Math 312, Intro. to Real Analysis: Homework #7 Solutions

Stephen G. Simpson

Wednesday, April 29, 2009

The assignment consists of Exercises 20.1, 20.18, 23.1, 23.4, 23.6, 24.1, 24.2, 24.6, 24.14, 24.17, 25.3, 25.6, 26.2, 26.6, 26.7 in the Ross textbook. Each exercise counts 5 points.

20.1. For f(x) = x/|x| = sgn x we have

lim_{x→∞} f(x) = lim_{x→0^{+}} f(x) = 1

and

lim_{x→−∞} f(x) = lim_{x→0^{−}} f(x) = −1.

20.18. For f(x) = \frac{\sqrt{1 + 3x^{2}} − 1}{x^{2}} and x ≠ 0 we have

f(x) = \frac{(1 + 3x^{2}) − 1}{x^{2}(\sqrt{1 + 3x^{2}} + 1)} = \frac{3}{\sqrt{1 + 3x^{2}} + 1}

hence

lim_{x→0} f(x) = \frac{3}{2}.

23.1. (a) R = 1, I = (−1, 1).

(b) R = ∞, I = (−∞, ∞).

(c) R = 1/2, I = [−1/2, 1/2].

(d) R = 3, I = (−3, 3).

(e) R = ∞, I = (−∞, ∞).

(f) R = 2, I = [−2, 2].

(g) R = 4/3, I = [−4/3, 4/3).

(h) R = 4, I = [−4, 4]. 23.4. Note that a_{n} = (6/5)^{n} if n is even, (2/5)^{n} if n is odd.

Moreover a_{n+1}/a_{n} = 2/(5 · 3^{n}) if n is even, (2 · 3^{n})/5 if n is odd.

(a) lim sup a_{n}^{1/n} = 6/5, lim inf a_{n}^{1/n} = 2/5,

lim sup a_{n+1}/a_{n} = ∞, lim inf a_{n+1}/a_{n} = 0.

(b) ∑ a_{n} and ∑(−1)^{n}a_{n} diverge, because a_{n} ↛ 0.

(c) For the power series ∑ a_{n}x^{n} we have R = 5/6 and I = (−5/6, 5/6).

23.6. Suppose a_{n} ≥ 0 and x ≥ 0. If ∑ a_{n}x^{n} converges, then ∑ a_{n}(−x)^{n} converges absolutely (because |a_{n}(−x)^{n}| = a_{n}x^{n}), hence ∑ a_{n}(−x)^{n} converges in view of Corollary 14.7.

24.1. For f_{n}(x) = \frac{1 + 2 cos^{2} nx}{\sqrt{n}} we have | cos^{2} nx| ≤ 1, hence |f_{n}(x)| ≤ \frac{3}{\sqrt{n}},

hence |f_{n}(x)| < ϵ whenever n > 9/ϵ^{2}. Since 9/ϵ^{2} is independent of x, we see that f_{n} converges uniformly to 0.

24.2. For f_{n}(x) = x/n we have:

(a) f(x) = lim f_{n}(x) = lim \frac{x}{n} = 0 (pointwise convergence).

(b) f_{n} → 0 uniformly on [0, 1]. This is because for all x ∈ [0, 1] we have

|f_{n}(x)| = \frac{|x|}{n} ≤ \frac{1}{n} < ϵ

whenever n > 1/ϵ.

(c) f_{n} ↛ 0 uniformly on [0, ∞). This is because for any n we can find x ∈ [0, ∞) such that |f_{n}(x)| ≥ 1. (An example of such an x is x = n.) Thus the definition of uniform convergence fails for ϵ = 1.

24.6. Let f_{n}(x) = (x − \frac{1}{n})^{2} and f(x) = x^{2}.

(a) Clearly for all x we have f_{n}(x) → f(x), i.e., pointwise convergence.

(b) For x ∈ [0, 1] we have

|f_{n}(x) − f(x)| = |(x − \frac{1}{n})^{2} − x^{2}| = |−\frac{2x}{n} + \frac{1}{n^{2}} | ≤ \frac{2}{n} + \frac{1}{n^{2}} → 0

independently of x ∈ [0, 1]. Thus f_{n} → f uniformly on [0, 1].

24.14. Let f_{n}(x) = \frac{nx}{1 + n^{2}x^{2}} .

(a) For x = 0 we have f_{n}(x) = 0 for all n, hence lim f_{n}(0) = 0. For x ≠ 0

we have

f_{n}(x) = \frac{1}{\frac{1}{nx} + nx} → 0

so again lim f_{n}(x) = 0. Thus f_{n} → 0 pointwise for all x.

(b) For all n we have f_{n}(1/n) = 1/2. Therefore, since 1/n ∈ [0, 1], we see that f_{n} ↛ 0 uniformly on [0, 1], with ϵ = 1/2.

(c) The derivative of f_{n}(x) is f′_{n}(x) = \frac{n − n^{3}x^{2}}{(1 + n^{2}x^{2})^{2}} which is clearly < 0

for all x ≥ 1. Thus f_{n}(x) is nonincreasing for all x ≥ 1, so in particular f_{n}(1) ≥ f_{n}(x) ≥ 0 for all x ≥ 1. But f_{n}(1) = n/(1+n^{2}) → 0, hence f_{n}(x) → 0 uniformly for all x ≥ 1.

