Mathematics 242 – Principles of Analysis Solutions – Final Examination December 16, 2004

I. We have

A = \{cos(x) : x ∈ [0, 3π/4]\} = [−\sqrt{2}/2, 1] ≐ [−.707, 1]

and

B = {x : 1 < x^{2} < 4} = (−2, −1) ∪ (1, 2)

so: A) A ∪ B = (−2, −1) ∪ [−\sqrt{2}/2, 2). B) For each x ∈ A, |x − 2| represents the distance along the number line from x to 2. So

C = \{|x − 2| : x ∈ A\} = [1, 2 + \sqrt{2}/2]

The least upper bound of C is 2 + \frac{\sqrt{2}}{2} ≐ 2.707.

II. A) We say lim_{n→∞} x_{n} = L if for all ε > 0, there exist N such that |x_{n} − L| < ε for all n ≥ N. (The N can be taken to be a natural number or an arbitrary real, as convenient.)

B) We have

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

To prove this, note that

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

There are several ways to proceed from here. For instance, using the triangle inequality in the numerator, we could say:

\frac{|n − 3|}{n^{2} + 1} ≤ \frac{n + 3}{n^{2} + 1},

which is

≤ \frac{2n}{n^{2} + 1} < \frac{2n}{n^{2}} = \frac{2}{n}

as soon as n ≥ 3. Here’s the finished proof based on this idea: Let ε > 0, and let N > max(3, \frac{2}{ε}) Then for all n ≥ N, we have

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

III. A) The sequence x_{n} = sin(n) is bounded since | sin(n)| ≤ 1 for all n. Hence the Bolzano-Weierstrass Theorem says that there exists a convergent subsequence x_{n_{k}} = sin(n_{k}) for some sequence of integers n_{k}.

B) Consider the sequence cos(n_{k}), using the n_{k} from part A. This is also a bounded sequence since | cos(n_{k})| ≤ 1 for all k. Hence the Bolzano-Weierstrass Theorem implies that there is a convergent subsequence of this sequence, say y_{n_{k_{ℓ}}} = cos(n_{k_{ℓ}}). Note that this sequence comes from a subsequence of the index sequence n_{k}. Hence x_{n_{k_{ℓ}}} is a subsequence of the convergent sequence from part A. So since any subsequence of a convergent sequence is convergent too, the x_{n_{k_{ℓ}}} is also convergent.

Note: A lot of people lost a lot of points on this one because they did not take into account that the n inside the sin and cos in the definitions of the sequences x_{n} and y_{n} is the same as the index for the sequences. You have to consider plugging integer values into the two trig functions. Things like nπ/2 are NOT INTEGERS!

IV. A) Let f : D → R and let c be an accumulation point of D. Then we say lim_{x→c} f(x) = L if for all ε > 0, there exist δ > 0 such that |f(x)−L| < ε for all x ∈ D with 0 < |x−c| < δ.

B) The limit here is 0. Proof: Let ε > 0 and let δ = \sqrt{ε}. Then for all x with 0 < |x − 0| < \sqrt{ε}, we have (using | cos(1/x)| ≤ 1 for all x ≠ 0)

|x^{2} cos(1/x) − 0| = |x|^{2}| cos(1/x)| ≤ |x^{2}| < (\sqrt{ε})^{2} = ε.

Note: It is not correct to evaluate the limit using the limit product rule here. The problem is that lim_{x→0} cos(1/x) does not exist. However, cos(1/x) is bounded and the other factor x^{2} goes to zero, so the product goes to zero.

V. A) f(0) = 0 and f(2) = 1 by computing.

B) The denominator x^{4} + 48 is nonzero for all x ∈ R. Hence f(x) = \frac{32x}{x^{4}+48} is continuous at all x. By the IVT on the interval [0, 2], for each k with 0 < k < 1, there is at least one x ∈ (0, 2) such that f(x) = k. To find a second x satisfying this condition, note that lim_{x→+∞} f(x) = 0. Hence by the IVT again, there is also at least one additional solution of f(x) = k for x ∈ (2, +∞).

C) (Note: We are interested in applying this statement to the f(x) from the problem, which is differentiable at all x. So it suffices to consider the contrapositive statement, assuming that f is differentiable: If for some k, f(x_{1}) = f(x_{2}) = k for some x_{1} ≠ x_{2}, then f′(x) = 0 for some x between x_{1} and x_{2}. This is exactly the situation of the special case of the MVT known as Rolle’s Theorem. So because the contrapositive statement is equivalent to this one, we have: If f is differentiable and f′(x) ≠ 0 for all x ∈ I, then for all k, the equation f(x) = k has at most one solution x ∈ I.

D) Our function f has derivative

f′(x) = \frac{1536 − 96x^{2}}{x^{4} + 48}

(quotient rule for derivatives!). This is zero for x > 0 only at x = 2. Hence by part C, on the intervals (0, 2) and (2, +∞), there can only be one solution of f(x) = k in each interval. This means that there are exactly two of them all together.

VI. The most economical way to show that f(x) = x^{2} + x − 1 is integrable on [0, 3] is to note that f(x) is monotone increasing on that interval (f′(x) = 2x + 1 > 0 for all x ∈ [0, 3].) Hence as in the proof of our general theorem, for the regular partition P_{n} (with Δx = 3/n),

|U(f, P_{n}) − L(f, P_{n})| = |f(3) − f(0)|\frac{3}{n} = \frac{36}{n}.

Since this goes to zero as n → ∞, we know f is integrable. The value is computed using the upper sum, for instance. As above, Δx = 3/n, and the points in the partition are P_{n} = {3i/n : i = 0, 1, …, n}. Then

U(f, P_{n}) = ∑_{i=1}^{n} ((\frac{3i}{i})^{2} + \frac{3i}{i} − 1) \frac{3}{n}

= \frac{27}{n^{3}} ∑_{i=1}^{n} i^{2} + \frac{9}{n^{2}} ∑_{i=1}^{n} i − \frac{3}{n} ∑_{i=1}^{n} 1

= \frac{27n(n + 1)(2n + 1)}{6n^{3}} + \frac{9n(n + 1)}{2n^{2}} − 3

= 9 + \frac{27}{2n} + \frac{9}{2n^{2}} + \frac{9}{2} + \frac{9}{2n} − 3

= \frac{21}{2} + \frac{18}{n} + \frac{9}{2n^{2}}

⇒ ∫_{0}^{3} x^{2} + x − 1 dx = lim_{n→∞} U(f, P_{n}) = \frac{21}{2}

B) See class notes and text.

VII. A) True. The sequence of partial sums is monotone increasing if a_{n} ≥ 0. Hence if s_{N} is bounded above, that sequence converges.

B) False. The series is absolutely convergent, since ∑_{n=1}^{∞} \frac{1}{n^{3}} is a p-series with p = 3 > 1. Conditionally convergent series are ones where ∑_{i=1}^{∞} a_{n} converges, but ∑_{n=1}^{∞} |a_{n}| diverges.

C) True. Let x_{n} be any sequence with lim_{n→∞} x_{n} = 0. Then whether or not x_{n} is rational, we have lim_{n→∞} f(x_{n}) = 0 = f(0). This shows lim_{x→0} f(x) = f(0), so f is continuous at x = 0.