PLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)

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1 Math 0560, Exam 3 November 6, 07 The Hoor Code is i effect for this examiatio. All work is to be your ow. No calculators. The exam lasts for hour ad 5 mi. Be sure that your ame is o every page i case pages become detached. Be sure that you have all 7 pages of the test. Trigoometric formulas are provided i the last page. PLEASE MARK YOUR ANSWERS WITH AN X, ot a circle!. (a) (b) (c) (d) (e). (a) (b) (c) (d) (e) (a) (b) (c) (d) (e) 4. (a) (b) (c) (d) (e) (a) (b) (c) (d) (e) 6. (a) (b) (c) (d) (e) (a) (b) (c) (d) (e) 8. (a) (b) (c) (d) (e) (a) (b) (c) (d) (e) 0. (a) (b) (c) (d) (e).... (a) (b) (c) (d) (e). (a) (b) (c) (d) (e) Please do NOT write i this box. Multiple Choice Total

2 Multiple Choice Name:.(6 pts) Fid ( ) lim ta (a) (b) (c) (d) 0 (e) 4 Solutio. This limit is idetermiate of type 0. Rewritig, we get ( ) ( ) ta lim ta = lim which is idetermiate of type 0. Usig L Hospital s Rule, 0 ( ) ( ) ta x x x lim x x sec L H = lim x x ( ) = lim sec x x =

3 .(6 pts) Fid the sum of the followig series. ( ) k+ + 3 k. 4 k k=0 (a) 3 8 (b) 6 5 (c) 4 5 (d) 8 3 (e) 0 Solutio. This is a sum of two coverget geometric series: ( ) k+ + 3 k ( ) k+ = + 3k 4 k 4 k 4 k k=0 k=0 ( = 4) k ( ) k 3 + = k=0 k=0 = + 4 = = 6 5 ( 4) k k=0 ( ) k 3 4 3

4 3.(6 pts) Fid the sum of the followig series, (a) e = [ e + e (c) the series diverges (d) (e) (b) 0 e ]. Solutio. This is a telescopig series. Let s look at some partial sums: ( s = e ) = 0 e e ( s = e ) ( + 0 e e 3 ) = 3 e e ( s 3 = e ) ( + 0 e e 3 ) ( 3 + e e 4 ) = 4 e 3 e 3 Thus the th partial sum is ad, by defiitio, = s = + e ( e + ) = lim s e = lim + =. e 4

5 4.(6 pts) Which oe of the followig statemet is TRUE? (a) (b) (c) (d) (e) = = = = ((si ) + ) ((si ) + ) ((si ) + ) ((si ) + ) oe of the above is absolutely coverget by ratio test. is diverget by compariso test. is diverget by ratio test. is absolutely coverget by root test. So Solutio. First, we observe that ( si () + ). ( si () + ) ad ( si () + ) The harmoic series diverges ad ( = si () + ), so diverges by the Compariso Test. = ( si () + ) 5

6 5.(6 pts) Which of the followig statemet is TRUE? (a) (b) (c) (d) (e) = = = = = ( ) 3 5 coverges coditioally. ( ) ( + ) ( ) ( + ) ( ) ( + ) ( ) 3 coverges absolutely. diverges. coverges coditioally. 5 diverges by divergece test. Solutio: Cosider the series = ( ) ( + ). Let b = ( + ) = +. We see that b b = ad clearly b 0 as, ( ) ( + ) so coverges by the Alteratig Series Test. To test for absolute = covergece, we cosider the series ( ) ( + ) = ( + ) = which diverges by compariso to the p-series is coditioal. = = = ( ) ( + ). Thus the covergece of 6

7 6.(6 pts) For what values of p is the followig series coverget? ( ) l. p (a) for all p (b) p > (c) p > 0 = (d) p < 0 (e) for ay p such that p 0 Solutio: Let us ote that for the values p 0 the geeral term does ot approaches zero ad hece the series diverges. So that we are left with p > 0. Thus we test our expressio for the alteratig series test to decide betwee p > 0 ad p >. Recall that we eed b = l() b b + b 0 as p 0 to satisfy To show it decreases it suffices to show f(x) := l(x) large values of x. I fact, x p for x [, ) is decreasig for f (x) = x xp l(x) px p = ( p l(x)) (x p ) xp+ So that f (x) < 0 p l(x) < 0 e < x p. Thus if we start our series at ay value such that p > e for p > 0 we have that b will decrease. Next, we compute the limit with p > 0, l(x) lim f(x) = lim x x = lim x = x p x px = p p lim x by L Hopital s x p = 0 Thus i particular, b always satisfies the limit coditio ad decreases for large for p > 0. Sice fiitely may terms does ot chage the covergece, we coclude that the series coverges for all p > 0. 7

8 7.(6 pts) Which of the followig statemets is true about the ifiite series? ( ) + + = (a) The series diverges, sice the limit of the -th term is. (b) The series diverges, by compariso with the harmoic series (c) The series coverges, sice the limit of the -th root of the -th term is 0. (d) The series coverges, sice the limit of the -th root of the -th term is. =. (e) The series coverges, by compariso with the geometric series = ( ). Solutio: Because of the structure of the series is atural to root test it. Deotig b the summad we have that + lim b = lim + = lim + / + / = / < Thus the Root test tells us that the series coverges absolutely ad hece it coverges. 8

