Download PHYSICS 116A Homework 2 Solutions I. [optional] Boas, Ch. 1, §6

PHYSICS 116A
Homework 2 Solutions
I. [optional]
Boas, Ch. 1, §6, Qu. 30 (proof of the ratio test). Just follow the hints.
If ρ, the ratio of succcessive terms for n → ∞ is less than 1, the hints show that the terms of
the series are less than those of a geometric series with common ratio σ where ρ < σ < 1. This
geometric series can be explicitly summed and seen to converge. Hence, by the comparison test,
the original series converges.
If ρ > 1, the terms of the series are greater than those of a geometric series with common ratio σ
where ρ > σ > 1. Since σ > 1, the individual terms in the geometric series to not tend to zero for
n → ∞, and the same must be true for the original series. Hence the series diverges according to
the preliminary test.
II. [optional]
Boas, Ch. 1, §16, Qu. 1 (a) (stacking books and relation to the harmonic series). For part (b),
when N is large approximate the sum by the corresponding integral.
Start at the top. The center of mass must be above the edge of the book below. Hence fraction
x1 = 1/2 sticks out, see the figure.
1
2
3
x1
x2
Next do the book below. The center of mass of the top two books must be above the edge of
the third book. Let the fraction of the second book which overlaps the book below be x2 . Then
x2 = 1/2 − x2 (since the center of mass of book 1 is x2 to the right, and the center of mass of
book 2 is 1/2 − x2 to the left). This gives x2 = 1/4.
Then do the third book and repeat the argument, which gives 2x3 = 1/2 − x3 since the center
of mass of books 1 and 2 (total mass 2) is x3 to the right, and the center of mass of book 3 is
1/2 − x3 to the left. This gives x3 = 1/6.
Repeating for the n-th book (counting from the top) gives (n − 1)xn = 1/2 − xn so
xn =
1
.
2n
Hence the total overhang is
1
2
1+
1 1 1
+ + ···
2 3 4
.
This is the harmonic series whose sum is infinite.
Note, however, that the sum to N terms is approximately (1/2) ln N (to see this compare with
RN
(1/2) 1 dn/n) and so the overhang only increases slowly with N .
1
1. (a)
a(n)
RN = aN +1 + aN +2 + aN +3 + · · ·
which is the area under the shaded rectangles in the
figure to the right. Hence
Z
RN <
∞
aN+1
a(n) dn ,
N
since the rectangles lie below the curve.
N
N+1
N+1
N+2
N+2
n
a(n)
Similarly the figure to the right shows that
RN >
Z
∞
aN+1
a(n) dn ,
N +1
because the rectangles like above the curve.
n
(b)
In the figure on the right, the regions where a(n) >
(height of shaded recatangle) is roughly balanced by
the region where a(n) < (height of shaded rectangle). This implies that the integral from N + 1/2
to infinity, is a better approxiation to the remainder
than the integral from N (which is too big) or the
integral from N + 1 (which is too small), i.e.
RN ≃
Z
∞
a(n)
aN+1
a(n) dn .
N +1/2
N
N+1/2
(c)
i.
S ≃ S10 =
10
X
1
= 1.5498 ,
n2
n=1
where we calculated the sum of the first 10 terms numerically.
ii.
Z
which gives
Now S = S10 + R10
∞
11
dn
< R10 <
n2
Z
∞
10
dn
n2
1
1
< R10 <
.
11
10
and so, using part (i), we have
1.6407 < S < 1.6498 .
2
N+1
N+2
n
iii.
R10 ≃
Z
10.5
1
dn
=
.
2
n
10.5
This gives
S = S10 + R10 ≃ 1.6450 .
iv. The exact result is
π2
= 1.6449.
6
Hence the error in S10 (part (i)) is 0.0952.
The bounds in part (ii) differ from the exact value by 0.0042 and −0.0049.
The error in part (iii) is −0.0001, which is indeed the most accurate estimate. It is
noteworthy that this simple correction increases the accuracy obtained from summing
the first ten terms by roughly one thousand fold.
√
2. Boas, problem 2.4–4. For the complex number z = − 3 + i, determine x, y, r, and θ, where
z = x + iy = r(cos θ + i sin θ) = reiθ . Plot the number in the complex plane and label it in five
ways as in Figure 3.3 on p. 48 of Boas. Also, plot the complex conjugate of z.
S=
p
√
√
Given z = − 3 + i, we have (x, y) = (− 3, 1) , and r = x2 + y 2 = 2 . Writing z = r(cos θ +
√
i sin θ), we identify cos θ = − 3/2 and sin θ = 1/2, which corresponds to an angle θ = 5π/6 = 150◦ .
In the complex plane, we can draw the point z and its complex conjugate point z:
√

(x, y) = (− 3, 1)


√



− 3 + i
(r, θ) = (2, 65 π)


2(cos 5 π + i sin 5 π)


6
6

2e5iπ/6
y
z
5π/6
x
√

(x, y) = (− 3, −1)


 √


− 3 − i
(r, θ) = (2, − 65 π)


