PHYSICS 116A Homework 2 Solutions

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Amanda Gabriella Lawson
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1 PHYSICS 6A Homework 2 Solutios I. [optioal] Boas, Ch., 6, Qu. 30 (proof of the ratio test). Just follow the hits. If ρ, the ratio of succcessive terms for is less tha, the hits show that the terms of the series are less tha those of a geometric series with commo ratio σ where ρ < σ <. This geometric series ca be explicitly summed ad see to coverge. Hece, by the compariso test, the origial series coverges. If ρ >, the terms of the series are greater tha those of a geometric series with commo ratio σ where ρ > σ >. Sice σ >, the idividual terms i the geometric series to ot ted to zero for, ad the same must be true for the origial series. Hece the series diverges accordig to the prelimiary test. II. [optioal] Boas, Ch., 6, Qu. (a) (stackig books ad relatio to the harmoic series). For part (b), whe N is large approximate the sum by the correspodig itegral. Start at the top. The ceter of mass must be above the edge of the book below. Hece fractio x = /2 sticks out, see the figure. 3 2 x 2 x Next do the book below. The ceter of mass of the top two books must be above the edge of the third book. Let the fractio of the secod book which overlaps the book below be x 2. The x 2 = /2 x 2 (sice the ceter of mass of book is x 2 to the right, ad the ceter of mass of book 2 is /2 x 2 to the left). This gives x 2 = /4. The do the third book ad repeat the argumet, which gives 2x 3 = /2 x 3 sice the ceter of mass of books ad 2 (total mass 2) is x 3 to the right, ad the ceter of mass of book 3 is /2 x 3 to the left. This gives x 3 = /6. Repeatig for the -th book (coutig from the top) gives ( )x = /2 x so x = 2. Hece the total overhag is (+ + + ). 2 This is the harmoic series whose sum is ifiite. Note, however, that the sum to N terms is approximately (/2)lN (to see this compare with (/2) N d/) ad so the overhag oly icreases slowly with N.

2 . (a) R N = a N+ +a N+2 +a N+3 + a() which is the area uder the shaded rectagles i the figure to the right. Hece a N+ R N < N a()d, sice the rectagles lie below the curve. N N+ N+2 Similarly the figure to the right shows that R N > N+ a()d, a N+ a() because the rectagles like above the curve. (b) I the figure o the right, the regios where a() > (height of shaded recatagle) is roughly balaced by the regio where a() < (height of shaded rectagle). This implies that the itegral from N + /2 to ifiity, is a better approxiatio to the remaider tha the itegral from N (which is too big) or the itegral from N + (which is too small), i.e. a N+ a() N+ N+2 (c) i. R N N+/2 a()d. S S 0 = 0 = 2 =.5498, where we calculated the sum of the first 0 terms umerically. ii. d 2 < R d 0 < 0 2 which gives < R 0 < 0. Now S = S 0 +R 0 ad so, usig part (i), we have.6407 < S < N N+/2 N+ N+2 2

3 iii. This gives iv. The exact result is R d 2 = 0.5. S = S 0 +R S = π2 6 = Hece the error i S 0 (part (i)) is The bouds i part (ii) differ from the exact value by ad The error i part (iii) is 0.000, which is ideed the most accurate estimate. It is oteworthy that this simple correctio icreases the accuracy obtaied from summig the first te terms by roughly oe thousad fold. 2. Boas, problem For the complex umber z = 3 + i, determie x, y, r, ad θ, where z = x+iy = r(cosθ +isiθ) = re iθ. Plot the umber i the complex plae ad label it i five ways as i Figure 3.3 o p. 48 of Boas. Also, plot the complex cojugate of z. Give z = 3+i, we have (x,y) = ( 3,), ad r = x 2 +y 2 = 2. Writig z = r(cosθ + isiθ),weidetifycosθ = 3/2adsiθ = /2,whichcorrespodstoaagle θ = 5π/6 = 50. I the complex plae, we ca draw the poit z ad its complex cojugate poit z: (x,y) = ( 3,) 3+i (r,θ) = (2, 5 6 π) 2(cos 5 6 π +isi 5 6 π) 2e 5iπ/6 z y 5π/6 x (x,y) = ( 3, ) 3 i (r,θ) = (2, 5 6 π) 2(cos 5 6 π isi 5 6 π) 2e 5iπ/6 z 3. Boas, problem First simplify the complex umber z = 3+i 2+i re iθ form. The, plot the umber i the complex plae. to the x+iy form or to the We ca simplify z by the followig steps: z 3+i 2+i = ( 3+i 2+i )( ) 2 i = 2 i 5 (7 i). The (x,y) = ( 7 5, 5 ), ad r = x 2 +y 2 = 2. Usig z = r(cosθ + isiθ) = re iθ, it follows that cosθ = 7 5 2, siθ = 5 2, = θ = 0.49 =

