Exercise 1. Sketch the graph of the following function. (x 2

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Gerald Wilkerson
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1 Writtn tst: Fbruary 9th, 06 Exrcis. Sktch th graph of th following function fx = x + x, spcifying: domain, possibl asymptots, monotonicity, continuity, local and global maxima or minima, and non-drivability points spcifying th typ, if any. Morovr, writ th quation of th tangnt lin to th graph of f in th point 0, f0. Solution It is asy to s that dom f = R so dom f = {, + }. Morovr f is continuous ovr its domain bing a composition of continuous functions. Obsrv that f = 0 x = or x = ±. Morovr f0 < 0 < x < and, according to th dnition of th absolut valu, w hav x + x if x < fx = x + x if x. Study at dom f: fx = x + x = x x + x + 6 = x + x x x 6x = x + x + o x = x + x + o = x + o +. x x Hnc y = x is th function of an obliqu asymptot for f, whn x. Similarly fx = x + x = x + x + x + o = x + x = x + + o x + x + o x and y = x + is th function of an obliqu asymptot for f, whn x +. Monotonicity and critical points f x = x + x, hnc x + x x x + x x + 6x if x < f x = x + x x = x + x + x + x x + 6x if x > x + x x + x if x < if x >. Th drivativ do not xist for x =, and in particular f is not boundd for ths valus and th sign of th drivativ dpnds only on th numrator in both cass and it is as follow: f x = 0 x = x = and x = x = +, i.. x, x ar th critical points of f; x + 6x > 0 x < x or x > x and x + 6x < 0 x < x < x ; vic vrsa for x + 6x.

2 Hnc w hav: for x <, f x < 0 and bcaus for x > f x > 0 thn, f is a local minimum for f; for x >, w hav f x > 0 for < x < x ; f x < 0 for x < x < x ; f x > 0 for x > x. Thn x, fx is a local maximum and x, fx is a local minimum for f. Obsrv that lim f x = 7 = ± so, f is a cusp point for f. x ± 0 ± Morovr lim x f x = = + and lim ± 0 + x f x = = so, f and ± 0 +, f ar vrtical tangnt points for f. Tangnt lin f 0 = 6 and f0 = 6 so th tangnt lin at 0, f0 is y = Exrcis. Considrd th function ft = arctan log t t log t, discuss th intgrability and calculat, if possibl, th intgrals:.a ft dt ;.b + 6 ft dt. x 6. Solution Th function ft admits primitivs as an lmntary functions, obtaind as follow. arctan log t arctan x t log t dt = ln t=x x dx = arctan x b.p. x + x x + dx. Hr th rst quality is obtaind by th indicatd substitution and th scond on by intgration by parts. Now th rational function in th last intgral dcomposs as x x + = A x + Bx + C x + whr,.g., w obtain th thr constants by: A = x + = x=, B : lim x x + x x + = A + B B = and C : x x + = = A + C C = x=0. In a pur algbraic way, it is possibl to obtain th constants A, B and C also by th rlation A x + Bx + C x + = Ax + + Bx + Cx x x = + x x +

3 quating th numrators in th last quation Ax + + Bx + Cx = x A + B + x B + C + A C = and solving th linar systm obtaind by comparing th polynomials at any powr of x A + B = 0 B = A A = B + C = 0 C = A A C = A = C = B =. Hnc for th last intgral abov w hav x x + dx = x x + x dx + and th primitivs ar givn by ft dt = arctan x x +.a.b + arctanln t = + ln t ln x arctan x lnx + + C ln ln t arctanln t lnln t + ftdt is a dnit Rimann intgral and, using th primitiv in t, w hav ft dt = arctan arctan + 4 ln. + C ft dt is impropr bcaus th domain of intgration is unboundd, at t, and bcaus ft is unboundd for t +. Using th ordr tst for impropr intgrals of positiv functions and fundamntal xampls, w hav π + o ft = t + t ln, hnc f has a convrgnt intgral t For t, obsrv that ln t = + t and a ot, so that ft = t + a ft dt for any a >. arctan + o t ft dt, for any ral a >, is divrgnt. Hnc, by additivity, th impropr intgral ft dt is divrgnt. Similarly, it is possibl to discuss. or.b using th substitution and primitiv in x = ln t. Exrcis. Discuss th absolut and simpl convrgnc of th following sris: n tan + n n x n n +.a ;.b ln, for x R. n + x + n + n= Solution.a n tan + n n For = a n w start discussing th Cauchy ncssary condition, and w n + n= n= hav n tan + n n n tan + o n a n = = = tan + o n n + n + + n

