Sequences and Series


 Kimberly Griffith
 2 years ago
 Views:
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
2 9. Covergece: Defiitio ad Examples 29 Example 9. The formulae a sequeces, respectively: (9.3) 2 3 ; ; b 2 ; c defie the The last sequece ca be defied recursively by: a 0 is give by the recursio a a. a ; , ad for 0 c 3 2 c 2. Similarly, the first The symbol! (read factorial ) is used to deote the product of the first itegers. This also has the recursive defiitio: a 0, ad for 0, a a. (Note that we have defied 0! ). Of the sequeces described i equatio 9.2, the first ad the third clearly grow without boud, but the secod is bouded; i fact, if we rewrite the geeral term as (9.4) b we see that the sequece b approaches. We say that b coverges to, as i the followig defiitio. Defiitio 9.2 A sequece a a 2 a coverges to a limit L, writte (9.5) lim a L if, for every ε 0, there is a such that for all we have a L ε. This just says that we ca be sure that a is as close to L as we eed it to be, just by takig the idex large eough. We will rarely have to actually use this defiitio, relyig more o uderstadig what it says, ad kow facts about limits. For example: Propositio 9. If the geeral term a of a sequece ca be expressed as f for a cotiuous fuctio f ad if we kow that lim x f x L, the we ca coclude that lim x a L. As a applicatio, usig results from the precedig chapter, we have Propositio 9.2 a) lim b) lim p for p 0 p 0 for p 0 c) lim A if A 0 Let p ad q be polyomials. d) lim p q 0 if deg p deg q lim p q if deg p deg q e) If the polyomials p ad q have the same degree, the lim p q a b leadig coefficiets of p ad q. p f) lim e for ay polyomial p. p g) lim for ay polyomial of positive degree ad ay positive c. l c where a ad b are the
3 Chapter 9 Sequeces ad Series 30 These ca all be derived by replacig by x, ad usig limit theorems already discussed (such as l Hôpital s rule). Example 9.2 Example lim 2 lim by e above 0 sice the umerator oscillates betwee  ad, ad the deomiator goes to ifiity. We should ot be perturbed by such oscillatio, so log as it remais bouded. For example we also have (9.6) lim si sice the term si remais bouded. The followig propositios state the geeral rule for hadlig such cases. Propositio 9.3 a) (Squeeze theorem) Give three sequeces a b c, if a b c for all adlim a lim c L the also lim b L b) If a b c, the sequece b is bouded, c 0 ad lim c Let s see why b) is true, usig a). Let M be the boud of the b. The (9.7) Mc b c Mc 0 0, the also lim a so a) applies ad the coclusio follows. I some cases where oe of the above rules apply, we have to retur to the defiitio of covergece. a Example 9.4 For ay a 0, lim! 0 To see why this is true, we thik of the sequece as recursively defied: the first term, a is a, ad each a is obtaied by multiplyig its predecessor by a. Now, evetually, that is, for large eough, a 2. Thus each term after that is less tha half its predecessor. This ow surely looks like a sequece covergig to zero. To be more precise, let N be the first iteger for which a N 2. The for ay k 0, (9.8) a N k N k! a N 2 k N! Now the sequece o the right is a fixed umber (a N N ) times a sequece ( 2 k ) which teds to zero. Thus our sequece coverges to zero, also by the squeeze theorem (propositio 9.3a). Note that i the above argumet, we oly had to show that the geeral term of our sequece is domiated by the geeral term of a sequece covergig to zero from some poit o. What happes to ay fiite collectio of terms of a sequece is ot relevat to the questio of covergece. We shall use the word evetually to mea from some poit o, or more precisely, for all greater tha some fixed iteger N. We restate propositio 9.3, usig the word evetually : 0.
