Size: px
Start display at page:

Transcription

1 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 limit process: ifiite series. The label series is just aother ame for a sum. A ifiite series is a sum with ifiitely may terms, such as (.) The idea of a ifiite series is familiar from decimal expasios, for istace the expasio ca be writte as π = π = , so π is a ifiite sum of fractios. Decimal expasios like this show that a ifiite series is ot a paradoxical idea, although it may ot be clear how to deal with o-decimal ifiite series like (.) at the momet. Ifiite series provide two coceptual isights ito the ature of the basic fuctios met i high school (ratioal fuctios, trigoometric ad iverse trigoometric fuctios, expoetial ad logarithmic fuctios). First of all, these fuctios ca be expressed i terms of ifiite series, ad i this way all these fuctios ca be approximated by polyomials, which are the simplest kids of fuctios. That simpler fuctios ca be used as approximatios to more complicated fuctios lies behid the method which calculators ad computers use to calculate approximate values of fuctios. The secod isight we will have usig ifiite series is the close relatioship betwee fuctios which seem at first to be quite differet, such as expoetial ad trigoometric fuctios. Two other applicatios we will meet are a proof by calculus that there are ifiitely may primes ad a proof that e is irratioal. 2. Defiitios ad basic examples Before discussig ifiite series we discuss fiite oes. A fiite series is a sum a + a 2 + a a N, where the a i s are real umbers. I terms of Σ-otatio, we write N a + a 2 + a a N = a. =

2 2 KEITH CONRAD For example, (2.) N = = N. The sum i (2.) is called a harmoic sum, for istace = A very importat class of fiite series, more importat tha the harmoic oes, are the geometric series N (2.2) + r + r r N = r. A example is = ( ) = 2 = = 2 = = The geometric series (2.2) ca be summed up exactly, as follows. Theorem 2.. Whe r, the series (2.2) is Proof. Let The + r + r r N = rn+ r S N = + r + + r N. rs N = r + r r N+. = rn+ r. These sums overlap i r + + r N, so subtractig rs N from S N gives ( r)s N = r N+. Whe r we ca divide ad get the idicated formula for S N. Example 2.2. For ay N, Example 2.3. For ay N, N = 2N N = (/2)N+ /2 = 2 N+. = 2 2 N. It is sometimes useful to adjust the idexig o a sum, for istace (2.3) = 2 = = 99 = ( + )2.

3 INFINITE SERIES 3 Comparig the two Σ s i (2.3), otice how the reumberig of the idices affects the expressio beig summed: if we subtract from the bouds of summatio o the Σ the we add to the idex o the iside to maitai the same overall sum. As aother example, = 2 = ( + 3) 2. =3 Wheever the idex is shifted dow (or up) by a certai amout o the Σ it has to be shifted up (or dow) by a compesatig amout iside the expressio beig summed so that we are still addig the same umbers. This is aalogous to the effect of a additive chage of variables i a itegral, e.g., (x + 5) 2 dx = 6 5 = u 2 du = 6 5 x 2 dx, where u = x + 5 (ad du = dx). Whe the variable iside the itegral is chaged by a additive costat, the bouds of itegratio are chaged additively i the opposite directio. Now we defie the meaig of ifiite series, such as (.). The basic idea is that we look at the sum of the first N terms, called a partial sum, ad see what happes i the limit as N. Defiitio 2.4. Let a, a 2, a 3,... be a ifiite sequece of real umbers. a is defied to be N a = lim a. N = The ifiite series If the limit exists i R the we say a is coverget. If the limit does ot exist or is ± the a is called diverget. Notice that we are ot really addig up all the terms i a ifiite series at oce. We oly add up a fiite umber of the terms ad the see how thigs behave i the limit as the (fiite) umber of terms teds to ifiity: a ifiite series is defied to be the limit of its sequece of partial sums. Example 2.5. Usig Example 2.3, So 2 = lim N N = is a coverget series ad its value is = lim N 2 2 N = 2. Buildig o this example we ca compute exactly the value of ay ifiite geometric series. Theorem 2.6. For x R, the (ifiite) geometric series x coverges if x < ad diverges if x. If x < the x = x. Proof. For N, Theorem 2. tells us { N x ( x N+ )/( x), if x, = N +, if x =. =

4 4 KEITH CONRAD Whe x <, x N+ as N, so N x = lim N = x x N+ = lim N x = x. Whe x > the umerator x N does ot coverge as N, so x diverges. (Specifically, the limit is if x > ad the partials sums oscillate betwee values tedig to ad if x <.) What if x =? If x = the the N-th partial sum is N + so the series diverges (to ). If x = the N = ( ) oscillates betwee ad, so agai the series ( ) is diverget. We will see that Theorem 2.6 is the fudametal example of a ifiite series. May importat ifiite series will be aalyzed by comparig them to a geometric series (for a suitable choice of x). If we start summig a geometric series ot at, but at a higher power of x, the we ca still get a simple closed formula for the series, as follows. Corollary 2.7. If x < ad m the x = x m + x m+ + x m+2 + = m Proof. The N-th partial sum (for N m) is xm x. x m + x m+ + + x N = x m ( + x + + x N m ). As N, the sum iside the paretheses becomes the stadard geometric series, whose value is /( x). Multiplyig by x m gives the asserted value. Example 2.8. If x < the x + x 2 + x 3 + = x/( x). A diverget geometric series ca diverge i differet ways: the partial sums may ted to or ted to both ad or oscillate betwee ad. The label diverget series does ot always mea the partial sums ted to. All diverget meas is ot coverget. Of course i a particular case we may kow the partial sums do ted to, ad the we would say the series diverges to. The covergece of a series is determied by the behavior of the terms a for large. If we chage (or omit) ay iitial set of terms i a series we do ot chage the covergece of divergece of the series. Here is the most basic geeral feature of coverget ifiite series. Theorem 2.9. If the series a coverges the a as. Proof. Let S N = a + a a N ad let S = a be the limit of the partial sums S N as N. The as N, a N = S N S N S S =. While Theorem 2.9 is formulated for coverget series, its mai importace is as a divergece test : if the geeral term i a ifiite series does ot ted to the the series diverges. For example, Theorem 2.9 gives aother reaso that a geometric series x diverges if x, because i this case x does ot ted to : x = x for all.

