Mathematics 242 – Principles of Analysis Solutions – Final Examination May 9, 2014

I. We have

A = {tan(x) : x ∈ [0, π/4]} = [0, 1]

and

B = {x : 1 < |x| < 3} = (−3, −1) ∪ (1, 3)

so: A) A ∪ B = [0, 1] ∪ ((−3, −1) ∪ (1, 3)) = (−3, −1) ∪ [0, 3). 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]

The least upper bound of C is 2.

II. A) We say lim_{n→∞} x_{n} = L if for all ε > 0, there exist n_{0} such that |x_{n} − L| < ε for all n ≥ n_{0}.

B) We have

lim_{n→∞} 1 + \frac{(−1)^{n}}{\sqrt{n}} = 1.

To prove this, note that

|1 + \frac{(−1)^{n}}{\sqrt{n}} − 1| = \frac{1}{\sqrt{n}}

For any given ε > 0, since N is not bounded in R, there exist

n_{0} > \frac{1}{ε^{2}} ⇔ \frac{1}{\sqrt{n_{0}}} < ε.

Hence if n ≥ n_{0}, then we have

|1 + \frac{(−1)^{n}}{\sqrt{n}} − 1| = \frac{1}{\sqrt{n}} ≤ \frac{1}{\sqrt{n_{0}}} < ε.

This shows the sequence converges to 1.

III. There is such an index sequence n_{k}. 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_{ℓ}} = sin(n_{ℓ}) for some strictly increasing index sequence of integers n_{ℓ}. Consider the sequence cos(n_{ℓ}), using these same indices n_{ℓ}. This is also a bounded sequence since | cos(n_{ℓ})| ≤ 1 for all ℓ. Hence the Bolzano-Weierstrass Theorem implies that there is a convergent subsequence of this sequence, say y_{n_{ℓ_{k}}} = cos(n_{ℓ_{k}}). Note that this sequence comes from a subsequence of the index sequence n_{ℓ}. Hence x_{n_{ℓ_{k}}} is a subsequence of a convergent sequence. Since any subsequence of a convergent sequence is convergent too, the x_{n_{ℓ_{k}}} is also convergent.

IV. A) Let f be defined on a deleted interval D about c (that is, at all points on some interval containing c, except possibly at c itself). 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. If we are considering the limit as x → +∞, then we should assume f(x) is defined for all x sufficiently large and the definition of lim_{x→+∞} f(x) = L becomes: For all ε > 0, there exist B > 0 such that |f(x) − L| < ε for all x > B.

B) The limit here is −1. Proof: Let ε > 0 and let δ = min(1, ε/3). Then for all x with 0 < |x − 0| < δ, we have

|x^{2} − 4x + 2 − (−1)| = |x^{2} − 4x + 3| = |x − 1||x − 3|.

Since |x − 1| < 1, we have 0 < x < 2. Hence |x − 3| < 3. But then:

|x^{2} − 4x + 2 − (−1)| = |x − 1||x − 3| < \frac{ε}{3} · 3 = ε.

This shows the limit equals L = −1.

C) As x → ∞, 1/x → 0. Hence the limit should be 1. Let ε > 0, and let B > \frac{1}{\sqrt{ε}}. Then x > B implies x > \frac{1}{\sqrt{ε}}, so \frac{1}{x^{2}} < ε. Hence if x > B, then

|\frac{1}{1 + \frac{1}{x^{2}}} − 1| = |\frac{\frac{−1}{x^{2}}}{1 + \frac{1}{x^{2}}} | = \frac{1}{x^{2} + 1} < \frac{1}{x^{2}} < ε.

This shows lim_{x→+∞} _{\frac{1}{1+ \frac{1}{x^{2}}}} = 1.

V. A) Statement of IVT: If f is continuous on [a, b] and y_{0} is any number between f(a) and f(b), then there exists c ∈ [a, b] with f(c) = y_{0}. See the class notes or the text for this proof.

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. We see f(0) = 0 and f(2) = \frac{64}{64} = 1. 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 given any 0 < k < 1, we will have f(b) < k for some b > 2 as well. By the IVT again, in the interval [2, b], there is also at least one additional solution of f(x) = k for x ∈ (2, +∞).

C) 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} by the special case of the MVT known as Rolle’s Theorem. However, our function f has derivative

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

(quotient rule for derivatives!). This is zero for x > 0 only at x = 2. Hence by Rolle’s Theorem, on the intervals (0, 2) and (2, +∞), there cannot be more than one solution of f(x) = k in each interval. This means that there are exactly two of them all together.

VI. We want to show that given any ε > 0, there exists a partition P of [0, 2] such that U(f, P) − L(f, P) < ε. Since the function changes from being decreasing to increasing (and is discontinuous) at x = 1, let’s chose a regular partition of [0, 1] with an even number n = 2q of subintervals, so that x_{q} = 1 is always one of the endpoints. We have x_{i} = 2i/(2q) = i/q for i = 0, …, 2q for each q ∈ N. On the first half of the interval, f is increasing. So for i = 1, …, q, we will have m_{i} = f((i − 1)/(2q)) = \frac{(i−1)}{2q} + 1 and for i = 1, …, q − 1, M_{i} = f(i/(2q)) = \frac{i}{2q} + 1, while M_{q} = 2. But then for i = q + 1, …, 2q, f

is decreasing so m_{i} = f(i/(2q)) = \frac{−i}{2q}, M_{i} = f((i − 1)/(2q)) = \frac{−(i−1)}{2q}. At this point, it is possible either to add up the upper and lower sums and subtract, or we can also be more clever. Following the proof that f monotone implies f integrable on [0, 1] and [1, 2], we can see that

