PHYSICS 116A Homework 3 Solutionsyoung.physics.ucsc.edu/116A/homework/sols/solutions3.pdf ·...
Transcript of PHYSICS 116A Homework 3 Solutionsyoung.physics.ucsc.edu/116A/homework/sols/solutions3.pdf ·...
PHYSICS 116A
Homework 3 Solutions
1. Boas, 2.14–9. Evaluate (−1)i in x+ iy form.
We write −1 = ei(π+2πn) (n = 0 , ±1 , ±2 , . . . , ) so
(−1)i = ei i(π+2πn) = e−π−2πn ,
which, rather surprisingly, is a set of real numbers. The principal value of (−1)i = e−π corre-sponding to the choice of n = 0 in the above equation.
2. Boas, 2.15–7. Write arctan(i√2) in x+ iy form.
Start with
z = tanw =sinw
cosw=
1
i
(
eiw − e−iw
eiw + e−iw
)
=1
i
(
e2iw − 1
e2iw + 1
)
.
Solving for e2iw,
iz =e2iw − 1
e2iw + 1=⇒ e2iw =
1 + iz
1− iz.
Hence,
w = arctan z =1
2iln
(
1 + iz
1− iz
)
. (1)
Setting z = i√2, we obtain:
arctan(i√2) =
1
2iln
(
1−√2
1 +√2
)
=1
2iln[
−(1−√2)2]
. (2)
after multiplying the numerator and denominator above by 1−√2 and noticing that (1−
√2)(1+√
2) = 1− 2 = −1. We shall make use of the definition of the multi-valued complex logarithm
ln z = Ln|z|+ i arg z = Ln|z|+ i(Argz + 2πn) , n = 0,±1,±2 . . . , (3)
with z = −(1−√2)2. In particular, we have Ln|z| = 2Ln(
√2−1) and Arg z = π. Hence, Eqs. (3)
and (2) yield:
arctan(i√2) =
(
n+ 12
)
π − iLn(√2− 1) , n = 0 , ±1 , ±2 , . . . .
3. Boas, 2.16–11. Prove that
cos θ + cos 3θ + cos 5θ + · · ·+ cos(2n− 1)θ =sin 2nθ
2 sin θ, (4)
sin θ + sin 3θ + sin 5θ + · · ·+ sin(2n− 1)θ =sin2 nθ
sin θ. (5)
Consider the geometric series:
S ≡ eiθ + e3iθ + e5iθ + · · ·+ e(2n−1)θ . (6)
1
Using the results of eq. (1.4) on p. 2 of Boas,
a+ ar + ar2 + · · ·+ arN−1 =a(1− rN )
1− r,
we identify a = eiθ, r = e2iθ and n = N . Hence,
S ≡ eiθ + e3iθ + e5iθ + · · ·+ e(2n−1)θ =eiθ(1− e2inθ)
1− e2iθ.
This result can be simplified by multiply both the numerator and denominator by −e−iθ. Theend result is:
S =e2inθ − 1
eiθ − e−iθ=
einθ(einθ − e−inθ)
eiθ − e−iθ=
einθ sinnθ
sin θ,
where we have used
sin θ ≡ eiθ − e−iθ
2i
to obtain the final result. In particular, using Re(einθ) = cosnθ and Im(einθ) = sinnθ, it followsthat:
Re S =cosnθ sinnθ
sin θ=
sin 2nθ
2 sin θ,
Im S =sin2 nθ
sin θ,
after employing the trigonometric identity sin 2nθ = 2 sinnθ cosnθ. Equating the real and imagi-nary parts of Eq. (6), we conclude that
Re S = cos θ + cos 3θ + cos 5θ + · · ·+ cos(2n− 1)θ =sin 2nθ
2 sin θ,
Im S = sin θ + sin 3θ + sin 5θ + · · ·+ sin(2n− 1)θ =sin2 nθ
sin θ,
which reproduces the results of Eqs. (4) and (5).
4. Boas Ch. 2, §17, Qu. 22. Show that tanh−1 z = 12 ln
(
1+z1−z
)
.
We write
z = tanhw =ew − e−w
ew + e−w=
e2w − 1
e2w + 1,
soz(
e2w + 1)
= e2w − 1 ,
which has solution
e2w =1 + z
1− z.
Taking the log gives
w ≡ tanh−1 z =1
2ln
(
1 + z
1− z
)
.