5 INFINITE SERIES 5 It is a ufortuate fact of life that the coverse of Theorem 2.9 is geerally false: if a we have o guaratee that a coverges. Here is the stadard illustratio. Example 2.. Cosider the harmoic series turs out that =.. Although the geeral term teds to it To show this we will compare N = / with N dt/t = log N. Whe t, /t /. Itegratig this iequality from to + gives + dt/t /, so N = N = + dt t = N+ dt t = log(n + ). Therefore the N-th partial sum of the harmoic series is bouded below by log(n + ). Sice log(n + ) as N, so does the N-th harmoic sum, so the harmoic series diverges. We will see later that the lower boud log(n + ) for the N-th harmoic sum is rather sharp, so the harmoic series diverges slowly sice the logarithm fuctio diverges slowly. The divergece of the harmoic series is ot just a couterexample to the coverse of Theorem 2.9, but ca be exploited i other cotexts. Let s use it to prove somethig about prime umbers! Theorem 2.. There are ifiitely may prime umbers. Proof. (Euler) We will argue by cotradictio: assumig there are oly fiitely may prime umbers we will cotradict the divergece of the harmoic series. Cosider the product of /( /p) as p rus through all prime umbers. Sums are deoted with a Σ ad products are deoted with a Π (capital pi), so our product is writte as p /p = /2 /3 /5. Sice we assume there are fiitely may primes, this is a fiite product. Now expad each factor ito a geometric series: /p = ( + p + p 2 + p ) 3 +. p p Each geometric series is greater tha ay of its partial sums. Pick ay iteger N 2 ad trucate each geometric series at the N-th term, givig the iequality (2.4) /p > ( + p + p 2 + p ) p N. p p O the right side of (2.4) we have a fiite product of fiite sums, so we ca compute this usig the distributive law ( super FOIL i the termiology of high school algebra). That is, take oe term from each factor, multiply them together, ad add up all these products. What umbers do we get i this way o the right side of (2.4)? A product of reciprocal itegers is a reciprocal iteger, so we obtai a sum of / as rus over certai positive itegers. Specifically, the s we meet are those whose prime factorizatio does t

6 6 KEITH CONRAD ivolve ay prime with expoet greater tha N. Ideed, for ay positive iteger >, if we factor ito primes as = p e pe 2 2 per r the e i < for all i. (Sice p i 2, 2 e i, ad 2 m > m for ay m so > e i.) Thus if N ay of the prime factors of appear i with expoet less tha N, so / is a umber which shows up whe multiplyig out (2.4). Here we eed to kow we are workig with a product over all primes. Sice / occurs i the expasio of the right side of (2.4) for N ad all terms i the expasio are positive, we have the lower boud estimate (2.5) p N /p >. Sice the left side is idepedet of N while the right side teds to as N, this iequality is a cotradictio so there must be ifiitely may primes. We ed this sectio with some algebraic properties of coverget series: termwise additio ad scalig. Theorem 2.2. If a ad b coverge the (a + b ) coverges ad + b ) = (a a + b. If a coverges the for ay umber c the series ca coverges ad a. = ca = c Proof. Let S N = N = a ad T N = N = b. Set S = a ad T = b, so S N S ad T N T as N. The by kow properties of limits of sequeces, S N + T N S + T ad cs N cs as N. These are the coclusios of the theorem. Example 2.3. We have ad ( 2 + ) 3 = /2 + /3 = = 5 4 = 5 /4 = Positive series I this sectio we describe several ways to show that a ifiite series a coverges whe a > for all. There is somethig special about series where all the terms are positive. What is it? The sequece of partial sums is icreasig: S N+ = S N + a N, so S N+ > S N for every N. A icreasig sequece has two possible limitig behaviors: it coverges or it teds to. (That is, divergece ca oly happe i oe way: the partial sums ted to.) We ca tell if a icreasig sequece coverges by showig it is bouded above. The it coverges ad its limit is a upper boud o the whole sequece. If the terms i a icreasig sequece are ot bouded above the the sequece

7 INFINITE SERIES 7 is ubouded ad teds to. Compare this to the sequece of partial sums N = ( ), which are bouded but do ot coverge (they oscillate betwee ad ). Let s summarize this property of positive series. If a > for every ad there is a costat C such that all partial sums satisfy N = a < C the a coverges ad a C. O the other had, if there is o upper boud o the partial sums the a =. We will use this over ad over agai to explai various covergece tests for positive series. Theorem 3. (Itegral Test). Suppose a = f() where f is a cotiuous decreasig positive fuctio. The a coverges if ad oly if f(x) dx coverges. Proof. We will compare the partial sum N = a ad the partial itegral N f(x) dx. Sice f(x) is a decreasig fuctio, x + = a + = f( + ) f(x) f() = a. Itegratig these iequalities betwee ad + gives (3.) a + + sice the iterval [, + ] has legth. Therefore (3.2) ad (3.3) N a = N a = a + = Combiiig (3.2) ad (3.3), (3.4) N = + N a a + =2 N+ f(x) dx N =2 N = f(x) dx a. f(x) dx = N+ f(x) dx N f(x) dx = f() + f(x) dx. N a f() + f(x) dx. Assume that f(x) dx coverges. Sice f(x) is a positive fuctio, N f(x) dx < f(x) dx. Thus by the secod iequality i (3.4) N a < f() + = f(x) dx, so the partial sums are bouded above. Therefore the series a coverges. Coversely, assume a coverges. The the partial sums N = a are bouded above (by the whole series a ). Every partial itegral of f(x) dx is bouded above by oe of these partial sums accordig to the first iequality of (3.4), so the partial itegrals are bouded above ad thus the improper itegral f(x) dx coverges. Example 3.2. We were implicitly usig the itegral test with f(x) = /x whe we proved the harmoic series diverged i Example 2.. For the harmoic series, (3.4) becomes (3.5) log(n + ) N = + log N.

