4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3

Transcription

1 Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x ( x = This meas that the series coverges absolutely as log as x < or x / < or / < x < / or / < x < 5/ ad the radius of covergece is /. It still remais to check the ed poits of the iterval though. Particularly at x = / we get ( ( =0 = ( ( =0 which is a diverget, harmoic series. At x = 5/ the series becomes: ( ( = (, =0 This coverges by the alteratig series test sice = 0. Thus we iclude x = 5/ but ot x = /. I iterval otatio, the values of x for which this power series coverges is ( /, 5/]. (x. Give the series, fid the values of x for which this power series coverges. Also state (! =0 clearly what the radius of covergece is. We start by settig up the Ratio Test: =0 ( x (! x (! ( x (! (! x (! x ( (! x ( ( ( = 0 This meas that the series coverges absolutely for all x by the ratio test. Thus the values of x are (, ad the radius of covergece is.

2 . (a Fid the third Taylor polyomial, T (x, for f(x = x cetered at a = 8. To fid the rd Taylor polyomial we first evaluate the first three derivatives ad the evaluate those fuctios at 8. This is summarized by the table below. Thus we have that: f ( (x f ( (8 f ( (8/! f(x = x f(8 = f (x = f (8 = x 6 6 f (x = f (8 = (x 08 6 f (x = 8(x 5 f (8 = T (x = (x (x 8 (x (b Use T (x foud i (a to approximate 9., (if your aswer is a decimal give 7 decimal places. 9. T (8. = 6 (0. 6 ( (0. = (c What is a upper boud for the maximum error for T (x x for 8 < x < 8.? The remaider of the Taylor series is give by: ad it is bouded i absolute value by: R = f ( (c (x 8 (! R M (! (x 8, where M is the maximum of f ( (x for the values of x that are of iterest. I this case we are usig = ad f (x = 5 6(x 7/. For 8 < x < 8. the ad Now this implies that, Thus Thus we have that 9 < x < / < (x 7/ < (9. 7/. (x < 7/ 9 7/ 5 6( x 7/ < 5 6(9 7/ x T (x 5!(6(9 x 5 7/ 8!(6(9 8. 7/

3 . For each of the followig series decide if they are absolutely coverget, coditioally coverget or diverget. Justify your aswers with the appropriate tests. (a = si( For this series you might attempt the ratio test at first, but that is ot goig to tell you much. However, you ca recogize that, si( <. Si is a coverget p-series the by the compariso test si( is also a coverget series by the compariso test. Cosequetly the origial series with out the absolute values is ot oly coverget but is absolutely coverget. ( (b (! = Usig the Ratio test, we have (c! (! Thus by the ratio test, the series coverges absolutely. 0 ( = We agai start with the ratio test, 0 ( 0 ( (!! = 0 0 ( ( ( 0 0 ( ( = 0 6 Thus by the Ratio Test, the series coverges absolutely.

4 (d ( ( l = This aother problem where the Ratio Test is icoclusive ad is hard to check ayway, if I set up the it for the ratio test it would look like ( ( l l ( ( ( (e Here I have applied L Hospitals rule sice the umerator ad the deomiator of the origial fractio are both goig to l( = 0. After some algebra we the have: ( ( l l ( ( ( = Cosequetly, the Ratio Test is icoclusive. We go back ad look closer ad recogize that sice 0 l ( = 0 the the origial series coverges by the alteratig series test. So we ow cosider the series ( = l, but we showed o a previous quiz that this is a diverget series by recogizig this as a collapsig series ad showig that the partial sum s diverge. Thus the origial series is coverget, but ot absolutely coverget. I other words this is a coditioally coverget series. ( ( = The ratio test i this case will produce a it of, so it will be icoclusive. However, if we look at the ( ( ( = 0 Cosequetly, by the Alteratig series test, the series coverges sice the it of the terms goes to 0. You also have to make sure that the series is decreasig, but that is also true after some value if you sketch a quick graph of the fuctio. O the other had if I look at the absolute value of the terms: (, = the it appears that the series is comparable to = =, which is a diverget harmoic series. To check that this compariso really works we set up the it compariso test givig, ( ( ( ( ( = =

5 (f! = First we ote that sice this is a series with positive terms the either the series coverges absolutely or it diverges. We the check the ratio test givig, (g (!! (!! = Thus by the ratio test, the series diverges. You could also make a argumet why the terms do ot go to zero ad thus is diverget by the divergece test. = e Agai sice the terms are positive, coditioal covergece is ot a optio. Lookig at this series we could use the compariso test with the coverget p-series. Either you could use the regular compariso test ad recogize that,. e e ad sice e is a coverget p-series, the the origial series must also coverge. You could also use the it compariso test ad compute the it: e e = e 0 = Sice this is a fiite it that is greater tha 0, the by the it compariso test either both series coverge or both diverge ad sice is a coverget p-series, the the origial series must also coverge. 5

