Notes: MA1102: Series and Matrices

Notes: MA: Series ad Matrices M.T. Nair (IIT Madras) Jauary - May 8 April 3, 8 Cotets Sequeces 3. Defiitio ad examples............................. 3. Covergece ad divergece........................... 3.3 Properties..................................... 6.4 Mootoe covergece theorem......................... Series. Some typical examples............................... Some tests for covergece............................ 5.. Itegral test................................ 5.. Compariso test.............................. 5..3 Ratio tests................................. 6..4 Root test................................. 7.3 Absolutely coverget series ad alteratig series............... 9.4 Rearragemets.................................. 3 Improper Itegrals 4 3. Various types of improper itegrals....................... 4 3. Examples..................................... 5 3.3 Gamma ad Beta fuctios........................... 7 4 Power series 9 4. Abel s theorem ad radius of covergece.................... 9 4. Termwise differetiatio ad termwise itegratio............... 3 4.3 Evaluatio at coverget edpoits....................... 33 4.4 Examples..................................... 33 4.5 Power series cetered at a poit......................... 35 5 Taylor s series ad Taylor s formulas 37 5. Taylor s theorem - Lagrage form........................ 38 5. Taylor s theorem - Cauchy form......................... 4

6 Fourier Series 4 6. Trigoometric series ad Trigoometric polyomials.............. 4 6. Fourier series of π-periodic fuctios...................... 43 6.3 Fourier series for eve ad odd fuctios.................... 46 6.4 Sie ad cosie series expasios........................ 49 6.5 Fourier Series of l-periodic Fuctios..................... 5 6.6 Fourier Series o Arbitrary Itervals...................... 54 6.7 Additioal Exercises............................... 56 7 Matrices 57 7. Various types of matrices............................. 57 7. Echelo ad row reduced echelo form (RREF)................ 59 7.3 Ier product ad orms of vectors....................... 6 7.4 Liear depedece ad idepedece...................... 63 7.5 RREF, Liear idepedece ad rak..................... 68 7.6 Determiats................................... 7 7.6. Properties................................. 7 7.6. Solutio of equatios by Cramer s rule................. 7 7.7 Iverse ad solutio of equatios usig RREF................. 7 7.8 Eigevalues ad eigevectors........................... 76 7.8. Eigevalues of hermitia, ormal ad uitary matrices........ 78 7.9 Eigevalue represetatios - diagoalizatio theorems............. 79

Sequeces. Defiitio ad examples List of umbers a, a,... is called a sequece. Such a sequece is deoted oe of the followig forms: (a, a,...), {a, a,...}, (a ), {a }. More precisely: Defiitio. A sequece of real umbers is a fuctio f : N R, ad it is usually represeted by it rage {a := f() : N}, which is also represeted by the way we have already metioed above. Example. The followig are examples of sequeces: (i) (.,.,.,...). (ii) (, 3, 4,...,,...). (iii) (, 3, 3 4,..., +,...). (iv) (,,,,...). (v) (, 5, 3,..., +,...).. Covergece ad divergece I Example 5 (), we have a =. = + + + + 3 = ( ) = ( ). 9 9 a = 9 <. Thus, first decimal places of ad a 9 are the same. Takig large eough, the quatity a 9 ca be made as small as we wat. For istace, give ay positive umber ε, 9 a < ε if < ε if > ( ) if > log ε ε Defiitio 3. A sequece (a ) of real umbers is said to coverge to a R if for every ε >, there is a positive iteger N such that a a < ε for all N. A sequece which does ot coverge is called a diverget sequece. 3

If (a ) coverges to a, the we write a a as or lim a = a, or sometimes a a whe the variable is uderstood from the cotext. Defiitio 4. A sequece (a ) of real umbers is said to diverge to if for every M >, there there is a positive iteger N such that a > M for all N. The sequece (a ) is said to diverge to if for every M >, there there is a positive iteger N such that a < M for all N. Note that If (a ) diverge to, the we write a as, ad if (a ) diverge to, the we write a as. a a ε >, N N such that a (a ε, a + ε) for all N. a a for every ope iterval I cotaiig x, N N such that a I for all N. (a ) diverges iff for every a R, there exists a ε >, such that a (a ε, a + ε) for ifiitely may s. Example 5. The sequece: (i) (.,.,.,...) coverges to /9. (ii) (,,,...,,...) coverges to. 3 4 (iii) (,, 3,...,,...) coverges to. 3 4 + (iv) ( + ) coverges to. (v) (,,, 3,...) diverges, ad diverges to. (vi) (, 5,,..., +,...) diverges. 3 4

Proofs. (i) We have already see that (.,.,.,...) coverges to /9. (ii) Let ε > be give. The (iii) Let ε > be give. The (iv) Let ε > be give. The = < ε + = + < ε [ ] +. ε [ ε + = + < = [ ] < ε +. ε For showig that the sequeces i (v) ad (vi) diverge, it is ecessary to show that they do ot coverge to ay umber a R. How ca we show this? Ca we verify with each ad every umber a R? There are ifiitely may such umber! So, we may adopt a particular logical method, called deductio ad absurdum or proof by cotradictio: Suppose there is some a R such that a a. Proceedig logically, if we arrive at a cotradictio, the we ca coclude that our assumptio that there is some a R such that a a is wrog. { I other words, there is NO a R such that a a. if odd, (v) I this case a = Suppose a if eve. a for some a R. The takig ε = /, we should be able to fid N N such that a a < / for all N. This is impossible: Note that a a < / a / < a < a + / so that / < a < 3/ & 3/ < a < 5/, which is impossible. Also, ote that for ay a R, if we take < ε < /, we see that either (a ε, a + ε) or ( (a ε, a + ε). Thus a (a ε, a + ε) for ifiitely may s. (vi) Note that + = + > N. Therefore, for ay a R ad for ay ε >, ]. + (a ε, a + ε) for all large eough N. I fact, the sequece ( +) diverges to, because, for ay M >, + = + > > M > M. 5

.3 Properties You must have heard about the Fiboacci sequece :,,, 3 5, 8, 3,.... More geerally, if a = ad a =, the a give by the recurrece relatio a + = a + + a. Of course, this sequece (a ) diverges to. However, the sequece (a + /a ) coverges, ad the limit is kow as golde ratio. Note that, if r = a + /a, the Suppose r r. The we have r + = a + a + = a + + a a + = + a a + = + r. so that r = +, ad hece r r + r ad r + + r + r r = + 5, the golde ratio. I the above discussio we used a few facts, which we have ot proved: () (r ) coverges. () r r ad r imply /r /r ad + /r + /r. Note that r = + 5 satisfies r = + r. Thus, so that r + r = ( + ) r From this, it ca be show that r + r = r r r r r + r = r r r r r r r ( + ) = r r r = r r r r r r. r r r r = r r r = ( 5 ). r You ca see that (we shall prove it soo!), sice r 3/ so that /r (/3). Hece, we ca deduce that r + r. Agai, we used the property that Also kow as Hemachadra sequece after the th cetury Jai scholar Hemachadra (89-73), who used this sequece about 5 years before Fiboacci. 6