8 8 KEITH CONRAD Thus the harmoic series diverges logarithmically. The followig table illustrates these bouds whe N is a power of. N log(n + ) N = / + log N Table Example 3.3. Geeralizig the harmoic series, cosider p = + 2 p + 3 p + 4 p + where p >. This is called a p-series. Its covergece is related to the itegral dx/x p. Sice the itegrad /x p has a ati-derivative that ca be computed explicitly, we ca compute the improper itegral: { dx x p = p, if p >,, if < p. Therefore /p coverges whe p > ad diverges whe < p. For example, /2 coverges ad / diverges. Example 3.4. The series 2 log 2 diverges by the itegral test sice dx x log x = log log x 2 =. (We stated the itegral test with a lower boud of but it obviously adapts to series where the first idex of summatio is somethig other tha, as i this example.) Example 3.5. The series, with terms just slightly smaller tha i the previous (log ) 2 2 example, coverges. Usig the itegral test, dx x(log x) = log x = log 2, which is fiite. 2 Because the covergece of a positive series is itimately tied up with the boudedess of its partial sums, we ca determie the covergece or divergece of oe positive series from aother if we have a iequality i oe directio betwee the terms of the two series. This leads to the followig covergece test. Theorem 3.6 (Compariso test). Let < a b for all. If b coverges the a coverges. If a diverges the b diverges. 2

9 INFINITE SERIES 9 Proof. Sice a b, (3.6) N a = N b. If b coverges the the partial sums o the right side of (3.6) are bouded, so the partial sums o the left side are bouded, ad therefore a coverges. If a diverges the N = a, so N = b, so b diverges. Example 3.7. If x < the x / coverges sice x / x so the series is bouded above by the geometric series x, which coverges. Notice that, ulike the geometric series, we are ot givig ay kid of closed form expressio for x /. All we are sayig is that the series coverges. The series x / does ot coverge for x : if x = it is the harmoic series ad if x > the the geeral term x / teds to (ot ) so the series diverges to. Example 3.8. The itegral test is a special case of the compariso test, where we compare a with + f(x) dx ad with f(x) dx. Just as covergece of a series does ot deped o the behavior of the iitial terms of the series, what is importat i the compariso test is a iequality a b for large. If this iequality happes ot to hold for small but does hold for all large the the compariso test still works. The followig two examples use this exteded compariso test whe the form of the compariso test i Theorem 3.6 does ot strictly apply (because the iequality a b does t hold for small ). Example 3.9. If x < the x coverges. This is obvious for x =. We ca t get covergece for < x < by a compariso with x because the iequality goes the wrog way: x > x, so covergece of x does t help us show covergece of x. However, let s write = x = x /2 x /2. Sice < x <, x /2 as, so x /2 for large (depedig o the specific value of x, e.g, for x = /2 the iequality is false at = 3 but is correct for 4). Thus for all large we have x x /2 so x coverges by compariso with the geometric series x/2 = ( x). We have ot obtaied ay kid of closed formula for x, however. If x the x diverges sice x for all, so the geeral term does ot ted to. Example 3.. If x the the series x /! coverges. This is obvious for x =. For x > we will use the compariso test ad the lower boud estimate! > /e. Let s recall that this lower boud o! comes from Euler s itegral formula The for positive x (3.7)! = t e t dt > t e t dt > x! < x ( ex ) /e =. e t dt = e.

10 KEITH CONRAD Here x is fixed ad we should thik about what happes as grows. The ratio ex/ teds to as, so for large we have ex (3.8) < 2 (ay positive umber less tha will do here; we just use /2 for cocreteess). Raisig both sides of (3.8) to the -th power, ( ex ) < 2. Comparig this to (3.7) gives x /! < /2 for large, so the terms of the series x /! drop off faster tha the terms of the coverget geometric series /2 whe we go out far eough. The exteded compariso test ow implies covergece of x /! if x >. Example 3.. The series Example coverges sice < 5 2 ad coverges by 52 Let s try to apply the compariso test to /(52 ). The rate of growth of the th term is like /(5 2 ), but the iequality ow goes the wrog way: 5 2 < 5 2 for all. So we ca t quite use the compariso test to show /(52 ) coverges (which it does). The followig limitig form of the compariso test will help here. Theorem 3.2 (Limit compariso test). Let a, b > ad suppose the ratio a /b has a positive limit. The a coverges if ad oly if b coverges. Proof. Let L = lim a /b. For all large, so L 2 < a b < 2L, L (3.9) 2 b < a < 2Lb for large. Sice L/2 ad 2L are positive, the covergece of b is the same as covergece of (L/2)b ad (2L)b. Therefore the first iequality i (3.9) ad the compariso test show covergece of a implies covergece of b. The secod iequality i (3.9) ad the compariso test show covergece of b implies covergece of a. (Although (3.9) may ot be true for all, it is true for large, ad that suffices to apply the exteded compariso test.) The way to use the limit compariso test is to replace terms i a series by simpler terms which grow at the same rate (at least up to a scalig factor). The aalyze the series with those simpler terms. Example 3.3. We retur to 5 2. As, the th term grows like /52 : /(5 2 ) /5 2.

11 INFINITE SERIES Sice /(52 ) coverges, so does /(52 ). I fact, if a is ay sequece whose growth is like / 2 up to a scalig factor the a coverges. I particular, this is a simpler way to settle the covergece i Example 3. tha by usig the iequalities i Example 3.. Example 3.4. To determie if coverges we replace the th term with the expressio = 2 3/2, which grows at the same rate. The series /(23/2 ) coverges, so the origial series coverges. The limit compariso test uderlies a importat poit: the covergece of a series a is ot assured just by kowig if a, but it is assured if a rapidly eough. That is, the rate at which a teds to is ofte (but ot always) a meas of explaiig the covergece of a. The differece betwee someoe who has a ituitive feelig for covergece of series ad someoe who ca oly determie covergece o a case-by-case basis usig memorized rules i a mechaical way is probably idicated by how well the perso uderstads the coectio betwee covergece of a series ad rates of growth (really, decay) of the terms i the series. Armed with the covergece of a few basic series (like geometric series ad p-series), the covergece of most other series ecoutered i a calculus course ca be determied by the limit compariso test if you uderstad well the rates of growth of the stadard fuctios. Oe exceptio is if the series has a log term, i which case it might be useful to apply the itegral test. 4. Series with mixed sigs All our previous covergece tests apply to series with positive terms. They also apply to series whose terms are evetually positive (just omit ay iitial egative terms to get a positive series) or evetually egative (egate the series, which does t chage the covergece status, ad ow we re reduced to a evetually positive series). What do ca we do for series whose terms are ot evetually all positive or evetually all egative? These are the series with ifiitely may positive terms ad ifiitely may egative terms. Example 4.. Cosider the alteratig harmoic series ( ) = The usual harmoic series diverges, but the alteratig sigs might have differet behavior. The followig table collects some of the partial sums. N N = ( ) / Table 2

