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) ...
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 corresponding 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
1 sin w = z = tan w = cos w i
Solving for e2iw , iz =
e2iw − 1 e2iw + 1
eiw − e−iw eiw + e−iw
=⇒
1 ln w = arctan z = 2i
√
1 arctan(i 2) = ln 2i
1 = i
e2iw =
Hence, √ Setting z = i 2, we obtain:
e2iw − 1 e2iw + 1
.
1 + iz . 1 − iz
1 + iz 1 − iz
.
(1)
√ ! √ i 1− 2 1 h √ ln −(1 − 2)2 . = 2i 1+ 2
(2)
√ √ 2 and noticing that (1 − 2)(1 + after multiplying the numerator and denominator above by 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) √ 2 √ with z = −(1 − 2) . In particular, we have Ln|z| = 2 Ln( 2 − 1) and Arg z = π. Hence, Eqs. (3) and (2) yield: √ √ arctan(i 2) = n + 21 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 . 1
(6)
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 . The end result is: e2inθ − 1 einθ (einθ − e−inθ ) einθ sin nθ S = iθ = = , e − e−iθ eiθ − e−iθ sin θ where we have used eiθ − e−iθ sin θ ≡ 2i to obtain the final result. In particular, using Re(einθ ) = cos nθ and Im(einθ ) = sin nθ, it follows that: Re S =
sin 2nθ cos nθ sin nθ = , sin θ 2 sin θ
Im S =
sin2 nθ , sin θ
after employing the trigonometric identity sin 2nθ = 2 sin nθ cos nθ. Equating the real and imaginary parts of Eq. (6), we conclude that Re S = cos θ + cos 3θ + cos 5θ + · · · + cos(2n − 1)θ = Im S = sin θ + sin 3θ + sin 5θ + · · · + sin(2n − 1)θ = which reproduces the results of Eqs. (4) and (5). 4. Boas Ch. 2, §17, Qu. 22. Show that tanh−1 z = We write z = tanh w =
1 2
ln
1+z 1−z
.
ew − e−w e2w − 1 = , ew + e−w e2w + 1
so which has solution
z e2w + 1 = e2w − 1 , e2w =
Taking the log gives w ≡ tanh
−1
1+z . 1−z
1 z = ln 2 2
1+z 1−z
.
sin 2nθ , 2 sin θ sin2 nθ , sin θ
5. The series for the principal value of the complex-valued logarithm, Ln(1 − z) = −
∞ X zn
n=1
n
,
(7)
converges for all |z| ≤ 1, z 6= 1. In particular, consider the conditionally convergent series, S≡
∞ inθ X e
n
n=1
,
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 , so S = − [LnR + iΘ] ,
(−π < Θ ≤ π) .
(11)
A simple way to do this is: h i 1 − eiθ = ReiΘ = eiθ/2 e−iθ/2 − eiθ/2 = −2ieiθ/2 sin(θ/2) = 2ei(θ−π)/2 sin(θ/2) , after using sin(θ/2) = follows that
1 iθ/2 2i (e
(12)
− e−iθ/2 ) and −i = e−iπ/2 . Since sin(θ/2) > 0 for 0 < θ < 2π, it 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 of the principal value of the argument function as defined in class. Hence, Eq. (11) simplifies to: θ S = −Ln(1 − e ) = −Ln 2 sin + 21 i(π − θ) , 2 iθ
(a) By taking the real part of Eq. (8), evaluate ∞ X cos nθ
n=1
n
,
where 0 < θ < 2π ,
3
for 0 < θ < 2π .
(14)
as a function of θ. Check that your answer has the right limit for θ = π. Noting that cos nθ = Re einθ , it follows from Eqs. (8) and (14) that ∞ X cos nθ
n
n=1
θ = Re S = −Ln 2 sin , 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, cos nπ = (−1)n and −(−1)n = (−1)n+1 , it follows that: Ln 2 =
∞ X (−1)n+1
n=1
n
=1−
1 2
+
1 3
−
1 4
+ ··· ,
a well-known result.
(b) By taking the imaginary part of Eq. (8), prove that ∞ X sin nθ
n
n=1
= 12 (π − θ) ,
where 0 < θ < 2π .
Noting that sin nθ = Im einθ , it follows from Eqs. (8) and (14) that ∞ X sin nθ
n
n=1
= Im S =
1 2 (π
− θ) ,
for 0 < θ < 2π .
(16)
Again, we can check the special case of θ = π. Since sin nπ = 0, Eq. (16) reduces in this limit to the trivial equation 0 = 0 .