2
5. The series for the principal value of the complex-valued logarithm,
Ln(1− z) = −∞∑
n=1
zn
n, (7)
converges for all |z| ≤ 1, z 6= 1. In particular, consider the conditionally convergent series,
S ≡∞∑
n=1
einθ
n, where 0 < θ < 2π . (8)
Using Eq. (7) with z = eiθ, it follows that
S = −Ln(1− eiθ) . (9)
The principal value of the complex logarithm is given by:
Ln(1− eiθ) = Ln|1− eiθ|+ iArg(1− eiθ) . (10)
To evaluate this expression we write the complex number z = 1− eiθ in polar form, i.e.
z = 1− eiθ = ReiΘ ,
soS = − [LnR+ iΘ] , (−π < Θ ≤ π) . (11)
A simple way to do this is:
1− eiθ = ReiΘ = eiθ/2[
e−iθ/2 − eiθ/2]
= −2ieiθ/2 sin(θ/2) = 2ei(θ−π)/2 sin(θ/2) , (12)
after using sin(θ/2) = 12i(e
iθ/2 − e−iθ/2) and −i = e−iπ/2. Since sin(θ/2) > 0 for 0 < θ < 2π, itfollows that
R ≡ |1− eiθ| = 2 sin(θ/2),
and we can then identify the principal value of the argument of z as
Θ ≡ Arg(1− eiθ) = 12(θ − π) . (13)
Note that when 0 < θ < 2π, it follows that Θ lies between −π and π, which is inside the range ofthe principal value of the argument function as defined in class. Hence, Eq. (11) simplifies to:
S = −Ln(1− eiθ) = −Ln
(
2 sinθ
2
)
+ 12 i(π − θ) , for 0 < θ < 2π . (14)
(a) By taking the real part of Eq. (8), evaluate
∞∑
n=1
cosnθ
n, where 0 < θ < 2π ,
3
as a function of θ. Check that your answer has the right limit for θ = π.
Noting that cosnθ = Re einθ, it follows from Eqs. (8) and (14) that
∞∑
n=1
cosnθ
n= ReS = −Ln
(
2 sinθ
2
)
, for 0 < θ < 2π . (15)
We can check Eq. (15) in the special case of θ = π. Using the results, sin(π/2) = 1, cosnπ = (−1)n
and −(−1)n = (−1)n+1, it follows that:
Ln 2 =∞∑
n=1
(−1)n+1
n= 1− 1
2 + 13 − 1
4 + · · · ,
a well-known result.
(b) By taking the imaginary part of Eq. (8), prove that
∞∑
n=1
sinnθ
n= 1
2(π − θ) , where 0 < θ < 2π .
Noting that sinnθ = Im einθ, it follows from Eqs. (8) and (14) that
∞∑
n=1
sinnθ
n= ImS = 1
2(π − θ) , for 0 < θ < 2π . (16)
Again, we can check the special case of θ = π. Since sinnπ = 0, Eq. (16) reduces in this limit tothe trivial equation 0 = 0 .
6. Evaluate the integral
In =
∫ ∞
0tne−kt2 dt ,
for k > 0 and n > −1.
Let us define a new variable, u = kt2 or equivalently t =√
u/k. Then,
du = 2ktdt =⇒ dt =du
2kt=
du
2k
(
k
u
)1/2
=du
2k1/2u1/2.
Hence,
In =1
kn/2
∫ ∞
0un/2e−u du
2k1/2u1/2du =
1
2k(n+1)/2
∫ ∞
0u(n−1)/2e−u du
4
Using the definition of the Gamma function,
Γ(x) =
∫ ∞
0tx−1e−t dt ,
we identify x = 12(n+ 1). Thus,
In =1
2k(n+1)/2Γ
(
n+ 1
2
)
As a check, let us set k = 1 and n = 0. Then, we obtain I0 =12Γ(
12) =
12
√π as expected. The factor
of k(n+1)/2 can be deduced using simple dimensional analysis. If t has dimensions of time, then kmust have dimensions of t−2, since the argument of the exponential must be dimensionless.∗ Sincedt also has dimensions of time, the entire integral has dimensions of tn+1, which must be respectedby the final result for In. Indeed k−(n+1)/2 has dimensions of tn+1, and so the dimensions of ourfinal result for In are consistent. In particular, it is possible to first carry out the computationwith k = 1, and then determine the k dependence of In strictly on dimensional grounds!
7. Boas, Ch. 11 §3, Qu. 5. Simplify Γ(12)Γ(4)/Γ(92).
Using xΓ(x) = Γ(x + 1) repeatedly, one obtains Γ(92) = 72 Γ(
72) = 7
2 · 52 Γ(
52), etc. until finally
obtaining Γ(92) =72 · 5
2 · 32 · 1
2 Γ(12) . Hence, using Γ(4) = 3! = 6, it follows that
Γ(12)Γ(4)
Γ(29)= 2
7 · 25 · 2
3 · 2 · 6 =32
35.
