Mathematics 242 – Principles of Analysis Final Exam Solutions

I. A) A = {sin(x) : 0 ≤ x ≤ 3π/4} = [0, 1] and B = {x : 1 < x^{2} < 4} = (−2, −1) ∪ (1, 2), so

A ∪ B = (−2, −1) ∪ [0, 2). B) C = {|x − 2| : x ∈ A} equals the set [1, 2], since we think of |x − 2| as the distance

from x to 2. Hence lub(C) = 2 (the largest distance to 2 of any element of C).

II. A) x_{n} converges to L if for every ϵ > 0, there exists some corresponding n_{0} ∈ N such

that |x_{n} − L| < ϵ for all n ≥ n_{0}. B) By the “Big Limit Theorem” (for instance),

lim_{n→∞} \frac{3n^{2} + n}{n^{2} + 1} = lim_{n→∞} \frac{(3n^{2} + n) · \frac{1}{n^{2}}}{(n^{2} + 1) · \frac{1}{n^{2}}}

= lim_{n→∞} \frac{3 + \frac{1}{n}}{1 + \frac{1}{n^{2}}}

= 3

To prove this by the definition, let n ≥ 3. Then

|\frac{3n^{2} + n}{n^{2} + 1} − 3| = |\frac{n − 3}{n^{2} + 1}| = \frac{n − 3}{n^{2} + 1} < \frac{n}{n^{2}} = \frac{1}{n}

Hence, given ϵ > 0, if n_{0} ≥ max{3, 1/ϵ} and n ≥ n_{0}, then

|\frac{3n^{2} + n}{n^{2} + 1} − 3| < \frac{1}{n} < ϵ

III. A) Since sin(x) takes values only between -1 and 1, x_{n} is a bounded sequence. Hence,

the Bolzano-Weierstrass Theorem implies that x_{n} has a convergent subsequence x_{n_{k}}. B) The answer is yes. Using the index sequence n_{k} found in part A, the sequence y_{n_{k}} is

also a bounded sequence since the cosine function is also bounded. By the Bolzano-Weierstrass Theorem again, there is a convergent subsequence y_{n_{k_{ℓ}}}. The corresponding subsequence x_{n_{k_{ℓ}}} is also convergent, since every subsequence of a convergent sequence (like x_{n_{k}}) is also convergent.

IV. A) lim_{x→c} f(x) = L if for every ϵ > 0 there exists a corresponding δ > 0 such that

|f(x) − L| < ϵ for all x with 0 < |x − c| < δ.

B) lim_{x→0} x^{2} sin(1/x) = 0 since for all x ≠ 0,

|x^{2} sin(1/x)| = |x|^{2}| sin(1/x)| ≤ |x|^{2}

(the last inequality follows since | sin(θ)| ≤ 1 for all θ). Hence, given ϵ > 0, let δ = \sqrt{ϵ}. Then if 0 < |x| < δ,

|x^{2} sin(1/x) − 0| ≤ |x|^{2} < δ^{2} = ϵ

V. A) The Intermediate Value Theorem: Let f be continuous on [a, b] and let m = min{f(x) :

x ∈ [a, b]}, M = max{f(x) : x ∈ [a, b]}. Then for all y_{0}, m ≤ y_{0} ≤ M, there exist x_{0} ∈ [a, b] such that f(x_{0}) = y_{0}.

B) For f(x) = a_{1} + a_{2} sin(x),

\{ \begin{matrix}f(−π/2) = a_{1} − a_{2} \\ f(π/2) = a_{1} + a_{2} \end{matrix}

By hypothesis, a_{2} > |a_{1}|. Hence a_{1} −a_{2} < 0 and a_{1} +a_{2} > 0. Since f is continuous everywhere, the Intermediate Value Theorem implies that there is some x_{0} ∈ [−π/2, π/2] such that f(x_{0}) = 0.

VI. The Fundamental Theorem of Calculus: Let f be continuous on [a, b], and let F (x) = ∫_{a}^{x} f(t) dt. Then:

1) F is differentiable on [a, b], and F′(x) = f(x).

2) If G is any antiderivative of f, then ∫_{a}^{b} f(x) dx = G(b) − G(a). See the class notes for the proof.

VII. A) False – f has a jump discontinuity at x = 1. If f is differentiable at c it is also

continuous at c. B) True – the sequence of partial sums is bounded above and monotone increasing since

a_{n} ≥ 0 for all n. Hence the sequence of partial sums converges to some real number. C) True – the conclusion of the MVT is that there exists some c ∈ (a, b) such that

f′(c) = \frac{f(b)−f(a)}{b−a}. This is true here (even though the hypotheses of the MVT are not satisfied!!) since

f′ (\frac{1}{8}) = 2 · \frac{1}{8} = \frac{1}{4} = \frac{f(2) − f(0)}{2 − 0}

D) False – this series is absolutely convergent, not conditionally convergent. The series ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{3}} itself converges by the Alternating Series Test. Then, the series of absolute values:

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

also converges since it is a p series with p = 3 > 1.

E) True – We proved in class that if f is continuous on [a, b] except for a finite number

of jump or removable discontinuities, then f is integrable. That is the case for f here. F) False – By the Ratio Test,

lim_{n→∞} |\frac{x^{n+1}}{(n + 1)2^{n+1}} \frac{n2^{n}}{x^{n}} | = lim_{n→∞} \frac{n}{n + 1} \frac{|x|}{2} = \frac{|x|}{2}

This is > 1 if |x| > 2. Hence the series diverges for all x with |x| > 2. (It converges absolutely by the Ratio Test for all |x| < 2. The series also converges conditionally for x = −2, but diverges for x = 2.)

G) True – This basically comes from the Integral Test calculation that shows the p-series with p > 1 is convergent. By thinking of the partial sum ∑_{n=1}^{N} \frac{1}{n^{p}} as the upper sum (left-hand Riemann sum) for the decreasing function f(x) = \frac{1}{x^{p}} on [1, N + 1] corresponding to the partition P = {1, 2, …, N + 1}, we get

∑_{n=1}^{N} \frac{1}{n^{p}} ≥ ∫_{1}^{N+1} \frac{1}{x^{p}} dx = \frac{(N + 1)^{1−p} − 1}{1 − p}.

As N → ∞, we get

∑_{n=1}^{∞} \frac{1}{n^{p}} ≥ lim_{n→∞} \frac{(N + 1)^{1−p} − 1}{1 − p} = \frac{1}{p − 1}

On the other hand, omitting the first term in the series gives ∑_{n=2}^{N} \frac{1}{n^{p}} as the lower sum (right-hand Riemann sum) for the decreasing function f(x) = \frac{1}{x^{p}} on [1, N] corresponding to the partition P = {1, 2, …, N}. So

∑_{n=2}^{N} \frac{1}{n^{p}} = ∑_{n=1}^{N} \frac{1}{n^{p}} − 1 ≤ ∫_{1}^{N} \frac{1}{x^{p}} dx = \frac{N^{1−p} − 1}{1 − p}

Again letting N → ∞, we get

∑_{n=1}^{∞} \frac{1}{n^{p}} ≤ 1 + lim_{N→∞} \frac{N^{1−p} − 1}{1 − p} = 1 + \frac{1}{p − 1} = \frac{p}{p − 1}