We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

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Sydney Small
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1 Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at a = -. We are maily goig to be cocered with power series i x, such as x =x+x +4x L = O... that are cetered at a = O. Importat facts about power series: Covergece: a power series c (x a) is coverget at a specified value of x if its sequece of partial sums {s (x)} coverges - that is, lims N (x)=lim c N (x a) exists. If the limit does ot exist at x, the the series is said to be diverget. Iterval of Covergece: every power series has a iterval of covergece. The iterval of covergece is the set of all real umbers x for which the series coverges. Radius of Covergece: every power has a radius of covergece R ad there are three possibilities: I. The series oly coverges at its ceter, R=O II. The series coverges for all x satisfyig x a < R ad diverges if x a >R III. The series coverges for all x, R= Recall that the absolute value iequality x a < R is equivalet to the simultaeous iequality a - R <x < a +R. A power series might or might ot coverge at the edpoits a - R ad a + R of this iterval. Absolute Covergece: withi its iterval of covergece a power series coverges absolutely. I other words, if x is a umber i the iterval of covergece ad is ot a edpoit of the iterval, the the series of absolute values c (x a) coverges. Ratio Test: Covergece of a power series ca ofte be determied by the ratio test. Suppose that c for all ad that lim c +(x a) + c (x a) = x a lim c + c =L If L<, the series coverges absolutely; if L> the series diverges; ad of L=, the test is icoclusive.

2 (x 3) Ex: Fid the iterval of covergece for = (may times Root Test ca also be applied) A Power Series Defies a Fuctio: A power series defies a fuctio f(x)= c (x a) whose domai is the iterval of covergece of the series. If the radius of covergece is R>O, the f is cotiuous, differetiable, ad itegrable o the iterval (a-r,a+r). Moreover, f (x) f(x)dx ca be foud by term-by-term differetiatio ad itegratio. Covergece at a edpoit may be either lost by differetiatio or gaied through itegratio. If y= c x is a power series i x, the the first two derivatives are y = x ad y = ( )x Notice that the first term i the first derivative ad the first two terms i the secod derivative are zero. We omit these zero terms ad write y = x ad y = ( )x = Idetity Property: If f(x)= c (x a) =R > for all umber x i the iterval of covergece, thec =O for all. Aalytic at a poit: A fuctio is aalytic at a poit a if it ca be represeted by a power series i x-a with positive or ifiite radius of covergece. I calculus it is see that the fuctios such that e x, cosx, six, l(-x), ad so o ca be represeted by the Taylor series. Recall e x - + x + x +..., e x =+ x! +x! +..., six=x x3 3! +x5 5! +...for x < cosx= x! +x4 4! x6 6! +... Taylor series cetered at O called Maclauri Series. Arithmetic of Power Series: Power series ca be combied through the operatios of additio, multiplicatio ad divisio. The procedures for power series are similar to those by which two polyomials are added, multiplied, ad divided - that is, we add

3 coefficiets of like powers of x, use the distributive law ad collect like terms, ad perform log divisio. Ex: e x six= +x+ x! +x3 3! +... x x3 3! +x5 5!... = ()x+()x x3 + 3! + 6 x4 + 5! + 4 x = x+x + x3 3 x Ex: Write ( )c x + c x + as a sigle power series whose geeral term ivolves x k. = Power Series Solutio of a Differetial Equatio We ca see that y=e x is a solutio to dx xy= By replacig x by x i the Maclauri series for e x we ca write the solutio as x y= which coverges everywhere! We ca the fid a way to fid this solutio directly usig a method very similar to the techique of udetermied coefficiets. Ex: Fid a solutio to dx xy= Assume a solutio of the form y= c x exists the y = c x = So dx xy= c x c x + dx = Now add the two series together dx xy= c x + c x c x + = Let k = - ad k = + i each respective sum

4 = c + (k+)c k+ x k c k x k k= k= = c + (k+)c k+ c k+ c k x k = Therefore c = ad (k+)c k+ c k =, k=,,3, This is a recurrece relatio k= c k+ = c k k+ k=,c = c k=,c 3 = 3 c = From this we fid k=3,c 4 = 4 c = c =! c k=4,c 5 = 5 c 3 = k=4,c 6 = 6 c 4 = 3 c = 3! c y= c x =c +c x+c x +c 3 x =c +c x +! c x4 + 3! c x x =c Ex: Fid solutios of 4 y + y = i the form of a power series i x!

5 dx xy= y= c x y = c x = dx xy= c x c x + = dx xy= c x + c x + = Solutio about Ordiary Poits Power Series Solutios Suppose the liear secod order DE a ( x) y + ay + a ( x) y= is put i stadard form y + P( x) y + Q( x) y= Defiitio: Ordiary ad Sigular Poits A poit xis said to be a ordiary poit of the secod order DE if both P(x) ad Q(x) i stadard form are aalytic (ca be writte as a power series i x - a with a positive radius of covergece) at x. A poit that is ot a ordiary poit is said to be a sigular poit of each equatio. Every fiite value of x is a ordiary poit of the differetial equatio x y + ( e ) y + (si x) y=. I particular, x = is a ordiary poit because the two fuctios are aalytic at this poit. y + ( e x ) y + (l x) y= has a sigular poit at x = because l x is discotiuous here. Polyomial Coefficiets We will maily be iterested i polyomial coefficiets because they are aalytic at every poit ad ratioal fuctios because they are aalytic at poits where the deomiator is ot zero. So if a (x), a (X) ad a (x) are polyomial with o commo factors the P(x) ad Q(x) are aalytic except where a (x) ad sigular whe a (x)=. Ex: ( x ) y + xy + 6y= Ex: ax y + bxy + cy= Ex: ( x + ) y + xy y=