9 8.(6 pts) Expad x x as a power series cetered aroud a =. Hit: Complete squares i the deomiator, ad use a well kow power series. (a) ( ) (x ) (b) (x ) (c) (x ) (d) (x ) (e) ( ) (x ) Solutio: Followig the hit we have x x = (x x + ) = (x ) = ( ) (x ) = (x ) 9

10 9.(6 pts) Use the MacLauri series to fid e x + x lim. x 0 x 4 (a) (b) 6 (c) 3 (d) 0 (e) Solutio: Recall that: Hece, for x < : ( x ) e x =! e y = = (y)! ( ) x!, for y < = x + x4! x6 3! +... e x + x = x4! x6 3! +... For x 0: e x + x x 4 =! x 3! +... e x lim x 0 + x x 4 =! = 0

11 5 0.(6 pts) Fid a power series represetatio for f(x) = ad determie its iterval of covergece. Hit: You ca obtai it by differetiatig power series represetatio ( x) for g(x) = 5 x. (a) (b) (c) (d) (e) = 5x, iterval of covergece: (, ) = 5x, iterval of covergece: [, ] = x, iterval of covergece: [, ) = 0x, iterval of covergece: [, ] = 5x, iterval of covergece: [, ) Solutio: Notice: f(x) = d ( 5 dx ) 5 =. Hece, for x < : x ( x) ( ) ( 5 = d ) 5x = x dx d dx = 5x Remark: f has the same redius of covergece as g. Now we must check (oe of) the edpoits to decide whether the aswer is (a) or (b): For x = : 5() = = The above sum clearly diverges. Hece, the aswer is (a). 5 = =

12 .(6 pts) Cosider the fuctio F (x) = = x What is F (3) (0)? Here F (3) represets the third derivative. (a) (b) 0 (c) 3 (d) 4 (e) 6 Solutio: The coefficiets of this power series are: c =, for, (c 0 = 0). Recall, the Maclauri series of F is give by: F (x) = But we have: Hece, F () (0)! I particular: F (3) (0) = F (x) = = c =, for. F () (0) = (!)c =! (3 )! 3 = 4 F () (0) x! c x = ( )! =, for

13 .(6 pts) The Taylor series for the fuctio f(x) = cos x cetered about a = π/ is give by (a) (x π/) + (b) x + x3 6 + (c) (x π/) + (x π/)3 6 + (d) + (x π/) + (e) x + Solutio: Takig derivatives, we have f (x) = si(x), f (x) = cos(x), f (x) = si(x). Pluggig i x = π/ gives f(π/) = 0, f (π/) =, f (π/) = 0, f (x) =. So, the Taylor series for f(x) cetered at π/ is which is optio (c). f(x) = (x π/) + (x π/)3 3!... 3

14 Partial Credit You must show your work o the partial credit problems to receive credit! 3. (4 pts.) Fid the radius ad iterval of covergece of the power series l x. Be sure to discuss the covergece at the two ed poits ad provide all details. particular, specify which tests you use. I Solutio: First, use the ratio test with a = l x : l( + ) lim a + a = lim + x+ l = x lim l( + ) x + = x lim l( + ) l L = H x lim + = x. l So, by the ratio test, the series coverges absolutely whe x <. So, the radius of covergece is. Now, we must check the edpoits of the iterval (, ). Lettig x =, we have l l. But, 0 for 3. Moreover, =3 diverges sice it is a p-series with p =. Therefore, l diverges as well by the compariso test. Lettig x =, we have l ( ). We wish to use the alteratig series test with b = l. It is clear that b is positive for > 0. So it remais to show (i) b is decreasig, ad (ii) lim b = 0. Takig a derivative gives ( ) l x = x x(/x) l x x 4 = l x x

15 which is egative whe l < 0. Solvig this iequality for gives > e. Therefore b is decreasig for 3. Now cosider lim b l = lim L H = lim = 0. So, by the alteratig series test, the series coverges. Thus, the iterval o covergece of our series is [, ). 5

16 4. (4 pts.) Give the power series represetatio cetered at x = 0 of the followig fuctio x arcta t f(x) = dt. 0 t by followig the steps below. You are required to use the summatio otatio. (a) First, write the power series expasio for cetered at t = 0. + t (b) Next, write the power series expasio for arcta t cetered at t = 0. (c) Fially, write the power series expasio for f(x) cetered at x = 0. Solutio: (a) Usig the formula for the sum of a geometric series, we have + t = ( t ) = ( t ) = ( ) t (b) Notice that d dt arcta(t) = + t = ( ) t from part (a). Takig atiderivatives shows that ( ) arcta(t) = + t+. (c) By the Fudametal Theorem of Calculus, we have f (x) = arcta(x) = ( ) ( ) x x + x+ = + x. So, takig atiderivatives oe last time, ( ) f(x) = ( + ) x+ 6