ENGI 344 5 Series Page 5-01 5. Series Cotets: 5.01 Sequeces; geeral term, limits, covergece 5.0 Series; summatio otatio, covergece, divergece test 5.03 Series; telescopig series, geometric series, p-series 5.04 Tests for Covergece: compariso ad limit compariso tests 5.05 Tests for Covergece: alteratig series; absolute ad coditioal covergece 5.06 Tests for Covergece: ratio ad root tests 5.07 Power Series, radius ad iterval of covergece 5.08 Taylor ad Maclauri Series, remaider term 5.09 Biomial Series 5.10 Itroductio to Fourier Series 5.11 Itroductio to Series Solutios of ODEs Appedix: 5.A Itegral Test [ot examiable; for referece oly]
ENGI 344 5.01 Sequeces Page 5-0 5.01 Sequeces; geeral term, limits, covergece A sequece is a set of related objects that all follow the same rule. progressio from curret ad previous elemets to the ext elemet. There is a logical Example 5.01.1 The alphabet: { a, b, c,..., y, z } The rule is: The th elemet a (the th letter of the alphabet) Example 5.01. The set of all atural umbers = { 1,, 3,... } Two ways to express the rule are: or the direct defiitio We are usually iterested i two features: A explicit form for the geeral term (ofte from a recurrece relatioship); The behaviour of the terms of a sequece as. Example 5.01.3 The terms of a sequece form a arithmetic progressio if cosecutive terms differ by a commo o-zero costat differece d : a a d a a d, with a iitial term a a 1 1 1 Writig dow the first few terms of the sequece, The geeral term is therefore a
ENGI 344 5.01 Sequeces Page 5-03 Example 5.01.4 The terms of a sequece form a geometric progressio if cosecutive terms chage by a commo costat ratio r : a a 1 1 r a a r r 0, r 1, with a iitial term a1 a 0 Writig dow the first few terms of the sequece, The geeral term is therefore a The sequece alterates i sig if r 0 The sequece coverges to a limit of zero if If r 1 the lim a
ENGI 344 5.01 Sequeces Page 5-04 The power sequece r 1 r, r, 3 r, 4 r, coverges to 0 for r < 0, coverges to 1 for r = 0 ad diverges for r > 0.
ENGI 344 5.01 Sequeces Page 5-05 Example 5.01.5 Fid the geeral term of this sequece ad determie whether it coverges. 1 4 7 10 13,,,,, 5 9 13 17 1 Algebra of Sequeces: Give two coverget sequeces a with limit value A ad / f x is a cotiuous fuctio with lim f x L ad if pa b with limit value B, qb is a coverget sequece with limit value (pa + qb) for all costats p, q; ab is a coverget sequece with limit value AB; a b is a coverget sequece with limit value A/B (provided that B 0 ); If x a f the a coverges with limit value L ; If a c b ad if c coverges with limit value C the A C B.
ENGI 344 5.01 Sequeces Page 5-06 Example 5.01.6 Fid the limit of the sequece si. The Race to Ifiity There is a hierarchy of fuctioal forms that all diverge to ifiity with icreasig : c! a a 1 c 0 log b 1 for all sufficietly large. b Example 5.01.7! lim 0 because goes to ifiity faster tha! OR the sequece is positive ad! 1 31 1 1 1 Therefore all terms are bouded above by the terms of a sequece that coverges to 0.
ENGI 344 5.01 Sequeces Page 5-07 If every term of a sequece is o less tha all of its predecessors a1 a a3 a4, the the sequece is a icreasig sequece. If every term of a sequece is o greater tha all of its predecessors a1 a a3, the the sequece is a decreasig sequece. A sequece is mootoic if ad oly if it is either icreasig or decreasig. If a U for some costat value U, the the sequece for some costat value L, the the sequece If a L A sequece is bouded if it is both upper ad lower bouded. a is upper-bouded. a is lower-bouded. Every upper-bouded icreasig sequece is coverget. Every lower-bouded decreasig sequece is coverget. Every bouded mootoic sequece is coverget.
