Lecture Appedix B: ome sample problems from Boas, Chapter Here are some solutios to the sample problems assiged for Chapter, 6 ad 9 : 5 olutio: We wat to use the geeral expressio for the form of a geometric series a r ar 0 To fid the fractio correspodig to 058333 To use this expressio we eed to look for a repeatig decimal i order the geometric series To this ed we rewrite the startig umber as 0583333 05 03 003 0003 00003 ow we ca rewrite the repeatig decimal as a sum ad use the geeral form to fid the desired fractio, 03 3 4 4 0 0 0 0 03 03 7 4 4 09 4 3 0 : olutio: Water purificatio as defied here is aother process described by a geometric series If oe-th of the curret impurity is removed at each stage, the total fractio of the impurity removed after stages of purificatio is give by Impurity, Physics 7 Lecture Appedix B Autum 008
o ow we cosider the limit for the two cases metioed, = ad = 3, too fid what fractio of the impurity we ca remove i the limit of a ifiite umber of stages, lim Impurity, lim lim, 3 3 lim Impurity 3, lim lim 3 3 3 Thus i the first case all of the impurity ca be removed, but i the secod case, with oly a third removed at each stage, the best we ca hope for is to remove oe half of the impurity 6: 3 olutio: We wat to practice usig the compariso test for covergece/divergece Let s proceed by comparig to the coverget geometric series with x C c c 4 8 6 0 0 ow we compare this to the series of iterest, a 4 9 6 As suggested i the problem we proceed by performig some maipulatios, ie, resumig, o the series We have, 4 9 6 5 36 49 64 5 ie, we look at the first term, the add together the ext terms, the add the ext 4 terms, the add the ext 8 terms, etc ow compare this to a similar redefiitio of the kow coverget series Physics 7 Lecture Appedix B Autum 008
C 4 8 4 8 4 6 64 8 terms of 4 4 6 6 6 6 64 Comparig these two forms for the two series shows that each parethesis i series is smaller tha the correspodig parethesis i series C Thus, sice C coverges, the compariso test tells us that must coverge also 6: 6 olutio: This exercise is iteded to ecourage us to thik about the (possibly icorrect) role of the lower limit i the itegral test I cosiderig the sum of terms a we might be tempted to iclude a lower limit of zero ad evaluate the itegral 0 d 0 0 This result is misleadig sice the ifiity arises etirely from the lower limit O the other had we kow that the covergece/divergece of the ifiite series caot deped o the first terms of the series, where is fiite By the same toke the covergece/divergece of the series caot deped, ie, be determied by, the behavior of the itegral at the lower limit Clearly choosig ay lower limit other tha the special value zero will lead to a fiite itegral ad the correct coclusio that the series coverges The lesso is that you should just igore the behavior of the itegral at the lower limit It tells us othig useful! 6: 0 olutio: Here we wat to practice usig the ratio test With the specified series we have Physics 7 Lecture Appedix B 3 Autum 008
0!!, a,!!!!!! a a lim lim 0 Thus the ratio test tells us that the series coverges (absolutely) * 9: 9 Here we cosider the series! First look at the prelimiary test, e lim a lim lim lim! e Thus the series must diverge To see this form of the asymptotic behavior of the factorial fuctio look ahead to Eq () i Chapter (tirlig s formula) * 9: 6 The series of iterest is defied by idividual terms, Physics 7 Lecture Appedix B 4 Autum 008 0 7 We first ote that the a, vaish as ad the prelimiary test is passed ext cosider the ratio test which says 7 lim 7 Due to the behavior of the first factor it is somewhat difficult to evaluate the large limit However, sice this factor is either 3 or, whe ca proceed to cosider the appropriate compassio series give by 0 0 0 3 7 7 7 ow apply the ratio test to the larger series to fid
3 7 7 lim lim lim 3 7 8 ow we ca easily apply the rule i Eq (9) to see that this series, ad thus the series of iterest, coverges (p = > ) Physics 7 Lecture Appedix B 5 Autum 008