6. Evaluate the integral In = for k > 0 and n > −1.
Z
∞
2
tn e−kt dt ,
0
Let us define a new variable, u = kt2 or equivalently t = du = 2ktdt Hence, In =
1 k n/2
Z
∞
u 0
=⇒
n/2 −u
e
du du = dt = 2kt 2k
p u/k. Then,
1/2 k du = 1/2 1/2 . u 2k u
du 1 du = (n+1)/2 1/2 1/2 2k u 2k
4
Z
∞ 0
u(n−1)/2 e−u du
Using the definition of the Gamma function, Γ(x) =
Z
∞
tx−1 e−t dt ,
0
we identify x = 12 (n + 1). Thus, In =
1 2k (n+1)/2
Γ
n+1 2
√ As a check, let us set k = 1 and n = 0. Then, we obtain I0 = 12 Γ( 21 ) = 21 π as expected. The factor of k (n+1)/2 can be deduced using simple dimensional analysis. If t has dimensions of time, then k must have dimensions of t−2 , since the argument of the exponential must be dimensionless.∗ Since dt also has dimensions of time, the entire integral has dimensions of tn+1 , which must be respected by the final result for In . Indeed k −(n+1)/2 has dimensions of tn+1 , and so the dimensions of our final result for In are consistent. In particular, it is possible to first carry out the computation with k = 1, and then determine the k dependence of In strictly on dimensional grounds! 7. Boas, Ch. 11 §3, Qu. 5. Simplify Γ( 21 )Γ(4)/Γ( 29 ). Using xΓ(x) = Γ(x + 1) repeatedly, one obtains Γ( 92 ) = 27 Γ( 27 ) = 27 · 52 Γ( 52 ), etc. until finally obtaining Γ( 29 ) = 27 · 25 · 32 · 21 Γ( 12 ) . Hence, using Γ(4) = 3! = 6, it follows that Γ( 21 )Γ(4) = Γ( 92 )
2 7
·
2 5
·
2 3
·2·6=
32 . 35
8. Boas, Ch. 11 §3, Qu. 13. Express as a Γ function Z
1 0
1 3 x ln dx . x 2
Introduce a new variable x = e−u . Then dx = −e−u du and ln(1/x) = ln eu = u. Noting that x = 0 =⇒ u = ∞ and x = 1 =⇒ u = 0, it follows that Z
1 0
Z Z ∞ 1 3 Γ(4) 2 1 ∞ 3 −v 3 −3u v e dv = = . u e du = 4 x ln dx = x 3 0 81 27 0 2
In the second step above, I used the overall minus sign to interchange the lower and upper limits of integration. At the third step, I changed the integration variable once more to v = 3u (the limits of integration are unchanged). Finally, I used the definition of the Gamma function [eq. (3.1) on p. 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 series expansion, ez = 1 + z + z 2 /2! + z 3 /3! + · · · . Suppose z had dimensions of length, for example. Then ez would be the sum of a dimensionless number plus a number with units of length plus a number with units of squared-length, and so on. But, one cannot consistently sum two quantities that possess different length dimensions. The only way to avoid this inconsistency is to conclude that z is dimensionless.
5
9. Boas, problem 11.5–3. Show that the binomial coefficient np can be written in terms of Gamma functions as: p Γ(p + 1) . (17) = n! Γ(p − n + 1) n The binomial coefficient is defined in eq. (13.6) on p. 28 of Boas as: p p(p − 1)(p − 2) · · · (p − n + 1) = . n n! To prove that it can be written as in Eq. (17), we first multiply numerator and denominator by Γ(p − n + 1), p p(p − 1)(p − 2) · · · (p − n + 1) Γ(p − n + 1) = . (18) n n! Γ(p − n + 1)
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 Γ(p + 1) = n 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
1 2
assuming that n is an integer.†
+ n = (−1)n π ,
(19)
We start from the reflection formula [eq. (5.4) on p. 541 of Boas]: Γ(p)Γ(1 − p) =
π . sin πp
(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 that for n = 0, Eq. (19) also yields a correct result, [Γ( 12 )]2 = π.
6
If we set p = Inserting p =
1 2
− n, then
1 2
− n into Eq. (20) yields:
1 − p = 1 − ( 12 − n) = n + 21 .
Γ
1 2
−n Γ
1 2
+n =
π
sin π
1 2
−n
Finally, we make use of the trigonometric identity, sin π 12 − n = sin 21 π − nπ = sin 12 π cos(nπ) − cos
. 1 2π
(21)
sin(nπ) = cos(nπ) ,
after using sin( 12 π) = 1 and cos( 12 π) = 0. Using the fact that cos(nπ) = (−1)n for any integer n, it follows that for any integer n . sin π 21 − n = (−1)n , Inserting this result back into Eq. (21) and using (−1)−n = (−1)n for any integer n, we obtain: Γ
1 2
−n Γ
which is the desired result.
1 2
+ n = (−1)n π ,
7
for any integer n ,