5
5


2(cos 6 π − i sin 6 π)

−5iπ/6
2e
z
3. Boas, problem 2.5–7. First simplify the complex number z =
reiθ form. Then, plot the number in the complex plane.
3+i
to the x + iy form or to the
2+i
We can simplify z by the following steps:
3+i
2−i
3+i
=
= 15 (7 − i) .
z≡
2+i
2+i
2−i
p
√
Then (x, y) = ( 75 , − 51 ), and r = x2 + y 2 = 2 . Using z = r(cos θ + i sin θ) = reiθ , it follows
that
7
1
cos θ = √ ,
sin θ = − √ ,
=⇒
θ = −0.1419 = −8.13◦ .
5 2
5 2
3
In the complex plane, z is located as follows
y
x
,−1
)
(7
5
5
4. Boas, problem 2.5–27. Find the absolute value of
2 + 3i
.
1−i
First, we write the number in the form of a + bi. Then we compute |a + bi| =
2 + 3i2 + 5i
1+i
2 + 3i
2 + 3i
=
=
= 12 (−1 + 5i) .
1−i
1−i
1+i
1+1
√
a 2 + b2 .
Hence,
q
2 + 3i q 1
13
=
4 1 + 25 =
2 .
1−i 5. Boas, problem 2.5–47. Solve for all possible values of the real numbers x and y in the equation:
(x + iy)3 = −1 .
(1)
To solve Eq. (1), we expand out this equation:
x2 − 3xy 2 + i(3yx2 − y 3 ) = −1 .
Equating real and imaginary parts yields:
x3 − 3xy 2 = −1 ,
3yx2 − y 3 = 0 .
(2)
One solution to the second equation of Eq. (2) is y = 0. Substituting this back into the first
equation of Eq. (2) yields x3 = −1 which implies that x = −1. Thus, we have found one solution.
To find the other solutions, we examine the case of y 6= 0. In this case, we can divide the second
equation of Eq. (2) above by y to obtain
√
3x2 − y 2 = 0 ,
=⇒
y = ±x 3 .
(3)
Inserting this equation back into the first equation of Eq. (2),√one obtains 8x3 = 1, which implies
that x = 12 . Inserting this back into Eq. (3) yields y = ± 3/2. Thus, we have found three
solutions:
√
√
(x, y) = (−1, 0) , ( 21 , 12 3) , ( 12 , − 21 3) .
4
In fact, there is an easier way to solve Eq. (1). Writing x + iy = reiθ , Eq. (1) becomes:
r3 e3iθ = −1 .
Taking the absolute value of both sides of the above equation yields r3 = 1, which implies that
r = 1. Hence, we must solve
e3iθ = cos 3θ + i sin 3θ = −1 .
Equating real and imaginary parts yields cos 3θ = −1 and sin 3θ = 0. The general solution to
these equations is θ = 31 (2n + 1)π for any integer n = 0, ±1, ±2, . . .. There are three solutions for
angles that lie in the range −π < θ ≤ π, namely θ = π , 13 π , − 13 π , which correspond precisely to
the three solutions found above.
6. Boas, problem 2.6–11. Test the following series for convergence:
∞ X
2 + i 2n
.
3 − 4i
(4)
n=0
P∞ n
This is a geometric series. By the ratio test,
n=0 z converges for |z| < 1 and diverges for
|z| > 1. (One can also prove separately that this geometric series also diverges for |z| = 1.) For
this problem, we identify:
2+i
.
z = w2 ,
where w =
3 − 4i
Noting that |z|2 = zz = w2 w2 = (ww)2 , it follows that |z| = (|z|2 )1/2 = ww. Hence,
2 + i 2
2−i
1
= 2+i
|z| = ww = = < 1,
3 − 4i 3 − 4i
3 + 4i
5
we conclude that the series given in Eq. (4) converges.
7. Boas, problem 2.7–13. Find the disk of convergence for the complex power series
∞
X
(z − i)n
n=0
n
.
Using the ratio test, we compute
(z − i)n+1
n
n
= lim
ρ ≡ lim |z − i| = |z − i| .
n→∞
n+1
(z − i)n n→∞ n + 1
The disk of convergence is defined to be the set of points in the complex plane such that ρ < 1,
which in this case is |z − i| < 1. This is a disk of radius 1 centered about the complex number i.
8. Boas, problem 2.9–9. Express the complex number 2e−iπ/2 in the x + iy form.
Using Euler’s formula,
2e−iπ/2 = 2 [cos(π/2) − i sin(π/2)] = −2i .
5
In the complex plane, this complex number lives on the circle of radius 2 centered at the origin,
at an angle −π/2 as measured in the counterclockwise direction from the x axis (or equivalently,
at an angle π/2 as measured in the clockwise direction from the x axis). That is, it lives on the
