Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

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1 Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial a =. To calculae he various quaniies: 3 f() = arcsin f() = f () = ( ) / f () = f () = ( ) 3/ f () = f () = ( ) 3/ + 3( ) 5/ f () = Therefore, he 3rd Taylor polynomial is + 3 /6. Problem. In his problem, you will calculae he n-h Taylor polynomial for / a =. To make he noaion easier, le f() = /. (a). Prove by inducion ha for all n, f (n) n 3 5 (n ) () = ( ) (n+)/ n Soluion. The base case for he inducion is when n =. In his case, f () = 3/ = ( ) ( +)/ so he formula works. For he inducion case, assume ha f (n) () has he form above and we show ha f (n+) () also has his form. Since f (n+) () = d/d f (n) (), we can ake he derivaive of he formula above o ge f (n+) n 3 5 (n ) () = ( ) n (n + ) ( (n+)/) I will leave you o check ha his epression simplifies in he correc form of f (n+) n+ 3 5 ((n + ) ) () = ( ) ((n+)+)/ n+ (b). Give he formula for he n-h Taylor polynomial o / a =.

2 Soluion. We have f() = and by Problem (a), we know f (k) k 3 5 (k ) () = ( ) k for k. Therefore, he n-h Taylor polynomial for f() a = is + k 3 5 (k ) ( ) ( ) k k k= or wrien wihou he sum noaion 3 ( ) +! ( ) ! ( )3 n 3 5 (n ) + + ( ) ( ) n n n! Problem 3. In his problem, you will develop he Taylor polynomials for ln( ). Le f() = ln( ). 3(a). Prove by inducion ha for all n, and hence for n, f (n) () = (n )!. f (n) () = (n )! ( ) n Soluion. For he base case when n =, we need o show ha f () =! ( ) By he Chain Rule, d/d(ln( )) = /( ). Since! =, he righ side of his above equaion is also /( ). This complees he base case. For he inducion case, assume ha for a fied n. We need o show ha f (n) () = f (n+) () = (n )! ( ) n ((n + ) )! ( ) n+ The righ side is n!/( ) n+. To calculae he lef side using he inducion hypohesis f (n+) () = d d f (n) () = d ( ) (n )! ( ) n = (n )!( n)( )( ) (n+) d The las epression simplifies o n!/( ) (n+) as required.

3 3(b). Give he n-h Taylor polynomial for f() a =. Since f() = ln =, we know ha he consan erm in he Taylor polynomial is. So, we can sar he indeing in our sum wih k = (because he k = erm is ). By Problem 3(a), we know ha f (n) () = (n )! for n. Therefore, he n-h Taylor polynomial for ln( ) a = is T n = k= f (k) () k = k= (k )! k = k= k k Problem 4. Prove ha if c is a consan and f is a (sufficienly differeniable) funcion, hen T n (cf()) = c T n (f()) where we ake he Taylor polynomial a = a. Soluion. We firs calculae he righ side as follows c T n (f()) = c k= f (k) (a) ( a) k = k= c f (k) (a) ( a) k To calculae he lef side, we firs noe ha because c is a consan, we have We now calculae he lef side as follows. T n (cf()) = k= d k dk cf = c dk d f = c f (k) k (d k /d k cf)(a) ( a) k = As hese formulas now mach, we have finished our proof. k= c f (k) (a) Problem 5. Evaluae he following (convergen) improper inegrals. For he firs inegral, d = d = (3 + ) 4 (3 + ) 3(3 + ) 4 For he second inegral, e / d = 4 ( a) k = = 39 e / d = e / = e / + e = /e

4 For he hird inegral, we need o spli he inegral ino wo improper inegrals by picking a convenien middle poin. I ll choose o spli i as e d = e d + e d We need o make sure ha boh of he inegrals on he righ converge and sum heir values. Firs, consider e d = e d = e = e + e = Noice ha we were able o remove he absolue value sign because is posiive on he inerval (, ]. You can approach he oher inegral in a couple of differen ways. One mehod is o noe ha by symmery, he area under e on he inerval (, ) is he same as on he inerval (, ). Therefore, we mus have e d = as well. Alernaely, you can wrie ou he definiion and calculae he inegral direcly. However, remember ha = on he inerval (, )! e d = e ( ) d = e d = e = e e = Problem 6. Use he definiion of he improper inegral o eplain why sin d diverges. Soluion. By definiion, sin d = sin d provided his i eiss. To show ha he i does no eis, we noe ha sin d = cos = cos + cos = cos + Recall ha when we plug an even muliple of π ino cos we ge. Tha is, cos nπ =. When we plug an odd muliple of π ino cos, we ge. Tha is, cos(n + )π =. This means ha nπ sin d = and (n+)π sin d = Therefore, for every M >, here are numbers > M such ha sin d = and > M such ha sin d =. Therefore, sin d canno approach a i as. Problem 7. Deermine wheher he following inegrals converge or diverge. You do no need o calculae he value of he convergen inegrals. For he firs inegral d