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
13 Chapter 9 Sequeces ad Series 40 if the limit L lim a a exists. I this case the series coverges absolutely for x c L, ad diverges for x c L. Thus, the domais of covergece ad divergece of the series are separated by the circle, cetered at c of radius L. It ca be show that, i geeral, there is a circle separatig these domais, eve if the limit of the ratio of successive coefficiets does t exist. 0 R Propositio 9.6 Give the power series represetatio f x a x c there is a umber R such that we get absolute covergece for all x x c R, ad divergece for all x x c R. a R is called the radius of covergece of the power series. We have this value of R: lim a R if the limit exists. The first example of a power series represetatio is that of the geometric series: Example 9.2 Example 9.22 x x successive coefficiets is k x coverges for for x has the radius of covergece R. k (9.6) k k as. Example 9.23 (9.62) so R (9.63) x! has radius of covergece R x for ay umber k. We use the ratio test. The ratio of!! Usig the ratio test:, ad the series coverges for all x. O the other had, the ratio test shows us that the series!x has radius of covergece R 0, so coverges oly for x 0. Newto thought of power series as geeralized polyomials  that is, as polyomials, oly loger. This is justified, because we ca operate with power series just as we operate with polyomials: we ca add, multiply, ad substitute i them by doig so term by term. Example 9.24 (9.64) x x x x x x x for R For x x x 2 x 3 0 x x 2 x 3 x 4 Example 9.25 x 2 x 2 x 2 x 2 for x
14 9.4 Power Series 4 To see the first, we ote that x 2 is obtaied from x by substitutig x 2 for x. Thus, the power series represetatio is obtaied i the same way. I the secod, we have substituted x 2 for x. Example 9.26 Fid a power series expasio for 5 2x cetered at the origi. What is its radius of covergece? To solve a problem like this, we have to relate the fuctio to aother fuctio, whose power series we kow. I this case that would be x. Now 5 2x x, so our fuctio is obtaied from x by first replacig x by 2 5 x, ad the dividig by 5. We follow the same istructios with the power series. Start with (9.65) Replace x by 2 5 x: (9.66) Divide by 5 ad clea up: (9.67) x 2 5 x 5 2x 5 x 2 5 x 2 5 x 2 x 5 We ca calculate the radius of covergece usig propositio 9.6, or we ca reaso as follows; sice the series we started with coverges for x, our fial series coverges for 2 5 x, or x 5 2. Fially, we ca also itegrate ad differetiate power series term by term: Propositio 9.7 Suppose that f x (9.68) x (9.69) f x a x has radius of covergece R. The 0 f t dt ad both have the same radius of covergece, R. a x a x Example 9.27 arctax 2 x2 We kow that the derivative of the arc taget is x 2. Now, i example 9.25, we have already foud the power series represetatio of that fuctio, so we obtai the power series represetatio of arctax by itegratig term by term. x Example 9.28 e x! fid (9.70) f x for all x Let f x x! 0 x! The, differetiatig term by term, we x! x!
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
Section 9.2 Series and Convergence
Sectio 9. Series ad Covergece Goals of Chapter 9 Approximate Pi Prove ifiite series are aother importat applicatio of limits, derivatives, approximatio, slope, ad cocavity of fuctios. Fid challegig atiderivatives
More informationIn 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
More information8.5 Alternating infinite series
65 8.5 Alteratig ifiite series I the previous two sectios we cosidered oly series with positive terms. I this sectio we cosider series with both positive ad egative terms which alterate: positive, egative,
More informationORDERS OF GROWTH KEITH CONRAD
ORDERS OF GROWTH KEITH CONRAD Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really wat to uderstad their behavior It also helps you better grasp topics i calculus
More informationSection IV.5: Recurrence Relations from Algorithms
Sectio IV.5: Recurrece Relatios from Algorithms Give a recursive algorithm with iput size, we wish to fid a Θ (best big O) estimate for its ru time T() either by obtaiig a explicit formula for T() or by
More informationINFINITE SERIES KEITH CONRAD
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
More informationSAMPLE 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
More information4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
SECTION 2.6 THE RATIO TEST 79 2.6. THE RATIO TEST We ow kow how to hadle series which we ca itegrate (the Itegral Test), ad series which are similar to geometric or pseries (the Compariso Test), but of
More information8.3 POLAR FORM AND DEMOIVRE S THEOREM
SECTION 8. POLAR FORM AND DEMOIVRE S THEOREM 48 8. POLAR FORM AND DEMOIVRE S THEOREM Figure 8.6 (a, b) b r a 0 θ Complex Number: a + bi Rectagular Form: (a, b) Polar Form: (r, θ) At this poit you ca add,
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More information3.2 Introduction to Infinite Series
3.2 Itroductio to Ifiite Series May of our ifiite sequeces, for the remaider of the course, will be defied by sums. For example, the sequece S m := 2. () is defied by a sum. Its terms (partial sums) are
More informationTAYLOR SERIES, POWER SERIES
TAYLOR SERIES, POWER SERIES The followig represets a (icomplete) collectio of thigs that we covered o the subject of Taylor series ad power series. Warig. Be prepared to prove ay of these thigs durig the
More informationOur aim is to show that under reasonable assumptions a given 2πperiodic 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
More informationwhen n = 1, 2, 3, 4, 5, 6, This list represents the amount of dollars you have after n days. Note: The use of is read as and so on.