a b for all N ad b imply a. This is a simple exercise. Thus, () is proved. For the property (), we prove the followig. Theorem 6. Suppose a a ad b b. The we have the followig.. a + b a + b.. a b ab. 3. If b for all N ad b, the a b a b. Remark 7. Note that, there ca exist sequeces (a ) ad (b ) such that a a ad b b with b for all N, but (a /b ) eed ot coverge. For example, take a = for all N ad b = /. I this case, a /b. Before provig the above, let us observe some simpler facts: Propositio 8. Let (a ) be a sequece of real umbers. The the followig results hold. () Suppose (a ) coverges. The there exists M > such that a M for all N. () Suppose a a ad a. The there exists α > ad N N such that a α for all N. Proof. Let a a, ad let ε > be give. The there exists N N such that a a < ε for all N. () Note that Therefore, a = (a a) + a a a + a < ε + N. () Suppose a. We kow that a M := max{ε +, a,..., a N } N. a = (a a) + a a a a a ε N. Takig ε < a /, we obtai a a / for all N. A defiitio is i order: Defiitio 9. A sequece (a ) is said to be a bouded sequece if there exists M > such that a M for all N. Propositio 8 () says that every coverget sequece is bouded. 7

Proof of Theorem 6. () Let ε > be give. Note that (a + b ) (a + b) = (a a) + (b b) a a + b b N. Let N, N N be such that Hece, () Note that a a < ε/ N ad b b < ε/ N. a a + b b < ε/ + ε/ N := max{n, N }. a b ab = a (b b) + (a a)b a b b + a a b Sice (a ) coverges, there exists M > such that a M for all N. Hece, a b ab M b b + a a b. Let ε > be give, ad let N, N N be such that The we have a a < ε ( b + ) N ad a a < ε M N. a b ab M b b + a a b < ε + ε = ε N := max{n, N }. Thus, a b ab. (3) Note that a a = a b b a b b b b = (a a)b a(b b) b b a a b + a b b b b Sice b b ad b, there exists α > ad N N such that b α for all N. Hece, a a a a b + a b b a a b + a b b b b b b α b Let N, N N be such that a a b α b < ε N ad a b b α b < ε N. The Thus, a b a b. a a a a b + a b b < ε N := max{n, N }. b b α b Theorem. (Sadwich theorem) Suppose a l ad b l. If a c b for all N, the c l. 8

Proof. Let ε > be give. The there exists N N such that a, b (l ε, l + ε) for all N. Sice a c b, we also have c (l ε, l + ε) for all N. We kow that every coverget sequece is bouded. But, a bouded sequece eed ot be coverget. A simple example to this effect is: a = ( ), N. However, we shall see that every bouded sequece (a ) which is either mootoically icreasig, that is, a a + for every N, or mootoically decreasig, that is, a a + for every N, is coverget. This is makig use of a importat property of the set of real umbers, called, least upper boud property. For statig this property we require a few defitios: Defiitio. Let S R. () S is said to be bouded above if there exists M > such that s M for all s S. () A umber M as i () is called a upper boud of S. (3) A upper boud M of S is called a least upper boud (lub) if M is least amog all upper bouds of S. (4) S is said to be bouded below if there exists M > such that s M for all s S. (5) A umber M as i () is called a lower boud of S. (6) A lower boud M of S is called a greatest lower boud (glb) if M is greatest amog all lower bouds of S. We observe the followig: Suppose S is bouded above. The M is the lub of S iff for every ε >, there exists s S such that M ε < s M. Suppose S is bouded below. The M is the glb of S iff for every ε >, there exists s S such that M s < M + ε. A atural questio is: If S is bouded above (resp. bouded below), does it have a least upper boud (resp. greatest lower boud)? Least Upper Boud (lub) Property: Every set S R which is bouded above has a least upper boud. Observe that: S is bouded above iff S := { s : s S} is bouded below. 9

Hece, from the Least Upper Boud Property, it follows that: If S R is bouded below, the it has a greatest lower boud. It ca be see easily that: Least upper boud of every set which is bouded above is is uique. Greatest lower boud of every set which is bouded below is uique. Notatio: Let S be a set which bouded above. The the least upper boud of S is called the supremum of S, ad it is deoted by sup(s). Let S be a set which bouded below. The the greatest lower boud of S is called the ifimum of S, ad it is deoted by if(s). It is to be metioed that: The supremum (resp. ifimum) of a set eed ot be a elemet of the set. If it is i the set, the it is called the maximum (resp. miimum) of the set. Example. S. (ii) The set S = { + (i) The set S = {x R : x < } is bouded above ad is the lub of (iii) The set S = N is ot bouded above. (iv) The set S = { + Example 3. S. () The set S = { + : N} is bouded above ad is the lub of S. : N} is ot bouded above. () The set S = {x R : x < } is bouded below ad is the glb of : N} is bouded below ad is the lub of S. (3) The set S = N is bouded below ad is the glb of N. (4) The set S = { + : N} is bouded below ad is the glb of S. (5) The set S = {,,,,...} is bouded above ad bouded below, ad ad, respectively, are the lub ad glb of S. (6) The set S = {,,,,...} is either bouded above or bouded below. Let S R. If (a ) is a sequece i R ad if the set S := {a : N} is bouded above, the we say that (a ) is bouded above, ad its supremum is deoted by sup a. If (a ) is a sequece i R ad if the set S := {a : N} is bouded below, the we say that (a ) is bouded below, ad the its ifimum is deoted by if a. Defiitio 4. A sequece (a ) is said to be mootoically icreasig if a a + for all N, ad it is called mootoically decreasig if a a + for all N.

.4 Mootoe covergece theorem Theorem 5. (Mootoe covergece theorem-) If (a ) is mootoically icresig ad bouded above, the it coverges to its supremum. Proof. Suppose (a ) is mootoically icresig ad bouded above ad let α := sup (a ). The for every ε >, there exists N such that M ε < a N M. Sice a a + α for all N, we also have Thus, a α. M ε < a α N. The followig theorem is a cosequece of the above (How?). Theorem 6. (Mootoe covergece theorem-) If (a ) is mootoically decreasig ad bouded below, the it coverges to its ifimum. Theorem 7. If < a <, the a. Proof. Let a = /( + b). The Sice b, it follows that a. a = Theorem 8. If c >, the c. ( + b) b. Proof. Let a = /c. The < a < so that a. Hece, c. Example 9. Let < a < ad s := + a + + a. The (s ) coverges ad its limit is /( a): s := + a + + a = a+ a a+ a. Example. / as, ad for ay a >, a /. Let / = + a. The Hece = ( + a ) a ( ) a!. ( ). Thus, / = + a. We may use similar trick to show that a /.