5 we can use he Comparison Tes as follows: = 4 for all [, ) because We know / 4 d converges because i has he form / p d wih p = 4 >. Therefore, he firs inegral converges. For he second inegral, we can use he Comparison Tes as follows: + e d + e because + e. We know / d diverges from class, and so he second inegral diverges. For he hird inegral, e d he problemaic inegral i is he lower i of. For [, ], we know ha e is beween e and. Therefore, for (, ), we have e e We know ha / d converges from Homework 3 since i has he form /p d where p = / <. Therefore, he second inequaliy above ells use ha e / d converges as well by he Comparison Tes. For he fourh inegral, we migh be emped o ry he Comparison Tes wih /. However, he inequaliy goes he wrong way. Tha is, bu / d converges, so he Comparison Tes doesn help using his paricular comparison and we need o hink a lile more. We know ha as, goes o infiniy more slowly han any posiive power of, so our inuiion says ha he facor of in he numeraor should no be enough o push he inegral ino divergence. d

6 To see his more formally, ry he Comparison Tes wih / p where p is a lile smaller han raher han wih /. For eample, i should be ha for large values of, we have. Since = we can apply he definiion of he i o fi an M such ha for all > M, / <. In oher words, < for all > M. By choosing M larger if necessary, we can assume ha M. We now have ha for all > M, = 3/ This comparison is useful since we know M /3/ d converges because p = 3/ >. To finish he problem, we wrie M d = d + M d The firs inegral on he righ side eiss because he funcion being inegraed is coninuous and he second inegral on he righ side converges by he Comparison Tes. Therefore, our iniial inegral converges as well. For he las inegral + d we migh be emped o ry a comparison wih /. However, + = Since / d diverges, his comparison doesn help us, and as above, we need o hink a lile more. Our inuiion should be ha for large values of, /( + ) is eremely close o / so adding he consan should make he fracion small enough o push he inegral ino convergence. One way o make his more formal is o noice ha for, we have and hence + + =. Since make he denominaor larger will make a fracion smaller, his insigh ells us ha + = for all. Now our comparison goes he righ way! We know / d diverges, and hence so does / d. So, he Comparison Tes ells us ha his final inegral diverges as well. Problem 8. Find a general formula o wrie (a + ib) in he form c + id. Tha is, find formulas for c and d in erms of a and b. (You can assume a + ib is no.)

7 Soluion. To find he muliplicaive inverse of a + ib we calculae a + ib a ib a ib = a ib a + b = a a + b i b a + b Problem 9. Prove ha e z is no equal o for any z C. Soluion. Wrie z = a + ib where a and b are real numbers. We know e z = e a+ib = e a cos b + i e a sin b To prove ha his value can never be, we need o show ha here are no real numbers a and b such ha boh e a cos b = and e a sin b =. Since e a (because a is real), i suffices o show ha we canno have boh sin b = and cos b =. If sin b =, hen b is a ineger muliple of π. However, he value of cos b when b is an ineger value of π is eiher or. Therefore, when sin b =, we have cos b, and hence hey canno boh be a he same ime. Problem. Find he following is. sin a sin b an sin 3 sin arcan /( ) 3 e Soluion. We consider he is one a a ime, ypically using L Hopial s rule. The firs hree is all have form /, so we immediaely apply L Hopial s rule. In all hree cases, we only need o apply i once. sin a sin b = a cos a b cos b = a b an sin 3 = sec 3 cos 3 = 3 sin arcan = cos /( + ) = There are a couple of ways you could approach he ne i. One mehod would be o simplify /3 o (/3). Since < /3 <, he value of (/3) goes o as he eponen goes o. Tha is, 3 = (/3) =.

8 Alernaely, you can rewrie as e ln and 3 as e ln 3. Then 3 = e ln ln 3) = e(ln e ln 3 Since ln ln 3 is negaive, he eponen of e goes o as goes o. Therefore, he i is. The ne i has for, so we sar by rewriing. = e = e We need o find he i of, which has he form, so we rewrie i in he form / as follows. = / = / / = = so = e =. The ne i has form, so we sar by rewriing again. /( ) = eln /( ) = e The i of he eponen has form /, so we apply L Hopial s rule: = / = / = So, he original i is e, or equivalenly /e. The ne i has form, so we combine i ino one fracion. e = e (e ) = e e The i now has he form / so we ry L Hopial s rule. e e = e e + e This i sill has form /, so we ry L Hopial s rule again. e e + e = e e + e + e =