4 I the complex plae, z is located as follows y x ( 7 5, 5 ) 4. Boas, problem Fid the absolute value of 2+3i i. First, we write the umber i the form of a+bi. The we compute a+bi = a 2 +b i i = ( 2+3i i )( ) +i = 2+3i2 +5i = +i + 2 ( +5i). Hece, 2+3i i = 4 ( +25 ) = Boas, problem Solve for all possible values of the real umbers x ad y i the equatio: (x+iy) 3 =. () To solve Eq. (), we expad out this equatio: Equatig real ad imagiary parts yields: x 2 3xy 2 +i(3yx 2 y 3 ) =. x 3 3xy 2 =, 3yx 2 y 3 = 0. (2) Oe solutio to the secod equatio of Eq. (2) is y = 0. Substitutig this back ito the first equatio of Eq. (2) yields x 3 = which implies that x =. Thus, we have foud oe solutio. To fid the other solutios, we examie the case of y 0. I this case, we ca divide the secod equatio of Eq. (2) above by y to obtai 3x 2 y 2 = 0, = y = ±x 3. (3) Isertig this equatio back ito the first equatio of Eq. (2), oe obtais 8x 3 =, which implies that x = 2. Isertig this back ito Eq. (3) yields y = ± 3/2. Thus, we have foud three solutios: (x,y) = (,0), ( 2, 2 3), ( 2, 2 3). 4

5 I fact, there is a easier way to solve Eq. (). Writig x+iy = re iθ, Eq. () becomes: r 3 e 3iθ =. Takig the absolute value of both sides of the above equatio yields r 3 =, which implies that r =. Hece, we must solve e 3iθ = cos3θ +isi3θ =. Equatig real ad imagiary parts yields cos3θ = ad si3θ = 0. The geeral solutio to these equatios is θ = 3 (2+)π for ay iteger = 0,±,±2,... There are three solutios for agles that lie i the rage π < θ π, amely θ = π, 3 π, 3π, which correspod precisely to the three solutios foud above. 6. Boas, problem 2.6. Test the followig series for covergece: ( ) 2+i 2. (4) 3 4i =0 This is a geometric series. By the ratio test, =0 z coverges for z < ad diverges for z >. (Oe ca also prove separately that this geometric series also diverges for z =.) For this problem, we idetify: z = w 2, where w = 2+i 3 4i. Notig that z 2 = zz = w 2 w 2 = (ww) 2, it follows that z = ( z 2 ) /2 = ww. Hece, z = ww = 2+i 2 ( )( ) 2+i 2 i 3 4i = = 3 4i 3+4i 5 <, we coclude that the series give i Eq. (4) coverges. 7. Boas, problem Fid the disk of covergece for the complex power series (z i). =0 Usig the ratio test, we compute ( )( ρ lim (z i) + + (z i) ) ( ) = lim z i = z i. + The disk of covergece is defied to be the set of poits i the complex plae such that ρ <, which i this case is z i <. This is a disk of radius cetered about the complex umber i. 8. Boas, problem Express the complex umber 2e iπ/2 i the x+iy form. Usig Euler s formula, 2e iπ/2 = 2[cos(π/2) isi(π/2)] = 2i. 5

