Mathematics 242 – Principles of Analysis Solutions for Midterm Exam 3 May 6, 2011

I. Both parts of this question refer to the function f : R → R defined by f(x) = x^{2} − 6x + 3.

A) (20) Consider the regular partitions P_{n} of the interval [0, 2] and show directly, using the upper and lower sums, that f is integrable on [0, 2].

Solution: Note that f is decreasing on [0, 2] since f′(x) = 2x − 6 < 0 for all x with 0 ≤ x ≤ 2. The partition is

P_{n} = {0, 2/n, 4/n, …, 2},

with x_{i} = 2i/n for i = 0, 1, …, n. Hence, since f is smallest at the right endpoint in each subinterval,

L_{P_{n}}(f) = ∑_{i=1}^{n} ((2i/n)^{2} − 6 · (2i/n) + 3) \frac{2}{n}

= \frac{8}{n^{3}} ∑_{i=1}^{n} i^{2} − \frac{24}{n^{2}} ∑_{i=1}^{n} i + \frac{6}{n} ∑_{i=1}^{n} 1

= \frac{8}{n^{3}} · \frac{n(n + 1)(2n + 1)}{6} − \frac{24}{n^{2}} · \frac{n(n + 1)}{2} + 6

= \frac{4}{3} + \frac{4}{n} + \frac{4}{3n^{2}} − 12 − \frac{12}{n} + 6

= \frac{−10}{3} − \frac{8}{n} + \frac{4}{3n^{2}}

.

Similarly, f is largest at the left endpoint in each subinterval, so

U_{P_{n}}(f) = ∑_{i=1}^{n} ((2(i − 1)/n)^{2} − 6 · (2(i − 1)/n) + 3) \frac{2}{n}

= \frac{8}{n^{3}} ∑_{i=1}^{n−1} i^{2} − \frac{24}{n^{2}} ∑_{i=1}^{n−1} i + \frac{6}{n} ∑_{i=1}^{n} 1

= \frac{8}{n^{3}} · \frac{(n − 1)n(2n − 1)}{6} − \frac{24}{n^{2}} · \frac{(n − 1)n}{2} + 6

= \frac{8}{3} − \frac{4}{n} + \frac{4}{3n^{2}} − 12 + \frac{12}{n} + 6

= \frac{−10}{3} + \frac{8}{n} + \frac{4}{3n^{2}}

.

Therefore, for any given ε > 0, if n > 16/ε,

U_{P_{n}}(f) − L_{P_{n}}(f) = \frac{16}{n} < ε.

This shows that f is integrable.

B) (15) Explain why the hypothesis of the Mean Value Theorem is satisfied for f on the interval [1, 5] and find the number c mentioned in the conclusion. Solution: f is a polynomial function, so it is differentiable, hence continuous everywhere. On the interval [1, 5], f(5) − f(1) = −2 − (−2) = 0. The MVT says that there is some c ∈ (1, 5) where f′(c) = 0 · (5 − 1) = 0. Since f′(c) = 2c − 6 = 0, this is true for c = 3.

II. (15) For which a ≥ 0 is

f(x) = \{ \begin{matrix}x^{a} sin(1/x) & if x ≠ 0 \\ 0 & if x = 0 \end{matrix}

f continuous at x = 0? For which a ≥ 0 is f differentiable at x = 0?

Solution: Note that this function is not defined for x < 0 for some a. Hence in the limits below, we will consider the case x → 0^{+} only. When x → 0^{−} also makes sense, the limits will be the same. For continuity at x = 0, we must have lim_{x→0^{+}} x^{a} sin(1/x) = f(0) = 0. This will be true by the squeeze theorem for limits as long as a > 0:

x^{a} · (−1) ≤ x^{a} sin(1/x) ≤ x^{a} · 1

for all x > 0. Hence since lim_{x→0^{+}} x^{a} = 0 if a > 0,

lim_{x→0^{+}} x^{a} sin(1/x) = 0 = f(0).

This is not true with a = 0 since then the limit does not exist. For differentiability at x = 0, the limit

lim_{x→0^{+}} \frac{f(x) − f(0)}{x − 0} = lim_{x→0^{+}} x^{a−1} sin(1/x)

must exist. This will be true (and the limit will equal zero) as long as a > 1 (squeeze theorem again).