ENGI 344 5.0 Series Page 5-08 5.0 Series The partial sum S of a sequece is the sum of the first terms of that sequece: S a a a a a 1 3 k k 1 This partial sum is also a fiite series. The sum of a ifiite series is the sum of all terms i the associated ifiite sequece, S a 1 a a a 3 k lim S. k 1 Example 5.0.1 1 ad fid the limit of the sequece of partial sums. Fid the first three partial sums of the sequece, 1,, 3,
ENGI 344 5.0 Series Page 5-09 Divergece Test If the geeral term of a series does ot coverge to zero, the the series diverges. lim a 0 a diverges (the sum does ot exist) 1 Example 5.0. Prove that all arithmetic series 1 1 a d diverge, (where d 0 ). However, the coverse of the divergece test is false. Some diverget series do have geeral terms that coverge to zero, or equivaletly: lim a 0 does ot guaratee that (except for alteratig series, sectio 5.05 below). a coverges 1
ENGI 344 5.0 Series Page 5-10 Example 5.0.3 Show that the harmoic series 1 diverges. 1 The divergece of the harmoic series is very slow. 4 terms are eeded for the partial sum to exceed. 31 terms are eeded for the partial sum to exceed 4. 1674 terms are eeded for the partial sum to exceed 8. The sum of the first billio (10 9 ) terms is still well uder 5. Yet the sum of the complete series does ot exist (diverges to ifiity)!
ENGI 344 5.03 Series Page 5-11 5.03 Series; telescopig series, geometric series, p-series Example 5.03.1 1 Fid the exact sum of the series S. 56 1 This is a example of a telescopig series. The simplest such series have geeral terms similar to a f f 1 Such series ofte ivolve the use of partial fractios..
ENGI 344 5.03 Series Page 5-1 Example 5.03. Fid the exact sum of the series S 8. 3 3 3
ENGI 344 5.03 Series Page 5-13 Example 5.03. (cotiued) S Collectig the survivig terms from this telescopig series, we fid that
ENGI 344 5.03 Series Page 5-14 Geometric Series Each term i a geometric series is a costat multiple r (the commo ratio) of the immediately precedig term, except for the [o-zero] first term a. Quotig the geeral term from the geometric progressio o page 5.03, the geometric series is Applyig the divergece test: 1 or 1 0 S ar ar
ENGI 344 5.03 Series Page 5-15 Example 5.03.3 (Example 5.0.1 revisited) Evaluate S 1. 1 Example 5.03.4 Fid the sum of the power series f x x x x x 3 4
ENGI 344 5.03 Series Page 5-16 p-series Aother importat family of stadard series is the p-series (also kow as hyperharmoic series): 1 1 The divergece test establishes divergece for p 0. It ca be show [sectio 5.A] that p-series coverge for all p 1 ad diverge for all p 1. The case p 1 is the harmoic series, for which Example 5.0.3 established divergece. p I 1741 Leohard Euler proved that 1. 1 6 The removal or isertio of a fiite umber of fiite terms does ot affect the covergece of a series. If S a coverges, the so does a a1 a a a1 a S 3 provided that a 1 ad a are fiite., 1 3
ENGI 344 5.04 Compariso Tests Page 5-17 5.04 Tests for Covergece: compariso ad limit compariso tests If all terms a i a series are positive, the the sequece of partial sums S is ecessarily icreasig. If aother series b is coverget with a sum B ad if a b, the 1 B is a upper boud to S Recall that a icreasig sequece that has a upper boud is coverget [page 5.07] S must coverge to a value B a coverges to some value A B. 1 This is the basis of the compariso test. Example 5.04.1 1 Is the series coverget? 1 56 [This is Example 5.03.1, for which the exact value of the sum was foud by telescopig the series.]
ENGI 344 5.04 Compariso Tests Page 5-18 Example 5.04. Is the series 1 l coverget? Put very briefly, the essece of the compariso test is: A coverget ceilig forces covergece: 0 a b ad b coverges a coverges. A diverget floor forces divergece: a b 0 ad b diverges a diverges.
