a f(x)dx is divergent.

Transcription

1 Mth 250 Exm 2 review. Thursdy Mrh 5. Brig TI 30 lultor but NO NOTES. Emphsis o setios 5.5, 6., 6.2, 6.3, 3.7, 6.6, 8., 8.2, 8.3, prt of 8.4; HW- 2; Q-. Kow for trig futios tht /2 d /2. From Mth 50, for derivtives kow power rule, produt rule, quotiet rule, hi rule d rules from referee p. 5. For itegrtio kow power rule, u-substitutio d rules -20 from referee p. 6. Kow everythig from Exm review, iludig F)-F2). The followig problems re very importt for exm 2 d the fil. The ottio F*** mes it ws o 3 out of 3 of the lst 3 fils. F3*) Osiolly you eed to fid it usig Mth 50 tehiques isted of L Hospitl s rule. F08 b A improper itegrl is f(x)dx where =, b = or where f(x) hs vertil symptote for some [, b]. Improper itegrls re defied s its. If the it exists s rel umber, the f(x)dx is overget. If the it does ot exist, is or is, the f(x)dx is diverget. t i) f(x)dx = f(x)dx, ii) f(x)dx = f(x)dx. t t t Assume the two itegrls o the RHS below re overget. Let be y ostt (hose so tht you evlute the two RHS itegrls), the iii) f(x)dx = f(x)dx + f(x)dx = I + I 2. For iii), if either I or I 2 is diverget, the f(x)dx is diverget. So stop s soo s oe of the two itegrls is show to be diverget. Now let <bbe rel d suppose f(x) is otiuous o [, b] exept for vertil symptote t, b or (, b). (Severl vertil symptotes be hdled i similr mer.) t iv) If the vertil symptote is t b, the f(x)dx = f(x)dx. v) If the vertil symptote is t, the vi) If the vertil symptote is t, the t f(x)dx = f(x)dx = t b t + t f(x)dx. f(x)dx + f(x)dx = f(x)dx + f(x)dx = I + I 2. Agi, if either I or I 2 is diverget, the t t + t f(x)dx is diverget. So stop s soo s oe of the two itegrls is show to be diverget. { Be ble to show tht x dx =, if p> p p is diverget, p. Kow tht dx overges iff p >, tht x p 0 dx overges iff p <, d x p 0 dx diverges for ll rel p. Hee x p dx diverges for ll rel p. But you x p eed to show overgee or divergee by tkig its for exm, quiz d homework problems.

2 Compriso Theorem: Suppose f d g re otiuous futios with f(x) g(x) 0 for x. i) If ii) If f(x)dx is overget, the g(x)dx is overget. g(x)dx is diverget, the f(x)dx is diverget. Kow wht the grphs of t t, se t (d other trig futios) look like o ref. p. 2. Kow wht the grph of l t, e t d e t look like (see p. 44, 46, 54 d ref. p. 4). Kow wht the grph of t t looks like o ref. p. 3. Memorize the followig its. t et =, t et =0 t e t =0, t e t = l t =, l t = t t 0+ t t =, t t = t π t π se t =, se t = t π t π t t t = π/2, t t t = π/2 t se t = π/2, t se t = π/2 F4**) Show f(x)dx or f(x)dx or f(x)dx is diverget beuse i) the it is ± or does ot exist, or ii) by usig the Compriso Theorem. F07 5, S08 5, dx Itegrls x d (p )dx re espeilly ommo where p>d ofte p x p =. F5***) Fid f(x)dx for improper itegrl where = d b = re llowed. F07 4, S08 4, F08 3. Ofte the itegrl eeds to be foud usig trig substitutio or t dx i) t 2 + x = 2 t [t ( t ) t ( )] = [π 2 t ( )] or t dx ii) t x x 2 = 2 t [se ( t ) se ( )] = [π 2 se ( )]. (Ofte =. For i), ofte = 0. For ii) eed > d ofte =2.) A sequee { } = { } = {, 2, 3...} where there is futio () = with domi the positive itegers {, 2, 3,...}. If = L, rel umber, the the sequee { } overges or is overget. If the it does ot exist or is ±, the the sequee { } diverges or is diverget. Nottio for sums is useful. = = Itegrls re its of sums: f(x)dx = f(x i ) b i= where x i = +i(b )/. Algebr of sums: =, =, i = i, ( i ± b i )= i ± b i. i= i= i= i= i= i= i= 2