III. (20) Show that if f is monotone increasing on [a, b], then f is integrable on [a, b].

Solution: If f is monotone increasing on [a, b] and P = {a = x_{0} < x_{1} < ⋯ < x_{n−1} < x_{n} = b} is a regular partition, then

U_{P}(f) − L_{P}(f) = ∑_{i=1} f(x_{i})\frac{b − a}{n} − ∑_{i=0}^{n−1} f(x_{i})\frac{b − a}{n}.

This is a telescoping sum, which cancels to

\frac{(f(b) − f(a))(b − a)}{n}.

Given ε > 0, this difference will be < ε as soon as n > \frac{(f(b)−f(a))(b−a)}{ε}. Therefore f is integrable on [a, b].

IV. True-False. Say whether each of the following statements is true or false. For true statements, give short proofs; for false ones give reasons or counterexamples. Do any three parts. If you submit solutions for all four, then I will consider the other one for Extra Credit.

A) (10) Let f(x) = e^{x} − e^{2x}. There exists some c ∈ (0, ln(3)) such that f(c) = −2.

Solution: The statement is TRUE. We apply the IVT. First, f is continuous everywhere since the exponentials e^{x} and e^{2x} are differentiable everywhere. On the interval [0, ln(3)], we have f(0) = 1 − 1 = 0 and f(ln(3)) = 3 − 9 = −6. Since −2 is in the range between the endpoint values, the (“weak form” of) the IVT implies that there exists c ∈ (0, ln(3)) such that f(c) = −2.

B) (10) If f is differentiable on (a, b) and there exists a real number B ≥ 0 such that |f′(x)| < B for all x ∈ (a, b), then f is uniformly continuous on (a, b).

Solution: This is TRUE. We apply the MVT to f on the interval [x, x′] where a < x < x′ < b are arbitary. Then there exists a c ∈ (x, x′) such that f(x) − f(x′) = f′(c)(x − x′). Taking absolute values, this implies

|f(x) − f(x′)| = |f′(c)||x − x′| ≤ B|x − x′|.

Given any ε > 0, let δ = ε/B. Then |x−x′| < ε/B implies |f(x)−f(x′)| < B·ε/B = ε. This shows uniform continuity of f on (a, b).

C) (10) If A ⊂ R is uncountably infinite, then A contains some nonempty interval (a, b).

Solution: This is FALSE. The Cantor set C is a counterexample. The “quick and dirty” approach for giving a reason is to recall that C is obtained by removing a collection of open intervals with total length equal to 1 from the interval [0, 1]. Hence what is left cannot contain a nonempty open interval (a, b). If so, then C would contain an interval of length b − a > 0. But that cannot be true since it would say C and its complement C′ could not fit together inside [0, 1].

Here is a sketch of a better proof as well. Recall that we showed the Cantor set was uncountably infinite in class using the base-3 description of the elements of the Cantor set – they are all the numbers in [0, 1] that have base-3 expansions using only the digits 0, 2. So the elements of the Cantor set are in correspondence with binary expansions of numbers in [0, 1], which we know is uncountably infinite by the Cantor diagonal argument. On the other hand, C cannot contain any interval. To see this, note that if c ∈ C and ε > 0, there will always be elements of the complement of the Cantor set in (c − ε, c + ε) – just go “far enough out” in the base three expansion

c = ∑_{n=1}^{∞} \frac{a_{n}}{3^{n}}

and change some of the digits. Namely, let n_{0} be sufficiently large that that the sum

∑_{n=n_{0}}^{∞} \frac{2}{3^{n}} = \frac{1}{3^{n_{0}−1}} < ε.

If we change a base-3 digit a_{n} in the expansion of c for some n ≥ n_{0} from a 0 or 2 to a 1, keeping at least one a_{n} = 2 to the right of that 1 so that the number will not be one of those in which the 1 can be replaced by an alternate expansion with a repeating 2 tail, we get a new number c′. The resulting number c′ will satisfy

|c − c′| < ∑_{n=n_{0}}^{∞} \frac{2}{3^{n}} < ε.

But it will be in the complement of C because of the digit equal to 1.

D) Let f be continuous on [a, b] and assume f(x) > 0 for all x ∈ [a, b]. Then there exists a constant k > 0 such that f(x) ≥ k > 0 for all x ∈ [a, b].

Solution: This is TRUE. By the Extreme Value Theorem, if k = glb{f(x) | x ∈ [a, b]}, then there exists a c ∈ [a, b] such that f(c) = k. But then by assumption k > 0 and f(x) ≥ k for all x ∈ [a, b].