8. Boas, Ch. 11 §3, Qu. 13. Express as a Γ function
∫ 1
0x2(
ln1
x
)3
dx .
Introduce a new variable x = e−u. Then dx = −e−udu and ln(1/x) = ln eu = u. Noting thatx = 0 =⇒ u = ∞ and x = 1 =⇒ u = 0, it follows that
∫ 1
0x2(
ln1
x
)3
dx =
∫ ∞
0u3e−3u du =
1
34
∫ ∞
0v3e−v dv =
Γ(4)
81=
2
27.
In the second step above, I used the overall minus sign to interchange the lower and upper limits ofintegration. At the third step, I changed the integration variable once more to v = 3u (the limitsof integration are unchanged). Finally, I used the definition of the Gamma function [eq. (3.1) onp. 538 of Boas] followed by Γ(4) = 3! = 6 to obtain the final result.
∗The simplest way to see that the argument of any exponential must be dimensionless is to consider the power seriesexpansion, ez = 1 + z + z2/2! + z3/3! + · · · . Suppose z had dimensions of length, for example. Then ez would be thesum of a dimensionless number plus a number with units of length plus a number with units of squared-length, and soon. But, one cannot consistently sum two quantities that possess different length dimensions. The only way to avoid thisinconsistency is to conclude that z is dimensionless.
5
9. Boas, problem 11.5–3. Show that the binomial coefficient(
pn
)
can be written in terms of Gammafunctions as:
(
p
n
)
=Γ(p+ 1)
n! Γ(p− n+ 1). (17)
The binomial coefficient is defined in eq. (13.6) on p. 28 of Boas as:
(
p
n
)
=p(p− 1)(p− 2) · · · (p− n+ 1)
n!.
To prove that it can be written as in Eq. (17), we first multiply numerator and denominator byΓ(p− n+ 1),
(
p
n
)
=p(p− 1)(p− 2) · · · (p− n+ 1)Γ(p− n+ 1)
n! Γ(p− n+ 1). (18)
By repeatedly employing xΓ(x) = Γ(x+ 1), the numerator of Eq. (18) can be written as:
p(p− 1)(p− 2) · · · (p− n+ 3)(p− n+ 2)(p− n+ 1)Γ(p− n+ 1)
= p(p− 1)(p− 2) · · · (p− n+ 3)(p− n+ 2)Γ(p− n+ 2)
= p(p− 1)(p− 2) · · · (p− n+ 3)Γ(p− n+ 3)
= · · ·
= p(p− 1)(p− 2)Γ(p− 2) = p(p− 1)Γ(p− 1) = pΓ(p) = Γ(p+ 1) .
Inserting this last result back into Eq. (18) yields
(
p
n
)
=Γ(p+ 1)
n! Γ(p− n+ 1)
as requested.
10. Boas, Ch. 11 §5, Qu. 5, part (a).
Use eq. (5.4) on p. 541 of Boas to show that
Γ(
12 − n
)
Γ(
12 + n
)
= (−1)n π , (19)
assuming that n is an integer.†
We start from the reflection formula [eq. (5.4) on p. 541 of Boas]:
Γ(p)Γ(1− p) =π
sinπp. (20)
†Boas restricts n to be a positive integer. However, Eq. (19) holds for negative integers as well. In fact, noting that(−1)n = (−1)−n for any integer n, it follows that Eq. (19) is unmodified under the interchange of n and −n. Note thatfor n = 0, Eq. (19) also yields a correct result, [Γ( 1
2)]2 = π.
6
If we set p = 12 − n, then
1− p = 1− (12 − n) = n+ 12 .
Inserting p = 12 − n into Eq. (20) yields:
Γ(
12 − n
)
Γ(
12 + n
)
=π
sin[
π(
12 − n
)] . (21)
Finally, we make use of the trigonometric identity,
sin[
π(
12 − n
)]
= sin(
12π − nπ
)
= sin(
12π)
cos(nπ)− cos(
12π)
sin(nπ) = cos(nπ) ,
after using sin(12π) = 1 and cos(12π) = 0. Using the fact that cos(nπ) = (−1)n for any integer n,it follows that
sin[
π(
12 − n
)]
= (−1)n , for any integer n .
Inserting this result back into Eq. (21) and using (−1)−n = (−1)n for any integer n, we obtain:
Γ(
12 − n
)
Γ(
12 + n
)
= (−1)n π , for any integer n ,
which is the desired result.
7