Series Give a sequece (a ) of real umbers, we ca forma ew sequece (s ) with s = a + + a = a k. k= Defiitio. Give a sequece (a ), a expressio of the form k= a k is called a series, ad s is called the -th partial sum of the series. We say that the series k= a k coverges to s R if s s, ad i that case we write k= a k = s, ad s is called the sum of the series k= a k. If (s ) diverges, the we say that the series k= a k diverges. Here is a importat property. Theorem. If = a coverges, the a. Proof. Suppose s l. The a + = s + s l l =.. Some typical examples Example 3. () The series + + coverges to /9. () For a R, cosider the series = a. If a =, the s = + a + a + + a = +. The Suppose a. The + a + a + + a = a+ a = a a+ a. If a <, the a + = a +, ad hece, i this case, s = a a+ a a. If a, the a + = a + for all N, ad hece, a, so that = a diverges. Thus, we have proved that + a + a + coverges iff a <. (3) The series + 3 + + + diverges to. + Note that for ay N, to. /. Hece, s + / for all N. Hece, (s ) diverges

(4) The series + + 3 + + + diverges to Note that s := + + 3 + + ca be writte as ( + ( + ) 3 + ( + 4) 5 + + ( + 8) 9 + 6) + + ( + + + + ). Note that so that Thus, ( + ) ( 3 + 4) ( 9 + + + 6) ( + + + ),, 3 4,, s + + + 3 3 + + = + 4 = +. s + N. This shows that (s ) is ubouded, ad hece it is diverget. I fact, it diverges to. (5) The series + + 3 + + + coverges. Clearly, the sequece (s ) of its partial sums is mootoically icresig. Hece, it is eough to show that it is bouded above. Note that + 3 =, 4 + 5 + 6 + 4 7 4 =, ( ) + ( + ) + + ( ) ( ) =. Hece s for all N. Thus, (s ) is mootoically icreasig ad bouded above, ad hece, it coverges. (6) For every p, the series + + p 3 + + + diverges to. p p 3

Let s (p) be its -th partial sum. Sice s (p) s () for all N ad (s () ) is ubouded, ) is also ubouded, ad hece diverges to. (s (p) (7) For every p, the series + p + 3 p + + p + coverges. Let s (p) be its -th partial sum. Note that s (p) s () for all N ad (s () ) is bouded. Sice (s (p) ) is mootoically icreasig, it follows that (s (p) ) mootoically icreasig ad bouded, ad hece it coverges. (8) For every p >, the series + p + 3 p + + p + coverges. But, Hece, Note that k = + + ( + ) = p k = + ( + ) dx dx + p x p k ( + ) p = + dx x p k = p. p dx = p dx k+ [ ] x = dx x p+ x=k+ p x = = [ ] p p + x= p (k + ) p p. k+ = p = k = ( + ) k+ p = + dx x p p. Thus, the sequece of partial sums,which is already mootoically icreasig, is bouded as well. Hece, the give series coverges. Thus, we have For p R, the series coverges p >. p = (9) The series +! +! + 3! + +! + coverges. Sice Hece,! =. s = +! +! + 3! + +! + + + + + 3. Sice (s ) is mootoically icreasig ad bouded above, it coverges. 4

The sum of the series =! is deoted by e. It ca be show that lim ( + ) exists ad it is equal to e (See the book ).. Some tests for covergece.. Itegral test The followig theorem ca be proved usig the techiques adopted i Example 3 (8). Theorem 4. (Itegral test) If f : [, ) [, ) is mootoically decreasig, the f() coverges =.. Compariso test ( ) f(x)dx coverges. Theorem 5. (Compariso test) Suppose (a ) ad (b ) are sequeces of o-egative terms. Suppose there exists k N such that a b for every k.. If = b coverges, the = a coverges.. If = a diverges, the = b diverges. Proof. Hit: Use mootoe covergece theorem, by observig that (i) the sequece of partial sums is mootoically icreasig ad (ii) every coverget sequece is bouded. Theorem 6. (Limit compariso test) Suppose (a ) ad (b ) are sequeces of positive a terms. Suppose lim = l. b. If < l <, the = a coverges = b coverges.. If l =, the = b coverges = a coverges. 3. If l =, the = a coverges = b coverges. Proof. Let ε >. The there exists N N such that l ε < a b < l + ε N. () Suppose α > ad < ε < α/. The l < a < 3l b N. See Example.5, page 3 i Calculus of Oe Variable by M.T. Nair 5

Hece l b < a < 3l b N. Therefore, the result follows from Theorem 5. () Suppose l =. The < a < ε N b so that < a < εb N. Therefore, the result follows from Theorem 5. Example 7. Cosider the series = + 3. Clearly, + 3 + ( ) ( + / ) 3 =. 3 /3 > for all N. Also, Note that (verify) +/ /3 for all k. Therefore, as. Hece, there exists k N such that +/ /3 < 3 Sice = ( 3..3 Ratio tests + ( ) ( + / ) 3 = 3 ) k. 3 /3 ( 3 ) coverges, it follows that + = 3 Theorem 8. Suppose (a ) is a sequece of positive terms. also coverges. (i) Suppose there exists k N ad < c < such that a + a = a coverges. c for all k. The (ii) Suppose there exists k N such that a + a for all k. The = a diverges. Proof. (i) a + c + k a k for all k. Sice =k c coverges, by compariso test, =k a also coverges. (ii) a + a k > for all k. Hece, a. Therefore, =k a diverges. Theorem 9. (d Alembert s atio test) Suppose (a ) is a sequece of positive terms, ad a ( + a+ ) lim = l, where either l is a o-egative real umber or l =, which meas a a diverges to. (i) If l <, the = a coverges. (ii) If l >, the = a diverges. 6

( ) If a+ a diverges to, the the coclusio i (ii) holds. Proof. Suppose l <. Let ε > ad let N N be such that l ε < a + a < l + ε for all N. (i) Tale ε > such that l + ε < ad apply Theorem 8 (). (ii) Tale ε > such that l ε ad apply Theorem 8 (). If l =, the there exists N N such that a + for all N so that a ; cosequetly, for all N. Hece, a a + a N > = a diverges to...4 Root test Theorem 3. (Cauchy s root test) Suppose (a ) is a sequece of positive terms, ( ad a+ ) lim a/ = l, where either l is a o-egative real umber or l =, which meas a diverges to. (i) If l <, the = a coverges. (ii) If l >, the = a diverges. Proof. Use similar techiques as i the proof of d Alembert s ratio test. Exercise 3. If l =, the d Alembert s ratio test ad Cauchy s ratio test fail. Oe may cosider the series = ad =. x Example 3. For ay x R, the series coverges. This follows from ratio test, sice! = x + /( + )! x /! = x +. The limit of the above series It ca be show that = x! is deoted by exp(x) or by the otatio ex. exp(x + y) = exp(x) exp(y); The fuctio f(x) := exp(x) is strictly mootoically icreasig, that is, x < y implies exp(x) < exp(y). The fuctio f(x) := exp(x) is cotiuous o R ad it is oe-oe ad oto [, ). a + Theorem 33. Let (a ) be a sequece of positive real umbers such that lim a < l <, the lim a / = l. = l. If 7