12 2 KEITH CONRAD These partial sums are ot clear evidece i favor of covergece: after, terms we oly get apparet stability i the first 2 decimal digits. The harmoic series diverges slowly, so perhaps the alteratig harmoic series diverges too (ad slower). Defiitio 4.2. A ifiite series where the terms alterate i sig is called a alteratig series. A alteratig series startig with a positive term ca be writte as ( ) a where a > for all. If the first term is egative the a reasoable geeral otatio is ( ) a where a > for all. Obviously egatig a alteratig series of oe type turs it ito the other type. Theorem 4.3 (Alteratig series test). A alteratig series whose -th term i absolute value teds mootoically to as is a coverget series. Proof. We will work with alteratig series which have a iitial positive term, so of the form ( ) a. Sice successive partial sums alterate addig ad subtractig a amout which, i absolute value, steadily teds to, the relatio betwee the partial sums is idicated by the followig iequalities: s 2 < s 4 < s 6 < < s 5 < s 3 < s. I particular, the eve-idexed partial sums are a icreasig sequece which is bouded above (by s, say), so the partial sums s 2m coverge. Sice s 2m+ s 2m = a 2m+, the odd-idexed partial sums coverge to the same limit as the eve-idexed partial sums, so the sequece of all partial sums has a sigle limit, which meas ( ) a coverges. Example 4.4. The alteratig harmoic series ( ) coverges. Example 4.5. While the p-series /p coverges oly for p >, the alteratig p-series ( ) / p coverges for the wider rage of expoets p > sice it is a alteratig series satisfyig the hypotheses of the alteratig series test. Example 4.6. We saw x / coverges for x < (compariso to the geometric series) i Example 3.7. If x < the x / coverges by the alteratig series test: the egativity of x makes the terms x / alterate i sig. To apply the alteratig series test we eed x + + < x for all, which is equivalet after some algebra to x < + for all, ad this is true sice x < ( + )/. Example 4.7. I Example 3. the series x /! was see to coverge for x. If x < the series is a alteratig series. Does it fit the hypotheses of the alteratig series test? We eed x + ( + )! < x! for all, which is equivalet to x < +

13 INFINITE SERIES 3 for all. For each particular x < this may ot be true for all (if x = 3 it fails for = ad 2), but it is defiitely true for all sufficietly large. Therefore the terms of x /! are a evetually alteratig series. Sice what matters for covergece is the log-term behavior ad ot the iitial terms, we ca apply the alteratig series test to see x /! coverges by just igorig a iitial piece of the series. Whe a series has ifiitely may positive ad egative terms which are ot strictly alteratig, covergece may ot be easy to check. For example, the trigoometric series si(x) = si x + si(2x) 2 + si(3x) 3 + si(4x) 4 turs out to be coverget for every umber x, but the arragemet of positive ad egative terms i this series ca be quite subtle i terms of x. The proof that this series coverges uses the techique of summatio by parts, which wo t be discussed here. While it is geerally a delicate matter to show a o-alteratig series with mixed sigs coverges, there is oe useful property such series might have which does imply covergece. It is this: if the series with all terms made positive coverges the so does the origial series. Let s give this idea a ame ad the look at some examples. + Defiitio 4.8. A series a is absolutely coverget if a coverges. Do t cofuse a with a ; absolute covergece refers to covergece if we drop the sigs from the terms i the series, ot from the series overall. The differece betwee a ad a is at most a sigle sig, while there is a substatial differece betwee a ad a if the a s have may mixed sigs; that is the differece which absolute covergece ivolves. Example 4.9. We saw x / coverges if x < i Example 3.7. Sice x / = x /, the series x / is absolutely coverget if x <. Note the alteratig harmoic series ( ) / is ot absolutely coverget, however. Example 4.. I Example 3. we saw x /! coverges for all x. Sice x /! = x /!, the series x /! is absolutely coverget for all x. Example 4.. By Example 3.9, x coverges absolutely if x <. Example 4.2. The two series si(x) 2, cos(x) 2, where the th term has deomiator 2, are absolutely coverget for ay x sice the absolute value of the th term is at most / 2 ad /2 coverges. The relevace of absolute covergece is two-fold: ) we ca use the may covergece tests for positive series to determie if a series is absolutely coverget, ad 2) absolute covergece implies ordiary covergece. We just illustrated the first poit several times. Let s show the secod poit. Theorem 4.3 (Absolute covergece test). Every absolutely coverget series is a coverget series.