U(f, P) − L(f, P) = (2 − 1)\frac{1}{2q} + ((−1) − (−2))\frac{1}{2q} = \frac{1}{q}

We can get this < ε for any q > 1/ε. The value of the integral is computed by taking the limit of the upper sum:

lim_{q→∞} [∑_{i=1}^{q} (\frac{i}{2q} + 1) + ∑_{i=q+1}^{2q} \frac{−(i − 1)}{2q}] \frac{1}{2q} = lim_{q→∞} [\frac{1}{4q^{2}} ∑_{i=1}^{q} i + \frac{1}{2q} ∑_{i=1}^{q} 1 + ∑_{i=1}^{q} (−1 − \frac{i}{2q} ) \frac{1}{2q} ]

= lim_{q→∞} [\frac{1}{4q^{2}} \frac{q(q + 1)}{2} + \frac{1}{2} − \frac{1}{2} − \frac{1}{4q^{2}} \frac{q(q + 1)}{2}]

= 0

(This can also be checked by the Fundamental Theorem:

∫_{0}^{2} f = ∫_{0}^{1} x + 1 dx + ∫_{1}^{2} −x dx = \frac{x^{2}}{2} + x|_{0}^{1} − \frac{x^{2}}{2} |_{1}^{2} = \frac{3}{2} − 2 + \frac{1}{2} = 0.)

VII. A) FALSE. Note that the problem does not say to assume the terms are all positive. For instance, the series ∑_{n=1}^{∞}(−1)^{n} has partial sums

S_{N} = \{ \begin{matrix}−1 & if N odd \\ 0 & if N even. \end{matrix}

So the set of partial sums is bounded with |S_{N}| ≤ 1 all N. However the sequence of partial sums does not converge, so the series does not converge.

B) TRUE. If f′(0) exists, then f must also be continuous at x = 0. This implies

lim_{x→0^{−}} cos(2x) = 1 = lim_{x→0^{+}} ax^{2} + bx + c.

Hence c = 1. Then to get f′(0) to exist, we must have

lim_{x→0^{−}} \frac{cos(2x) − 1}{x} = 0 = lim_{x→0^{+}} \frac{ax^{2} + bx + 1 − 1}{x}

This implies b = 0. Finally, and similarly, to get f′′(0) to exist we must have a = −2. Here’s why: We have f′(x) = −2 sin(2x) for all x < 0 and f′(x) = 2ax for x > 0.

lim_{x→0^{−}} \frac{−2 sin(2x) − 0}{x − 0} = −4 = 2a = lim_{x→0^{+}} \frac{2ax − 0}{x}

C) FALSE. The Ratio Test gives that the power series converges absolutely on the open interval (−2, 2). But in fact, note that with x = 2 we get ∑_{n=1}^{∞} \frac{1}{\sqrt{n}}. This series diverges by the integral test:

lim_{b→∞} ∫_{1}^{b} \frac{1}{\sqrt{x}} dx = lim_{b→∞} 2\sqrt{b} − 2

is not finite. (Or: It’s a p-series with p = 1/2 < 1, so it must diverge.) Hence the power series does not converge absolutely on the closed interval [−2, 2].

D) TRUE. The idea is the same as in V C above. If there were distinct x_{1} < x_{2} with f(x_{1}) = f(x_{2}), then Rolle’s Theorem would imply that f′(c) = 0 for some c ∈ (x_{1}, x_{2}).

E) FALSE. This can be seen by experimenting a bit with the formulas. For instance with n = 3, k = 1, we get \overline{x} = \frac{5·4}{6·5} = \frac{2}{3}, and \overline{y} = \frac{5·4}{9·7} = \frac{20}{63}. However (\overline{x})^{3} = \frac{8}{27}, which is clearly less than \overline{y}. Hence \overline{y} > (\overline{x})^{3}.

In fact, we claim that for each fixed positive integer k,

lim_{n→∞} \overline{y} − (\overline{x})^{n} = \frac{1}{4} − \frac{1}{e^{2}} > 0.

It follows that there exists an n_{0} (depending on k) such that \overline{y} > \overline{x}^{n} for all n ≥ n_{0}. (This says that the centroid of the region R lies above the upper boundary of R when n is sufficiently large(!)) Here’s how to see this: First

\overline{y} = \frac{(n + k + 1)(n + 1)}{(2n + 2k + 1)(2n + 1)} = \frac{n^{2} + (k + 2)n + 1}{4n^{2} + (4k + 4)n + 1}

and it follows easily that lim_{n→∞} \overline{y} = \frac{1}{4}. Now consider

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

This is a 1^{∞} indeterminate form. So we proceed as in a problem on Problem Set 9. We take the logarithm and apply L’Hopital’s Rule to evaluate this limit:

lim_{n→∞} ln (\frac{(n + k + 1)(n + 1)}{(n + k + 2)(n + 2)})^{n} = lim_{n→∞} n ln (\frac{(n + k + 1)(n + 1)}{(n + k + 2)(n + 2)})

= lim_{n→∞} \frac{ln (\frac{(n^{2}+(k+2)n+1)}{(n^{2}+(k+4)n+4)})}{\frac{1}{n}}

= lim_{n→∞} \frac{\frac{n^{2}+(k+4)n+4}{n^{2}+(k+2)n+1} · \frac{2n^{2}+6n+3k+4}{(n^{2}+(k+4)n+4)^{2}}}{\frac{−1}{n^{2}}}

= −2

Hence

lim_{n→∞} \overline{x}^{n} = lim_{n→∞} (\frac{(n + k + 1)(n + 1)}{(n + k + 2)(n + 2)})^{n} = e^{−2}.

This concludes the proof.