Carleto College, Witer 207 Math 2, Practice Fial Prof. Joes Note: the exam will have a sectio of true-false questios, like the oe below.. True or False. Briefly explai your aswer. A icorrectly justified aswer may ot receive full (or ay) credit. (a) It s possible for a power series to coverge for all x i (, 2] ad for x = 0 but ot for ay other value of x. This is false. Every power series has a radius of covergece, ad so the set of x-values where the series coverges will be a iterval (though oe that may or may ot iclude its edpoits). The set give i this problem is ot a iterval. (b) Deote by g(x) the twetieth derivative of f(x) = xe x4. The g(0) = 0. A power series for f(x) ca be foud by substitutio ad multiplicatio by x. We have that e x4 = +( x 4 )+ ( x4 ) 2 So the 2! + ( x4 ) 3 3! + ( x4 ) 4 4! + ( x4 ) 5 + = x 4 + x8 5! 2 x2 6 + x6 24 x20 20 + xe x4 = x x 5 + x9 2 x3 6 + x7 24 x2 20 + Sice this is a power series represetatio for f(x), it is also the Taylor series for xe x4. This meas that the coefficiet of the x 20 term is the twetieth derivative of f(x) evaluated at x = 0, divided by 20!. This is the same as g(0)/20!. But the x 20 term is zero i our power series, so we coclude that g(0) = 0. (c) Suppose that for the series = a, the sequece of terms a satisfies lim a = 0. The = a might coverge, or might diverge; there is ot eough iformatio to tell. This is true. The th term test for divergece oly tells us that if lim a is ot 0, the = a diverges. Whe lim a = 0, we ca t coclude aythig. For istace, the series = ad = both satisfy lim 2 a = 0, yet the first diverges ad the secod coverges. (d) To evaluate 2x 2 2 dx, you must use partial fractios. x 3 3x False. You ca use the substitutio u = x 3 3x. The du = 3x 2 3 dx = (2/3)(2x 2 2) dx. (e) The Taylor series of ay fuctio f(x) at a = 0 coverges for all values of x. False. For istace, the Taylor series for l( + x) at a = 0 coverges oly for x i the iterval (, ]. 2. Fid the followig itegrals: (a) l x dx
Use Itegratio by parts: set u = l x ad dv = dx. Thus du = /x dx ad v = x. The itegral the becomes x l x dx, which is x l x x + C. (b) l x dx x This itegral diverges. You ca show this by compariso to dx, or you ca calculate it x directly. To calculate directly, first let s fid l x dx, which we ca do usig the substitutio x u = l x. The du = (/x) dx, so the itegral becomes u du. This gives u 2 /2 + C, or (l x) 2 /2 + C. So ow we have l x x [ dx = lim (l x) 2 /2 ] t = lim (l t t t)2 /2 0, ad this last limit is ifiity, meaig that the itegral diverges. (c) x 3 x 2 dx This oe is readymade for a trig substitutio. Use x = si θ. The dx = cos θ dθ, ad we get si 3 θ si 3 si 2 θ cos θ dθ = θ cos θ dθ = si 3 θ dθ. cos θ Sice the itegral is of a odd power of sie, we covert all but oe power to cosies, ad the substitute u = cos θ: si 3 θ dθ = ( cos 2 θ) si θ dθ = ( u 2 ) du. So we ow get or 3 u3 u + C, 3 cos3 θ cos θ + C. Sice x = si θ, by drawig our triagle (or usig si 2 θ + cos 2 θ = ), we get cos θ = x 2, ad so our fial aswer is 3 ( x2 ) 3/2 ( x 2 ) /2 + C. 3. Solve the iitial value problem dy dx = y2 x 2 +, y(0) = Separate to get y 2 dy = dx. The itegrate to get /y = x 2 + ta (x) + C. Solve for y to get y = ta (x) + C. Sice y(0) =, we have = /C, so C =, ad the fial aswer is y = ta (x). 2
4. The itegral 0 e x3 dx caot be evaluated exactly. Use a method from the course this term to approximate this itegral to withi a error of at most /60. Leave your aswer as a fractio. (So select a method where fidig the error boud wo t be to difficult) The most coveiet error estimate we saw this term is the oe for alteratig series. So would t it be great if we could express this itegral with a alteratig series. Well we ca! Substitute x 3 ito the Taylor series for e x at x = 0, ad we get e x3 = x 3 + x6 2! x9 3! + x2, 4! ad this equality holds for all x. Therefore e x3 dx = x x4 4 + x7 7 2! x 0 0 3! + x3 3 4!, Ad so 0 e x3 dx = ( 4 + 7 2! 0 3! + ) 3 4! 0 = 4 + 4 60 + 32 So we ve expressed the defiite itegral we wat as a alteratig series. The alteratig series error estimate the tells us that 0 e x3 dx = 4 + 4, with a error of at most /60. We we express /4 + /4 as a fractio, we get 23/28. 