14 4 KEITH CONRAD Proof. For all we have so a a a, a + a 2 a. Sice a coverges, so does 2 a (to twice the value). Therefore by the compariso test we obtai covergece of (a + a ). Subtractig a from this shows a coverges. Thus the series x / ad x coverge for x < ad x /! coverges for all x by our treatmet of these series for positive x aloe i Sectio 3. The argumet we gave i Examples 4.6 ad 4.7 for the covergece of x / ad x /! whe x <, usig the alteratig series test, ca be avoided. (But we do eed the alteratig series test to show x / coverges at x =.) A series which coverges but is ot absolutely coverget is called coditioally coverget. A example of a coditioally coverget series is the alteratig harmoic series ( ) /. The distictio betwee absolutely coverget series ad coditioally coverget series might seem kid of mior, sice it ivolves whether or ot a particular covergece test (the absolute covergece test) works o that series. But the distictio is actually profoud, ad is icely illustrated by the followig example of Dirichlet (837). Example 4.4. Let L = /2 + /3 /4 + /5 /6 + be the alteratig harmoic series. Cosider the rearragemet of the terms where we follow a positive term by two egative terms: If we add each positive term to the egative term followig it, we obtai , which is precisely half the alteratig harmoic series (multiply L by /2 ad see what you get termwise). Therefore, by rearragig the terms of the alteratig harmoic series we obtaied a ew series whose sum is (/2)L istead of L. If (/2)L = L, does t that mea = 2? What happeed? This example shows additio of terms i a ifiite series is ot geerally commutative, ulike with fiite series. I fact, this feature is characteristic of the coditioally coverget series, as see i the followig amazig theorem. Theorem 4.5 (Riema, 854). If a ifiite series is coditioally coverget the it ca be rearraged to sum up to ay desired value. If a ifiite series is absolutely coverget the all of its rearragemets coverge to the same sum. For istace, the alteratig harmoic series ca be rearraged to sum up to, to 5.673, to π, or to ay other umber you wish. That we rearraged it i Example 4.4 to sum up to half its usual value was special oly i the sese that we could make the rearragemet achievig that effect quite explicit. Theorem 4.5, whose proof we omit, is called Riema s rearragemet theorem. It idicates that absolutely coverget series are better behaved tha coditioally coverget oes. The ordiary rules of algebra for fiite series geerally exted to absolutely coverget series but ot to coditioally coverget series.

15 INFINITE SERIES 5 5. Power series The simplest fuctios are polyomials: c + c x + + c d x d. I differetial calculus we lear how to locally approximate may fuctios by liear fuctios (the taget lie approximatio). We ow itroduce a idea which will let us give a exact represetatio of fuctios i terms of polyomials of ifiite degree. The precise ame for this idea is a power series. Defiitio 5.. A power series is a fuctio of the form f(x) = c x. For istace, a polyomial is a power series where c = for large. The geometric series x is a power series. It oly coverges for x <. We have also met other power series, like x, x x,!. These three series coverge o the respective itervals [, ), (, ) ad (, ). For ay power series c x it is a basic task to fid those x where the series coverges. It turs out i geeral, as i all of our examples, that a power series coverges o a iterval. Let s see why. Theorem 5.2. If a power series c x coverges at x = b the it coverges absolutely for x < b. Proof. Sice c b coverges the geeral term teds to, so c b for large. If x < b the c x = c b x, b which is at most x/b for large. The series x/b coverges sice it s a geometric series ad x/b <. Therefore by the (exteded) compariso test c x coverges, so c x is absolutely coverget. Example 5.3. If a power series coverges at 7 the it coverges o the iterval ( 7, 7). We ca t say for sure how it coverges at 7. Example 5.4. If a power series diverges at 7 the it diverges for x > 7: if it were to coverge at a umber x where x > 7 the it would coverge i the iterval ( x, x ) so i particular at 7, a cotradictio. What these two examples illustrate, as cosequeces of Theorem 5.2, is that the values of x where a power series c x coverges is a iterval cetered at, so of the form ( r, r), [ r, r), ( r, r], [ r, r] for some r. We call r the radius of covergece of the power series. The oly differece betwee these differet itervals is the presece or absece of the edpoits. All possible types of iterval of covergece ca occur: x has iterval of covergece (, ), x / has iterval of covergece [, ), ( ) x / has iterval of covergece (, ] ad x / 2 (a ew example we have ot met before) has iterval of covergece [, ]. The radius of covergece ad the iterval of covergece are closely related but should ot be cofused. The iterval is the actual set where the power series coverges. The radius is simply the half-legth of this set (ad does t tell us whether or ot the edpoits are icluded). If we do t care about covergece behavior o the boudary of the iterval of covergece tha we ca get by just kowig the radius of covergece: the series always coverges (absolutely) o the iside of the iterval of covergece.

16 6 KEITH CONRAD For the practical computatio of the radius of covergece i basic examples it is coveiet to use a ew covergece test for positive series. Theorem 5.5 (Ratio test). If a > for all, assume a + q = lim a exists. If q < the a coverges. If q > the a diverges. If q = the o coclusio ca be draw. Proof. Assume q <. Pick s betwee q ad : q < s <. Sice a + /a q, we have a + /a < s for all large, say for. The a + < sa whe, so a + < sa, a +2 < sa + < s 2 a, a +3 < sa +2 < s 3 a, ad more geerally a +k < s k a for ay k. Writig + k as we have = a < s a = a s s. Therefore whe the umber a is bouded above by a costat multiple of s. Hece a coverges by compariso to a costat multiple of the coverget geometric series s. I the other directio, if q > the pick s with < s < q. A argumet similar to the previous oe shows a grows at least as quickly as a costat multiple of s, but this time s diverges sice s >. So a diverges too. Whe q = we ca t make a defiite coclusio sice both / ad /2 have q =. Example 5.6. We give a proof that x has radius of covergece which is shorter tha Examples 3.9 ad 4.. Take a = x = x. The a + = + x x a as. Thus if x < the series x is absolutely coverget by the ratio test. If x > this series is diverget. Notice the ratio test does ot tell us what happes to x whe x = ; we have to check x = ad x = idividually. Example 5.7. A similar argumet shows x / has radius of covergece sice as. x + /( + ) x / = x x + Example 5.8. We show the series x /! coverges absolutely for all x. Usig the ratio test we look at x + /( + )! x /! = x + as. Sice this limit is less tha for all x, we are doe by the ratio test. The radius of covergece is ifiite. Power series are importat for two reasos: they give us much greater flexibility to defie ew kids of fuctios ad may stadard fuctios ca be expressed i terms of a power series.