Geometric eries Before we defie what is meat by a series, we eed to itroduce a related topic, that of sequeces. Formally, a sequece is a fuctio that computes a ordered list. uppose that o day 1, you have
More informationInfinite 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...
More informationSection 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
More informationTheorems About Power Series
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 oegative umber R, called the radius
More information4.3. 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
More informationSoving 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
More informationLesson 12. Sequences and Series
Retur to List of Lessos Lesso. Sequeces ad Series A ifiite sequece { a, a, a,... a,...} ca be thought of as a list of umbers writte i defiite order ad certai patter. It is usually deoted by { a } =, or
More information1. a n = 2. a n = 3. a n = 4. a n = 5. a n = 6. a n =
Versio PREVIEW Homework Berg (5860 This pritout should have 9 questios. Multiplechoice questios may cotiue o the ext colum or page fid all choices before aswerig. CalCb0b 00 0.0 poits Rewrite the fiite
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0.,
More informationExample 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
More informationTrigonometric 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
More information8.1 Arithmetic Sequences
MCR3U Uit 8: Sequeces & Series Page 1 of 1 8.1 Arithmetic Sequeces Defiitio: A sequece is a comma separated list of ordered terms that follow a patter. Examples: 1, 2, 3, 4, 5 : a sequece of the first
More information7 b) 0. Guided Notes for lesson P.2 Properties of Exponents. If a, b, x, y and a, b, 0, and m, n Z then the following properties hold: 1 n b
Guided Notes for lesso P. Properties of Expoets If a, b, x, y ad a, b, 0, ad m, Z the the followig properties hold:. Negative Expoet Rule: b ad b b b Aswers must ever cotai egative expoets. Examples: 5
More information7) an = 7 n 7n. Solve the problem. Answer the question. n=1. Solve the problem. Answer the question. 16) an =
Eam Name MULTIPLE CHOICE. Choose the oe alterative that best comletes the statemet or aswers the questio. ) Use series to estimate the itegral's value to withi a error of magitude less tha .. l( + )d..79.9.77
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, 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
More informationContents. 7 Sequences and Series. 7.1 Sequences and Convergence. Calculus II (part 3): Sequences and Series (by Evan Dummit, 2015, v. 2.
Calculus II (part 3): Sequeces ad Series (by Eva Dummit, 05, v..00) Cotets 7 Sequeces ad Series 7. Sequeces ad Covergece......................................... 7. Iite Series.................................................
