SOLUTIONS TO EXAM 3 Problem Fid the sum of the followig series 2 + ( ) 5 5 2 5 3 25 2 2 This series diverges Solutio: Note that this defies two coverget geometric series with respective radii r 2/5 < ad r 2 /5 < So we get 2 + ( ) 5 ( ) 2 + 5 ( ) 5 2/5 + ( /5) 5 2
2 SOLUTIONS TO EXAM 3 Problem 2 Use Compariso Tests (or tests of your ow choosig) to determie which oe of the followig series is diverget l() 2 / 3/2 2 +2+2 4 +8 2 3 2 + Solutio: Note that for ay 4, l() for atural log is icreasig So l() o that domai ad hece the first series is diverget for it boud from above a p-series with p /2 <
SOLUTIONS TO EXAM 3 3 Problem 3 Which oe of the followig series coverges coditioally? (a) ( ) + (b) ( ) 2 5 (c) ( ) 2 + (d) ( ) 4 (e) ( ) l() Solutio: Recall that a series is coditioally coverget if it coverges but ot i absolute value Takig a look at the first oe we see that ( + ) + + which gives b + ( + ) + + b ad we kow that b 0 as Thus the series coverges by the Alteratig Series Test Takig absolute value we see that ( ) + + + 2 /2 which defies a p-series with p <, ad hece it diverges Thus the series is ot absolutely coverget
4 SOLUTIONS TO EXAM 3 Problem 4 Cosider the followig series ( 2 + ) (I) (II) 3 2 Which of the followig is true? (a) (I) coverges ad (II) diverges! 2 (b) The ratio test is icoclusive o (II) (c) (I) diverges ad (II) coverges (d) Both the series coverge (e) Both the series diverge Solutio: Computig a i the first series we get 2 + < ad thus it 3 2 3 coverges absolutely For the secod oe, let us ratio test it We compute a + ( + )! a 2 + + 2 ( + )! 2 ( + ) + 2 ( + ) + 2 2 which is ubouded ad hece the secod series diverges
SOLUTIONS TO EXAM 3 5 Problem 5 Cosider the followig series ( ) l() e (I) (II) e + 2 2 Which of the followig statemets is true? Oly I ad III coverge Oly III coverges Oly I ad II coverge All three coverge All three diverge 2 (III) si( 2 ) + 2 Solutio: Let f(x) : l(x) x o [, ) Note that f (x) x x l(x) 2 x ( x) 2 x x ( l(x) ) < 0 2 for all x > e 2 Thus the terms of the first sum are evetually decreasig Note also that applyig L Hospital s rule oce we get ( l(x) lim lim ( x) ) 2 0 as x x x x x 2 x Thus the Alteratig test tell us that a tail of this series coverges, hece it coverges For the secod oe, ote that e e + 2 + (2/e) + 0 0 for 2/e < Thus it diverges by the divergece test For the last oe, ote that upo takig absolute value we get si( 2 ) + 2 si(2 ) + 2 + 2 2 Thus it coverges absolutely via compariso test with a p-series with p 2 >
6 SOLUTIONS TO EXAM 3 Problem 6 Fid the sum of the followig series ( )! e e e l(2) e Solutio: Recallig that e x x for all x, we ca evaluate the above series by substitutig for x, yieldig e!
SOLUTIONS TO EXAM 3 7 Problem 7 Fid a power series represetatio for the the fuctio f(x) l ( x 2 ) d Hit: l ( dx x2 ) 2x x 2 ( 2)x 2+2 2+2 ( 2) x 2 ( 2) x 2+2 2+2 ( 2)(2 + )x2 ( 2) x 2+ 2+ Solutio: Usig the hit as a startig place, we ca fid the expasio for the derivative ad the itegrate term by term to arrive at a power series for the iitial fuctio From our kowledge of the geometric series, we ca write x x 2, so locally d 2 dx l ( x2 ) 2x 2+ The we itegrate ad solve for our costat of itegratio, f(0) l() 0, so i the ed we fid our power series about 0 is l( x 2 ( 2)x 2+2 ) 2 + 2
8 SOLUTIONS TO EXAM 3 Problem 8 Which of the power series give below is the McLauri series (ie Taylor series at a 0) of the fuctio x 2+! ( ) x 4+ (2)! x 2 (2)! ( ) x 4+3 (2+)! x 2+ (2)! Solutio: f(x) xe x2? We use the Maclauri series for e x, ie e x x! Substitutig i x2 (for x), we ( (x 2 ) x 2 ) obtai e x2!! Thus, f(x) x 2 x 2+ xex2 x!!