17 INFINITE SERIES 7 Ituitively, if a kow fuctio f(x) has a power series represetatio o some iterval aroud, say f(x) = c x = c + c x + c 2 x 2 + c 3 x 3 + c 4 x 4 + for x < r, the we ca guess a formula for c i terms of the behavior of f(x). To begi we have f() = c. Now we thik about power series as somethig like ifiite degree polyomials. We kow how to differetiate a polyomial by differetiatig each power of x separately. Let s assume this works for power series too i the iterval of covergece: so we expect Now let s differetiate agai: so f (x) = c x = c + 2c 2 x + 3c 3 x 2 + 4c 4 x 3 +, f () = c. f (x) = 2 ( )c x 2 = 2c 2 + 6c 3 x + 2c 4 x 2 +, f () = 2c 2. Differetiatig a third time (formally) ad settig x = gives The geeral rule appears to be so we should have f (3) () = 6c 3. f () () =!c, c = f () ().! This procedure really is valid, accordig to the followig theorem whose log proof is omitted. Theorem 5.9. Ay fuctio represeted by a power series i a ope iterval ( r, r) is ifiitely differetiable i ( r, r) ad its derivatives ca be computed by termwise differetiatio of the power series. This meas our previous calculatios are justified: if a fuctio f(x) ca be writte i the form c x i a iterval aroud the we must have (5.) c = f () ().! I particular, a fuctio has at most oe expressio as a power series c x aroud. Ad a fuctio which is ot ifiitely differetiable aroud will defiitely ot have a power series represetatio c x. For istace, x has o power series represetatio of this form sice it is ot differetiable at. Remark 5.. We ca also cosider power series f(x) = c (x a), whose iterval of covergece is cetered at a. I this case the coefficiets are give by the formula c = f () (a)/!. For simplicity we will focus o power series cetered at oly.

18 8 KEITH CONRAD Remark 5.. Eve if a power series coverges o a closed iterval [ r, r], the power series for its derivative eed ot coverge at the edpoits. Cosider f(x) = ( ) x 2 /. It coverges for x but the derivative series is 2( ) x 2, which coverges for x <. Remark 5.2. Because a power series ca be differetiated (iside its iterval of covergece) termwise, we ca discover solutios to a differetial equatio by isertig a power series with ukow coefficiets ito the differetial equatio to get relatios betwee the coefficiets. The a few iitial coefficiets, oce chose, should determie all the higher oes. After we compute the radius of covergece of this ew power series we will have foud a solutio to the differetial equatio i a specific iterval. This idea goes back to Newto. It does ot ecessarily provide us with all solutios to a differetial equatio, but it is oe of the stadard methods to fid some solutios. Example 5.3. If x < the x =. Differetiatig both sides, for x < we have x Multiplyig by x gives us x = x = ( x) 2. x ( x) 2, so we have fially foud a closed form expressio for a ifiite series we first met back i Example 3.9. Example 5.4. If there is a power series represetatio c x for e x the (5.) shows c = /!. The oly possible way to write e x as a power series is x! = + x + x2 2 + x3 3! + x4 4! +. Is this really a valid formula for e x? Well, we checked before that this power series coverges for all x. Moreover, calculatig the derivative of this series reproduces the series agai, so this series is a fuctio satisfyig y (x) = y(x). The oly solutios to this differetial equatio are Ce x ad we ca recover C by settig x =. Sice the power series has costat term (so its value at x = is ), its C is, so this power series is e x : e x = x!. Loosely speakig, this meas the polyomials, + x, + x + x2 2, + x + x2 2 + x3 6,... are good approximatios to e x. (Where they are good will deped o how far x is from.) I particular, settig x = gives a ifiite series represetatio of the umber e: (5.2) e =! = ! + 4! + Formula (5.2) ca be used to verify a iterestig fact about e. Theorem 5.5. The umber e is irratioal.

19 Proof. (Fourier) For ay, ( e = + 2! + 3! + + )! ( = + 2! + 3! + +! INFINITE SERIES 9 ( + ) +! ) ( + 2)! + ( + )! + ( + + ( + 2)( + ) + The secod term i paretheses is positive ad bouded above by the geometric series + + ( + ) 2 + ( + ) 3 + =. Therefore ( < e + 2! + 3! + + )!!. Write the sum + /2! + + /! as a fractio with commo deomiator!, say as p /!. Clear the deomiator! to get (5.3) <!e p. So far everythig we have doe ivolves o uproved assumptios. Now we itroduce the ratioality assumptio. If e is ratioal, the!e is a iteger whe is large (sice ay iteger is a factor by! for large ). But that makes!e p a iteger located i the ope iterval (, ), which is absurd. We have a cotradictio, so e is irratioal. We retur to the geeral task of represetig fuctios by their power series. Eve whe a fuctio is ifiitely differetiable for all x, its power series could have a fiite radius of covergece. Example 5.6. Viewig /( + x 2 ) as /( ( x 2 )) we have the geometric series formula + x 2 = ( x 2 ) = ( ) x 2 whe x 2 <, or equivaletly x <. The series has a fiite iterval of covergece (, ) eve though the fuctio /( + x 2 ) has o bad behavior at the edpoits: it is ifiitely differetiable at every real umber x. Example 5.7. Let f(x) = x /. This coverges for x <. For x < we have f (x) = x = x = x. Whe x <, a atiderivative of /( x) is log( x). Sice f(x) ad log( x) have the same value (zero) at x = they are equal: log( x) = x for x <. Replacig x with x ad egatig gives (5.4) log( + x) = ( ) for x <. The right side coverges at x = (alteratig series) ad diverges at x =. It seems plausible, sice the series equals log( + x) o (, ), that this should remai true at the x ).

20 2 KEITH CONRAD boudary: does ( ) / equal log 2? (Notice this is the alteratig harmoic series.) We will retur to this later. I practice, if we wat to show a ifiitely differetiable fuctio equals its associated power series (f () ()/!)x o some iterval aroud we eed some kid of error estimate for the differece betwee f(x) ad the polyomial N = (f () ()/!)x. Whe the estimate goes to as N we will have proved f(x) is equal to its power series. Theorem 5.8. If f(x) is ifiitely differetiable the for all N N f () () f(x) = x + R N (x),! where R N (x) = N! Proof. Whe N = the desired result says = x f(x) = f() + f (N+) (t)(x t) N dt. x f (t) dt. This is precisely the fudametal theorem of calculus! We obtai the N = case from this by itegratio by parts. Set u = f (t) ad dv = dt. The du = f (t) dt ad we (cleverly!) take v = t x (rather tha just t). The x f (t) dt = f t=x x (t)(t x) f (t)(t x) dt so = f ()x + f(x) = f() + f ()x + t= x f (t)(x t) dt, x f (t)(x t) dt. Now apply itegratio by parts to this ew itegral with u = f (t) ad dv = (x t) dt. The du = f (3) (t) dt ad (cleverly) use v = (/2)(x t) 2. The result is so x f (t)(x t) dt = f () x x x f (3) (t)(x t) 2 dt, f(x) = f() + f ()x + f () x 2 + f (3) (t)(x t) 2 dt. 2! 2 Itegratig by parts several more times gives the desired result. We call R N (x) a remaider term. Corollary 5.9. If f is ifiitely differetiable the the followig coditios at the umber x are equivalet: f(x) = (f () ()/!)x, R N (x) as N. Proof. The differece betwee f(x) ad the Nth partial sum of its power series is R N (x), so f(x) equals its power series precisely whe R N (x) as N.