ENGI 344 5.04 Compariso Tests Page 5-19 Limit Compariso Test All terms a ad b must be o-egative ad L is some fiite costat. a lim L 0 ad b coverges a coverges. b a lim 0 ad b diverges a diverges. b The referece series b 1 is ofte the geometric series or a p-series. Example 5.04.3 l Is the series coverget? 3 8 3
ENGI 344 5.04 Compariso Tests Page 5-0 Example 5.04.4 4 Is the series coverget? 3 9 1
ENGI 344 5.05 Alteratig Series Test Page 5-1 5.05 Tests for Covergece: alteratig series; absolute ad coditioal covergece Alteratig Series Test If cosecutive terms of a series alterate i sig, the a simpler test for covergece is available. a If 1 0 a ad lim a 0 the a coverges. 1 Example 5.05.1 Is the alteratig harmoic series 1 1 1 1 1 a 1 1 3 4 coverget? 1 1 Note that if we take the absolute values of the terms i the alteratig harmoic series, 1 1 1 the we obtai the harmoic series 1, which diverges. 3 4
ENGI 344 5.05 Alteratig Series Test Page 5- Absolute Covergece If ad oly if the series the the series 1 a If ad oly if the series the the series 1 a a coverges, 1 is absolutely coverget. a coverges but the series a diverges, 1 1 is coditioally coverget. The alteratig harmoic series is a example of a coditioally coverget series. If a series is absolutely coverget, the the alteratig series test is a waste of time. Aother test to establish the covergece of 1 a is absolutely coverget i oe step. 1 a automatically establishes that If a series is coditioally coverget, the the alteratig series test is ofte required to establish covergece of a, but aother test is required o a to complete the 1 1 proof of coditioal covergece. Therefore test the covergece of a first! 1 Absolute covergece allows the algebra of fiite series to be exteded to ifiite series. I particular, the order of a ifiite umber of terms may be chaged without affectig the sum of the series. The derivative of a absolutely coverget series is the sum of the derivatives of the idividual terms. This is o loger ecessarily true for coditioally coverget series. For example, 1 1 1 1 1 1 1 1 1 1 1 3 1 l but 1 l. 3 4 5 6 7 3 5 7 4
ENGI 344 5.05 Alteratig Series Test Page 5-3 Example 5.05. Ivestigate the covergece of the alteratig p-series 1 1 1 1 1 a 1 p 1 1 1 p 3 p 4 p This is obviously a alteratig series. Example 5.05.3 Is 1 1 1 1 1 1 1 absolutely coverget, coditioally 3 4 coverget or diverget?
ENGI 344 5.05 Alteratig Series Test Page 5-4 Approximatio Errors For ay coverget alteratig series, the error i approximatig the sum S of the etire series caused by evaluatig the partial sum S of just the first terms is S S a 1 Example 5.05.4 Estimate S 0 1 correct to four decimal places.! The sigs of the terms of this series are clearly alteratig ad this series coverges (by the alteratig series test). 1 lim 0! The error caused by usig S to estimate S is
ENGI 344 5.06 Ratio ad Root Tests Page 5-5 5.06 Tests for Covergece: ratio ad root tests Amog the most useful tests for series covergece is the ratio test: a If lim 1 1 the a a If lim 1 1 the a 1 1 a a is absolutely coverget. is diverget. a If lim 1 1 the the test fails ad there is o iformatio o series covergece. a The ratio test is most useful whe the geeral term icludes a expoetial or factorial factor. The test fails whe the geeral term is a algebraic fuctio (terms of the form costat i the umerator ad/or deomiator) oly. Example 5.06.1 Is the series diverget? a 1 1! absolutely coverget, coditioally coverget or
ENGI 344 5.06 Ratio ad Root Tests Page 5-6 Note: Proof that 1 lim 1 e :
ENGI 344 5.06 Ratio ad Root Tests Page 5-7 Root Test: If lim a 1 the If lim a 1 the 1 1 a a is absolutely coverget. is diverget. If lim a 1 the the root test fails (provides o iformatio). Like the ratio test, the root test fails for purely algebraic fuctios. The root test fails wheever the ratio test fails. The root test works best for geeral terms of the form a f. Usually the ratio test ca be applied more easily wheever the root test provides a result. Example 5.06. Is the series diverget? x a l absolutely coverget, coditioally coverget or
ENGI 344 5.07 Power Series Page 5-8 5.07 Power series, radius ad iterval of covergece A power series with cetre x c is of the form bd d b b 3b f x a x c x c a a x c a x c a x c 0 where b ad d are real costats ad b 0. 0 1 3 A commo choice of parameters for power series is b 1 ad d 0 : 0 0 1 3 3 f x a x c a a x c a x c a x c u ow represet the geeral term b d Let a x c ad let us apply the ratio test for covergece of a power series: b b d 1 1 1 bd a x c a u a x c a x c u u lim 1 b x c lim 1 u a The series is absolutely coverget wheever a b where R is the radius of covergece The iterval of covergece I icludes c R x c R ad may iclude oe or both edpoits, which must be tested separately. I is also the domai of the fuctio f x. f c a. Also ote that 0 All power series are absolutely coverget at the cetre, eve if 0 R. If R, the the power series is absolutely coverget for all x.