circle of radius 2 on the negative y-axis—which is the complex number −2i as obtained above.
9. Boas, problem 2.9–28. Show that the absolute value of a product of two complex numbers is equal
to the product of the two absolute values. Also, show that the absolute value of the quotient of
two complex numbers is the quotient of the absolute values.
We shall prove that for any two complex numbers z1 and z2 , the following two results hold:
z1 |z1 |
.
(i) |z1 z2 | = |z1 ||z2 | ,
(ii) =
z2
|z2 |
These results can be established by writing the complex numbers in polar form, z1 = r1 eiθ1 and
z2 = r2 eiθ2 . Then,
|z1 z2 | = |r1 r2 ei(θ1 +θ2 ) | = r1 r2 |ei(θ1 +θ2 ) | = r1 r2 ,
where we have used the fact that r1 and r2 are non-negative real numbers and eiθ is a complex
number whose magnitude is one, for any real angle θ. Similarly,
|z1 ||z2 | = |r1 eiθ1 ||r2 eiθ2 | = r1 r2 |eiθ1 ||eiθ2 | = r1 r2 .
Thus, |z1 z2 | = |z1 ||z2 | = r1 r2 , which establishes (i) above. Next,
iθ z1 r1 e 1 r1 eiθ1 r1 i(θ −θ ) r1
1
2
=
,
=
z2 r2 eiθ2 = r2 eiθ2 = r2 e
r2
and
|z1 |
|r1 eiθ1 |
r1
=
=
|z2 |
|r2 eiθ2 |
r2
|eiθ1 |
|eiθ2 |
=
r1
.
r2
Thus |z1 /z2 | = |z1 |/|z2 | = r1 /r2 , which establishes (ii) above.
√
3
10. Boas, problem 2.10–20. Find all the values of −8i.
The complex power function is defined as
z c = ec ln z = ec[Ln|z|+i arg z] = ec[Ln|z|+i(Arg z+2πn)] ,
n = 0 , ±1 , ±2 , . . . ,
where c is a complex number. In this problem, z = −8i and c = 13 , in which case Ln|z| = Ln 8 =
3 Ln 2, and Arg z = − 21 π. Hence,
√
3
−8i = eLn 2−iπ/6+2πn/3 = 2e−iπ/6+2πn/3 .
We obtain distinct values for n = 0, ±1. Other values of n simply
yield one of the three possible
√
3
values for the cube root of −8i. Thus, the possible values of −8i are:
√
3 − i,
2e−iπ/6 = 2[cos(π/6) − i sin(π/6] =
2eiπ/2 = 2[cos(π/2) + i sin(π/2)] = 2i ,
√
2e−5iπ/6 = 2[cos(5π/6) − i sin(5π/6)] = − 3 − i .
6
By taking the cube of each of the three numbers above, one can check that the result is indeed
−8i for each case.
√
3
As an aside, note that the principal value of −8i√is defined to be the cube root that corresponds
to n = 0. Thus the principal cube root of −8i is 3 − i (and not 2i as you might have guessed).
Of course this result is a consequence of our convention for the principal value of the argument,
where −π < Arg z ≤ π.
11. Boas, problem 2.10–28. Find formulas for sin 3θ and cos 3θ.
We make use of De Moivre’s theorem [eq. (10.2) on p. 64 of Boas]:
(eiθ )n = (cos θ + i sin θ)n = cos nθ + i sin nθ .
Setting n = 3 and expanding out (cos θ + i sin θ)3 , one obtains
cos 3θ + i sin 3θ = (cos θ + i sin θ)3
= cos3 θ + 3i cos2 θ sin θ + 3i2 cos θ sin2 θ + i3 sin3 θ
= cos3 θ − 3 cos θ sin2 θ + i(3 cos2 θ sin θ − sin3 θ) ,
after using i2 = −1 and i3 = −i. Equating the real and imaginary parts of the above equation
then yields:
cos 3θ = cos3 θ − 3 cos θ sin2 θ ,
sin 3θ = 3 cos2 θ sin θ − sin3 θ .
12. Boas, problem 2.11–18. Evaluate the real indefinite integral
Z
eax sin bx dx ,
for real a and b ,
by first evaluating
Z
e(a+ib)x dx
and then taking the imaginary part of the resulting expression.
First, we evaluate
Z
e(a+ib)x dx =
=
1
a − ib ax
e(a+ib)x = 2
e (cos bx + i sin bx)
a + ib
a + b2
eax (a cos bx + b sin bx)
eax (a sin bx − b cos bx)
+
i
,
a 2 + b2
a 2 + b2
(5)
where we have used Euler’s formula to write eibx = cos bx + i sin bx in the first line of the above
equation. If we equate the imaginary parts of Eq. (5), and note that
Im e(a+ib)x = Im ea eibx = ea sin bx ,
then it follows that
Z
eax sin bx dx =
eax (a sin bx − b cos bx)
,
a 2 + b2
7
for real a and b .
As a bonus, one can also equate the real parts of Eq. (5) and make use of
Re e(a+ib)x = Re ea eibx = ea cos bx ,
to obtain:
Z
eax cos bx dx =
eax (a cos bx + b sin bx)
,
a 2 + b2
8
for real a and b .