3 - A ifiite series is sum = The th prtil sum s = i = i= If the sequee {s } is overget with s = s, rel umber, the is overget or overges with = s. If {s } is diverget, the is diverget or diverges. Suppose 0. The geometri series the geometri series r diverges. r = r if r <. If r, the A overget ifiite series remis overget if the st k terms re disrded or if k + dditiol terms 0,,..., k re dded. A diverget ifiite series remis diverget if the st k terms re disrded or if k + dditiol terms re dded. Note tht the geometri series r = 0 + r. =0 Note tht dr = dr r is geometri series with = dr. Suppose is overget (diverget). The = remis overget (diverget) for y 0.. The p-series = p + p overges for p> d diverges for p. p 3p The diverget series is lso lled the hrmoi series. Let d d be itegers with d. The telesopig series kd ( + )( + d + ) = k d ( + )( + d + ) = k ( i= i + i + d + )= k [ d d + ]= k[ d + ]. Similrly, the telesopig series kd =j+ ( + )( + d + ) = k d =j+ ( + )( + d + ) = k ( i=j+ i + i + d + )= k [ j ++ + j j + d d + ] = k[ j j + d + ]. Typilly k =ork =/d, j = d d 3. To see this, ote tht 3

4 the prtil sum s = A i + + B i + d + i= kd (i + )(i + d + ). The term i hs prtil frtio expsio or kd = A(i + + d)+b(i + ). If i =, th kd = Ad or A = k. If k i = ( + d), the kd = B( d) orb = k. Thus s = ( i= i + k ). Thus i + d + kd ( + )( + d + ) = s = k ( i= i + ). All but the st d terms i + d + of d the lst d terms of i + i + d + el whe s is writte s telesopig sum. kd Thus ( + )( + d + ) = k [ d d + ]. kd Whe fidig, write out the prtil frtio expsio ( + )( + d + ) to fid A d B d fid the it of the prtil sum k ( i= i + i + d + )= k [ d d + ]. You should be ble to fid the 2d terms of s = i=j+ ( ) quikly sie i+ i+d+ they re the st d terms of d the lst d terms of i + i + d +. Ex. ( i=2 i i + )= + 2 sie d = 2 with i+ = i +d = i Ex. ( i= i i +3 )= sie d = The th term test for divergee: If it is ot true th =0, the is diverget. So diverges if = L 0 or if does ot exist. Note tht =0 is iolusive: the series ould overge or diverge. The itegrl test: Suppose f is otiuous, positive, deresig futio o [, ). Let = f(). ) If f(x)dx is overget, the is overget. = b) If f(x)dx is diverget, the is diverget. = Tips: i) Usully = d the sum eed ot strt t sie overgee or divergee of ifiite series i=d does ot deped o the st k terms. 4

5 ii) Be ble to show tht overges for p>. x dx = p p if p> to show tht the p series iii) Be ble to show tht x dx = if p to show tht the p series p p diverges for p. iv) If f(x) dx = L, the series overges, but L i geerl. v) The itegrl test is ofte used where f(x) =/g(x) orf(x) = where g(x) g(x)h(x) d h(x) re iresig. vi) Rell tht otiuous f(x) is deresig o [, ) iff (x) < 0o(, ). vii) Futio g(x) =x p is positive d iresig for x>0 d p>0, h(x) =[l(x)] p is positive d iresig iresig for x>d p>0, g(x) =e x is positive d iresig o (, ). viii) The futio g(x) = l(x)/x is positive d deresig for x > eso for x > 3. Kow tht i) xp 0sx for ll p. ex So ii) ex s x for ll p. xp iii) l x 0sx. x x So iv) s x. l x Let t (θ) =w. The t[t (θ)] = θ = t w d w is the gle (i rdis) whose t is θ. So fid w ( π/2,π/2) suh tht t(w) =θ. The t (θ) =w. Kow tht t(0) = 0 d t(π/4) =. Thus t (0) = 0 d t () = π/4. Let si (θ) =w. The si[si (θ)] = θ = si w d w is the gle (i rdis) whose sie is θ. So fid w [ π/2,π/2] suh tht si(w) =θ. The si (θ) =w. Kow tht si(0) = 0, si(π/2) = d si( π/2) =. Thus si (0) = 0, si () = π/2, d si ( ) = π/2. Let se (θ) =w. The se[se (θ)] = θ = se w =/ os(w). So fid w (0,π/2) (π/2,π) suh tht se(w) =θ or os(w) =/θ. The se (θ) =w. Kow tht os( π )=/2 so 2 = se(π/3). Thus 3 se (2) = π/3. - The Compriso Test: Suppose d b re series with positive terms. ) If b is overget d 0 < b for ll, the is overget. b) If b is diverget d 0 <b for ll, the is diverget. Tips: For the ompriso test, usully b = is p series whih overges for p p> d diverges for p, or b = r is geometri series whih overges for r < d diverges for r. p 5

6 where the first term = b is egtive. The Limit Compriso Test: Suppose d b re series with positive terms. If = >0 where is rel, the either both series diverge or both series b overge. - Note:! is red ftoril d! =( )( 2)( 3) 3 2. So! =[( )!] = ( )[( 2)!] = ( )( 2)[( 3)!] et eter. Here is iteger d 0! =. Kow tht domites! whih domites the expoetil for y ostt. Also the expoetil domites p! for y >, p > 0. Thus = 2 0 = 3 2. Similrly, = 0 = p d = 0 =. Now = 0 = p! 2 2. By p L Hospitl s rule, = p p l() = p(p ) p 2 p(p )(p 2) p 3 = [l()] 2 [l()] 3 p! p = = =0if> d p is iteger. Thus [l()] p =0if> for y rel p by the ompriso theorem.! p To summrize, = 0, = 0 for y ostt, = 0 d p = 0 for y ostt p, = 0 for y ostt, d!! p = 0 for y > d y rel p. - ( Also, kow tht + = e. ) A ltertig series hs terms tht re ltertely positive d egtive. If b > 0, the ( ) b = b b 2 + b 3 b 4 + is ltertig series where the first term = b is positive, while ( ) b = b + b 2 b 3 + b 4 is ltertig series Altertig Series Test: Suppose b > 0 d ( ) b (or ( ) b ) is ltertig series. i) If b + b for ll (or if b is deresig), d ii) b =0, the the series is overget. Tip: For ltertig series, try the th term test for divergee d the Altertig Series Test for overgee. Note tht if b 0 the = ( ) b 0 (does ot exist), d diverges by the th term test. - 6

7 A series is bsolutely overget if is overget. A series is oditiolly overget if = is diverget but is overget. Absolute Covergee Theorem: If is bsolutely overget, the is overget. Thus if is diverget the is ot bsolutely overget d = is diverget. If is overget, the is overget. Compriso Test for Absolute Covergee: If b for ll d b overges, the is bsolutely overget d so overget. Tip: This test is useful for showig bsolute overgee, d is ofte useful if the umertor equls (or otis) si(), os() or( ) sie the the si(), os() d ( ) =. The Rtio Test: i) If + = L<, the is bsolutely overget d so overget. ii) If + = L> where L = is llowed, the diverget d so ot bsolutely overget. iii) If + =, the the test is iolusive (so it is possible tht is diverget, oditiolly overget or bsolutely overget). Tips: i) If the idex of is ivoved s expoetil or ftoril ( or ( )! where is rel), the the rtio test is ofte useful for showig overgee or divergee of. ii) If is oly ivolved lgebrilly or logrithmilly (eg, or p (4 3)(4 ) 2 l(4+) ), the ofte the rtio test will fil (be iolusive). (+) The Root Test: i) If = L<, the is bsolutely overget d so overget. ii) If = L> where L = is llowed, the diverget d so ot bsolutely overget. iii) If =, the the test is iolusive (so it is possible tht is diverget, oditiolly overget or bsolutely overget). Tips: i) The root test is useful whe th powers our. ii) / = d =. / iii) = /. 7

8 F6***) Show tht series overges or diverges usig series tests. F08 4, S08 7, F07 6. F7***) Show tht series is bsolutely overget, oditiolly overget or diverget. (Usully the series is oditiolly overget ltertig series, so you hve to show tht is diverget, but is overget.) F08 5, S08 8, F07 7. Tips: i) The geometri series r d the sled geometri series =j+ overge if r < d diverge if r. ii) The p-series =j+ d the sled p-series p diverge for p. =j+ p r +k =j+ overge for p> d iii) The th term (divergee) test: If 0, the diverges. This test is ofte useful for ltertig series d for series where is rtio of polyomils d roots. iv) The itegrl test: Suppose f is otiuous, positive, deresig futio o [, ). Let = f(). The overges iff f(x)dx is overget. = The itegrl test is useful for p-series d d sometimes for futios p [l()] p f(x) = /g(x) orf(x) = /[g(x)h(x)] where g d h re iresig futios. v) Absolute Covergee Theorem: If overges, the overges. =j+ =j+ vi) The Compriso Test: ) If b overges d 0 b for ll, the overges. b) If b is diverges d 0 b for ll, the is diverges. vii) Compriso Test for Absolute Covergee: If b for ll d b overges, the is bsolutely overget d so overget. viii) The Limit Compriso Test: Let d b be series with positive terms. If = >0 where is rel, the either both series diverge or both series overge. b Tests vi) viii) re useful if is rtio of polyomils d roots. For the it ompriso test, let b =/ p = lrgest power i umertor/lrgest power i deomitor of where p>0 (so tht the series overges). ix) Altertig Series Test: Suppose b > 0 d ( ) b (or ( ) b )is ltertig series. If ) b + b for ll (or if b is deresig), d b) b =0, the the series is overget. x) The Rtio Test: Let + = L. ) If L<, the is bsolutely overget. b) If L>, the diverges. ) If L =, the the test fils. This test is useful if otis terms like ( + j)! d. xi) The Root Test: Let = L. ) If L<, the is bsolutely overget. b) If L>, the diverges. ) If L =, the the test fils. This test is useful if =( ), = () or p = p ( ). 8