More informationConvexity, 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
More informationThe second difference is the sequence of differences of the first difference sequence, 2
Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for
More information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More informationProperties 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
More informationThe geometric series and the ratio test
The geometric series ad the ratio test Today we are goig to develop aother test for covergece based o the iterplay betwee the it compariso test we developed last time ad the geometric series. A ote about
More informationSECTION 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,
More information1 n. n > dt. t < n 1 + n=1
Math 05 otes C. Pomerace The harmoic sum The harmoic sum is the sum of recirocals of the ositive itegers. We kow from calculus that it diverges, this is usually doe by the itegral test. There s a more
More informationFIBONACCI 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
More information1 The Binomial Theorem: Another Approach
The Biomial Theorem: Aother Approach Pascal s Triagle I class (ad i our text we saw that, for iteger, the biomial theorem ca be stated (a + b = c a + c a b + c a b + + c ab + c b, where the coefficiets
More informationBasic Elements of Arithmetic Sequences and Series
MA40S PRECALCULUS 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
More information5. SEQUENCES AND SERIES
5. SEQUENCES AND SERIES 5.. Limits of Sequeces Let N = {0,,,... } be the set of atural umbers ad let R be the set of real umbers. A ifiite real sequece u 0, u, u, is a fuctio from N to R, where we write
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid
More informationPage 2 of 14 = T(2) + 2 = [ T(3)+1 ] + 2 Substitute T(3)+1 for T(2) = T(3) + 3 = [ T(4)+1 ] + 3 Substitute T(4)+1 for T(3) = T(4) + 4 After i
Page 1 of 14 Search C455 Chapter 4  Recursio Tree Documet last modified: 02/09/2012 18:42:34 Uses: Use recursio tree to determie a good asymptotic boud o the recurrece T() = Sum the costs withi each level
More informationThe Euler Totient, the Möbius and the Divisor Functions
The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA 01075 1 Ackowledgemets This work was supported by the Mout Holyoke College fellowship
More informationTaylor Series and Polynomials
Taylor Series ad Polyomials Motivatios The purpose of Taylor series is to approimate a fuctio with a polyomial; ot oly we wat to be able to approimate, but we also wat to kow how good the approimatio is.
More informationTHE 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,
More informationChapter Gaussian Elimination
Chapter 04.06 Gaussia Elimiatio After readig this chapter, you should be able to:. solve a set of simultaeous liear equatios usig Naïve Gauss elimiatio,. lear the pitfalls of the Naïve Gauss elimiatio
More informationa 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
More informationChapter 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
More information2.3. GEOMETRIC SERIES
6 CHAPTER INFINITE SERIES GEOMETRIC SERIES Oe of the most importat types of ifiite series are geometric series A geometric series is simply the sum of a geometric sequece, Fortuately, geometric series
More informationReview for College Algebra Final Exam
Review for College Algebra Fial Exam (Please remember that half of the fial exam will cover chapters 14. This review sheet covers oly the ew material, from chapters 5 ad 7.) 5.1 Systems of equatios i
More informationDivide and Conquer. Maximum/minimum. Integer Multiplication. CS125 Lecture 4 Fall 2015
CS125 Lecture 4 Fall 2015 Divide ad Coquer We have see oe geeral paradigm for fidig algorithms: the greedy approach. We ow cosider aother geeral paradigm, kow as divide ad coquer. We have already see a
More information23 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
More informationDivide and Conquer, Solving Recurrences, Integer Multiplication Scribe: Juliana Cook (2015), V. Williams Date: April 6, 2016
CS 6, Lecture 3 Divide ad Coquer, Solvig Recurreces, Iteger Multiplicatio Scribe: Juliaa Cook (05, V Williams Date: April 6, 06 Itroductio Today we will cotiue to talk about divide ad coquer, ad go ito
More information2.7 Sequences, Sequences of Sets
2.7. SEQUENCES, SEQUENCES OF SETS 67 2.7 Sequeces, Sequeces of Sets 2.7.1 Sequeces Defiitio 190 (sequece Let S be some set. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For
More informationThe Field of Complex Numbers
The Field of Complex Numbers S. F. Ellermeyer The costructio of the system of complex umbers begis by appedig to the system of real umbers a umber which we call i with the property that i = 1. (Note that
More informationNPTEL STRUCTURAL RELIABILITY
NPTEL Course O STRUCTURAL RELIABILITY Module # 0 Lecture 1 Course Format: Web Istructor: Dr. Aruasis Chakraborty Departmet of Civil Egieerig Idia Istitute of Techology Guwahati 1. Lecture 01: Basic Statistics
More informationARITHMETIC AND GEOMETRIC PROGRESSIONS
Arithmetic Ad Geometric Progressios Sequeces Ad ARITHMETIC AND GEOMETRIC PROGRESSIONS Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives
More informationCS103A 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
More informationMeasurable Functions
Measurable Fuctios Dug Le 1 1 Defiitio It is ecessary to determie the class of fuctios that will be cosidered for the Lebesgue itegratio. We wat to guaratee that the sets which arise whe workig with these
More informationDivergence of p 1/p. Adrian Dudek. adrian.dudek[at]anu.edu.au
Divergece of / Adria Dudek adria.dudek[at]au.edu.au Whe I was i high school, my maths teacher cheekily told me that it s ossible to add u ifiitely may umbers ad get a fiite umber. She the illustrated this
More informationf(x + T ) = f(x), for all x. The period of the function f(t) is the interval between two successive repetitions.