6 Existece of Power Series Solutios If x= x is a ordiary poit of the secod order DE, we ca always fid two liear idepedet solutios i the form of a power series cetered at x : f( x) = c ( x x) = A series solutio coverges at least for x x < R, where R is the distace from x to the closest sigular poit (real or complex). A solutio of the form y= c ( ) x a is said to be a solutio about the ordiary poit = x. The distace R i the theorem is the miimum value or the lower boud for the radius of covergece of the series solutios of the DE about x. Fidig a Power Series Solutio Fidig a power series solutio is very similar to fidig a particular solutio of a ohomogeeous equatio by the method of udetermied coefficiets. We substitute y=cx ito the DE, combie series, ad the equate all coefficiets to = the right-had side of the equatio to determie the coefficiets c. Because the righthad side is zero, by the idetity property, all coefficiets of x must be equal to zero, this does ot mea all coefficiets are zero. The geeral solutio of the DE is y cy x c y x = ( ) + ( ) where c c = ad c = cy Ex: Solve y + xy= Sice there are o sigular poit The previous theorem guaratees two power series solutios cetered at, coverget for x <. Substitutig =, = y cx =, = y cx y= ( ) cx = ( ) = = y + xy= c x + x cx = ( ) + c x + cx = = Add the two together lettig k = - i the first series ad k = + i the secod series

7 Therefore [ k+ k ] k y + xy= c + ( k+ )( k+ ) c + c x = k= ( k )( k ) c c,,, k+ + k = k = ad c = the c = ad the previous expressio is called a recurrece relatio, we ca solve for c k + i terms ck : ck ck+ =, k =,,3... ( k+ )( k+ ) This geerates cosecutive coefficiets of the assumed solutio oe at a time as we let k take o the cosecutive itegers: c k =, c3 = 3 c k =, c4 = 3 4 c k =, c5 = = 4 5 c3 k =, c6 = = c c k =, c = = c c5 k =, c8 = = Ad so o. Now substitutig the coefficiets just obtaied ito the origial assumptio y= cx = c + cx+ cx + cx + cx + cx +... = We get c 3 c 4 c 6 y= c+ cx + x x + + x By groupig the terms cotaiig co ad c we obtai y= c y( x) + cy ( x) ( ) y( x) = x + x x +... = + x k k ( ) y( x) = x + x x +... = x+ x (3 k)(3k+ ) k 4 7 3k+ k= k= Ex: Solve ( x + ) y + xy y=

8 Ex: Solve y ( + x) y= Solutios about Sigular Poits For the DE a (x) y +a (x) y +a (x)y= we saw that if x=x is a ordiary poit we ca fid a solutio of the form y= c (x x ) What if x=x is a sigular poit? It turs out we may be able to fid a solutio of the form y= c (x x ) +r where r is a costat that must be determied. Regular ad Irregular Sigular Poits Workig with stadard form y +P(x) y +P(x) y +Q(x)y= A sigular poit xo is said to be a regular sigular poit of a secod order DE if the fuctios (x x )P(x) ad (x x ) Q(x) are both aalytic at x. A sigular poit that is ot regular is said to be a irregular sigular poit of the equatio. For polyomial coefficiets of a (x x ) y +a (x) y +a (x)y= where a,a,a have o commo factors, let a (x)y=. Form P(x) ad Q(x) by reducig a a ad a a to lowest terms, respectively. If the factor (x=x ) appears at most to the first power i the deomiator of P(x) ad at most the secod power i the deomiator of Q(x), the x=x is a regular sigular poit. Ex: Fid the sigular poits ad determie whether regular or irregular for a. (x 4) y +3(x ) y +5y= b. x (x+) y +(x ) y +y= c. ( x ) y x y +3y= Frobeius' Theorem If x=x is a regular sigular poit of the DE, the there exists at least oe solutio of the form y=(x x ) r c (x x ) = c (x x ) +r where the umber r is a costat to be determied. The series will coverge at least o some iterval <x x <R

9 The trick to this is that we will eed to fid r. If r is foud to be a umber that is ot a o-egative iteger the the correspodig solutio is ot a power series. Ex: Because x = is a regular sigular poit of the differetial equatio 3x y + y = we try to fid a solutio of the form y= c x +r Now y = (+r)c x +r ad y = (+r)(+r )c x +r Idicial Equatio r(3r-) is called the idicial equatio of the problem ad the r values r = /3 ad r = are called the idicial roots, or expoets, of the sigularity x =. I geeral after substitutig ad simplifyig, the idicial equatio is a quadratic i r that results from equatig the total coefficiet of the lowest power of x to zero. We solve for the two values of r ad substitute them ito the recurrece relatio. Furbeius's Theorem guaratees that at least oe solutio of the assumed series form ca be foud. Ex: Solve x y +(+x) y +y= Ex: Solve x y +3 y y=

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will