SOLUTIONS TO EXAM 3 9 Problem 9 The followig is the fifth order Taylor polyomial of the fuctio f(x) at a 2 T 5 (x) 2 2 (x 2) + 3 (x 2)2 2(x 2) 3 + 2 (x 2)4 + 0(x 2) 5 What is f (4) (2)? 2 2 48 2 24 Solutio: The coefficiet of the term (x 2) 4 i the fifth order Taylor polyomial is f (4) (2) 4! 2 Thus, f (4) (2) 4! 24 2 2 2
0 SOLUTIONS TO EXAM 3 Problem 0 Which of the followig is a graph of the parametric curve defied by x t si(t), y t cos(t) π π for 0 t 3π? ( 2 Solutio: Plottig a few poits i a table, we get t x y 0 0 0 π 0 2 2 3π 0 3 So, the curve should start at (0, 0), ed at (0, 3), ad pass through the poit, 0) The oly choice that does this is
Problem Cosider the series SOLUTIONS TO EXAM 3 [ ] ( ) l(+) l() Fill i the followig blaks ad be sure to show your work I each case idicate which test you are usig ad show how it is applied Is the series absolutely coverget? (YES or NO) Solutio: To check if the series is aboslutely coverget, we take the absolute value of the [ ] iside to get l( + ) l() We otice that this series is telescopig, ad the partial sums are s (l(2) l()) + (l(3) l(2)) + (l(4) l(3)) + + (l( + ) l()) l() + l( + ) l( + ) Therefore, lim s lim l(+) So, the series [ ] l(+) l() is diverget, which shows that the origial series is ot absolutely coverget Is the series coverget? (YES or NO) Hit: Laws of logarithms may help here Solutio: We use the alteratig series test with ( ) + b l( + ) l() l We have ( ( d l + )) dx x + x ( l + ) x 2 x 2 + x, which is egative for all x Therefore, the sequece {b } is decreasig Also, ( lim b lim l + ) l() 0 Therefore, by the alteratig series test, the series coverget [ ] ( ) l( + ) l() is
2 SOLUTIONS TO EXAM 3 Problem 2 Fid the radius of covergece ad iterval of covergece of the followig power series: ( ) (x + ) 2 Show all of your work for credit Solutio: We recall the geeral form of a power series cetred at a: c (x a) Now, let s express the series we have i this form: ( ) (x + ) 2 ( ) 2 (x ( )) I this form, we see a ad c ( ) 2 So, before we do ay tests, we kow our series coverges at x a Now, for x, we apply the Ratio Test: lim c + c lim ( ) + (x + ) + 2 2 + + ( ) (x + ) lim (x + ) x + 2 + 2 So, the Ratio Test guaratees us (abs) covergece if x+ <, which is equivalet to x ( ) < 2 ie, we have covergece o ( 3, ), ad R 2 Recall, that 2 the Ratio Test is icoclusive whe we get the above limit ie whe x + 2, which occurs whe x 3 ad x (boudary terms) We must check these idividually: x 3: ( ) ( 3 + ) ( ) ( 2) 2 2 /2 Which we kow diverges by the p-test (p ) 2 x : ( ) ( + ) ( ) (2) 2 2 ( ) Which we recogize as a alteratig series with b We see: lim 0 ad, + > + < Hece, by the Alteratig Series Test, the series coverges We coclude, the Iterval of Covergece is ( 3, ]
SOLUTIONS TO EXAM 3 3 Problem 3 (a) Give the Taylor series expasio for si (x 5 ) about 0 (Use Σ otatio ad the write out the first three o-zero terms) Solutio: Recall the Maclauri Series (Taylor Series aroud 0) expasio for si(y): si(y) Hece, si(x 5 ) ( ) (2+)! y 2+ ( ) (2+)! (x 5 ) 2+ y y3 3! + y5 5! ( ) (2+)! x 0+5 x 5 x5 3! + x25 5! (b) Use your aswer from part (a) to write the defiite itegral 0 0 si(x 5 ) dx as the sum of a series (usig Σ otatio) Solutio: Usig (a), ad recallig that we may itegrate power series term by term iside their radius of covergece, we have: 0 0 si(x 5 )dx 0 0 ( ( ) ( ( ) (2 + )! (2 + )! ( 0 0 ) (0) 0+6 0 + 6 ) ( ) (2 + )! x0+5 dx ) x 0+5 dx ( ( ) (2 + )! ( 0 ( ( ) x 0+6 (2 + )! (0 + 6) ( ) 0 )0+6 0 + 6 0 ) ( ) (2 + )! x0+5 dx 0 0 ) ( ) (2 + )! 0 + 6 ( ) 0+6 0 (c) Use your aswer from part (a) to compute the followig limit Solutio: Usig part (a): si (x 5 ) x 5 lim x 0 x 5 si(x 5 ) x 5 x5 3! si(x5 ) x 5 x 5 3! lim x 0 si(x 5 ) x 5 x 5 3! + x25 5! + x0 5! + 0 + 6