4 x qx and x qx. π bcaus tan > tan =. Morovr, a n > 0 for any n >, hnc w hav absolut and 4 simpl divrgnc..b Lt q = x x + so that w can writ th sris as n + q n ln and obsrv that for x = n + w hav q = 0 so th sris is trivial and for x = w hav that q and th sris ar not wll dnd. Morovr w hav n + ln = ln < 0 for any n 0. n + n + Absolut convrgnc W discuss x x + n n + ln = n + q n ln n + ln = n + n ln n + n + for x R. Hr w obsrv that n + = n Thn, for q, that may b obtaind, for xampl, by studying th graph of th functions x qx as in th gur, w hav dirnt charactrs of th sris. For < x and x w hav 0 < q < so that q n ln is absolutly convrgnt n + bcaus,.g. by comparison, 0 q n ln < vntually for n. n + n For x < w hav q > and qn ln > n vntually for n, so w hav n + absolut divrgnc. For x = w hav that q = and th sris rducs to ln that is absolutly n + divrgnt bcaus of th innitsimal tst, bing ln quivalnt to for n. n + n So th sris is absolutly convrgnt for < x and for this valus of th paramtr x is also simply convrgnt by th thorm of absolut convrgnc. Simpl convrgnc By th thorm of absolut convrgnc, th sris n. n + q n ln is also simply convrgnt n + for < x, i.. q < but w hav to discuss th convrgnc for th cas q. 4

5 For x = w hav q = and dning b n := ln = ln + > 0 for n + n + vry n 0, th sris is n ln = n+ ln n + n + = n+ ln + = n b n n + i.. an altrnating sign sris that w discuss using Libniz tst: i b n = + o 0, for n, so th rst condition is satisd; n + ii b n is dcrasing for n, in fact: n is strictly incrasing w us this symbol, thn n +, thn is dcrasing, thn + and ln +. n + n + n + W can also pass to th function x ln + and s that its drivativ x + x + x + is ngativ for x. For x < w hav q < and th sris is n q n ln, with q >, and n + w us Libniz tst again. Th squnc do not satisfy th rst condition, in fact b n + for n, so th Libniz critrium may not b usd. Howvr, it is asy to s that, bcaus b n +, th altrnating sign maks th sris indtrminat. Exrcis 4. Find th solution of th following Cauchy problm: y + y + y = x x y0 = 0. y 0 = W shall writ th gnral intgral of th quation as yx = y 0 x+y p x whr y 0 is th gnral intgral of th associatd homognous quation y + y + y = 0 and y p is a particular solution of th non-homognous quation y + y + y = g, with sourc gx = x x. Th charactristic polynomial of th homognous quation is λ + λ + = 0, whos solutions ar λ, = ± i. Hnc w hav y 0 x = x A cos x + B sin x, with constants A, B R. Thn, bcaus th sourc trm is of th form polynomialx xpαx, w may sarch for a particular solution of th form y p x = ax + b x, bing α not a root of th associatd charactristic polynomial. So w hav y = a x + y and y = a x + y = a x + y. Substituting in th quation y + y + y = x x and quating w obtain a = and b = 4, i.. x 4 x. y p x = Hnc th gnral intgral of th quation is yx = x A cos x + B sin x + A, B R. Now w can x th constant A and B imposing th givn initial conditions: y0 = A 4 = 0 A = 4 y 0 = yx + x A sin x + B cos x + x + and th solution of th Cauchy problm is givn by yx = x 4 cos x + 9 sin x + x 4 x=0 x 4 x. x 4 x with = B 4 = B = 9,

6 Exrcis. Comput th following limit: lim n + n4 + ln n ln lim n + n4 + ln n ln + n n sin n = lim n + n4 + ln n n n 4 + o n 4 n = lim n + n4 + ln n n o 6 n 4 = lim ln n n + n 4 =. + n n sin. n n 6n + o n 4 = lim n + n4 + ln n n 4 + o 6

BSc Engineering Sciences A. Y. 2017/18 Written exam of the course Mathematical Analysis 2 August 30, x n, ) n 2

BSc Engineering Sciences A. Y. 2017/18 Written exam of the course Mathematical Analysis 2 August 30, x n, ) n 2 BSc Enginring Scincs A. Y. 27/8 Writtn xam of th cours Mathmatical Analysis 2 August, 28. Givn th powr sris + n + n 2 x n, n n dtrmin its radius of convrgnc r, and study th convrgnc for x ±r. By th root