24.17. Assume that f_{n} → f uniformly and f_{n} is continuous. By Theorem 24.3 it follows that f is continuous. Assuming x_{n} → x, we are asked to prove that lim f_{n}(x_{n}) = f(x). In other words, given ϵ > 0, we must show that |f_{n}(x_{n}) − f(x)| < ϵ for all sufficiently large n. By the Triangle Inequality we have

|f_{n}(x_{n}) − f(x)| ≤ |f_{n}(x_{n}) − f(x_{n})| + |f(x_{n}) − f(x)|  (1)

for all n. Since f_{n} → f uniformly, we can find N_{1} such that |f_{n}(z)−f(z)| < ϵ/2 holds for all n > N_{1} and all z. In particular |f_{n}(x_{n}) − f(x_{n})| < ϵ/2 whenever n > N_{1}. On the other hand, since f is continuous and x_{n} → x, we can find N_{2} such that |f(x_{n}) − f(x)| < ϵ/2 whenever n > N_{2}. Letting N = max(N_{1}, N_{2}), we see from (1) that |f_{n}(x_{n}) − f(x)| < ϵ whenever n > N. This completes the proof.

25.3. Let f_{n}(x) = \frac{n + cos x}{2n + sin^{2} x}.

(a) Clearly lim f_{n}(x) = 1/2 for all x, so the limit function is f(x) = 1/2. Since | cos x| ≤ 1 and 0 ≤ sin^{2} x ≤ 1, we have

|f_{n}(x) − \frac{1}{2}| = |\frac{n + cos x}{2n + sin^{2} x} − \frac{1}{2}| = |\frac{2 cos x − sin^{2} x}{2(2n + sin^{2} x)}| ≤ \frac{3}{4n} → 0

and this is independent of x. Thus f_{n} → 1/2 uniformly.

(b) By part (a) and Theorem 25.2 we have

lim_{n→∞} ∫_{2}^{7} f_{n}(x) dx = ∫_{2}^{7} lim_{n→∞} f_{n}(x) dx = ∫_{2}^{7} \frac{1}{2} dx = \frac{5}{2}.

25.6. (a) Assume that ∑_{k=0}^{∞} |a_{k}| < ∞. Then ∑_{k=0}^{∞} a_{k}x^{k} converges uniformly for all x ∈ [−1, 1], by the Weierstrass M-test. (This is because |a_{k}x^{k}| ≤ |a_{k}| for all x ∈ [−1, 1].) Moreover, the partial sums ∑_{k=0}^{n} a_{k}x^{k} are polynomials, hence they are continuous on [−1, 1], so by Theorem 24.3 the limit function ∑_{k=0}^{∞} a_{k}x^{k} is continuous on [−1, 1].

(b) For example, we know that ∑_{k=1}^{∞} \frac{1}{k^{2}} < ∞ (this is the p-series with p = 2). Therefore, it follows by part (a) that the power series ∑_{k=1}^{∞} \frac{1}{k^{2}} x^{k} is continuous on [−1, 1].

26.2. (a) We have

∑_{n=0}^{∞} x^{n} = \frac{1}{1 − x}

for all x ∈ (−1, 1) (geometric series). Differentiating term by term according to Theorem 26.5, we see that

∑_{n=1}^{∞} nx^{n−1} = \frac{1}{(1 − x)^{2}}

for all x ∈ (−1, 1). Multiplying through by x, we see that

∑_{n=1}^{∞} nx^{n} = \frac{x}{(1 − x)^{2}}

for all x ∈ (−1, 1).

(b) Plugging x = 1/2 into the result of part (a), we see that

∑_{n=1}^{∞} \frac{n}{2^{n}} = ∑_{n=1}^{∞} n(\frac{1}{2})^{n} = \frac{\frac{1}{2}}{(1 − \frac{1}{2})^{2}} = 2.

(c) Plugging x = 1/3 and x = −1/3 into the result of part (a), we see that

∑_{n=1}^{∞} \frac{n}{3^{n}} = ∑_{n=1}^{∞} n(\frac{1}{3})^{n} = \frac{\frac{1}{3}}{(1 − \frac{1}{3})^{2}} = \frac{3}{4}

and

∑_{n=1}^{∞} \frac{(−1)^{n}n}{3^{n}} = ∑_{n=1}^{∞} n(−\frac{1}{3})^{n} = \frac{−\frac{1}{3}}{(1 − (−\frac{1}{3}))^{2}} = −\frac{3}{16}.

26.6. Let

s(x) = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − ⋯

and

c(x) = 1 − \frac{x^{2}}{2!} + \frac{x^{4}}{4!} − ⋯

and note that the interval of convergence for these series is (−∞, ∞).

(a) Differentiating term by term according to Theorem 26.5, we get

s′(x) = 1 − \frac{3x^{2}}{3!} + \frac{5x^{4}}{5!} − ⋯ = c(x)

and

c′(x) = 0 − \frac{2x}{2!} + \frac{4x^{3}}{4!} − \frac{6x^{5}}{6!} + ⋯ = −s(x).

(b) By part (a) plus the usual laws of differentiation, we have (s^{2} +c^{2})′ = 2ss′ + 2cc′ = 2sc − 2cs = 0. From this it follows that s^{2} + c^{2} is a constant function.

(c) By part (b) s^{2} + c^{2} is a constant function, say s^{2} + c^{2} = K for some constant K. Plugging in x = 0 we get s(0)^{2} + c(0)^{2} = 1^{2} + 0^{2} = 1, hence K = 1. This proves that s(x)^{2} + c(x)^{2} = 1 for all x.

26.7. Let f(x) = |x|. This function is not differentiable at x = 0, because the left-hand derivative is −1 and the right-hand derivative is 1. Therefore, this function is not representable as a power series ∑_{n=0}^{∞} a_{n}x^{n}. If it were representable in this way, it would be be differentiable, by Theorem 26.5.