Solutions to selected problems from the midterm exam Math 222 Winter 2015

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Transcription:

Soluios o seleced problems from he miderm eam Mah Wier 5. Derive he Maclauri series for he followig fucios. (cf. Pracice Problem 4 log( + (a L( d. Soluio: We have he Maclauri series log( + + 3 3 4 4 +..., ( so log( + The a ermwise iegraio yields + 3 3 4 +.... ( (b C( L( cos d. log( + Soluio: We have he Maclauri series so d + 3 3 3 4 4 4 +.... (3 cos! + 4 4! 6 6! +..., (4 cos The a ermwise iegraio yields C( e e (c S( d. Soluio: We have he wo Maclauri series which imply ha e e The a ermwise iegraio yields S(! 4! + 4 6! 6 8! +.... (5 cos d! 3 3 4! + 5 5 6! 7 7 8! +.... (6 e + +! + 3 3! + 4 4! + 5 5! +..., (7 e +! 3 3! + 4 4! 5 5! +..., (8 e e + 3! + 4 5! + 6 7! +.... (9 d + 3 3 3! + 5 5 5! + 7 7 7! +.... (

. Decide if he followig series coverge. (To be compared wih Pracice Problem 5 (a log e + e. Soluio: For large, we have which imply ha e + +, e +, ( log e + e log( +, ( where i he las sep we have used he firs erm of he Maclauri series log( + +.... This suggess ha he origial series coverges, ha i should be comparable o he series. The he i compariso es gives us he i I order o compue his i, we replace i by log e + e. (3 maipulae i as log e +e log(e + e log, (4 log(e + e log ( e e e +e e + e e e e e e + e, where we have applied L Hôpial s rule wice, oce i he firs sep, oce i he peulimae sep. The coclusio is ha (5 (b hece he series coverges. log ( arca. Soluio: For large, we have, (6 arca 3 3, so log ( arca ( log 3 3. (7 This suggess ha he origial series coverges, ha i should be comparable o Page

he series. The he i compariso es gives us he i I order o compue his i, we replace i by log arca apply L Hôpial s rule wice, as log ( arca. (8 log f(, wih f( arca, (9 log f( f ( f( f (f( [f (] [f(] f (f( [f (] [f(]. ( Now, from he Maclauri series f( arca 3 3 +... 3 +..., ( we have f(, f (, f ( 3, which yields I oher words, we have log f( f (f( [f (] [f(] f (f( [f (] [f(] 3 ( 3. log ( arca so he series coverges. 3. Deermie he covergece radius of he power series b + L b, or 3 3, (3 ( +, where a. Soluio: This should be compared wih Pracice Problem 7. Wih he -h erm of he give series deoed by b ( +, a applicaio of he raio es leads o + + a +. (4 + ( + + + + ( + I order o compue he i, we eed o have some idea o how fas he idividual erms i + grows wih as. I paricular, oe ha grows faser ha if a >, ha does o grow a all if a. Therefore we spli he problem io wo cases. Firs, assume ha a >. The we have + + a + + + a + a+ + a + a a. (5 + Page 3

Now assume ha a. I his case, we have + + a + + + + a+ + + a +. (6 Based o hese compuaios, we coclude ha b + L b { a if a >, if a. (7 I he firs case, we have covergece for a < divergece for a >, hece he covergece radius is R a. I he secod case, we have covergece for < divergece for >, hece he covergece radius is R. We ca also wrie i i a sigle formula, as R mi {, a}. (8 4. Compue he ui age T (, he pricipal ui ormal N(, he curvaure κ(, for each of he followig plae curves. (a X( ( si, cos, < < π. Soluio: By a direc compuaio, we have which yield X ( ( cos, si (9 X ( ( cos + si + cos cos + si cos, (3 T ( X ( X ( ( cos cos, Goig furher, we compue leadig o X ( X ( cos si si ( cos, cos si cos. (3 X ( (si, cos, (3 κ( X ( X ( X ( 3 cos ( cos 3 si cos ( cos cos si cos, (33 cos, (34 where we have ake io accou he fac ha cos < for < < π, hus cos cos. The laer fac also implies ha X ( X ( cos <, so ha he curve is bedig o he righ, as he miimal ur from X o X is clockwise. Hece he pricipal ormal N( is equal o T (, roaed 9 degree clockwise: ( si cos N(, (35 cos Page 4

Noe: By usig he double-agle formula cos cos ( si (, we ge cos cos ( + si ( si ( + cos ( cos ( + si ( si (. (36 This gives simplificaios o may of he formulas above. For eample, we have where we have also used si si( cos(, T ( X ( X ( ( si(, cos(, (37 κ( 4 si( (38. T N π Figure : Cycloid, give by X( ( si, cos. The blue par correspods o he parameer values < < π. (b X( (cos 3, si 3, < < π. Soluio: A direc compuaio yields X ( ( 3 cos si, 3 si cos, (39 X ( 9 cos 4 si + 9 si 4 cos 9 si cos (cos + si 9 si cos. (4 Sice boh si cos are posiive for < < π, we have herefore Furhermore, we have X ( 3 si cos 3 si cos, (4 T ( X ( ( X ( 3 cos si 3 si cos, 3 si cos ( cos, si. (4 3 si cos X ( (6 cos si 3 cos 3, 6 si cos 3 si 3, (43 Page 5

X ( X ( 3 cos si 3 si cos 6 cos si 3 cos 3 6 si cos 3 si 3 8 cos 4 si + 9 cos si 4 8 si 4 cos + 9 si cos 4 9 si 4 cos 9 si cos 4 9 si cos. (44 Because X ( X ( <, we see ha he curve is bedig o he righ, which meas ha he pricipal ormal N( is equal o T (, roaed 9 degree clockwise: Fially, le us compue he curvaure, as N( (si, cos. (45 κ( X ( X ( X ( 3 9 si cos 7 si 3 cos 3 3 si cos. (46 T N T N (a Asroid, give by X( (cos 3, si 3. The blue par correspods o he parameer values < < π. (b Hyperbolic spiral, give by X( ( cos, si, >. Figure : Curves from Quesio 4(b Quesio 4(c. (c X( ( cos, si, >. Soluio: Le us sar wih ( si cos + ( cos si X ( ( si cos, cos si si + cos + si cos + cos + si si cos +, (47 (48 Page 6

which imply ha Furhermore, we have X ( +, (49 T ( X ( X ( ( si cos, cos si. (5 + ( X ( cos ( si cos ( 4, ( si ( cos si 4 ( 3 cos + si + cos 4, 3 si cos + si 4. From he preiary compuaio si cos 3 cos + si + cos cos si 3 si cos + si ( si cos ( 3 si cos + si ( cos si ( 3 cos + si + cos (5 4 + 4, (5 we ifer ha which i is ur gives X ( X ( 4 4, (53 κ( X ( X ( X ( 3 6 ( + 3 4. (54 ( + 3 Fially, sice X ( X ( >, he curve is bedig o he lef, as he miimal ur from X o X is couer-clockwise, hece he pricipal ormal N( is equal o T (, roaed 9 degree couer-clockwise: N( ( cos + si, si cos. (55 + Page 7