Proof. Let ε > be give ad let k N be such that that is, l ε < a + a < l + ε k, (l ε)a < a + < (l + ε)a k. Takig ε small eough such that l ε >, we obtai Writig k + j =, we have (l ε) j a k < a k+j < (l + ε) j a k j N. (l ε) k a k < a < (l + ε) k a k > k, that is, a k (l ε) k (l ε) < a < (l + ε) a k (l ε) k > k, so that [ a ] / [ k (l ε) < a / a ] / k (l ε) k < (l + ε) (l ε) k > k, ] / Recall that for ay x >, lim x / =. Hece, lim =. Let k N be (l ε) k k k ad [ a ] / k ε < < + ε k. (l ε) k Thus, that is, [ ( ε)(l ε) < a / < (l + ε)( + ε) k, l ε(l + + ε) < a / < l + ε(l + + ε) k. Hece, lim a / exists ad it is equal to l. It ca happe that root test ca be applied whereas ratio test may ot be applied. Example 34. Cosider the series Hece, i this case, + + + 4 + 3 + 6 + + 5 + + + + +3 +. + a a = ad a + a =, whereas a / {, } so that lim + a/ =. Thus, ratio test caot be applied, whereas root test shows that the series coverges. a k 8

.3 Absolutely coverget series ad alteratig series Theorem 35. Let (a ) be a sequece of real umbers such that + a coverges. The + a coverges. For its proof we shall make use of the followig result, whose proof we omit: Theorem 36. (Cauchy s criterio of covergece of sequeces) Let (a ) be a sequece of real umbers. The (a ) coverges iff for every ε >, there exists N N such that a a m < ε, m N. Defiitio 37. A sequece (a ) of real umbers is called a Cauchy sequece if for every ε >, there exists N N such that Thus, Theorem 36 says that a a m < ε, m N. (a ) coverges (a ) is a Cauchy sequece. It is to be observed that: Every coverget sequece is a Cauchy sequece. Ideed, if a a, the for ay, m N, a a m a a + a a m. For ay give ε >, let N N be such that a a < ε/ for all N. The from the above, a a m a a + a a m < ε + ε = ε, m N. Defiitio 38. A series + a is said to be absolutely coverget if + a coverges. Thus, Theorem 35 says that: Every absolutely coverget series is coverget. Proof of Theorem 35. Let s ad s be the -th partial sums of = a ad = a, respectively. The for > m, we have s s m s s m. ( ) Sice = a assumed to be coverget, (s ) is a Cauchy sequece, ad by the relatio ( ), (s ) is also a Cauchy sequece. Hece, by Theorem 36, (s ) coverges. 9

Example 39. For ay x R, the series as well. = x! Defiitio 4. The fuctio exp : R R defied by exp(x) = = is absolutely coverget, ad hece coverges x!, x R, is called the expoetial fuctio. It ca be show that 3 exp(x + y) = exp(x) exp(y) x, y R. Note that exp(x) > for x >, ad exp() =. Hece, so that exp(x) exp( x) = exp() = exp(x) ad exp( x) = exp(x) x R. I particular, exp(x) > for every x R. It ca also be proved that 4 the fuctio x exp(x) is cotiuous, strictly icreasig, exp(x) as x ad exp(x) as x. Hece, for every y >, there exists a uique x R such that exp(x) = y. This x is called the logarithm of y, deoted by log(y). Defiitio 4. A series of the form = ( )+ a, where a > for all N, is called a alteratig series. By Theorem 35, for ay p >, the series ( ) is absolutely coverget, ad hece = coverges as well. Does the above series coverge for < p? The aswer is i affirmative, due to the followig theorem. Theorem 4. (Leibitz test for covergece of alteratig series) If (a ) is a strictly decreasig sequece of positive real umbers such that a, the = ( )+ a coverges. 3 See page, Theorem.9, i Calculus of Oe variable, Ae Books, Pvt. Ltd, 4, by M.T. Nair. 4 See page, Theorem.9, i Calculus of Oe variable, Ae Books, Pvt. Ltd, 4, by M.T. Nair. p

Proof. Let (a ) be a strictly decreasig sequece of positive real umbers such that a. Let s be the -th partial sum of the series = ( ) a. The for every N, s = a a + a 3 a 4 + + a a = (a a ) + (a 3 a 4 ) + + (a a ) = a (a a 3 ) (a 4 a 5 ) (a a ) a. From the above it follows that (s is a mootoically icreasig ad bouded above by a. Hece, it coverges, say a s. The we have s + = s + a + s + = s. Now, let ε > be give ad let N, N N be such that s s < ε N ad s + s < ε N. ( ) By the above theorem, the series is coverget. I fact: = ( ) For ay p >, coverges. Remark 43. Does the series + 3 4 + 5 + 6 7 8 + = p coverge? Note that the above series ca be writte as (a + b ), where = a = ( )+, b = ( )+. ( ) + Sice ad ( ) + coverge, it follows that = = coverges..4 Rearragemets = [ ( ) + ] + ( )+ also Suppose, for a give sequece (a ) of real umbers, the series = a coverges. If (b ) is a sequece obtaied from (a ) by rearragig its terms, the is it true that the series = b is also coverget? I other words, if = a coverges ad if σ : N N is a bijective fuctio, the is it true that the series = a σ() is also coverget? The aswer is i affirmative if the terms of the series are all positive (Exercise). I fact, i this case, the sums of all the rearraged series are the same. From the above discussio, it follows that:

If a is absolutely coverget, the = rearraged series are coverget. a = s ad all the However, if the series is coditioally coverget, the differet rearragemets of the series ca have differet sums. Example 44. Cosider the series = + 3 4 + 5 6 + 7 8 +. () We kow that this series coverge. Now cosider the series 4 + 3 6 8 + + 4 4 +. () If s ad s are the -th partial sums of the series i () ad (), respectively, ad if s is the sum of the series i (), the it ca be show that 5 so that s 3 = s s 3 = s s, N, s 3+ = s 3 + + s, s 3+ = s 3 + + 4 s. Therefore (why?) s s/. Thus, the series i () ad () have differet sums. Questio: If = a is coverget, the is it true that every series obtaied from = a by rearragig the terms also coverget? Not ecessarily. Look at the followig example. Example 45. Cosider the coverget alteratig series + 3 4 + 5 6 + 7 8 +. Cosider the followig rearraged series: ( + 3 ) ( + 5 + 7 + 9 + 4) ( + 3 + 5 + 7 + 9 + + 3 6). + +. ( ( ) + + ( + ) ) 5 See page 59, Example 53 i Calculus of Oe variable, Ae Books, Pvt. Ltd, 4, by M.T. Nair.