ENGI 344 5.07 Power Series Page 5-9 Example 5.07.1 Determie the radius of covergece ad iterval of covergece for the power series f x 0 x 3
ENGI 344 5.07 Power Series Page 5-30 Example 5.07. Determie the radius of covergece ad iterval of covergece for the power series x 1 f x 3 1
ENGI 344 5.08 Taylor ad Maclauri series Page 5-31 5.08 Taylor ad Maclauri Series; Remaider Term A fuctio f x that is represeted by a power series (usig oly o-egative iteger powers of x c ) is a Taylor series: f x a x c 0 0 0 f c a 0 0 a f c Assumig that the series exists ad is absolutely coverget, the f x a x c 0 1 f c 0 1a 0 0 a 1 1 f c 1 1 f x a x c 0 f c 0 0 1a 0 a f c 1 1 f x a x c 0 3 f c f c 0 0 0 31a3 a3 ad so o, so that the geeral term is a 0 3 1 f c ad the Taylor series for f! c f f x x c! The ratio test ca determie the radius of covergece R. x is
ENGI 344 5.08 Taylor ad Maclauri series Page 5-3 Example 5.08.1 Fid the Taylor series for f x l x about x 1.
ENGI 344 5.08 Taylor ad Maclauri series Page 5-33 Example 5.08. Prove L Hôpital s rule for the case of a (0/0) idetermiacy. Let fuctios f (x) ad g (x) be such that f (a) = g (a) = 0. f a The is a 0 type of idetermiacy. g a 0 Represet both fuctios by their Taylor series about x a, the
ENGI 344 5.08 Taylor ad Maclauri series Page 5-34 Maclauri Series A Taylor series with a cetre at xc 0 is a Maclauri series. Example 5.08.3 x Fid the Maclauri series for f x e ad fid its iterval of covergece.
ENGI 344 5.08 Taylor ad Maclauri series Page 5-35 Example 5.08.4 Fid the Maclauri series for f x cos x ad fid its iterval of covergece.
ENGI 344 5.08 Taylor ad Maclauri series Page 5-36 Example 5.08.4 (Alterative Solutio) Use the Euler expressio for the cosie fuctio i terms of the expoetial fuctio: j e cos jsi ad
ENGI 344 5.08 Taylor ad Maclauri series Page 5-37 Example 5.08.5 Use the Maclauri series for the cosie fuctio to fid the Maclauri series for the sie fuctio. The Maclauri series for f x si x may also be foud directly, through repeated differetiatios, or via the Euler equatio ad the Maclauri series for the expoetial fuctio (as i Example 5.08.4 for the cosie fuctio) or by differetiatio of cos x.
ENGI 344 5.08 Taylor ad Maclauri series Page 5-38 Remaider Term The practical use of Taylor (ad Maclauri) series arises whe the partial sum of the first few terms of the series T x is used to approximate the value of a o-liear fuctio. The error caused by trucatig the series at the th term is the th remaider term R x : 1 1! f R x f x T x x c where is some umber betwee x ad c. 1 The Taylor series is well-defied oly if lim R x 0 c R x c R. Example 5.08.6 x The Maclauri series for the expoetial fuctio f x e is e x 3 x x x 1 x k!! 3! k 0 k 1 1 e x f R x x 0 0 x 1! 1! Because factorial fuctios diverge to ifiity faster tha expoetial fuctios, 1 x lim R x e lim 0 1! ad the Maclauri series is therefore well-defied. x The error caused by approximatig f x e by the quadratic 3 a 1 x T x 1 x o! ae the iterval 0 x a is at most. 6 For a 0.1, this maximum error is less tha 10 4, with a maximum relative error of uder 0.03%. However, for a 10, the maximum error is substatial: more tha 3 10 6. May more terms eed to be take to maitai accuracy, the further away from the cetre oe goes. Note that is the equatio of the taget lie to y f x y T x 1 at x c.
ENGI 344 5.09 Biomial Series Page 5-39 5.09 Biomial Series A special case of Maclauri series arises for f x 1 x.
ENGI 344 5.09 Biomial Series Page 5-40 Summary: x 1 1 k 1 1 1 k k! x k Iterval of covergece: I for 0 or [ the series termiates at k = ] I 1, 1 for 0 I 1, 1 for 1 0 I 1, 1 for 1 ad ot a iteger, with absolute covergece at x 1. with coditioal covergece at x 1. with divergece at x 1.
