Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing

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1 Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals from the frot ito their ow fractio, so we have: = We ca rewrite this as: = which is equivalet to: = = = Boas, problem. 6 Suppose a large umber of particles are boucig back ad forth betwee x = 0 ad x =, except that at each edpoit some escape. Let r be the fractio reflected each time; the r is the fractio escapig. Suppose the particles start at x = 0 headig toward x = ; evetually all particles will escape. Write a ifiite series for the fractio that escape at x = ad similarly for the fractio that escape at x = 0. Sum both the series. What is the largest fractio of the particles that ca escape at x = 0? (Remember that r must be betwee 0 ad.) Suppose we start off with N particles at x = 0 headig toward x =.. The first time the particles reach x =, ( r)n particles escape ad rn particles are reflected.. The first time the particles retur to x = 0, ( r)rn particles escape ad r N particles are reflected. 3. The secod time the particles reach x =, ( r)r N particles escape ad r 3 N particles are reflected. 4. The secod time the particles retur to x = 0, ( r)r 3 N particles escape ad r 4 N particles are reflected.

2 ad so o. The patter should be clear. At x =, the umber of particles that escape is equal to ( r)n + ( r)r N + ( r)r 4 N + = ( r)n( + r + r 4 + ) = ( r)n r = N + r, () after factorig ( r ) = ( r)( + r). I the peultimate step, we summed the geometric series: r = + r + r 4 + =, r <. () r Similarly, at x = 0, the umber of particles that escape is equal to ( r)rn + ( r)r 3 N + ( r)r 5 N + = ( r)rn( + r + r 4 + ) = ( r)rn r = Nr + r. (3) Thus, the total umber of particles that escape (either at x = or at x = 0) is give by the sum of eqs. () ad (3): N + r + Nr + r = N, which cofirms that all the origial particles evetually escape. I the above aalysis, it was assumed that r is fixed to lie betwee zero ad oe. If r = 0, the eqs. () ad (3) imply that all N particles escape at x =. This makes sese, sice if r = 0 the the fractio of particles escapig at oe of the edpoits is 00%. If r =, eqs. () ad (3) seem to suggest that N particles escape at x = 0 ad N particles escape at x =. This result caot possibly be correct, sice if r =, the the fractio of particles reflectig off a edpoit is 00%. Thus, if r =, o particles escape. How is this apparet paradox resolved? The aswer is that the summatio of the geometric series give i eq. () is valid oly whe r <. A closer look at the formulae give i eqs. () ad (3) before eq. () is applied show that the umber of particles that escape at either edpoit is ideed equal to zero if r =. Fially, to determie the largest fractio of the particles that ca escape at x = 0, we maximize eq. (3), subject to the iequality 0 < r <. The maximum value of eq. (3) occurs i the limit as r,i which case N particles would escape at x = 0. As oted at the ed of the previous paragraph, o particles actually escape if r =. But, I ca take r arbitrarily close to (with r < ), i which case the umber of particles escapig at x = 0 ca be arbitrarily close to N. Hece, we coclude that at most half of the particles ca escape at x = 0. If you wated to get picky ad say that the umber of particles must be a iteger, the you would coclude that at most N particles ca escape at x = 0. If N is extremely large (e.g. Avogadro s umber, ), the the error that you make by statig that half the particles ca escape at x = 0 becomes egligible.

3 3. Boas, problem. 6 Fid the limit of /! as. The simplest way to compute this limit is to write:! = ( ) 3 = ( 3 ) ( ), (4) Note that each of the factors ( ) (, ) ( 3,, ) appearig i eq. (4) is larger tha. Hece, it immediately follows that:! >. Takig, oe sees that the iequality above yields lim! =. 4. Boas, problem.5 8 Use prelimiary test to see if the series l() = possibly coverges. We kow that the series = diverges, ad we kow that for ay >, that l() >, so this series must be greater for all terms of >, so this must also diverge. 5. Boas, problem.6 5 Test the followig series for covergece usig the compariso test: (a) =, (b) = l. For ay positive iteger, we have < so that >. Hece, usig the compariso test, = > = =, 3

4 sice the harmoic series diverges. We coclude that series (a) also diverges. Likewise, for ay positive iteger, we have l < so that Hece, usig the compariso test, l >. = l > = =, ad we coclude that series (b) also diverges. 6. Boas, problem.6 6. The studet has eglected to ote that the divergece i the itegral dx stems from the behavior of the itegral at x=0, ot at x=, 0 x ote that the itegral dx coverges, ad that the sum may start at =, ad x we ote that the sum may also coverge (it does, i fact). 7. Boas, problem.6 8. Use the ratio test to fid whether coverges or diverges.!()! (3)! Usig the ratio test, we evaluate: ρ = ( + )!( + )! (3)! ( + )( + )( + ) (3 + 3)!!()! = (3 + 3)(3 + )(3 + ) 4 7, as, after usig the defiitio of the factorial to obtai (+)! = (+)!, ad similarly, ( + )! = ( + )( + )()!, (3 + 3)! = (3 + 3)(3 + )(3 + )(3)!. Sice ρ ρ 4 7 < as, we coclude that the series coverges. 8. Boas, problem Use the special compariso test to fid whether coverges or diverges. =5 (5) 4

5 We shall use the special compariso test (p. 5 of Boas) with b = ad a = /( ) 0. Oe ca easily check that: a lim = lim b = lim =. The last step is a cosequece of the fact that expoetial growth is always faster tha power law growth. I particular, It follows that the series coverges. lim = lim = Boas, problem Test the series ( ) for covergece. + Note first that each term i this series is smaller tha the correspodig term i the series ( ), which is equal to ( ). So by the compariso test, if this series coverges, so does the series i the origial questio. The ote that we have a alteratig series with terms, where each term gets smaller tha the last, ad the limit evetually goes to zero (from sectio ), so this coverges. 0. Boas, problem Test the series (!) for covergece. ()! Use the ratio test (this is ofte a very good idea for factorials), ote that we ca write a + as a sum over terms [(+)!], which ca be rewritte as (+)(+)(!). (+)! (+)(+)[()!] I the limit that goes to ifiity, the leader order terms give us, which is 4 equal to. This is less tha, so it coverges by the ratio test. 4. Boas, problem Test the series ( ) = for covergece. l() Sice this is a alteratig series, we eed to kow oly that each term has a absolute value smaller tha the previous term, ad that the fial limit is 0 (as goes to ifiity). For the first part, we ca ivestigate the terms (otig l(+) that the absolute value makes the umerator positive i every case), ad ote that this is smaller tha the th term, as the atural log is a mootoically icreasig fuctio. To derive the limit as goes to ifiity, we ote that the l( ) is ifiite, ad thus this series coverges. 5

6 . Boas, problem.0 8. Fid he iterval of covergece the power series = ( ) x, (6) ad check the edpoits of the iterval for covergece or divergece. Usig the ratio test, we compute ρ = x + / + x / = x + x, as. The iterval of covergece is determied by the coditio ρ lim ρ <. Hece, the power series above coverges for x < ad diverges for x >. We ow examie the edpoits x = ad x = separately. At x =, eq. (6) is (coditioally) coverget, sice it is a alteratig series, = ( ) /, whose terms mootoically approach zero as. At x =, eq. (6) reduces to =, which is diverget due to the p-series test [cf. problem 7 above]. Hece, it follows that the iterval of covergece for the power series give i eq. (6) is: < x. 6

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