Uniform Convergence of Power Series - MavDISK

Math 316-01 Intermediate Analysis
Questions for Section 37: Uniform Convergence of Power Series
1. Theorem 37.10: If a power series anx n converges at x = R for some
R > 0 then it converges uniformly on [0, R].
Proof: Assume that anR n converges for some R > 0. By definition,
this means that (sn(R)) converges to a real number, where sn(x) = a0 +
a1x + · · · + anx n . To say that anx n converges uniformly on [0, R] is to say
that the sequence of functions (sn(x)) converges uniformly on this interval.
By the Cauchy Convergence Criterion for function sequences, this requires
that we demonstrate that there exists N such that n > m > N implies
|sn(x) − sm(x)| < ǫ for all x ∈ [0, R]. Let ǫ > 0 be given. Since (sn(R))
converges by hypothesis, we know that we can find N so that n > m > N
implies |sn(R) − sm(R)| < ǫ. We will show that this choice of N works for
all x ∈ [0, R].
Let x ∈ [0, R] be given. We can assume without loss of generality that x < R,
because we already know what happens when x = R. Let bm = amRm for all
m and let β = x
R . Since (sn(R)) converges, we know that |bm+1+· · ·+bn| < ǫ
whenever n > m > R, and we wish to show |sn(x) − sm(x)| = |bm+1βm+1 +
· · · + bnβn | < ǫ. If each bi ≥ 0 this would be no problem because β < 1, but
we don’t know this. Mimicking the algebra at the bottom of page 342, let
s1 = bm+1, s2 = bm+1 + bm+2, . . ., sn−m = bm+1 + · · · + bn. Then we know
that |sk| < ǫ for all k. Moreover, we have
bm+1β m+1 + · · · + bnβ n =
s1β m+1 + (s2 − s1)β m+2 + (s3 − s2)β m+2 + · · · + (sn−m − sn−m−1)β n =
s1(β m+1 − β m+2 ) + s2(β m+2 − β m+3 ) + · · · + sn−m−1(β n−1 − β n ) + sn−mβ n =
β m+1 (1 − β)(s1 + s2β + · · · + sn−m−1β n−m−2 ) + sn−mβ n .
Since 0 ≤ β < 1 and |sk| < ǫ for all k, after taking the absolute value and
using the triangle inequality and canceling positive and negative terms we
obtain
|bm+1 + · · · + bn| ≤ β m+1 (1 − β)(ǫ + ǫβ + · · · + ǫβ n−m−2 ) + ǫβ n = ǫβ m+1 < ǫ.
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2. A similar proof shows that if anx n converges at x = −R for some R > 0
then it converges uniformly on [−R, 0].
3. Combining comments 1 and 2, if f(x) = anx n has radius of convergence
R, then it is uniformly convergent on [−K, K] for any 0 ≤ K < R. Reason:
it converges at x = K and x = −K, hence it is uniformly convergent on
[0, K] and [−K, 0], hence on [−K, K]. This is Theorem 37.1.
4. An important corollary of Theorem 37.1 is Corollary 37.11: If f(x) =
anx n converges at x = K, then f is continuous on [0, K], and in particular
at x = K.
Proof: By Corollary 37.11, the sequence of functions (sn(x)) is uniformly
convergent on [0, K]. Moreover, each function sn(x) = a0 + a1x + · · · + anx n
is a polynomial, hence continuous on [0, K]. By Theorem 36.1 the limit
function f(x) must be continuous on [0, K].
5. Theorem 37.2: The series anx n and the series nanx n−1 have the
same radius of convergence R. If f(x) = anx n for all x ∈ (−R, R) then
f ′ (x) = nanx n−1 for all x ∈ (−R, R).
Proof: Let x = 0. The root test applied to nanx n−1 yields
lim sup |nanx n−1 | 1
n = lim sup n 1
1
n 1
n |anx
|x|
n | 1
n = lim sup |anx n | 1
n,
which is the result of applying the root test to anxn . Hence both power
series have the same radius of convergence. Now let c be an number within
the radius of convergence of both power series, assuming R > 0 (a function
defined on a single point cannot be differentiated). We will prove that f ′
(c) =
nancn−1 . Let sn(x) = a0 + a1x + · · · + anxn . Then (sn(x)) converges
pointwise to f(x) and each s ′ n(x) is continuous on [0, c]. If we can show that
(x)) is uniformly convergent on [0, c], then by Corollary 36.8 we will know
(s ′ n
that f ′ (x) = nanxn−1 on [0, c]. In particular, f ′ (c) = nancn−1 . To
prove that (s ′ n(x)) is uniformly convergent on [0, c] it will suffice by Theorem
37.10 to show that (s ′ n (c)) converges. It does since c is within the radius of
convergence of both power series.