5. Describe aother method for approximatig the itegral from the previous problem, ad write dow (but do ot evaluate) a sum of five umbers that is a approximatio of this itegral. Aother way to approximate the itegral is to use Simpso s rule. Whe we use it with four sub-itervals (or i other words = 4), we get a approximatio of ) (e 0 + 4e ( /4)3 + 2e ( /2)3 + 4e ( 3/4)3 + e. 2 The error estimate here is much harder to fid tha i the previous problem. 6. Use ideas from the course to approximate e 2 to withi 0.. Leave you aswer as a fractio, but explai why your approximatios are correct to withi 0.. Oe way we ca approximate e is to recogize that it s the value of the fuctio f(x) = e x whe x =. So we approximate the fuctio e x usig its Taylor series at x = 0: e x = + x + x2 2! + x3 3! + x4 4! +, 3
ad so if we trucate at = 4, we get that the degree-4 Taylor polyomial for f(x) = e x at x = 0 is + x + x2 2 + x3 6 + x4 24. So this tells us that e = e + + 2 + 6 + 24 = 65/24, which is approximately 2.70833, though you do t eed that for this problem). How good is this approximatio? Well, we ca use the Taylor series error estimate. Sice = 4 here, we eed to take K to be a umber such that f (5) (x) < K for x i [0, ]. But f (5) (x) = e x, ad e x e for x i [0, ], ad e < 3. So we take K = 3. The error estimate is K b a + ( + )! = 3 5! = 3 20 = 0.025. So we ve foud e to withi the desired degree of accuracy. If you wat to play more with this, fid the error i the approximatio to e that you get by pluggig ito the degree-0 Taylor polyomial for e x. (Note that you do t eed to fid the actual approximatio i order to fid the error, though of course you re welcome to fid the approximatio). The do the same for the degree-00 Taylor polyomial. 7. Determie whether these series coverge. If a series coverges ad is geometric, fid its sum. 2 a) 2 Use the ratio test. The limit you get works out to 2, ad so the series diverges. b) 2 + = The expressio for the th term of this series is very close to / 2 whe is large, ad / 2 = /2 / 2 = / 3/2. So use either direct (easier) or limit compariso to the coverget p-series = /3/2 to show that this series coverges.! c) [Hit: it s helpful to write ( + )+ as ( + ) ( + ).] = This oe is a good cadidate for the ratio test, sice it ivolves a!. Whe you apply the ratio test, the limit you get is ( ) ( + )! lim ( + ) +! = lim ( + ) ( + ) = lim ( + ) + ( + ) ( + ) = lim. + To fid this limit, use the techique from class: let b = ( ), + so that l b = l. + Rewrite this as l b = l + /, 4
so that ow lim l b has the form 0/0, so we ca apply L Hopital s rule. After doig that, we get + lim l b /( + )2 = lim / 2 + = lim 2 /( + ) 2 = lim + =. So we ve show that lim l b =, ad so lim b = e = /e. Sice this is less tha oe, the origial series coverges by the ratio test. d) l Here it looks like the l will be isigificat relative to the i the deomiator, ad so it s temptig to compare to the diverget p-series /. But direct compariso does t work, sice the iequality goes the wrog way. Use limit compariso with the series from the previous setece: lim l = lim /, by L Hopital s rule, ad so the limit comes out to. Thus the origial series behaves the same as /, ad so diverges. 8. Determie whether this series coverges absolutely, coverges coditioally, or diverges: ( ) l. The series coverges by the alteratig series test: it s easy to check that for 2, / l is decreasig ad teds to 0 as goes to ifiity. However, whe we take absolute values, we get the series. Use the itegral test to show that this series diverges. l (x + ) 9. Determie the iterval of covergece of the power series. 2 3 Usig the ratio test, you should get that the series coverges whe 3 < x+ < 3, or i other words whe 4 < x < 2. The problem At the edpoit x = 2, we get the series =, 2 which is a coverget p-series (p = 2). At the edpoit x = 4, we get the series = = ( ) 2, which coverges by the alteratig series test (or coverges absolutely by the x = 2 case, ad so coverges). So the iterval of covergece is [ 4.2]. 5