ENGI 344 5.09 Biomial Series Page 5-41 Example 5.09.1 Fid the Maclauri series for f x 1 x ad its iterval of covergece. I the biomial expasio, for "x" read "x" ad for "" read "½".
ENGI 344 5.10 Fourier Series Page 5-4 5.10 Fourier Series This sectio offers a very brief overview of [discrete] Fourier series. Some majors will explore Fourier series ad trasforms i much greater detail i subsequet semesters. The Fourier series of f x o the iterval LL, is where ad a0 x x f x acos bsi 1 L L L 1 x a f x dx L L L L cos, 0,1,, 3, 1 x b f x dx L L L si, 1,, 3, The a, b are the Fourier coefficiets of f x. Note that the cosie fuctios (ad the fuctio 1) are eve, while the sie fuctios are odd. If f x f x for all x), the b 0 f x is eve ( for all, leavig a Fourier cosie series (ad perhaps a costat term) oly for f x. If f x is odd ( f x f x for all x), the a 0 for all, leavig a Fourier sie series oly for f x. Note that a a g x dx 0 a 0 if g x is odd g x dx if g x is eve with the multiplicatio table odd fuctio eve fuctio odd fuctio eve odd eve fuctio odd eve
ENGI 344 5.10 Fourier Series Page 5-43 Example 5.10.1 Expad f x 0 x 0 x 0 x i a Fourier series.
ENGI 344 5.10 Fourier Series Page 5-44 Example 5.10.1 (Additioal Notes also see "www.egr.mu.ca/~ggeorge/344/demos/") The first few partial sums i the Fourier series 1 1 1 f x cos x si x x 4 1 are S0 4 S 1 cos si 4 x x S 1 cos si si 4 x x x S 1 1 3 cos si si cos3 si 3 4 x x x 9 x 3 x ad so o. The graphs of successive partial sums approach f (x) more closely, except i the viciity of ay discotiuities, (where a systematic overshoot occurs, the Gibbs pheomeo).
ENGI 344 5.10 Fourier Series Page 5-45 Example 5.10. Fid the Fourier series expasio for the stadard square wave, 1 1 x 0 f x 1 0 x 1 The graphs of the third ad ith partial sums (cotaiig two ad five o-zero terms respectively) are displayed here, together with the exact form for f (x), with a periodic extesio beyod the iterval ( 1, +1) that is appropriate for the square wave. y S3 x
ENGI 344 5.10 Fourier Series Page 5-46 Example 5.10. (cotiued) y S9 x Covergece At all poits limits lim x x o x x i ( L, L) where f x is cotiuous ad is either differetiable or the f o x ad lim f x both exist, the Fourier series coverges to x x o f x. At fiite discotiuities, (where the limits lim x x o f x ad lim f x x x o both exist), the f xo f xo Fourier series coverges to, f x lim f x ad f x lim f x ). (usig the abbreviatios o o xx o xx o f (x) ot cotiuous cotiuous but cotiuous ad at x x ot differetiable differetiable I all cases, the Fourier series at x x coverges to o o f x f x o o (the red dot).
ENGI 344 5.10 Fourier Series Page 5-47 Half-Rage Fourier Series A Fourier series for domai to LL,. f x, valid o 0, L, may be costructed by extesio of the A odd extesio leads to a Fourier sie series: where L 0 x f x b si L 1 x b f x dx L L si, 1,, 3, A eve extesio leads to a Fourier cosie series: where a0 x f x a cos L L 0 1 x a f x dx L L cos, 0,1,, 3, ad there is automatic cotiuity of the Fourier cosie series at x 0 ad at x L.
