Physics 451
Fall 2004 Homework Assignment #7 — Solutions
Textbook problems: Ch. 5: 5.4.1, 5.4.2, 5.4.3, 5.5.2, 5.5.4 Chapter 5 5.4.1 Given the series ln(1 + x) = x −
x3 x4 x2 + − + ···, 2 3 4
−1 < x ≤ 1
show that ln
1+x 1−x
x3 x5 =2 x+ + + ··· , 3 5
−1 < x < 1
We use the property ln(a/b) = ln a − ln b to write ln
1+x 1−x
= ln(1 + x) − ln(1 − x) = =
∞ X n=1 ∞ X n=1
∞
(−1)n+1
xn X (−x)n (−1)n+1 − n n n=1
((−1)n+1 + 1)
X xn xn =2 n n n odd
Note that, since we use the ln(1 + x) series for both +x and −x, the common range of convergence is the intersection of −1 < x ≤ 1 and −1 < −x ≤ 1, namely |x| < 1. 5.4.2 Determine the values of the coefficients a1 , a2 , and a3 that will make (1 + a1 x + a2 x2 + a3 x3 ) ln(1 + x) converge as n−4 . Find the resulting series. Using the expansion for ln(1 + x), we write (1 + a1 x + a2 x2 +a3 x3 ) ln(1 + x) n ∞ X a1 xn+1 a2 xn+2 a3 xn+3 x n+1 + + + = (−1) n n n n n=1 We want to collect identical powers of x on the right hand side. To do this, we must shift the index n according to n → n − 1, n → n − 2 and n → n − 3 for the second, third and last terms on the right hand side, respectively. After doing
so, we may combine terms with powers x4 and higher. The first few terms (x, x2 and x3 ) may be treated as exceptions. The result is (1 + a1 x + a2 x2 +a3 x3 ) ln(1 + x) = (x − 12 x2 + 13 x3 ) + a1 (x2 − 21 x3 ) + a2 x3 n ∞ X a1 xn a2 xn a3 xn x n+1 + − + − (−1) n n − 1 n − 2 n−3 n=4
(1)
= x + (a1 − 12 )x2 + (a2 − 21 a1 + 31 )x3 ∞ X 1 a1 a2 a3 n+1 − + − + (−1) xn n n − 1 n − 2 n − 3 n=4 Combining the terms over a common denominator yields
1 a1 a2 a3 − + − n n−1 n−2 n−3 (n − 1)(n − 2)(n − 3) − a1 n(n − 2)(n − 3) + a2 n(n − 1)(n − 3) −a3 n(n − 1)(n − 2) = n(n − 1)(n − 2)(n − 3) 3 (1 − a1 + a2 − a3 )n + (−6 + 5a1 − 4a2 + 3a3 )n2 +(11 − 6a1 + 3a2 − 2a3 )n − 6 = n(n − 1)(n − 2)(n − 3)
To make this converge as n−4 , we need to cancel the coefficients of n3 , n2 and n in the numerator. Solving for 1 − a1 + a2 − a3 = 0,
−6 + 5a1 − 4a2 + 3a3 = 0,
11 − 6a1 + 3a2 − 2a3 = 0
yields the solution a1 = 3,
a2 = 3,
a3 = 1
Finally, inserting this back into (1), we obtain 2
3
(1 + 3x + 3x + x ) ln(1 + x) = x +
11 3 5 2 2x + 6 x +6
∞ X n=4
or ln(1 + x) =
x + 25 x2 +
11 3 6 x
+6
(−1)n
xn n(n − 1)(n − 2)(n − 3)
xn n n=4 (−1) n(n−1)(n−2)(n−3) (1 + x)3
P∞
5.4.3 Show that ∞ X a) [ζ(n) − 1] = 1 n=2
Using the sum formula for the Riemann zeta function, we have ∞ X
[ζ(n) − 1] =
n=2
∞ X
"
n=2
∞ X 1 pn p=1
!
"∞ # "∞ # ∞ ∞ X X 1 X X 1 −1 = = pn pn n=2 p=2 p=2 n=2 #
where in the last step we have rearranged the order of summation. In doing so, we have now changed this to a geometric series, with sum ∞ X
p−2 1 = −1 1−p p(p − 1)
p−n =
n=2
In this case ∞ X
∞
∞ X
X 1 = [ζ(n) − 1] = p(p − 1) p=2 n=2 p=2
1 1 − p−1 p
=1
since this is a telescoping series. b)
∞ X
(−1)n [ζ(n) − 1] =
n=2
1 2
The solution to this is similar to that of part a). The addition of (−1)n yields ∞ X
n
(−1) [ζ(n) − 1] =
n=2
∞ X n=2
" n
(−1)
∞ X 1 pn p=2
# =
"∞ ∞ X X n=2
1 (−p)n p=2
# =
"∞ ∞ X X p=2
1 (−p)n n=2
The sum over n is still a geometric series, this time with ∞ X n=2
(−p)−n =
1 (−p)−2 = −1 1 − (−p) p(p + 1)
In this case ∞ X
∞ X
∞
X 1 (−1) [ζ(n) − 1] = = p(p + 1) p=2 n=2 p=2 n
1 1 − p p+1
=
1 2
#
5.5.2 For what range of x is the geometric series
P∞
n=0
xn uniformly convergent?
P∞ n We use the Weierstrass M test. We first note that the geometric series P∞ n=0n x is absolutely convergent for |x| < 1. This means that the series n=0 s is convergent for 0 ≤ s < 1. While this is all very obvious, the introduction of this convergent series in s allows us to bound the x series by an x-independent convergent one. This is precisely the setup of the Weierstrass M test. We simply choose Mn = sn . Then, so long as |x|n ≤ Mn (ie |x| ≤Ps), the ∞ geometric series is uniformly convergent. Therefore we have shown that n=0 xn is uniformly convergent provided |x| ≤ s < 1. P P 5.5.4 If the series of the coefficients an and bn are absolutely convergent, show that the Fourier series X (an cos nx + bn sin nx) is uniformly convergent for −∞ < x < ∞. This is also a case for the Weierstrass M test. Note that, if we let α(x) = an cos nx + bn sin nx denote the n-th element of the series, then |α(x)| = |an cos nx + bn sin nx| ≤ |an cos nx| + |bn sin nx| ≤ |an | + |bn | P for the entire domain x ∈ (−∞, ∞). Since the problem states that an and P b are absolutely convergent, we now take simply M = |a | + |b |. Clearly, n n P n Pn Mn converges, and since |α(x)| ≤ Mn , we conclude that α(x) is uniformly convergent for x ∈ (−∞, ∞).
