Math 311 - hw 3 solutions
Tues, Sept 16, 2014
1. Consider the infinite series of functions given below.
∞
X
n sin nx
2n
n=1
a. Prove that the infinite series converges uniformly on R to a continuous function f .
We apply the Weierstrass M -Test with Mn := 2nn . Since | sin nx| ≤ 1 for all n ∈ N and x ∈ R we have
n sin nx n
for all n ∈ N, x ∈ R.
2n ≤ 2n = Mn
P∞
We show that n=1 Mn converges by the ratio test. We have
Mn+1 n + 1 2n
1
1
1
=
=
1+
→ < 1;
Mn n+1
2
n
2
n
2
P∞
nx
hence, n=1 Mn converges and therefore the infinite series converges uniformly. Since the function n sin
is
2n
continuous for each n, the limit must be continuous by Theorem 6.4.2.
b. Show that f is integrable on [0, π] and evaluate
Z π
f (x) dx.
0
Since f is continuous on [0, π] it is integrable on [0, π]. By Theorem 6.4.3 we have
Z π
∞
∞
∞
∞ Z π
π
X
X
X
X
1
1
1
4
n sin nx
n
dx =
− n cos nx = −
((−1) − 1) =
= ;
f (x) dx =
n
n
k
2
2
2
4
3
0
0
n=1
n=1
n=0
n=1 0
since only the odd terms of the sum are nonzero, we have reindexed.
2. Use the methods of 6.4 (and induction) to prove that for all x ∈ (−1, 1)
∞ X
n+k n
1
x
for all x ∈ (−1, 1) and k ≥ 0.
=
k
(1 − x)k+1
n=0
We use induction on on k = 0, 1, 2, . . . . We first check the case k = 0. We have
∞
∞ X
X
1
1
n+0 n
n
=
=
x =
x
for all x ∈ (−1, 1).
(1 − x)0+1
1 − x n=0
0
n=0
We now suppose that the formula holds for some k ≥ 0. Then differentiating termwise (see Theorem 6.4.12) we
have
∞ ∞
k+1
d
1
d X n + k n X (n + k) · · · (n + 1) n−1
=
=
x
=
nx
(1 − x)k+2
dx (1 − x)k+1
dx n=0
k
k!
n=1
∞
X
(n + k + 1) · · · (n + 2)(n + 1) n
x
k!
n=0
∞ X
n+k+1 n
= (k + 1)
x
k+1
n=0
=
by reindexing
for all x ∈ (−1, 1). Dividing through by k + 1 yields the desired result.
3. Consider the following power series. Find its radius of convergence and the set of points x at which the series
converges.
∞
X
(−1)n (x + 1)n
√
2n n
n=1
We first find the radius of convergence. Setting cn :=
(−1)n
√
2n n
we have
ρ := lim sup |cn |1/n = lim sup
n
n
1
1
= .
2
2n1/2n
Hence, R = 1/ρ = 2 is the radius of convergence. Thus the series converges for all x such that |x + 1| < 2. We
check the endpoints x = −3, 1. For x = −3 we have
∞
∞
∞
X
X
X
(−1)n (x + 1)n
(−1)n (−3 + 1)n
1
√
√
√ .
=
=
n n
n n
2
2
n
n=1
n=1
n=1
This series diverges since it is a p-series with p = 1/2 ≤ 1. For x = 1 we have
∞
∞
∞
X
X
X
(−1)n (x + 1)n
(−1)n (−1 + 1)n
1
√
√
=
=
(−1)n √ .
n n
n n
2
2
n
n=1
n=1
n=1
√
This series converges by the Alternating Series Test since {1/ n}n is a nonnegative nonincreasing sequence which
converges to 0. Therefore, D := (−3, 1] is the set of points x at which the series converges.
4. Prove that
∞
X
(−1)n−1 (x − 1)n
ln x =
n
n=1
for all x ∈ (0, 2).
For |x − 1| < 1 we have
∞
∞
X
X
1
1
=
=
(1 − x)n =
(−1)n (x − 1)n .
x
1 − (1 − x) n=0
n=0
Then by Theorem 6.4.10 we obtain
Z x
∞
∞
X
X
(−1)n
(−1)n−1 (x − 1)n
1
n+1
dt =
(x − 1)
=
ln x =
n+1
n
1 t
n=0
n=1
for all x ∈ (1 − 1, 1 + 1) = (0, 2).
5. Use Taylor’s Formula to prove that
cos x =
∞
X
(−1)n x2n
(2n)!
n=0
for all x ∈ R.
We first use Taylor’s Formula (Theorem 6.5.3) to prove that for all x ∈ R, Rn (x) → 0. Observe that f (x) := cos x
is infinitely differentiable, that is, all higher order derivatives exist. Moreover, the nth derivative (for n ≥ 0) is
given by

 cos x if n = 4k,


− sin x if n = 4k + 1,
(n)
f (x) =

− cos x if n = 4k + 2,



sin x if n = 4k + 3.
Hence |f (n) (x)| ≤ 1 for all x ∈ R. By Taylor’s formula, given x 6= 0 and n ≥ 0 we have
Rn (x) =
f (n+1) (c)xn+1
(n + 1)!
for some c between 0 and x. Therefore,
|Rn (x)| =
n+1
|f (n+1) (c)xn+1 |
|x|n+1
≤
.
(n + 1)!
(n + 1)!
|x|
Since (n+1)!
→ 0 as n → ∞ (see exercise 6.5.1), Rn (x) → 0 by the Squeeze Theorem. Hence, the Taylor Series for
f converges for all x ∈ R.
We must now show that the given series is in fact the Taylor series.
Note that for n = 2m + 1 we have f (n) (x) = f (2m+1) (x) = (−1)m+1 sin x and that for n = 2m we have
(n)
f (x) = f (2m) (x) = (−1)m cos x. Hence
(
(−1)m
if n = 2m,
f (n) (0)
= (2m)!
n!
0
otherwise.
P∞ (−1)n x2n
Therefore, the Taylor series is given by n=0 (2n)! and the required result holds.