Fourier Series. Itroductio Whe the Frech mathematicia Joseph Fourier (76883) was tryig to study the flow of heat i a metal plate, he had the idea of expressig the heat source as a ifiite series of sie
More informationApproximating the Sum of a Convergent Series
Approximatig the Sum of a Coverget Series Larry Riddle Ages Scott College Decatur, GA 30030 lriddle@agesscott.edu The BC Calculus Course Descriptio metios how techology ca be used to explore covergece
More informationFourier Series and the Wave Equation Part 2
Fourier Series ad the Wave Equatio Part There are two big ideas i our work this week. The first is the use of liearity to break complicated problems ito simple pieces. The secod is the use of the symmetries
More informationAsymptotic 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
More informationA function f whose domain is the set of positive integers is called a sequence. The values
EQUENCE: A fuctio f whose domi is the set of positive itegers is clled sequece The vlues f ( ), f (), f (),, f (), re clled the terms of the sequece; f() is the first term, f() is the secod term, f() is
More informationThe Limit of a Sequence
3 The Limit of a Sequece 3. Defiitio of limit. I Chapter we discussed the limit of sequeces that were mootoe; this restrictio allowed some shortcuts ad gave a quick itroductio to the cocept. But may importat
More informationMATH 140A  HW 5 SOLUTIONS
MATH 40A  HW 5 SOLUTIONS Problem WR Ch 3 #8. If a coverges, ad if {b } is mootoic ad bouded, rove that a b coverges. Solutio. Theorem 3.4 states that if a the artial sums of a form a bouded sequece; b
More information.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,
More informationBINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients
652 (1226) 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
More informationif A S, then X \ A S, and if (A n ) n is a sequence of sets in S, then n A n S,
Lecture 5: Borel Sets Topologically, the Borel sets i a topological space are the σalgebra geerated by the ope sets. Oe ca build up the Borel sets from the ope sets by iteratig the operatios of complemetatio
More informationx(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More informationGeometric Sequences and Series. Geometric Sequences. Definition of Geometric Sequence. such that. a2 4
3330_0903qxd /5/05 :3 AM Page 663 Sectio 93 93 Geometric Sequeces ad Series 663 Geometric Sequeces ad Series What you should lear Recogize, write, ad fid the th terms of geometric sequeces Fid th partial
More informationHandout: How to calculate time complexity? CSE 101 Winter 2014
Hadout: How to calculate time complexity? CSE 101 Witer 014 Recipe (a) Kow algorithm If you are usig a modied versio of a kow algorithm, you ca piggyback your aalysis o the complexity of the origial algorithm
More informationTHE LEAST SQUARES REGRESSION LINE and R 2
THE LEAST SQUARES REGRESSION LINE ad R M358K I. Recall from p. 36 that the least squares regressio lie of y o x is the lie that makes the sum of the squares of the vertical distaces of the data poits from
More information13 Fast Fourier Transform (FFT)
13 Fast Fourier Trasform FFT) The fast Fourier trasform FFT) is a algorithm for the efficiet implemetatio of the discrete Fourier trasform. We begi our discussio oce more with the cotiuous Fourier trasform.
More informationLecture Notes CMSC 251
We have this messy summatio to solve though First observe that the value remais costat throughout the sum, ad so we ca pull it out frot Also ote that we ca write 3 i / i ad (3/) i T () = log 3 (log ) 1
More informationThe Harmonic Series Diverges Again and Again
The Harmoic Series Diverges Agai ad Agai Steve J. Kifowit Prairie State College Terra A. Stamps Prairie State College The harmoic series, = = 3 4 5, is oe of the most celebrated ifiite series of mathematics.