If s is the -th partial sum of the above series, the we see that s (+) = ( + 3 ) ( + 5 + 7 + 9 + 4) ( + 3 + 5 + 7 + 9 + + 3 6). ( + ( ) + + ( + ) Note that ( + ) = [( ) ] + 4 so that there are umber of terms i the last bracket ad, sice we have (+) (+) ( ( ) + + ) ( + ) ). ( + ) = +. Thus, s (+) = = ( + 3 + + ) ( + + 4 + + ) ( + 3 + + ) ( + + ) + + ( ) + + + + Hece, s (+) as, so that the rearraged series diverges to. Questio: If = a is absolutely coverget, the is it true that all the series obtaied from = a with terms rearraged coverge to the same sum? 3

3 Improper Itegrals 3. Various types of improper itegrals Recall that defiite itegral b f(x)dx is usually defied for a bouded fuctio f : [a, b] R, a which is either cotiuous or piecewise cotiuous. We would like to see if we ca extet the usual type of itegral to the cases to have itegrals of the forms a f(x)dx, b f(x)dx, f(x)dx b a f(x)dx eve whe f is ot ecessarily bouded i a iterval cotaiig the ed-poits. It is atural to defie them as follows:. Suppose f is defied ad cotiuous o [a, ). If lim the improper itegral a f(x)dx as a f(x)dx = lim t t a t t a f(x)dx.. Suppose f is defied ad cotiuous o (, b]. If lim t defie the improper itegral b f(x)dx as b f(x)dx = lim t b 3. Suppose f is defied ad cotiuous o [a, b). If lim the improper itegral b f(x)dx as a b a f(x)dx = lim t b t a t f(x)dx. t t b a 4. Suppose f is defied ad cotiuous o (a, b]. If lim the improper itegral b f(x)dx as a b a f(x)dx = lim t a b t f(x)dx. b t a t f(x)dx. f(x)dx exists, the we defie b f(x)dx exists, the we t f(x)dx exists, the we defie f(x)dx exists, the we defie 5. If f is defied ad cotiuous o [a, c) ad (c, b], ad if the improper itegrals c f(x)dx a ad b f(x)dx exist, the the we defie the improper itegral b f(x)dx as c a b a f(x)dx = c a f(x)dx + b c f(x)dx. 4

6. If f is cotiuous o R ad if the improper itegrals c f(x)dx ad f(x)dx exist c for some c R, the the we defie the improper itegral f(x)dx = c f(x)dx + c f(x)dx. f(x)dx as If a improper itegral exist, the we also say that the improper itegral coverges. Here are a few results o improper itegrals over a iterval J which is of fiite or ifiite legth. If f(x) g(x) for all x J, ad if g(x)dx exists, the f(x)dx also exists, J J ad f(x)dx f(x)dx. if J 3. Examples J f(x) dx exists, the J J f(x)dx also exists. Example 46. We show that for ever p R, dx x p dx coverges p > ad i that case x = p p. Clearly, if p, the q := p ad t dx t [ x x = x q q+ dx = p q + ] t = q + (tq+ ) as t. Now, assume that p >. First cosider the case of p =. I this case, t Next, let p > with p. The Note that Hece, < p < p > dx x p dx x = [log(x)]t = log(t) as t. t dx x p = [ x p+ p + ] t = p + (t p+ ) p + (t p+ ) = p (t p ) as t, p + (t p+ ) = p ( t p ) p coverges p > ad i that case 5 dx x = p p. as t.

Example 47. We show that for ever p R, dx x p dx coverges p < ad i that case x = p p. Clearly, if p, the q := p ad dx [ x x = x q q+ dx = p q + ] = q +. Now, assume that p >. First cosider the case of p =. I this case, t dx x = [log(x)] t Next, let p > with p. The Note that Hece, < p < p > dx x p t = log(t) = log(/t) as t. [ ] dx x p+ x = = p p + t p + ( t p+ ) p + ( t p+ ) = p ( t p ) p as t, p + ( t p+ ) = p (t p ) as t. coverges p < ad i that case Example 48. For p R ad for ay a >, dx x = p p. a dx coverges p >. xp Note that, Hece, t > a > t > > a t lim t a t a t Example 49. For p R ad for ay b >, a dx t x = p dx x p = a dx a x p dx x p + dx t dx exists lim xp t x p 6 t dx x p, dx x p. exists.

b dx coverges p <. xp Note that, Hece, Example 5. We have b > > t > > b > t > b lim t t b t b t dx x = p dx x p = t t dx b x + p dx x p dx dx exists lim xp t t x p b exists. dx x p, dx x p. si x dx ad xp cos x dx coverge for p > xp This follows from the fact that si x x p dx dx x p, cos x dx x p dx x p ad I fact, dx coverges for p >. xp si x dx ad xp cos x dx coverge for all p > xp This ca be see by usig itegratio by parts for the proper itegrals ad takig limits (Exercise). For more examples, see Sectio 4. i the book by Nair 6 3.3 Gamma ad Beta fuctios The gamma fuctio Γ(x) for x > is defied as Γ(x) = 6 Calculus of Oe variable, Ae Books, Pvt. Ltd, 4. t x e t dt. 7

Sice x > ad e t, we have so that tx e t dt coverges. Also, sice ad dt t coverges, t x e t t x t x e t /t as t, t x e t dt also coverges. Hece, Γ(x) is well-defied for x >. It ca be see that so that Γ() =, Γ(x + ) = xγ(x) for x Γ( + ) =! N. The beta fuctio β(x, y) for x >, y > is defied as Note that Hece, / β(x, y) = ( t) y = t x ( t) y dt t x ( t) y dt. ( ) y y t t. / t x y dt / t x dt. Sice / t x dt coverges, / t x ( t) y dt also coverges. Also, / t x ( t) y dt = / s y ( s) x ds. Thus, / tx ( t) y dt coverges, ad hece β(x, y) is well-defied for x, y >. β(x, y) = Γ(x)Γ(y) Γ(x + y). 8

4 Power series Power series is a geeralizatio of polyomials. Recall that, by a polyomial we mea a expressio of the form a + a x + + a x, where a, a, a,..., a are real umbers, ad x is a real variable. Give real umbers a, a, a,..., a, a polyomial ca also be thought of as a fuctio f : R R defied by defied by f(x) = a + a x + + a x, x R. Istead of fiite umber of umbers a, a, a,..., a, if we are give a sequece (a, a, a,..., ) of real umbers, the we may cosider a series or more compactly as a + a x + + a x + a x. = Such a series is called a power series. We say that the power series = a x coverges at at poit c if the the series = a c coverges. 4. Abel s theorem ad radius of covergece Note that partial sums of a power series are polyomials. A atural questio is: For what values of x the above series coverges? I this regard, we have the followig importat result. Theorem 5. (Abel s theorem) If the series = a x coverges at a poit x = c, the it coverges absolutely for all x with x < c. Proof. Suppose = a c coverges. Let x be such that x < c. The ( a x x ) ( = a c x ) a c N. c c As = a c coverges, a c. I particular, ( a c ) is bouded, that is, there exists M > such that a c M for all N. Thus, a x M ( x ) N. c Note that x c <. Hece, by compariso test, = a x coverges. Corollary 5. If the series = a x coverges at a poit x = d, the it coverges absolutely for all x with x > d. 9

