Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

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1 Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc, I may be able o clarify wha you need o fi.. Deermine he value of Soluion. Consider he power series f() (e ) k k(k + ) k k(k + ). This converges for < by he raio es: a k+ k k n k+ /((k + )(k + 3)) k k /(k(k + )) k(k + ) k (k + )(k + 3). We re ineresed in f(e ), which eiss, because e 0.78 <. Per he hin, we break he coefficien up ino parial fracions: k(k + ) A k + B k +. A(k + ) + Bk Se k 0, see ha A. Se k, see ha B. So and f() k(k + ) / k + / k + k k(k + ) ( / k k / ) k + ( k k ) k k k + k k k +.

2 So we re concerned wih he series k k and k k Per he hin, we also noe ha so inegraing erm by erm, 0 d + + +, ln( ) which is one of he wo series we care abou. The oher is So we have formulas for k k f() k k ( ) ( ) ln( ) /. and k. Combining hem, we ge k+ k k + ln( ) + ln( ) + + /. Subsiuing e, and using he fac ha (e ) 3 e, we ge he final value: (e ) k k(k + ) ln(3 e) + ln(3 e) + (e ) + (e ) / (e ).. Suppose ha f() is a funcion which is differeniable for all. Show ha here is some such ha f () e f() an

3 Proof. Suppose no. Then for every, Muliplying boh sides by e f(), we ge f () e f() an. e f() f () an. By he chain rule, he lef hand side is he derivaive of e f(). Inegraing boh sides, we ge ha e f() an d To inegrae arcangen, use inegraion by pars, wih u an, dv d, v, and du d. So + an d u dv uv v du an d + The laer inegral can be evaluaed by doing a subsiuion, +, so d d and d + d ln ln( + ) so an d an ln( + ) + C. So, we see ha e f() an ln( + ) + C for all values of. Now, we ll ge a conradicion by choosing sufficienly large. We claim ha he righ hand side looks asympoically like π, for 0. Indeed, an ln( + ) + C ( an ln( + ) + C ) an ln( + ) C + π ln( + ) + 0 ln(+ assuming ha ) eiss. And indeed, we can evaluae i by L Hospial s rule: ln( + ) + + 0, so ha an ln( + ) + C π π. 3

4 So, here is an N such ha if > N, hen an ln( + ) + C e f() By aking o be bigger han boh N and 0, we see ha is wihin 0.00 of π/. so ha which is false. e f() > π 0.00 > 0 e f() > 0, 3. Suppose n c n n and n d n n are wo power series in. Suppose he radius of convergence of he firs one is, and he radius of convergence of he second one is 3. Wha is he radius of convergence of he sum n (c n + d n ) n? Soluion. I is. If <, hen boh series converge, so heir sum converges. If is sricly beween and 3, hen he firs series diverges, and he second series converges, so heir sum diverges. In paricular, we know ha he power series n (c n + d n ) n Converges for all such ha < Diverges for all such ha < < 3. In paricular, i diverges on he inervals ( 3, ) and (, 3). The firs fac implies ha he radius of convergence R mus be a leas, since he inerval of convergence conains (, ). The second fac implies ha he radius of convergence R can be any bigger han. Indeed, if R >, hen ( R, R) would inersec (, 3) and ( 3, ), which is false, since he series converges on ( R, R), and diverges on (, 3) and ( 3, ). So he radius is a mos and a leas. Therefore i is. 4. Wrie down a series which converges o.5.5 ln d We have used he fac ha he sum n (a n + b n ) of a divergen series n a n and a convergen series n b n is divergen. To see his, suppose for he sake of conradicion ha he sum converged. Then since differences of convergen series are convergen, and n (a n + b n ) and n b n are boh convergen, heir difference n [(a n + b n ) b n ] n a n is convergen, conradicing he assumpion ha n a n was no convergen. 4

5 Soluion. Following he hin, we do a change of variables, so + and.5.5 ln d 0.5 ln( + ) The power series for ln( + ) is goen by inegraing he power series d. ln( + ) C Seing 0 on boh sides reveals C 0. So ln( + ) + ( ) n+ n n. We can divide by, o ge a series for ln( + )/: ln( + ) n + 3 n0 ( ) n n n +. This series has radius of convergence, by using he raio es in he usual fashion. Since and 0.5 are wihin he radius of convergence, we can inegrae erm by erm 0.5 ln( + ) d 0.5 ( ) n n0 n0 0.5 d 0.5 d n d n + [ ( ) n n+ (n + ) n0 ( ) 0.5 ( ) n n+ (n + ) n+. (n + ) 3 d ] If we le m n +, we can rewrie his in he slighly simpler form ( ) m 0.5m m m m 5. Suppose f() is a differeniable funcion, wih f (0), and suppose ha for every and y, f( + y) f()f(y). Show ha f() e for all. The raio beween consecuive erms is n+ n+, whose absolue value approaches in he i. So if <, he series converges. 5

6 Proof. For any consan a, we have he ideniy f( + a) f()f(a). Differeniaing boh sides wih respec o, f ( + a) f ()f(a), since f(a) is a consan. Now, seing 0, we ge As a was arbirary, i follows ha for all. f (a) f (0)f(a) f(a) f(a). f () f() This could alernaively be seen as follows: Firs noe ha so f(0) f(0 + 0) f(0)f(0) 0 f(0) f(0) f(0)(f(0) ), and so f(0) is 0 or. If i were 0, hen for any, f(0 + ) f()f(0) f() 0 0, so f would be he consan funcion 0, conradicing he assumpion ha f (0). So f(0). Now for any, f f( + ) f() () 0 Since f() is a consan, his is f( ) f() 0 So f () f() for any. f()f( ) f() 0 f( + 0) f(0) f() 0 f()f( ). 0 f()f (0) f(). Regardless of how we go ha, we now know ha f solves he linear differenial equaion f () f() 0 Muliplying by he facor of inegraion e d e, we ge e f () e f() 0 The lef hand side is he derivaive of e f(), so e f() mus be a consan. Thus e f() k f() ke. So f () ke and f (0) ke 0 k, so k. Thus f() ke e for all. 6