6. Corollary 37.3: The series anx n and the series n!
(n−k)! anx n−k have
the same radius of convergence R for each k ∈ N, and for all x ∈ (−R, R) we
have f (k) (x) = n!
(n−k)! anx n−k .
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Proof: Just apply Theorem 37.2 k times.
7. Corollary 37.4: If anx n = bnx n for all x ∈ (−R, R) then an = bn
for all n.
Proof: Let f(x) = cnxn on (−R, R). Then f (k) (x) = n!
(n−k)! cnx n−k on
(−R, R). Therefore f (k) (0) = k!ck. This implies ak = f(k) (0)
k! = bk. Repeat
for all k ≥ 0.
8. As an application, we will evaluate the integral x
arctant dt for any
0
x ∈ (0, 1). Our goal is to expand arctant as a power series and apply
Corollary 36.5, page 331. First, some scratch work, not worrying about
1
details: Fix x ∈ (0, 1). We learned in Calc II that an antiderivative of 1+t2 is arctant. We have
1
1 − t =
∞
t n ,
1
1 + t =
1
=
1 + t2 arctan t =
∞
n=0
∞
n=0
∞
n=0
n=0
(−1) n t n ,
(−1) n t 2n ,
(−1) n
2n + 1 t2n+1 .
This is not a proof that arctant = ∞ (−1)
n=0
n
radius of convergence of the series ∞
(−1, 1) → R by f(t) = ∞
n=0
n=0
2n+1 t2n+1 , just a conjecture. The
(−1) n
2n+1 t2n+1 is R = 1. Defining f :
(−1) n
2n+1 t2n+1 , we know that f ′ (t) = ∞
n=0 (−1)n t 2n
on (−1, 1) by Theorem 37.2. However, we know that 1
1+t
valid for −1 < t < 1, therefore 1
1+t 2 =
=
n=0 (−t)n is
n=0 (−1)n t 2n is valid for −1 < t < 1,
therefore f ′ (t) = 1
1+t 2. Since arctan t has the same derivative as f(t), we
have f(t) = arctan t + c for some c ∈ R. But 0 = f(0) = arctan(0) + c =
c, therefore c = 0 and f(t) = arctant. Moreover, since ∞ (−1)
n=0
n
2n+1 x2n+1
converges, the series ∞ (−1)
n=0
n
2n+1 t2n+1 converges uniformly to arctant on the
interval [0, x] by Theorem 37.10. Now we are in business. By Corollary 36.5
(with fn(t) = (−1)n
2n+1 t2n+1 for each n and [a, b] = [0, x]) we have
x
arctan t dt =
0
∞
x
n=0
0
(−1) n
2n + 1 t2n+1 =
3
∞
n=0
(−1) n
(2n + 1)(2n + 2) x2n+1 .

9. In Comment 8 above we proved that
x
arctant dt =
0
∞
n=0
(−1) n
(2n + 1)(2n + 2) x2n+1
for all 0 < x < 1. We will show that this formula is also valid for x = 1.
The arctangent function is continuous on [0, 1], hence the function F(x) =
arctant dt is uniformly continuous on [0, 1] by Theorem 31.1. Therefore
x
0
1
The power series ∞
n=0
0
arctan t dt = F(1) = lim F(x) =
x→1− lim
x→1 −
∞
n=0
(−1) n
(2n + 1)(2n + 2) x2n+1 .
(−1) n
(2n+1)(2n+2) x2n+1 converges at x = 1 by the alter-
nating series test, hence the function G : [0, 1] → R defined by G(x) =
∞
n=0
hence
lim
x→1 −
(−1) n
(2n+1)(2n+2) x2n+1 is uniformly continuous on [0, 1] by Theorem 37.10,
∞ (−1) n
(2n + 1)(2n + 2) x2n+1 ∞ (−1)
= lim G(x) = G(1) =
x→1− n
(2n + 1)(2n + 2) .
n=0
Putting everything together we have
1
arctant dt =
0
∞
n=0
n=0
(−1) n
(2n + 1)(2n + 2) .