ENGI 344 5.10 Fourier Series Page 5-48 Example 5.10.3 Fid the Fourier sie series ad the Fourier cosie series for f x x o 0,1. f x x happes to be a odd fuctio of x for ay domai cetred o x 0. The odd extesio of f x to the iterval 1,1 is f Evaluatig the Fourier sie coefficiets, x itself. This fuctio happes to be cotiuous ad differetiable at x 0, but is clearly discotiuous at the edpoits of the iterval x 1. Fifth order partial sum of the Fourier sie series for f x x o [0, 1]
ENGI 344 5.10 Fourier Series Page 5-49 Example 5.10.3 (cotiued) The eve extesio of f x to the iterval [ 1, 1] is f x x. Evaluatig the Fourier cosie coefficiets,
ENGI 344 5.10 Fourier Series Page 5-50 Example 5.10.3 (cotiued) Third order partial sum of the Fourier cosie series for f x x o 0,1 Note how rapid the covergece is for the cosie series compared to the sie series. for cosie series ad y S x for sie series for f x x o 0,1 y S x 3 5
ENGI 344 5.11 Series Solutios of ODEs Page 5-51 5.11 Itroductio to Series Solutios of ODEs If the fuctios px, qx ad r x i the ODE d y dy p x qx y r x dx dx are all aalytic i some iterval xo h x xo h (ad possess Taylor series expasios aroud x o with radii of covergece of at least h), the a series solutio to the ODE aroud x o with a radius of covergece of at least h exists: 0 y x y x a x xo, a! o Example 5.11.1 Fid a series solutio as far as the term i x 3, to the iitial value problem d y dy x x e y 4 ; y 0 1, y0 4 dx dx Noe of our previous methods apply to this problem. The fuctios x, e x ad 4 are all aalytic everywhere. The solutio of this ODE, expressed as a power series, is
ENGI 344 5.11 Series Solutios of ODEs Page 5-5 Example 5.11. Fid the geeral solutio (as a power series about x 0 ) to the ordiary differetial equatio d y xy 0 dx Let the geeral solutio be The yx a 0 x. 1 1 0 0 1 1 0 0 1 0 y x a x a x ad y x a x a x Substitute ito the ODE: ax x ax ax ax 0 0 1 0 1 0 We wat the expoet of x to be the same i both summatios, so that we ca brig the two summatios together. 0 1 1 1 3 43 1 a x a x a x a x a x 3 4 0 ad 3 4 0 1 0 1 ax ax 0 0 a x a x a x a x a x Brig the two summatios together for all terms from = owards: 0 1 3 1 a x 3 a x 1 a a x 0 But this equatio must be true regardless of the choice of x. Therefore the coefficiet of each power of x must be zero. a a a3 0 ad a, 3, 4, 1
ENGI 344 5.11 Series Solutios of ODEs Page 5-53 Example 5.11. (cotiued) ad a y ad a y 0 0 are arbitrary. 1 0 Therefore the geeral solutio is a0 a1 y x a0 a1x 0x 0x x x 1 0 0x 0x a0 8 a1 9 10 11 x x 0x 0x 67 1440 or 3 4 5 6 7 01 1 4 1 8 1 1 5 1 9 y x a x x a x x x 1 67 0 1440 The arbitrary costats ca be determied from iitial coditios: a y 0 ad a y 0 0 1 The Maple commads > with(detools): > ode := diff(y(x),x,x) + x*x*y(x) = 0; > Order := 15; > dsolve(ode, y(x), series); produce the output 1 4 1 5 1 y x y 0 D y 0 x y 0 x D y 0 x y 0 x 8 1 0 67 1 1 1 D y 0 x y 0 x D y 0 x O x 9 1 13 15 1440 88704 4640 Follow the lik for this example o the course web site, at "http://www.egr.mu.ca/~ggeorge/344/demos/idex.html".
ENGI 344 5.11 Series Solutios of ODEs Page 5-54 The two pricipal methods of series solutio of ODEs of the form d y dy p x qx y r x dx dx usig a Maclauri series are y 0 y 0 3 yx y0 y0 x x x! 3! 1) use the ODE to fid y 0 r 0 p0 y0 q0 y0 ad differetiate y x r x px yx qx yx y to fid the higher derivatives y x ad hece 0 a (as i Example 5.11.1) ad successively! ) Substitute y x the values of the 0 a x ito the ODE ad match coefficiets of a (as i Example 5.11.). x to fid Method is ot advised for Example 5.11.1, as that would ivolve the product of two ifiite series, x x e y ax! 0 0 Method 1 could be used for Example 5.11., as show o the ext two pages.