Proof. Follows from Theorem 66 (How?). Remark 53. Note that for the proof of Abels s theorem (Theorem 66) we used oly the boudedess of the sequece (a c ). Thus, we actually have, i place of Theorem 66, the followig: If (a c ) is bouded for some c, the the series = a x coverges absolutely for all x with x < c. From this it follows that = x coverges for x <. Defiitio 54. Give a power series = a x, R := sup{ x : a x coverges at x} = is called the radius of covergece of = a x, ad D := sup{x R : a x coverges at x} = is called the domai of covergece of = a x. Note that R ; if = a x diverges at a poit, the R < ; if = a x coverges at every poit i R, the R = ; {x R : x < R} D ad if R <, the {x R : x > R} D c. Note that the domai of covergece of a power series is a iterval. Therefore, it is also called the iterval of covergece of the power series. Example 55. (i) The = x coverges at every x with x <, ad diverges at every x with x. The radius of covergece of x = The domai of covergece of x = is. is (, ). (ii) The x = coverges at every x with x < ad also at x =, ad diverges at every x with x > ad also at x =. The radius of covergece of x = The domai of covergece of x = is. is [, ). 3

(iii) The x = with x >. coverges (absolutely) at every x with x ad diverges at every x The radius of covergece of x = is. The domai of covergece of x = is [, ]. (iv) The x = coverges (absolutely) at every x R.! The radius of covergece of x = is.! The domai of covergece of x = is R := (, ).! How to determie the radius of covergece? Theorem 56. Suppose l := lim a / exists or a / l :=. The the radius of covergece of {, l =, = a x is R = /l, with the covetio that R =, l =. Proof. Follows from Cauchy s root test. a + Theorem 57. Suppose there exists k N such that a for all k, l := lim ( a a ) + exists or l :=. The the radius of covergece of = a a x is R = /l, {, l =, with the covetio that R =, l =. Proof. Follows from d Alembert s ratio test. The procedure i the above two theorems ca be used for determiig the radius of covergece of the series = a (x x ) as well. 4. Termwise differetiatio ad termwise itegratio Recall that every polyomial ca be differetiated as may times as we wat. Moreover, for a polyomial, f(x) = a + a x + + a x, we have, ad i geeral, a = f(), a = f () (), a = f () () a k = f (k) (), k =,,,...,. k! This shows that the coefficiets of a polyomial are uiquely determied by the values of the polyomial ad its derivatives at. Do we have similar result for power series? I this regard we have the followig theorem 7. 7 Calculus of Oe variable, Ae Books, Pvt. Ltd, 4. 3

Theorem 58. Let R be the radius of covergece of the power series = a x. The the fuctio f defied by f(x) = a x, R < x < R = is differetiable i the iterval ( R, R), ad f (x) = a x, = R < x < R ad radius of covergece of = a x also equal to R. Usig the above theorem, repeatedly, we have the followig. Corollary 59. Let R be the radius of covergece of the power series = a x. The for every k N, the k-th derivative of f, amely f (k), exists at every x ( R, R) ad f (k) (x) = =k I particular, a k = f (k) (), k =,,,.... k!! ( k)! a x k, R < x < R. If the power series = a x coverges i a ope iterval I cotaiig, the the fuctio f(x) := = a x is ifiitely differetiable i I ad a k = f (k) (), k =,,,.... k! From the above corollary we ca deduce the followig. Corollary 6. (Uiqueess) Suppose = a x ad = b x have radii of covergece R ad R, respectively, ad suppose that = a x = = b x i some iterval cotaiig. The R = R ad a = b for every. Not oly that a power series ca be differetiated term by term, it ca also be itegrated term by term. Theorem 6. Let R be the radius of covergece of the power series = a x. Let The for every a, b ( R, R), b a f(x)dx = f(x) = a x, R < x < R. = b a x dx = = a = a + [b+ a + ], R < x < R. 3

4.3 Evaluatio at coverget edpoits Let R be the radius of covergece of = a x ad let f(x) = a x, R < x < R. = Suppose the series = a x coverges at x = R, that is, = a R coverges. What ca we say about its value? Here is a theorem regardig this 8. Theorem 6. (Abel) Let R be the radius of covergece of = a x ad let f(x) = a x, R < x < R. If = a x coverges at x = R, the f is left cotiuous at x = R ad = lim f(x) = a R. x R Similarly, if = a x coverges at x = R, the the f is right cotiuous at x = R ad lim f(x) = a ( ) R. x R + = = 4.4 Examples Example 63. Cosider the power series ( ) x for x <. Note that = + x = ( ) x, x <. = Itegratig term by term: log( + t) = t dx + x = = ( ) t+, x <. + Sice the above series coverges at the right edpoit, by Theorem 69, log() = = ( ) +. 8 G. PedrikA First Course i Aalysis, Spriger, 9, Page 47 33

Example 64. Cosider the series expasio Itegratig term by term: + x = ( ) x, x <. ta x = = = ( ) x+, x <. + Sice the above series coverges at the right edpoit, by Theorem 69, we obtai 9 π 4 = ( ) +. = Remark 65. For ay α R ad x R, we have Therefore, if α, the Takig α = x with x, Itegratig ( α)( + α + + α k ) = α k+ k N. k α = α + α+ α. = = k + x = ( ) x k+ xk+ + ( ) + x. log( + x) = k = Note that for < x, x t k+ ( )k+ + t dt x Hece, k log( + x) ( ) x+ + + ( )k+ = x t k+ + t dt. t k+ dt xk+ k + k +. ( ) x+ + k +. This is true for all x with x. I particular, we have k log ( ) + k +. Lettig k, we have = log = = ( ) +. Thus, we prove the coclusio i Example 63 without usig Theorem 69. Aalogously, we obtai the coclusio i Example 64 without usig Theorem 69. 9 Value for the Madhava series 34

4.5 Power series cetered at a poit So far we have bee oly cocered with power series of the form = a x. For a give x R ad a real umbers a, a, a,..., the series a (x x ) = is called a power series cetered at x. The aalogue of Abel s theorem is the followig: Theorem 66. (Abel s theorem) If the series = a (x x ) coverges at a poit x = c, the it coverges absolutely for all x with x x < c x. The radius of covergece ad domai of covergece of this series ca be defied by writig it as = a y with y = x x. Thus, R := sup{ x x : a x coverges at x} = is the radius of covergece of = a (x x ), ad D := sup{x R : a (x x ) coverges at x} = is the domai of covergece of = a x. Note that = a (x x ) coverges for every x with x x < R. The iterval of covergece, i this case, is (x R, x + R). The geeral power series = a (x x ) ca be viewed as the earlier oe i terms of a ew variable y := x x, i.e., a y with y = x x. Thus, = a y coverges at y = c = I this geeral case we have the followig: a (x x ) coverges at x = x + c, = y < r x x < r. Theorem 67. Let R be the radius of covergece of the power series = a (x x ). The the fuctio f defied by f(x) = a (x x ), = R < x < R is ifiitely differetiable i the iterval ( R, R), ad for each k N, f (k) (x) = =k! ( k)! a x k, R < x < R 35

ad radius of covergece of! =k a ( k)! (x x ) k is also equal to R. I particular, a k = f (k) (x ), k =,,,.... k! Theorem 68. Let R be the radius of covergece of the power series = a (x x ). Let f(x) = The for every a, b ( R, R), b a f(x)dx = a (x x ), R < x < R. = b a (x x ) dx, R < x < R. = a Theorem 69. (Abel) Let R be the radius of covergece of = a (x x ) ad let f(x) = a (x x ), R < x < R. = If = a x coverges at x = R, the lim f(x) = a (R x ). x R = Similarly, if = a x coverges at x = R, the lim f(x) = a ( ) (R + x ). x R + = 36