More informationB1. Fourier Analysis of Discrete Time Signals
B. Fourier Aalysis of Discrete Time Sigals Objectives Itroduce discrete time periodic sigals Defie the Discrete Fourier Series (DFS) expasio of periodic sigals Defie the Discrete Fourier Trasform (DFT)
More informationMath 152 Final Exam Review
Math 5 Fial Eam Review Problems Math 5 Fial Eam Review Problems appearig o your iclass fial will be similar to those here but will have umbers ad fuctios chaged. Here is a eample of the way problems selected
More informationS. 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,
More informationMath 475, Problem Set #6: Solutions
Math 475, Problem Set #6: Solutios A (a) For each poit (a, b) with a, b oegative itegers satisfyig ab 8, cout the paths from (0,0) to (a, b) where the legal steps from (i, j) are to (i 2, j), (i, j 2),
More informationApproximating 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.
More informationNumerical Solution of Equations
School of Mechaical Aerospace ad Civil Egieerig Numerical Solutio of Equatios T J Craft George Begg Buildig, C4 TPFE MSc CFD Readig: J Ferziger, M Peric, Computatioal Methods for Fluid Dyamics HK Versteeg,
More informationSection 6.1. x n n! = 1 + x + x2. n=0
Differece Equatios to Differetial Equatios Sectio 6.1 The Expoetial Fuctio At this poit we have see all the major cocepts of calculus: erivatives, itegrals, a power series. For the rest of the book we
More informationLecture 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 BruMikowski iequality for boxes. Today we ll go over the
More informationBuilding 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
More information7. Sample Covariance and Correlation
1 of 8 7/16/2009 6:06 AM Virtual Laboratories > 6. Radom Samples > 1 2 3 4 5 6 7 7. Sample Covariace ad Correlatio The Bivariate Model Suppose agai that we have a basic radom experimet, ad that X ad Y
More informationRecursion and Recurrences
Chapter 5 Recursio ad Recurreces 5.1 Growth Rates of Solutios to Recurreces Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer. Cosider, for example,
More informationHere 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
More informationClass 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,
More information7.1 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?
More informationWHEN 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
More informationNUMBERS COMMON TO TWO POLYGONAL SEQUENCES
NUMBERS COMMON TO TWO POLYGONAL SEQUENCES DIANNE SMITH LUCAS Chia Lake, Califoria a iteger, The polygoal sequece (or sequeces of polygoal umbers) of order r (where r is r > 3) may be defied recursively
More informationChapter One BASIC MATHEMATICAL TOOLS
Chapter Oe BAIC MATHEMATICAL TOOL As the reader will see, the study of the time value of moey ivolves substatial use of variables ad umbers that are raised to a power. The power to which a variable is
More informationEquation of a line. Line in coordinate geometry. Slopeintercept form ( 斜 截 式 ) Intercept form ( 截 距 式 ) Pointslope form ( 點 斜 式 )
Chapter : Liear Equatios Chapter Liear Equatios Lie i coordiate geometr I Cartesia coordiate sstems ( 卡 笛 兒 坐 標 系 統 ), a lie ca be represeted b a liear equatio, i.e., a polomial with degree. But before
More informationMath 105: Review for Final Exam, Part II  SOLUTIONS
Math 5: Review for Fial Exam, Part II  SOLUTIONS. Cosider the fuctio fx) =x 3 l x o the iterval [/e, e ]. a) Fid the x ad ycoordiates of ay ad all local extrema ad classify each as a local maximum or
More information0,1 is an accumulation
Sectio 5.4 1 Accumulatio Poits Sectio 5.4 BolzaoWeierstrass ad HeieBorel Theorems Purpose of Sectio: To itroduce the cocept of a accumulatio poit of a set, ad state ad prove two major theorems of real
More informationArithmetic Sequences and Partial Sums. Arithmetic Sequences. Definition of Arithmetic Sequence. Example 1. 7, 11, 15, 19,..., 4n 3,...
3330_090.qxd 1/5/05 11:9 AM Page 653 Sectio 9. Arithmetic Sequeces ad Partial Sums 653 9. Arithmetic Sequeces ad Partial Sums What you should lear Recogize,write, ad fid the th terms of arithmetic sequeces.
More informationTests for Convergence of Series. a n > 1 n. 0 < a n < 1 n 2. 0 < a n <.
Tests for Covergece of Series ) Use the compari test to cofirm the statemets i the followig eercises.. 4 diverges, 4 3 diverges. Aswer: Let a / 3), for 4. Sice 3 /, a >. The harmoic series
More information