ENGI 344 5.11 Series Solutios of ODEs Page 5-55 Example 5.11. Alterative Solutio to d y xy 0 dx y0 1 y 0 y 0 3 yx y 0 x x x 1!! 3! Let A y0 ad B y0 y x x y x y 0 0 y x xy x x y x y 0 0 4 4 IV 4 0 IV y x y x xy x xy x x y x y x xy x x x y x IV y x x y x xy x y A 3 4 4 4 4 V V y x x y x x y x y x xy x 4x 3 y x x 4 6 y x 4x x y x 8x 3 y x x 4 6 y x y 0 6B 3 3 4 4 8 4 6 x yx x 3 yx x 4 x yx x x 6 yx x 3 yx VI y x x y x x y x x y x x y x 4 1 6 30 1 VI y 0 0 5 6 3 60 6 30 36 1 5 6 3 5 6 60x 6x yx 66x x yx 1x x yx 60x 18x yx 66x x yx VII y x x x y x x x y x x y x x y x VII y 0 0 4 5 5 6 60 90 60 18 13 6 66 VIII y x x y x x x y x x x y x x x y x 4 8 5 VIII 60 156x x y x 19x 4x y x y 0 60A IX y x 3 7 4 8 4 5 64x 8x yx 60 156x x yx 19 10x yx 19x 4x x y x 3 7 4 8 IX 816x 3x yx 5 76x x yx y 0 5B etc. 3 A 4 6B 5 6 7 60A 8 5B 9 yx A Bx 0x 0x x x 0x 0x x x 4! 5! 8! 9! 1 1 1 1 1 1 67 0 1440 4 8 5 9 yx A x x B x x x
ENGI 344 5.11 Series Solutios of ODEs Page 5-56 Example 5.11. Alterative Solutio (cotiued) However, the above does ot allow us to deduce a geeral patter for a. We require a geeralized form of the product rule of differetiatio for higher order derivatives. The product rule of differetiatio is uv uv uv du dv where u, v dx dx Extedig the product rule to higher orders of differetiatio, uv uv uv uv uv uv uv uv uv uv uv uv uv uv uv u v uv u v uv uv uv u v 3u v 3uv uv k d u d v The coefficiets of are clearly the etries k k dx dx k 0,1,,, i row of Pascal s triagle, which! are also the biomial coefficiets Ck k! k! It is straightforward to verify that 4 d 4 4 4 uv u v 4u v 6u v 4uv uv dx The geeral form for ay atural umber is 1 k k k uv Ck k k k 0 d d u d v dx dx dx which is also 1 1 d d u d u dv d u d v uv v dx dx 1 dx 1 dx dx dx 3 3 d u d v d v u 3 1 3 3 dx dx dx
ENGI 344 5.11 Series Solutios of ODEs Page 5-57 Example 5.11. Alterative Solutio (cotiued) Startig over: y x x y x k k d d d d y x y x x y x C x y x dx dx dx dx k k k k 0 3 4 d d d d d C0 x y x C1 x y 3 x C x y 4 x dx dx dx dx dx 3 3 4 x y x x y x y x 0 0 Therefore, for 4, 3 4 y x x y x x y x 3 y x 4 y 0 0 0 3 y 0 4 0 3 4 y y 0 0 a4 4! 3 3 4! 4!! 1 3 4! y a But!! a a a 4 4 1 a a 1 4, which is the recurrece relatio for 4 1 1 1 a,, 4 3 a a 5 4 a a 65 a We also have a0 A, a1 B, a a3 0 The same coclusio follows, 4 0 5 1 6, etc. 1 4 1 8 1 1 y x A1 x x x 4 3 8 7 4 3 1 11 8 7 4 3 1 5 1 9 1 13 Bx x x x 5 4 985 4 131985 4
ENGI 344 5.11 Series Solutios of ODEs Page 5-58 Example 5.11.3 Fid the complete solutio (as a power series about x = 0) to the iitial value problem 1 d y dy x 5x 3y 0, y 0 0, y0 1 dx dx Let the geeral solutio be yx a 0 x. 1 1. The ad 1 y x a x y x a x Substitute ito the ODE: 1 1 a x ax ax ax 1 0 1 1 5 3 0 We prefer all summatios to ivolve the same expoet x i the geeral term. Substitute m = (ad thus = m + ) ito the first term: 1 1 m a x m m a x m0 Start each summatio from = : m 0 1 m m 1 amx 1 ax 3 a3x 1 ax, m0 1 5 ax 51a1x ax 5a1x 5 ax 1 ad 3 3 0 1 ax a0x a1x ax 3 a0 3 a1x 3 ax 0 ad combie the summatios for all from owards: 1 a 1 a 5a 3a x 3 1 0 1 m a 6a x 5a x 3a 3a x 0 0 3 1 1 a 5 3 a x a 3a 6a 8a x 0 3 6 8 1 3 1 a a a a x a a x 0 0 3 1 This method ca be difficult to follow. A slower alterative is o the ext page.