21 INFINITE SERIES 2 Example 5.2. I (5.4) we showed log( + x) equals ( ) x / for x <. What about at x =? For this purpose we wat to show R N () as N whe f(x) = log( + x). To estimate R N () we eed to compute the higher derivatives of f(x): f (x) = + x, f (x) = ( + x) 2, f (3) 2 (x) = ( + x) 3, f (4) 6 (x) = ( + x) 4, ad i geeral for. Therefore so R N () = N! f () (x) = ( ) ( )! ( + x) ( ) N N! ( + t) N+ ( t)n dt = ( ) N ( t) N dt, ( + t) N+ R N () dt ( + t) N+ = N N2 N as N. Sice the remaider term teds to, log 2 = ( ) /. Example 5.2. Let f(x) = si x. Its power series is ( ) x2+ (2 + )! = x x3 3! + x5 5! x7 7! +, which coverges for all x by the ratio test. We will show si x equals its power series for all x. This is obvious if x =. Sice si x has derivatives ± cos x ad ± si x, which are bouded by, for x > R N (x) = x N! f (N+) (t)(x t) N dt N! N! x x = xn+ x t N dt x N dt. N! This teds to as N, so si x does equal its power series for x >. If x < the we ca similarly estimate R N (x) ad show it teds to as N, but we ca also use a little trick because we kow si x is a odd fuctio: if x < the x > so si x = si( x) = ( ) ( x)2+ (2 + )! from the power series represetatio of the sie fuctio at positive umbers. Sice ( x) 2+ = x 2+, si x = ( ) x2+ x2+ (2 + )! = ( ) (2 + )!,

22 22 KEITH CONRAD which shows si x equals its power series for x < too. Example By similar work we ca show cos x equals its power series represetatio everywhere: cos x = ( ) x2 (2)! = x2 2! + x4 4! x6 6! +. The power series for si x ad cos x look like the odd ad eve degree terms i the power series for e x. Is the expoetial fuctio related to the trigoometric fuctios? This idea is suggested because of their power series. Sice e x has a series without the ( ) factors, we ca get a match betwee these series by workig ot with e x but with e ix, where i is a square root of. We have ot defied ifiite series (or eve the expoetial fuctio) usig complex umbers. However, assumig we could make sese of this the we would expect to have e ix = (ix)! = + ix x2 2! ix3 = x2 2! + x4 4! + i = cos x + i si x. For example, settig x = π would say e iπ =. Replacig x with x i the formula for e ix gives 3! + x4 4! + ix5 5! + (x x3 3! + x5 5! e ix = cos x i si x. Addig ad subtractig the formulas for e ix ad e ix lets us express the real trigoometric fuctios si x ad cos x i terms of (complex) expoetial fuctios: cos x = eix + e ix, si x = eix e ix. 2 2i Evidetly a deeper study of ifiite series should make systematic use of complex umbers! The followig questio is atural: who eeds all the remaider estimate busiess from Theorem 5.8 ad Corollary 5.9? After all, why ot just compute the power series for f(x) ad fid its radius of covergece? That has to be where f(x) equals its power series. Alas, this is false. Example Let f(x) = { e /x2, if x,, if x =. It ca be show that f(x) is ifiitely differetiable at x = ad f () () = for all, so the power series for f(x) has every coefficiet equal to, which meas the power series is the zero fuctio. But f(x) if x, so f(x) is ot equal to its power series aywhere except at x =. Example 5.23 shows that if a fuctio f(x) is ifiitely differetiable for all x ad the power series (f () ()/!)x coverges for all x, f(x) eed ot equal its power series aywhere except at x = (where they must agree sice the costat term of the power series is f()). This is why there is o-trivial cotet i sayig a fuctio ca be represeted by its power series o some iterval. The eed for remaider estimates i geeral is importat. )

### Sequences and Series

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

### 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

### 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

### 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

### 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

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

### 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

### 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

### 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

### 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...

### 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

### 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

### 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,

### 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

### 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

### 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,

### 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,

### 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

### 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

### 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.

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### .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,

### 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

### 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

### 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

### 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

### 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

### 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

### 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.

### 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

### Output Analysis (2, Chapters 10 &11 Law)

B. Maddah ENMG 6 Simulatio 05/0/07 Output Aalysis (, Chapters 10 &11 Law) Comparig alterative system cofiguratio Sice the output of a simulatio is radom, the comparig differet systems via simulatio should

### 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

### 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

### 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

### 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

### 5: Introduction to Estimation

5: Itroductio to Estimatio Cotets Acroyms ad symbols... 1 Statistical iferece... Estimatig µ with cofidece... 3 Samplig distributio of the mea... 3 Cofidece Iterval for μ whe σ is kow before had... 4 Sample

### CHAPTER 3 DIGITAL CODING OF SIGNALS

CHAPTER 3 DIGITAL CODING OF SIGNALS Computers are ofte used to automate the recordig of measuremets. The trasducers ad sigal coditioig circuits produce a voltage sigal that is proportioal to a quatity

### 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

### 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

### 5 Boolean Decision Trees (February 11)

5 Boolea Decisio Trees (February 11) 5.1 Graph Coectivity Suppose we are give a udirected graph G, represeted as a boolea adjacecy matrix = (a ij ), where a ij = 1 if ad oly if vertices i ad j are coected

### 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

### Tradigms of Astundithi and Toyota

Tradig the radomess - Desigig a optimal tradig strategy uder a drifted radom walk price model Yuao Wu Math 20 Project Paper Professor Zachary Hamaker Abstract: I this paper the author iteds to explore