5 Taylor s series ad Taylor s formulas f () (x ) Questio: Does the series (x x ) coverge to f(x) for x I?! = We have see that if a fuctio f ca be represeted as a power series = coverges i a iterval (x r, x + r), the a = f () (x ) for every N. Thus, i this case we do! have f () (x ) f(x) = (x x ) for x (x r, x + r).! = Defiitio 7. Suppose f is ifiitely differetiable at i a iterval cotaiig x. The the series f () (x ) = (x x! ) is called the Taylor s series of f cetered at x, ad we write f () (x ) f(x) (x x ).! = If f is a polyomial, the there is a k N such that f () (x ) = for all > k, ad f(x) = Example 7. We kow that Therefore, k = f () (x ) (x x ) =! = x = x, x <, = + x = ( ) x, x <, + x = = ( ) x, x <. = d ( ) =!, dx x x= d ( = dx + x )x= d ( dx Example 7. () Let f(x) = si x. The we have { + x ) f () (x ) (x x ).! x= = ( )!,, odd, ( ) / ()!, eve f () (x) = ( ) si x, f (+) (x) = ( ) cos x, =,,,.... Thus the Taylor s series of f(x) := si x is [ ] ( ) si x si x (x x ) + ( ) cos x (x x ) +. ()! ( + )! = 37

I particular, takig x =, si x = ( ) ( + )! x+. Similarly, cos x = ( ) ()! x. 5. Taylor s theorem - Lagrage form Questio: Uder what coditio ca we say that f(x) = = f () (x ) (x x )! for x i a iterval cotaiig x? I this cotext, first we prove the followig theorem. Theorem 73. (Taylor s theorem - Lagrage form) Suppose f is + times differetiable i a ope iterval (a, b) ad f (+) is cotiuous i [a, b]. Let x (a, b). The for ay x [a, b], there exists c x betwee x ad x such that f(x) = Proof. For y [a, b], let ad where Note that k= f (k) (x ) (x x ) k + f (+) (c x ) k! ( + )! (x x ) +. P (y) = k= f (k) (x ) (y x ) k k! g(y) = f(y) P (y) ϕ(x)(y x ) +, ϕ(x) = f(x) P (x) (x x ) +. g(x ) = f(x ) P (x ) =, g(x) = f(x) P (x) ϕ(x)(x x ) + =. Also, sice we have P (k) (x ) = f (k) (x ), k =,,,..., +, g (k) (x ) =, k =,,,..., +. Now, we apply Roll s theorem repeatedly: 38

g(x ) = = g(x) x betwee x ad x such that g (x ) =. g (x ) = = g (x ) x betwee x ad x such that g () (x ) =. g () (x ) = = g () (x ) x 3 betwee x ad x such that g (3) (x 3 ) =. Cotiuig this, we obtai x, x,... x such that g (k) (x k ) = for k =,,...,. Fially, g () (x ) = = g () (x ) x + betwee x ad x such that g (+) (x + ) =. Let c := x +. Sice g (+) (y) = f (+) (y) ϕ(x)( + )!, we obtai so that f (+) (c) ϕ(x)( + )! = f(x) P (x) (x x ) + = ϕ(x) = f (+) (c) ( + )!. That is, f(x) = P (x) + f (+) (c) ( + )! (x x ) +. Corollary 74. Suppose f is ifiitely differetiable i a ope iterval (a, b). If there exists M > such that f (+) (x) M for all x (a, b), the for ay x, x (a, b), f(x) = = f () (x ) (x x ).! Example 75. Takig f(x) = si x, we have P + (x) = so that si x = = Similarly, takig f(x) = cos x, we have P + (x) = cos x = k= ( ) ( + )! x+. = Example 76. Recall that f(x) = log( + x) := k= ( ) ()! x. x dx + t ( ) k (k + )! xk+ ad f (x), ( ) k (k)! xk ad f (x), so that for x, so that f (x) = + x, f (x) = ( + x), f (3) (x) = ( ) ( + x) 3, f (4) (x) = 3( )3 ( + x) 4, ad more geerally, f (+) (x) =!( ) ( + x) +. 39

I particular, f (+) () ( + )! = ( ) +. Note that, for each fixed x, f (+)!( )! (x) = = ( + x) ( + x) ( + x) ( + x) as. Thus, Corollary 74 caot be applied. However, we kow that log( + x) = = ( ) x+ +, x, that is, log( + x) has the Taylor series expasio for x. 5. Taylor s theorem - Cauchy form Theorem 77. (Taylor s theorem - Cauchy form) Suppose f is + times differetiable i a ope iterval (a, b) ad f (+) is cotiuous i [a, b]. Let x (a, b). The for ay x [a, b], f (k) (x ) f(x) = (x x ) k + x f (+) (t)(x t) dt. k!! x k= Proof. By fudametal theorem of calculus, f(x) = f(x ) + x x f (t)dt. Thus, the theorem is true if =. If =, the by itegratio by parts, takig first fuctio as f (t) ad the secod fuctio as (x t), x f (t)dt x = [ f (t)(x t)] x x + x f (t)(x t)dt x = f (x )(x x ) + x x f (t)(x t)dt. Now, assume that the theorem is true for some = m N, that is, f(x) = m k= f (k) (x ) (x x ) k + k! m! x x f (m+) (t)(x t) m dt, ad the assumptios of the theorem are satisfied for = m +. The, by itegratio by parts, x x f (m+) (t)(x t) m dt = [ f (m+) (t) ] x (x t)m+ m + = f (m+) (x ) (x x ) m+ m + 4 x + x x + x f (m+) (t) x f (m+) (t) (x t)m+ dt m + (x t)m+ dt. m +

Thus, f(x) = m+ k= f (k) (x ) (x x ) k + k! Hece, by iductio, we obtai the required formula., x f (m+) (t)(x t) m+ dt. (m + )! x Corollary 78. Suppose f is + times differetiable i a ope iterval (a, b) ad f (+) is cotiuous i [a, b]. Let x (a, b). If M > is such that the for every x [a, b], f(x) f (+) (t) M t [a, b], k= f (k) (x ) k! (x x ) k (x t) + M. ( + )! I particular, if (M ) is bouded, the f (k) (x ) k= (x x k! ) k coverges ad f(x) = k= f (k) (x ) (x x ) k. k! 4

