEMAS Trabalho 02 (MATLAB) Prof. Dr. José

### 5.4 Amortization. Question 1: How do you find the present value of an annuity? Question 2: How is a loan amortized?

5.4 Amortizatio Questio 1: How do you fid the preset value of a auity? Questio 2: How is a loa amortized? Questio 3: How do you make a amortizatio table? Oe of the most commo fiacial istrumets a perso

### NATIONAL SENIOR CERTIFICATE GRADE 12

NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 04 MARKS: 50 TIME: 3 hours This questio paper cosists of 8 pages ad iformatio sheet. Please tur over Mathematics/P DBE/04 NSC Grade Eemplar INSTRUCTIONS

### Analysis Notes (only a draft, and the first one!)

Aalysis Notes (oly a draft, ad the first oe!) Ali Nesi Mathematics Departmet Istabul Bilgi Uiversity Kuştepe Şişli Istabul Turkey aesi@bilgi.edu.tr Jue 22, 2004 2 Cotets 1 Prelimiaries 9 1.1 Biary Operatio...........................

### 1 Correlation and Regression Analysis

1 Correlatio ad Regressio Aalysis I this sectio we will be ivestigatig the relatioship betwee two cotiuous variable, such as height ad weight, the cocetratio of a ijected drug ad heart rate, or the cosumptio

### NEW HIGH PERFORMANCE COMPUTATIONAL METHODS FOR MORTGAGES AND ANNUITIES. Yuri Shestopaloff,

NEW HIGH PERFORMNCE COMPUTTIONL METHODS FOR MORTGGES ND NNUITIES Yuri Shestopaloff, Geerally, mortgage ad auity equatios do ot have aalytical solutios for ukow iterest rate, which has to be foud usig umerical

### CHAPTER 7: Central Limit Theorem: CLT for Averages (Means)

CHAPTER 7: Cetral Limit Theorem: CLT for Averages (Meas) X = the umber obtaied whe rollig oe six sided die oce. If we roll a six sided die oce, the mea of the probability distributio is X P(X = x) Simulatio:

### SOME GEOMETRY IN HIGH-DIMENSIONAL SPACES

SOME GEOMETRY IN HIGH-DIMENSIONAL SPACES MATH 57A. Itroductio Our geometric ituitio is derived from three-dimesioal space. Three coordiates suffice. May objects of iterest i aalysis, however, require far

### Lesson 15 ANOVA (analysis of variance)

Outlie Variability -betwee group variability -withi group variability -total variability -F-ratio Computatio -sums of squares (betwee/withi/total -degrees of freedom (betwee/withi/total -mea square (betwee/withi

### Math 114- Intermediate Algebra Integral Exponents & Fractional Exponents (10 )

Math 4 Math 4- Itermediate Algebra Itegral Epoets & Fractioal Epoets (0 ) Epoetial Fuctios Epoetial Fuctios ad Graphs I. Epoetial Fuctios The fuctio f ( ) a, where is a real umber, a 0, ad a, is called

### Chapter 14 Nonparametric Statistics

Chapter 14 Noparametric Statistics A.K.A. distributio-free statistics! Does ot deped o the populatio fittig ay particular type of distributio (e.g, ormal). Sice these methods make fewer assumptios, they

### How To Solve The Homewor Problem Beautifully

Egieerig 33 eautiful Homewor et 3 of 7 Kuszmar roblem.5.5 large departmet store sells sport shirts i three sizes small, medium, ad large, three patters plaid, prit, ad stripe, ad two sleeve legths log

### Listing terms of a finite sequence List all of the terms of each finite sequence. a) a n n 2 for 1 n 5 1 b) a n for 1 n 4 n 2

74 (4 ) Chapter 4 Sequeces ad Series 4. SEQUENCES I this sectio Defiitio Fidig a Formula for the th Term The word sequece is a familiar word. We may speak of a sequece of evets or say that somethig is

### Sequences and Series Using the TI-89 Calculator

RIT Calculator Site Sequeces ad Series Usig the TI-89 Calculator Norecursively Defied Sequeces A orecursively defied sequece is oe i which the formula for the terms of the sequece is give explicitly. For

### Lesson 17 Pearson s Correlation Coefficient

Outlie Measures of Relatioships Pearso s Correlatio Coefficiet (r) -types of data -scatter plots -measure of directio -measure of stregth Computatio -covariatio of X ad Y -uique variatio i X ad Y -measurig

### Week 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable

Week 3 Coditioal probabilities, Bayes formula, WEEK 3 page 1 Expected value of a radom variable We recall our discussio of 5 card poker hads. Example 13 : a) What is the probability of evet A that a 5

### Mathematical goals. Starting points. Materials required. Time needed

Level A1 of challege: C A1 Mathematical goals Startig poits Materials required Time eeded Iterpretig algebraic expressios To help learers to: traslate betwee words, symbols, tables, ad area represetatios

### Math C067 Sampling Distributions

Math C067 Samplig Distributios Sample Mea ad Sample Proportio Richard Beigel Some time betwee April 16, 2007 ad April 16, 2007 Examples of Samplig A pollster may try to estimate the proportio of voters

### 0.7 0.6 0.2 0 0 96 96.5 97 97.5 98 98.5 99 99.5 100 100.5 96.5 97 97.5 98 98.5 99 99.5 100 100.5

Sectio 13 Kolmogorov-Smirov test. Suppose that we have a i.i.d. sample X 1,..., X with some ukow distributio P ad we would like to test the hypothesis that P is equal to a particular distributio P 0, i.e.