10. If we want to approximate the value of 1
arctant dt with an error
0
no greater than 0.001, we can exploit the fact that it can be expressed as
the limit of partial sums of the alternating series ∞ . Setting
an =
1
(2n+1)(2n+2) and sn = n k=0
(−1) k
, we have
(2k+1)(2k+2)
sn
1
− arctant dt
< an+1 <
0
4
n=0
1
(2n + 3) 2.
(−1) n
(2n+1)(2n+2)

Therefore
1
s15 − arctant dt
Hence
0
0
1 1
< = < 0.001.
332 1089
1 15
(−1)
arctan t dt ≈
k
(2k + 1)(2k + 2)
k=0
= 0.438 · · ·
with error < 0.001. In fact, a Mathematica calculation shows us that
1
arctant dt = 1
(π − ln 4) = 0.438825 · · · .
4
0
11. We proved in Comment 8 above that arctanx = ∞ (−1)
n=0
n
2n+1 x2n+1 for
0 ≤ x < 1. Since the power series converges at x = 1 by the alternating
series test the function f(x) = ∞
n=0
(−1) n
2n+1 x2n+1 is uniformly convergent [0, 1].
Since each sn(x) is continuous, the limit function f(x) is continuous on [0, 1].
So is the function arctan x. Therefore
∞ (−1)
arctan1 = lim arctanx = lim f(x) = f(1) =
x→1− x→1− n
2n + 1 .
On the other hand, arctan 1 = π,
therefore we have a proof that
4
Since we have π
∞
n=0
(−1) n π
=
2n + 1 4 .
4
4
imate π by 4sn with error no greater than 2n+3 .
n=0
realized as the limit of an alternating series, we can approx-
Homework for Section 37, due ??? (only the starred problems will
be graded):
Hints:
3 ∗ (a, b), 9 ∗ (a, b, c), 10 ∗ (a, b, c)
3. (a) Conjecture the right formula for ∞
n=1 n2 x n , then prove that is valid for
|x| < 1 (see the first few lines of Comment 8 above). Note that 1
1
1−x
′ = nx n−1 , x 1
1−x
′ = nx n , therefore x
(1−x) 2 = nx n (conjec-
1−x = x n ,
tured, not proved). Repeat this to conjecture the right formula f(x) for
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2 n 1
n x . The prove that the second derivative of this function is , similar
1−x
to the logic used in Comment 8 above, citing the appropriate theorems. This
proves that n2xn = f(x)+ax+b, where f(x) is your conjectured formula.
Determine b by evaluating at x = 0. Determine a by differentiating then and
evaluating at x = 0. (b) Evaluate f( 1).
Keep track of the range of the index
2
of summation.
9. (a) In Example 36.6, pp. 331–332, it is proved that ln(1+x) = ∞ (−1)
n=1
n+1
is valid for −1 < x < 1. Now mimic the derivation of x
arctant dt in Com-
0
ment 8 above. (b) Mimic the limit arguments in Comment 8, citing the
appropriate theorems. (c)Use the fact that x ln x − x is an antiderivative of
ln x. Evaluate the formula in (b) at x = 1.
10. (a) In Example 37.5, p. 340, it is shown that ∞ x
n=0
n
n!
n x n
converges to the
limit function e x for all x. Use this to conjecture the power series expansion
of (e x − 1)/x, then prove that the sequence of partial sums of this series
converges to (e x − 1)/x for x = 0 using limit properties. Your answer will be
in the form ∞
n=1 bnx n−1 , but you can also express it in the form ∞
n=0 bn+1x n .
(b) Your function f(x) = (e x −1)/x = ∞
n=0 bn+1x n is technically not defined
at x = 0, but you might as well assume that it is by setting f(0) = b1 = 1.
(Note that (e x − 1)/x is not defined at x = 0, but it does have a limit of
1 as x → 0 using L’Hospital’s rule.) Use Theorem 37.2 to find a power
series expansion of f ′ (x) which is valid for (−R, R) for an arbitrary value of
R > 0. Since this is valid for all R, it is valid on (−∞, ∞). (c) Evaluate
the appropriate limit, using continuity of f(x) and its derivatives, citing the
appropriate theorems.
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