6 Fourier Series The material for this sectio is take from the book Calculus of Oe Variable 6. Trigoometric series ad Trigoometric polyomials Defiitio 79. Let (a ) ad (b ) be sequeces of real umbers. The a series of the form c + (a cos x + b si x) is called a trigoometric series. = The trigoometric series c + = (a cos x + b si x) is also writte as c + a cos x + b si x + a cos x + b si x +. As i the case of power series, if there exists a k N {} such that a = ad b = for all k, the the resultig series ca be represeted as k c + (a cos x + b si x). = Defiitio 8. A fuctio of the form k (a cos x + b si x). = where a, b R are called a trigoometric polyomials. We observe that trigoometric polyomials are π-periodic o R, i.e., if f(x) is a trigoometric polyomial, the f(x + π) = f(x) x R. From this, we ca ifer that, if the trigoometric series c + (a cos x + b si x) = coverges at a poit x R, the it has to coverge at x+π as well; ad hece at x+π for all itegers. This shows that we ca restrict the discussio of covergece of a trigoometric series to a iterval of legth π. I particular, if the series c + = (a cos x + b si x) coverges o a iterval I, the we have a π-periodic fuctio f defied by f(x) = c + (a cos x + b si x) = for x i the set {x + π : x I, Z}. Hece, we caot expect to have a trigoometric series expasio for a fuctio f : R R if it is ot a π-periodic fuctio. Defiitio 8. A fuctio f : R R is said to have period T for some T > if f(x+t ) = f(x) for all x R. A fuctio f : R R with period T is called a T -periodic fuctio. M.T. Nair, Calculus of Oe Variable, Ae Publishers, 4. 4

6. Fourier series of π-periodic fuctios We kow that a coverget trigoometric series is π-periodic. What about the coverse. That is, suppose that f is a π-periodic fuctio. Is it possible to represet f as a trigoometric series? Suppose, for a momet, that we ca write f(x) = c + (a cos x + b si x) = for all x R. The what should be the coefficiets c, a, b? To aswer this questio, let us further assume that f is itegrable o [ π, π] ad the series ca be itegrated term by term. For istace if the above series is uiformly coverget to f i [ π, π], the term by term itegratio is possible. By Weierstrass test, we have the followig result: If = ( a + b ) coverges, the c + = (a cos x + b si x) is a domiated series o R ad hece it is uiformly coverget. For, m N {}, we observe the followig orthogoality relatios: π, if m cos x cos mxdx = π, if = m, π π π π π si x si mxdx = cos x si mxdx =. π, if = m =, {, if m π, if = m, Thus, uder the assumptio that f is itegrable o [ π, π] ad the series ca be itegrated term by term, we obtai c = π f(x)dx, π π π a = f(x) cos xdx, b = f(x) si xdx. π π π π Defiitio 8. The Fourier series of a π-periodic fuctio f is the trigoometric series where a = π π a π + (a cos x + b si x), = π f(x) cos xdx ad b = π π π f(x) a + (a cos x + b si x). The umbers a ad b are called the Fourier coefficiets of f. = See M.T. Nair, Calculus of Oe Variable, Ae Books, 4. See M.T. Nair, Calculus of Oe Variable, Ae Books, 4. f(x) si xdx ad this fact is writte as 43

If f is a trigoometric polyomial, the its Fourier series is itself. The followig two theorems show that there is a large class of fuctios which ca be represeted by their Fourier series. Iterested readers may look for their proofs i books o Fourier series; for example Bhatia 3. Theorem 83. Suppose f is a mootoic fuctio o [ π, π]. The the Fourier series of f coverges, ad the limit fuctio f(x) is give by { f(x) if f is cotiuous at x, f(x) = [f(x ) + f(x+)] if f is ot cotiuous at x. Theorem 84. (Dirichlet s theorem) Suppose f : R R is a π-periodic fuctio which is piecewie cotiuous, piecewise differetiable ad f is piece wise cotiuous o [ π, π]. The the Fourier series of f coverges, ad the limit fuctio f(x) is give by { f(x) if f is cotiuous at x, f(x) = [f(x ) + f(x+)] if f is ot cotiuous at x. I the above theorem we used the terms, piecewie cotiuous ad piecewise differetiable. They are i the followig sese: Defiitio 85. Let f : [a, b] R is said to be. piecewie cotiuous if there exist a fiite umber of poits x, x,..., x i [a, b] with a = x < x < < x = b such that f is cotiuous i each ope iterval (x i, x i ) ad the limits f(x i +) := lim h + f(x i + h) ad f(x i ) := lim h + f(x i h) exist;. piecewie differetiable if there exist a fiite umber of poits x, x,..., x i [a, b] with a = x < x < < x = b such that f is differetiale i each ope iterval (x i, x i ) ad the limits f f(x i + h) f(x i +) (x i +) := lim h + h ad f f(x i ) f(x i h) (x i ) := lim h + h exist. {, x, Example 86. Let f(x) = I this case, take we take the poits x, < x. x =, x =, x =. We see that f is differetiable i the ope itervals (, ), (, ), ad f( +) =, f( ) =, f(+) =, f( ) =, f ( +) =, f ( ) =, f (+) =, f( ) =. 3 R. Bhatia, Forrier Series, Trim Series, 44

Ideed, f ( +) = f( + h) f( +) lim h + h f( ) f( h) f ( ) = lim h + f (+) = lim h + f ( ) = lim h + h f( + h) f(+) h f( ) f( h) h = lim h + h = lim h + = lim h + = lim h + =, =, h h =, h [ ( h)] h =. Remark 87. It is kow that there are cotiuous fuctios f defied o [ π, π] whose Fourier series does ot coverge poitwise, eve o a dese subset of [ π, π]. Its proof relies o cocepts from advaced mathematics. Recall that, i the case of a power series, the partial sums are polyomials ad the limit fuctio is ifiitely differetiable. I the case of a Fourier series, the partial sums are ifiitely differetiable, but, the limit fuctio, if exists, eed ot be eve cotiuous at certai poits. This fact is best illustrated by the followig example. {, π x, Example 88. Let f(x) = Note that this fuctio satisfies the coditios i Dirichlet s theorem (Theorem 84). Hece, its Fourier series coverges to f(x) for, < x π. every x, ad at the poit, the series coverges to /. Note that a = π {, =, cos xdx = π,, ad for N, Thus, b = π π Thus, Fourier series of f is ad by Theorem 84, I particular, for x = π/, = + π si xdx = [ ] cos π = [ ] ( ). π π b = + π f(x) = + π = π, odd,, eve. = = si[( + )π/] ( + ) si( + )x, ( + ) si( + )x, x ( + ) 45 = + π = ( ) ( + )

which leads to the Madhava Nilakatha series 6.3 Fourier series for eve ad odd fuctios π 4 = ( ) ( + ). = Recall that a fuctio f : R R is called a (i) eve fuctio, if f( x) = f(x) for all x R, (ii) odd fuctio, if f( x) = f(x) for all x R. Cosider the cases (i) ad (ii) separately: Case(i): Suppose f is a eve fuctio. I this case, f(x) cos x is a eve fuctio ad f(x) si x is a odd fuctio. Hece b = for all N. Thus: If f is a eve fuctio, the the Fourier series of f is a + a cos x with a := π f(x) cos xdx. π = Note that at the poits x = ad x = π, the above series takes the forms respectively. a + a ad a + ( ) a, = Case(ii): Suppose f is a odd fuctio. I this case, f(x) si x is a eve fuctio. Hece a = for all N {}. Thus: = If f is a odd fuctio, the the Fourier series of f is b si x with b := π f(x) si xdx. π = At the poit x = π/, the series takes the form ( ) b +. Let us illustrate the above observatios by some examples. Example 89. Cosider the fuctio f defied by = f(x) = x, x [ π, π]. 46