### THE HEIGHT OF q-binary SEARCH TREES

THE HEIGHT OF q-binary SEARCH TREES MICHAEL DRMOTA AND HELMUT PRODINGER Abstract. q biary search trees are obtaied from words, equipped with the geometric distributio istead of permutatios. The average

### Factors of sums of powers of binomial coefficients

ACTA ARITHMETICA LXXXVI.1 (1998) Factors of sums of powers of biomial coefficiets by Neil J. Cali (Clemso, S.C.) Dedicated to the memory of Paul Erdős 1. Itroductio. It is well ow that if ( ) a f,a = the

### Solutions to Exercises Chapter 4: Recurrence relations and generating functions

Solutios to Exercises Chapter 4: Recurrece relatios ad geeratig fuctios 1 (a) There are seatig positios arraged i a lie. Prove that the umber of ways of choosig a subset of these positios, with o two chose

### FOUNDATIONS OF MATHEMATICS AND PRE-CALCULUS GRADE 10

FOUNDATIONS OF MATHEMATICS AND PRE-CALCULUS GRADE 10 [C] Commuicatio Measuremet A1. Solve problems that ivolve liear measuremet, usig: SI ad imperial uits of measure estimatio strategies measuremet strategies.

### Annuities Under Random Rates of Interest II By Abraham Zaks. Technion I.I.T. Haifa ISRAEL and Haifa University Haifa ISRAEL.

Auities Uder Radom Rates of Iterest II By Abraham Zas Techio I.I.T. Haifa ISRAEL ad Haifa Uiversity Haifa ISRAEL Departmet of Mathematics, Techio - Israel Istitute of Techology, 3000, Haifa, Israel I memory

### Solutions to Selected Problems In: Pattern Classification by Duda, Hart, Stork

Solutios to Selected Problems I: Patter Classificatio by Duda, Hart, Stork Joh L. Weatherwax February 4, 008 Problem Solutios Chapter Bayesia Decisio Theory Problem radomized rules Part a: Let Rx be the

### Lecture 4: Cheeger s Inequality

Spectral Graph Theory ad Applicatios WS 0/0 Lecture 4: Cheeger s Iequality Lecturer: Thomas Sauerwald & He Su Statemet of Cheeger s Iequality I this lecture we assume for simplicity that G is a d-regular

### MARTINGALES AND A BASIC APPLICATION

MARTINGALES AND A BASIC APPLICATION TURNER SMITH Abstract. This paper will develop the measure-theoretic approach to probability i order to preset the defiitio of martigales. From there we will apply this

### Time Value of Money. First some technical stuff. HP10B II users

Time Value of Moey Basis for the course Power of compoud iterest \$3,600 each year ito a 401(k) pla yields \$2,390,000 i 40 years First some techical stuff You will use your fiacial calculator i every sigle

### Maximum Likelihood Estimators.

Lecture 2 Maximum Likelihood Estimators. Matlab example. As a motivatio, let us look at oe Matlab example. Let us geerate a radom sample of size 00 from beta distributio Beta(5, 2). We will lear the defiitio

### Factoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett <garrett@math.umn.edu>

(March 16, 004) Factorig x 1: cyclotomic ad Aurifeuillia polyomials Paul Garrett Polyomials of the form x 1, x 3 1, x 4 1 have at least oe systematic factorizatio x 1 = (x 1)(x 1

### Permutations, the Parity Theorem, and Determinants

1 Permutatios, the Parity Theorem, ad Determiats Joh A. Guber Departmet of Electrical ad Computer Egieerig Uiversity of Wiscosi Madiso Cotets 1 What is a Permutatio 1 2 Cycles 2 2.1 Traspositios 4 3 Orbits

### UC Berkeley Department of Electrical Engineering and Computer Science. EE 126: Probablity and Random Processes. Solutions 9 Spring 2006

Exam format UC Bereley Departmet of Electrical Egieerig ad Computer Sciece EE 6: Probablity ad Radom Processes Solutios 9 Sprig 006 The secod midterm will be held o Wedesday May 7; CHECK the fial exam

### Solving equations. Pre-test. Warm-up

Solvig equatios 8 Pre-test Warm-up We ca thik of a algebraic equatio as beig like a set of scales. The two sides of the equatio are equal, so the scales are balaced. If we add somethig to oe side of the