6 5. Determie if the followig series coverge or diverge ad if they coverge give the exact value of the series. (a = This is clearly a telescopig series, so we compute the partial sum: s = = = = 9 5 ( 5... (... ( ( It is the easy to see that s = 9 ad thus the series coverges ad the sum is equal to 9 If you have trouble seeig how the cacellatio works above write out a few if the partial sums. s = s = s = s 8 = ( ( = = = 9 0 (b (7 = ( First we rewrite the series as: = (7 ( = = ( 7 = = ( 7 = = 7 ( 7 This is a geometric series with r = 7 >. Thus this series diverges. You could also show taht this series diverges by the divergece test as well. 6

7 6. Use the Maclauri series for si(x to compute a Maclauri series for F (x which satisfies, ( si x F (x = dx, F (0 = x We begi with the Maclauri series for si(x which is: si(x = ( x (! = x x! x5 5! x7 7!..., R = =0 Thus, Thus, Thus, si ( x = ( x (! = x x6! x0 5! =0 x 7!..., R = si ( x = ( x x (! = x x5! x9 5! x..., R = 7! =0 F (x = ( si x dx = C ( x x (! dx =0 = C ( x ( (! =0 = C x x6 6(! x0 0(5! x (7!..., R = Now sice F (0 =, the C = ad we have the Mclauri series for F (x is give by, F (x = ad has a ifiite radius of covergece. ( x ( (! =0 7

8 7. Determie if each of the followig sequeces coverge or diverge. If they coverge give their it. Justify your aswers. ( (a b = si This sequece is of the form 0, ad we ca apply L Hospital s rule after applyig some algebra to make it a 0/0 type of problem. ( si si ( cos cos ( = by L Hospital s Rule Cosequetly, this sequece does coverge to the value. (b c = e e e Here we use some it laws. (c d = e e e Thus the it coverges to the value 0. (!! (e e (e (e e e e e = 0 (!! (!!!!!! = Thus this series diverges to positive ifiity. 8

9 8. True or False: Decide if each of the followig statemets are true or false, ad discuss briefly why. (a If a = 0, the a coverges. = This is false, this statemet is oly true i the special case of alteratig series with decreasig terms. A couter example to this statemet is that the harmoic series has terms that go to zero, but the sum = does ot coverge. (b If a power series c (x coverges at x = 5 but diverges at x = 7 the it must coverge =0 at x =. This is true, the implicatio is that the radius of covergece is at least but o more tha 5, this implies that the series must coverge for x <, must diverge for x > 5 ad without more iformatio we do t kow what happes for < x < 5. The value x = though is i the first case, ad thus the series must coverge here. (c If the partial sum s 5 = k= bouded above by the itegral 5 ( k is used to approximate s = ( k the the error s s 5 is 5 ( x dx. This is false, the statemet that is give is the boud that is used for fuctios that coverge via the itegral test. The coditios eeded to apply this boud are that the terms are positive, the fuctio is decreasig after some poit, ad that the correspodig fuctio is cotiuous. The first requiremet, that the terms are positive, is ot satisfied here so we ca t apply this boud. There is a boud though i this case ad it would be give by b = b 6 = 6, by the error boud for alteratig series. k= 9

10 9. Cosider the followig coverget series, s = k= k(l(k Fid the smallest so that the th partial sum s approximates s to withi.0. Do your calculatios exactly istead of guessig ad checkig. This is a case whe we ca apply the error bouds for a series that coverges via the itegral test. Here the fuctio is certaily cotiuous ad o-egative where it eeds to be ad a quick glace at the graph shows that it is decreasig as well. It is of ote that this series starts at k = rather tha k =, which is good because the fuctio is cotiuous o [, but ot o [,. I ay case we defie s = k= the we kow that: x(l(x dx s s x(l(x dx k(l(k Sice we oly wat to get withi.0 we are really oly iterested i the right had side of the previous equatio. If the problem asked to bouds o the error or to fid a iterval for which the error was i, the we would use both the lower ad the upper boud, but here we oly eed the upper boud. Now we compute the itegral o the right, s s x(l(x dx = l( b b = l( x(l(x dx u du, u = l(x, du = x [ ] b u l( [ b ] l( Now i order to get the desired level of accuracy eeded you eed to fid the first where /(l( <.0, so we set up the iequality, or or (sice l(x is a icreasig fuctio l(.0 00 l( e 00 So the first iteger that works is the first iteger greater tha or equal to (e 00 e So the first umber that works is

11 0. Compute the it below by usig a Taylor series for x cetered at a =. x ( x x You should oly eed the first three terms of the series. First let us write out the first terms of the Taylor series: f ( (x f ( (0 f ( (0/! f(x = x f( = f (x = x x f ( = / -6 f (x = (9 x ( x x f ( = ( = This gives that the series is of the form: /x = 6(x (x... So ( /x x = 6 (x... Cosequetly, the it we seek is 6 sice all of the terms ivolvig (x go to 0 as x.