ENGI 344 5.11 Series Solutios of ODEs Page 5-59 From before, ax ax ax ax 1 0 1 1 5 3 0 Write out eough terms from each series to spot a patter: 0 1 3 4 1 a x 3 a x 43 a x 54 a x 65 a x 3 4 5 6 3 4 1 ax 3a3x 43a4x 51a 1 3 4 1x ax 3a3 x 4a4x 0 1 3 4 a0x a1x ax a3x a4x 3 0 Collectig the coefficiets of 0 x ad 1 x together, the series begis with Clearly a geeral term ca be costructed for = owards. The geeral term is of the form The series ca be re-writte as 3 6 8 1 3 a a a a x a a x 0 0 3 1
ENGI 344 5.11 Series Solutios of ODEs Page 5-60 Example 5.11.3 (cotiued) Equatig coefficiets of x : x 0 : x 1 : All higher powers of x : The Maple commads > with(detools): > ode := (1-x^)*diff(y(x),x,x) - 5*x*diff(y(x),x) - 3*y(x) = 0; > Order := 11; > dsolve(ode, y(x), series); produce the output 3 4 3 15 4 8 y x y 0 D y 0 x y 0 x D y 0 x y 0 x D y0 x 5 3 8 5 35 6 64 7 315 8 18 9 693 y 0 x D y 0 x y 0 x D y 0 x y 0 x 10 Ox 11 16 35 18 63 56 Follow the lik for this example o the course web site, at "http://www.egr.mu.ca/~ggeorge/344/demos/idex.html".
ENGI 344 5.11 Series Solutios of ODEs Page 5-61 Example 5.11.4 Fid the geeral solutio (as a power series about x 0 ) to the ordiary differetial equatio d y xy dx d y dx d y dx xy xy 0 Let the geeral solutio be yx a 0 x. 1 1. The ad 1 y x a x y x a x Substitute ito the ODE: 1 ax ax 0 1 0 Writig out the two series, term by term: Equatig coefficiets of 0 x : Equatig coefficiets of 1 x : Equatig coefficiets of x : For all 1, equatig coefficiets of x :
ENGI 344 5.11 Series Solutios of ODEs Page 5-6 Example 5.11.4 (cotiued)
ENGI 344 5.A Itegral Test Page 5-63 5.A Itegral Test [for referece oly ot examiable i this course] If f x is a cotiuous fuctio with positive values that are decreasig for all x k,, 1,, ad the improper itegral lim a f k k k exists as a fiite real umber, the the series k k k k a coverges ad k f x dx a a f x dx If the improper itegral does ot have a fiite value, the the series diverges. L L f x dx x k Proof: Examie the area betwee xi 1 ad x i uder the curve y f x : Both rectagles have width oe uit. The smaller rectagle has height f i ad area smaller tha that uder the curve. The larger rectagle has height f i 1 ad area larger tha that uder the curve. Therefore i 1 f i f x dx 1 f i 1 i1 This iequality is true for all areas from i1 k owards. Addig all of these areas together, we fid The left side ca be re-arraged as k 1 f i f x dx f i k i k1 i k1 f x dx f i f k f i f k f x dx i k i k The right side ca be re-arraged as 1 f i f i f x dx f x dx f i k k i k1 i k i k k
ENGI 344 5.A Itegral Test Page 5-64 Also ote that the i th term of the series is a i f i k k k. It the follows that k f x dx a a f x dx The covergece or divergece of the series is therefore liked completely to the covergece or divergece of the improper itegral. The double iequality also allows upper ad lower bouds to be evaluated for the sum of a coverget series whose terms form a sequece draw from a positive, cotiuous ad decreasig fuctio f x. The covergece of the p-series ca be established by a combiatio of the divergece ad itegral tests: p 0 1 lim p 1 p 0 0 p 0 the p-series diverges for p 0. 1 Whe 0 x so that we may use the itegral test. 1 x p p the fuctio f x is positive, cotiuous ad decreasig o 1, dx Therefore the p-series p p1 x lim p 1 1 p 1 p 1 1 p 1 0 p 1 lim l x p 1 1 1 1 p coverges for p 1 ad diverges otherwise. Aother example of the itegral test arises i showig that 1 p coverges if ad oly if 1 l p [ details omitted]. END OF CHAPTER 5 END OF ENGI 344!