### 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

### Math 113 HW #11 Solutions

Math 3 HW # Solutios 5. 4. (a) Estimate the area uder the graph of f(x) = x from x = to x = 4 usig four approximatig rectagles ad right edpoits. Sketch the graph ad the rectagles. Is your estimate a uderestimate

### Case Study. Normal and t Distributions. Density Plot. Normal Distributions

Case Study Normal ad t Distributios Bret Halo ad Bret Larget Departmet of Statistics Uiversity of Wiscosi Madiso October 11 13, 2011 Case Study Body temperature varies withi idividuals over time (it ca

### Measures of Spread and Boxplots Discrete Math, Section 9.4

Measures of Spread ad Boxplots Discrete Math, Sectio 9.4 We start with a example: Example 1: Comparig Mea ad Media Compute the mea ad media of each data set: S 1 = {4, 6, 8, 10, 1, 14, 16} S = {4, 7, 9,

### 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

### 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.

### 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

### 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

### Definition. A variable X that takes on values X 1, X 2, X 3,...X k with respective frequencies f 1, f 2, f 3,...f k has mean

1 Social Studies 201 October 13, 2004 Note: The examples i these otes may be differet tha used i class. However, the examples are similar ad the methods used are idetical to what was preseted i class.

### http://www.webassign.net/v4cgijeff.downs@wnc/control.pl

Assigmet Previewer http://www.webassig.et/vcgijeff.dows@wc/cotrol.pl of // : PM Practice Eam () Questio Descriptio Eam over chapter.. Questio DetailsLarCalc... [] Fid the geeral solutio of the differetial

### Center, Spread, and Shape in Inference: Claims, Caveats, and Insights

Ceter, Spread, ad Shape i Iferece: Claims, Caveats, ad Isights Dr. Nacy Pfeig (Uiversity of Pittsburgh) AMATYC November 2008 Prelimiary Activities 1. I would like to produce a iterval estimate for the

### Irreducible polynomials with consecutive zero coefficients

Irreducible polyomials with cosecutive zero coefficiets Theodoulos Garefalakis Departmet of Mathematics, Uiversity of Crete, 71409 Heraklio, Greece Abstract Let q be a prime power. We cosider the problem

### 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

### 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

### 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

### 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?

### Notes on exponential generating functions and structures.

Notes o expoetial geeratig fuctios ad structures. 1. The cocept of a structure. Cosider the followig coutig problems: (1) to fid for each the umber of partitios of a -elemet set, (2) to fid for each the

### 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

### Modified Line Search Method for Global Optimization

Modified Lie Search Method for Global Optimizatio Cria Grosa ad Ajith Abraham Ceter of Excellece for Quatifiable Quality of Service Norwegia Uiversity of Sciece ad Techology Trodheim, Norway {cria, ajith}@q2s.tu.o

### 19 Another Look at Differentiability in Quadratic Mean

19 Aother Look at Differetiability i Quadratic Mea David Pollard 1 ABSTRACT This ote revisits the delightfully subtle itercoectios betwee three ideas: differetiability, i a L 2 sese, of the square-root

### 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

### 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...........................

### 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

### COMPUTER LABORATORY IMPLEMENTATION ISSUES AT A SMALL LIBERAL ARTS COLLEGE. Richard A. Weida Lycoming College Williamsport, PA 17701 weida@lycoming.

COMPUTER LABORATORY IMPLEMENTATION ISSUES AT A SMALL LIBERAL ARTS COLLEGE Richard A. Weida Lycomig College Williamsport, PA 17701 weida@lycomig.edu Abstract: Lycomig College is a small, private, liberal

### 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:

### 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.

### Chapter 5 Unit 1. IET 350 Engineering Economics. Learning Objectives Chapter 5. Learning Objectives Unit 1. Annual Amount and Gradient Functions

Chapter 5 Uit Aual Amout ad Gradiet Fuctios IET 350 Egieerig Ecoomics Learig Objectives Chapter 5 Upo completio of this chapter you should uderstad: Calculatig future values from aual amouts. Calculatig

### The following example will help us understand The Sampling Distribution of the Mean. C1 C2 C3 C4 C5 50 miles 84 miles 38 miles 120 miles 48 miles

The followig eample will help us uderstad The Samplig Distributio of the Mea Review: The populatio is the etire collectio of all idividuals or objects of iterest The sample is the portio of the populatio

### Vladimir N. Burkov, Dmitri A. Novikov MODELS AND METHODS OF MULTIPROJECTS MANAGEMENT

Keywords: project maagemet, resource allocatio, etwork plaig Vladimir N Burkov, Dmitri A Novikov MODELS AND METHODS OF MULTIPROJECTS MANAGEMENT The paper deals with the problems of resource allocatio betwee

### 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

### 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

### 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

### 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

### 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

### 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,

### GCSE STATISTICS. 4) How to calculate the range: The difference between the biggest number and the smallest number.

GCSE STATISTICS You should kow: 1) How to draw a frequecy diagram: e.g. NUMBER TALLY FREQUENCY 1 3 5 ) How to draw a bar chart, a pictogram, ad a pie chart. 3) How to use averages: a) Mea - add up all

### Ekkehart Schlicht: Economic Surplus and Derived Demand

Ekkehart Schlicht: Ecoomic Surplus ad Derived Demad Muich Discussio Paper No. 2006-17 Departmet of Ecoomics Uiversity of Muich Volkswirtschaftliche Fakultät Ludwig-Maximilias-Uiversität Müche Olie at http://epub.ub.ui-mueche.de/940/

### 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.

THE ABRACADABRA PROBLEM FRANCESCO CARAVENNA Abstract. We preset a detailed solutio of Exercise E0.6 i [Wil9]: i a radom sequece of letters, draw idepedetly ad uiformly from the Eglish alphabet, the expected

### 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

### 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

### Metric, Normed, and Topological Spaces

Chapter 13 Metric, Normed, ad Topological Spaces A metric space is a set X that has a otio of the distace d(x, y) betwee every pair of poits x, y X. A fudametal example is R with the absolute-value metric

### 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