6 I the complex plae, this complex umber lives o the circle of radius 2 cetered at the origi, at a agle π/2 as measured i the couterclockwise directio from the x axis (or equivaletly, at a agle π/2 as measured i the clockwise directio from the x axis). That is, it lives o the circle of radius 2 o the egative y-axis which is the complex umber 2i as obtaied above. 9. Boas, problem Show that the absolute value of a product of two complex umbers is equal to the product of the two absolute values. Also, show that the absolute value of the quotiet of two complex umbers is the quotiet of the absolute values. We shall prove that for ay two complex umbers z ad z 2, the followig two results hold: (i) z z 2 = z z 2, (ii) z = z z 2. These results ca be established by writig the complex umbers i polar form, z = r e iθ ad z 2 = r 2 e iθ 2. The, z z 2 = r r 2 e i(θ +θ 2 ) = r r 2 e i(θ +θ 2 ) = r r 2, where we have used the fact that r ad r 2 are o-egative real umbers ad e iθ is a complex umber whose magitude is oe, for ay real agle θ. Similarly, z z 2 = r e iθ r 2 e iθ 2 = r r 2 e iθ e iθ 2 = r r 2. Thus, z z 2 = z z 2 = r r 2, which establishes (i) above. Next, ad z z 2 = r e iθ r 2 e iθ 2 = r r 2 e iθ e iθ 2 z z 2 = r e iθ r 2 e iθ 2 = r r 2 z 2 = r e i(θ θ 2 ) = r, r 2 r 2 ( e iθ ) e iθ = r. 2 r 2 Thus z /z 2 = z / z 2 = r /r 2, which establishes (ii) above. 0. Boas, problem Fid all the values of 3 8i. The complex power fuctio is defied as z c = e clz = e c[l z +iargz] = e c[l z +i(argz+2π)], = 0, ±, ±2,..., where c is a complex umber. I this problem, z = 8i ad c = 3, i which case L z = L8 = 3L2, ad Argz = 2π. Hece, 3 8i = e L 2 iπ/6+2π/3 = 2e iπ/6+2π/3. We obtai distict values for = 0,±. Other values of simply yield oe of the three possible 3 values for the cube root of 8i. Thus, the possible values of 8i are: 2e iπ/6 = 2[cos(π/6) isi(π/6] = 3 i, 2e iπ/2 = 2[cos(π/2)+isi(π/2)] = 2i, 2e 5iπ/6 = 2[cos(5π/6) isi(5π/6)] = 3 i. 6

7 By takig the cube of each of the three umbers above, oe ca check that the result is ideed 8i for each case. As a aside, ote that the pricipal value of 3 8i is defied to be the cube root that correspods to = 0. Thus the pricipal cube root of 8i is 3 i (ad ot 2i as you might have guessed). Of course this result is a cosequece of our covetio for the pricipal value of the argumet, where π < Argz π.. Boas, problem Fid formulas for si 3θ ad cos 3θ. We make use of De Moivre s theorem [eq. (0.2) o p. 64 of Boas]: (e iθ ) = (cosθ +isiθ) = cosθ +isiθ. Settig = 3 ad expadig out (cosθ +isiθ) 3, oe obtais cos3θ +isi3θ = (cosθ +isiθ) 3 = cos 3 θ +3icos 2 θsiθ +3i 2 cosθsi 2 θ +i 3 si 3 θ = cos 3 θ 3cosθsi 2 θ +i(3cos 2 θsiθ si 3 θ), after usig i 2 = ad i 3 = i. Equatig the real ad imagiary parts of the above equatio the yields: cos3θ = cos 3 θ 3cosθsi 2 θ, si3θ = 3cos 2 θsiθ si 3 θ. 2. Boas, problem Evaluate the real idefiite itegral e ax sibxdx, for real a ad b, by first evaluatig e (a+ib)x dx ad the takig the imagiary part of the resultig expressio. First, we evaluate e (a+ib)x dx = a+ib e(a+ib)x = a ib a 2 +b 2 eax (cosbx+isibx) = eax (acosbx+bsibx) a 2 +b 2 +i eax (asibx bcosbx) a 2 +b 2, (5) where we have used Euler s formula to write e ibx = cosbx+isibx i the first lie of the above equatio. If we equate the imagiary parts of Eq. (5), ad ote that the it follows that Im e (a+ib)x = Im e a e ibx = e a sibx, e ax sibxdx = eax (asibx bcosbx) a 2 +b 2, for real a ad b. 7

8 As a bous, oe ca also equate the real parts of Eq. (5) ad make use of Re e (a+ib)x = Re e a e ibx = e a cosbx, to obtai: e ax cosbxdx = eax (acosbx+bsibx) a 2 +b 2, for real a ad b. 8

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

$Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing$ Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig 0.694444444... Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals