Mah 6: Review for Final Exam, Par II. Use a second-degree Taylor polynomial o esimae 8. We choose f(x) x and x 7 because 7 is he perfec cube closes o 8. f(x) x / f(7) f (x) x / f (7) x / 7 / 7 f (x) 9 x 5/ f (7) 9x 5/ 9 7 5/ 87 Now plug in o he Taylor polynomial formula wih x 7. P (x) f(x ) + f (x )(x x ) + f (x ) (x x ) + (x 7) (x 7)! 7 87 Finally, evaluae a x 8. 8 P (8) + (8 7) 7 87 (8 7) 664 87.65797. Wha is he larges possible error ha could have occurred in your previous esimae? We now ha f(x) P n (x) K n+ (n + )! x x n+. In his case, n, x 7, and x 8. K max of f (x) on [7, 8] max of 7x 8/ on [7, 8] 7 78/ 7747 Puing his all ogeher, we have f(x) P (x) 7747! 8 7 5 544.94.. Use a comparison o show wheher each of he following converges or diverges. If an inegral converges, give a good upper bound for is value. (a) (b) 7 + 5 sin x x For all x, we have 7 + 5 sinx 7 + 5() because he maximum of sin x is. x x x x x x [ ] [ ( )] Therefore, he original inegral in quesion mus converge o a value less han. + x + x x + 7x For x, we have + x + x x + 7x x x + 7x. (We ve made he numeraor smaller and he denominaor larger, so he new fracion is smaller.)
x Bu x + 7x x x 7x x 4 x and we now ha yourself or noice ha p ). Therefore he original inegral mus also diverge. x diverges (compue for 4. Decide if each of he following sequences {a } converges or diverges. If a sequence converges, compue is i. (a) a + Terms are.,.,.,.,... ( + ), so he sequence converges o. (b) a ( ) Terms are,,,,... ( ) doesn exis, so he sequence diverges. (c) a + 5 Terms are 8/9, /, 8/, /5,... 7 + + 5 7 + 5 (by L Hopial s Rule or by inspecion), so he sequence converges o 5. 5. Decide if each of he following series converges or diverges. If a series converges, find is value. (a). +. +. +. +... a, so he series diverges by he nh Term Tes. (We eep adding s forever.) [Compare his wih he firs sequence of he previous problem.] (b) + / + / + /4 +... This is he famous Harmonic Series, which diverges even hough he erms do approach. We can use he Inegral Tes: (c) 5 5/ + 5/9 5/7 +... x diverges, which means ha This is a geomeric series wih r a, so i converges o mus diverge oo. r 5 ( /) 5 4. 6. Decide if each of he following series converges or diverges. If a series converges, find upper and lower bounds for is value. (a) ( ) + The erms of his series alernae in sign. And,.... 4 And, +. [Alernaing Series Tes] Therefore, by he Alernaing Series Tes, he series mus converge. We now ha any wo consecuive parial sums will provide upper and lower bounds: lower bound S [To ge beer bounds, use laer parial sums, such as S 5 and S 6.] upper bound S +
(b) ()! (!) a + a [(+)]! + [(+)!] ()! (!) ( + )! (!) ()! + [( + )!] ( + )( + ) ( + ) 4 + 6 + + 6 + [Raio Tes] Use L Hopial or divide each erm by. (c) (d) 4 Since he i of he raio is greaer han, he series diverges. ( + ) 5 5) [nh Term Tes] ( + 5) 98 + 5 6, so, by he nh Term Tes, he series diverges. 5 + 98 + 5 For, we have 6 9 8 5 > + 5 + 5 >. 9 8 Bu, 5 + 4 5 5 5 5 and we now ha p ). Therefore, he original series, which has larger erms, mus diverge also. [Comparison Tes] diverges (use Inegral Tes or noe ha (e) (ln()) [Inegral Tes] x(lnx) x x(lnx) x u x x lnx ln ln u du Subsiue u lnx, so du x. [ ln ln The inegral converges, so he series mus converge oo. ]
Furher, we now ha Therefore, our lower bound is And our upper bound is a + x(lnx) x(lnx) (ln()) a + ln. x(lnx) (ln) + ln. x(lnx). 7. Does he firs series from he previous problem converge absoluely or condiionally? ( ) +, which diverges by he Inegral Tes (chec for yourself). + Therefore, he firs series from he previous problem converges condiionally. 8. Compue he radius and inerval (including endpoins) of convergence for (x+) + (+) 5 + (x+) 5 (x + ) + (x + ) + (x + ) 5 x + 5 So, we are guaraneed convergence when x + 5 5 5 + (x + ) 5. Use L Hopial on he middle fracion. <. Bu his is equivalen o he following. < x + < 5 5 < x + < 5 8 < x < To chec convergence a he endpoins (where he Raio Tes is inconclusive), we plug in o he series iself. ( + ) 5 A x, we have 5 5, which is he Harmonic Series and hus diverges. A x 8, we have Series Tes. ( 8 + ) 5 ( 5) 5 ( ), which converges by he Alernaing Thus, he inerval of convergence is 8 x < and he radius of convergence is 5. 9. Find he complee Taylor series (in summaion noaion) for f(x) ln( x) abou x and deermine is inerval of convergence. f(x) ln( x) f() f (x) x f () f (x) ( x) f () f (x) ( x) f () f (4) (x) 6 ( x) 4 f () 6
Now plug in o he Taylor series formula wih x. f(x ) + f (x )x + f (x )! (x x ) + f (x ) (x x ) +... (x) +! x + 6 x +... We now find he inerval of convergence as in he previous problem. x x x x4 4... x x + (+) x x x x + x + Use L Hopial on he second fracion. So, we are guaraneed convergence for x <, which is equivalen o < x <. Now chec he endpoins. A x, we have, which is he negaive of he Harmonic Series and hus diverges. ( ) ( ) + A x, we have, which converges by he Alernaing Series Tes. So, he inerval of convergence is x <.. Wrie he complee series equal o e x and show ha i converges. We now e w + w + w! + w +..., so by subsiuion we obain he following.! Thus, e x e x + ( x ) + ( x )! + ( x )! ] [ x + x4! x6! +... ] [x x + x5 5! x7 7! +... + 5 5! 7 7! +... ( ) ( + )! The erms of his series alernae in sign. And, 5 5! 7.... 7! And, ( + )!. Therefore, by he Alernaing Series Tes, he series mus converge. +... x + x4! x6! +...
. (Secions A and B may omi his quesion.) The probabiliy densiy funcion (pdf) of he weighs of newborn oads in a cerain pond is given by f(x) (x + ) 4 4, where x is he weigh (in ounces). Noe ha he domain is x since no oad can have a negaive weigh. (a) Wha mus be he value of? We now ha he oal area under any pdf mus be (because i mus accoun for % of evens.) (x + ) 4 (x + ) 4 (x + ) (x + ) ( + ) ( + ) So, we have / or. (b) Wha fracion of he newborn oads weigh more han one ounce? (x + ) 4 (x + ) 4 (x + ) ( + ) ( + ) 8 8 Noe ha we could insead have compued and goen he same answer. (x + ) 4