Mathematics 242, section 1 – Principles of Analysis Solutions For Final Examination – May 15, 2013

I. (20) Let A = {x^{2} − 2 : −1 < x < 2} and B = {x : |x − 1| < 4}. Find lub(A ∪ B) and glb(A ∩ B). Solution: By the definitions, we see A = [−2, 2) and B = (−3, 5). Hence, since A ⊂ B, A∪B = B = (−3, 5) and A∩B = A = [−2, 2). Hence lub(A∪B) = 5 and glb(A∩B) = −2. II. A) (10) State the ε, n_{0} definition of convergence for a sequence of real numbers.

Solution: The sequence x_{n} converges to c if for all ε > 0 there exist n_{0} ∈ N such that |x_{n} − c| < ε for all n ≥ n_{0}.

B) (10) Identify lim_{n→∞} \frac{5n^{2}+1}{n^{2}+n+4}.

Solution: Using the limit theorems, we have

lim_{n→∞} \frac{5n^{2} + 1}{n^{2} + n + 4} = lim_{n→∞} \frac{(5n^{2} + 1) · \frac{1}{n^{2}}}{(n^{2} + n + 4) · \frac{1}{n^{2}}}

= lim_{n→∞} \frac{5 + \frac{1}{n^{2}}}{1 + \frac{1}{n} + \frac{4}{n^{2}}}

= \frac{5 + lim_{n→∞} \frac{1}{n^{2}}}{1 + lim_{n→∞} \frac{1}{n} + lim_{n→∞} \frac{4}{n^{2}}}

= \frac{5 + 0}{1 + 0 + 0}

= 5.

C) (10) Show that your result in part B is correct using the definition. Solution: Given ε > 0, let n_{0} > max(20, \frac{6}{ε}). (No matter how small ε is, this is possible because N is not bounded above.) Then for all n ≥ n_{0}, we have

|\frac{5n^{2} + 1}{n^{2} + n + 4} − 5| = |\frac{−5n − 19}{n^{2} + n + 4}| < \frac{5n + 19}{n^{2}}

Since n ≥ 20, we have 5n + 19 < 6n, so this is

< \frac{6n}{n^{2}} = \frac{6}{n} ≤ \frac{6}{n_{0}} < ε.

Hence by the definition, \frac{5n^{2}+1}{n^{2}+n+4} converges to 5.

A) (15) Show that if x_{n} is a monotone increasing sequence of real numbers that is bounded above, then x_{n} converges to some real number.

Solution: The set X = {x_{n}} is bounded above in R. Hence the LUB axiom implies that it has a least upper bound c. Given any ε > 0, c − ε is not an unpper bound for X, therefore there exists some n_{0} such that c − ε < x_{n_{0}} ≤ c. But the sequence is monotone increasing so c − ε < x_{n_{0}} ≤ x_{n} ≤ c for all n ≥ n_{0}. That implies |x_{n} − c| < ε for all n ≥ n_{0}, and hence x_{n} → c by the definition.

B) (10) True/False and prove/give a reason: The sequence

x_{n} = \{ \begin{matrix}−1 + \frac{1}{n^{2}} & if n is a prime integer > 1 \\ 1 − \frac{1}{n^{2}} & if n is not prime \end{matrix}

has convergent subsequences.

Solution: This is TRUE. From the definition, x_{n} < 1 for all n and x_{n} > −1 for all n. Since {x_{n}} is a bounded sequence, the Bolzano-Weierstrass theorem implies it must have convergent subsequences. (In fact, we can also see directly from the above that the subsequences for n_{k} = kth prime number and n_{k} = 2k, k ≥ 1 are convergent.)

C) True/False and prove/give a reason: The infinite “continued radical”

\sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + ⋯}}}}}

represents a finite real number. (Hint: If so, that number would be the limit of a sequence defined by x_{1} = \sqrt{3} and x_{n} = \sqrt{3 + x_{n−1}} for all n ≥ 2.)

Solution: This is TRUE. First we notice that in the sequence described in the Hint,

\begin{aligned}x_{2} & = \sqrt{3 + \sqrt{3}} \\ x_{3} & = \sqrt{3 + \sqrt{3 + \sqrt{3}}}, \end{aligned}

etc. So as n → ∞, if the sequence x_{n} converges, it converges to the expression defined by the infinite “continued radical.” We claim next the sequence is bounded below by 0 (clear) and above by 3. The second part requires some proof, but we can argue that x_{n} < 3 for all n by induction as follows. The base case is x_{1} = \sqrt{3} < 3. Assuming x_{k} < 3, then we have x_{k+1} = \sqrt{3 + x_{k}} < \sqrt{3 + 3} = \sqrt{6} < 3. Hence x_{n} < 3 for all n. Finally, we show by induction again that x_{n} is monotone increasing. Note that x_{2} = \sqrt{3 + \sqrt{3}} > \sqrt{3} = x_{1} since x_{2}^{2} = 3 + \sqrt{3} > 3 = x_{1}^{2}. Assuming x_{k+1} > x_{k}, then x_{k+2} = \sqrt{3 + x_{k+1}} > \sqrt{3 + x_{k}} = x_{k+1} also. Hence the monotone convergence theorem (part A) implies that x_{n} → a for some real number a < 3. In fact, taking the limit on both sides of x_{n+1} = \sqrt{3 + x_{n}}, we can see that a must satisfy

a = \sqrt{3 + a} ⇒ a^{2} − a − 3 = 0

By the quadratic formula,

a = \frac{1 + \sqrt{13}}{2} ≐ 2.302775638.

IV. A) (15) Let

f(x) = \{ \begin{matrix}x + 3 & if x is a rational number \\ −x^{2} + 3 & if x is an irrational number \end{matrix}

Is f continuous at x = 0? Why or why not?

Solution: Yes, f(x) is continuous at 0. To see this, let ε > 0, and δ < min(ε, \sqrt{ε}). If |x| < δ and x is rational then we have

|f(x) − f(0)| = |x + 3 − 3| = |x| < ε,

while if x is irrational we have

|f(x) − f(0)| = | − x^{2} + 3 − 3| = |x|^{2} < (\sqrt{ε})^{2} = ε.

Therefore f is continuous at 0.

B) (25) State and prove the Intermediate Value Theorem. (You may assume as known the theorem that if f is continuous at c and x_{n} → c is a sequence contained in the domain of f, then f(x_{n}) → f(c)).

Solution: The IVT in the form we proved first says that if f is continuous on [a, b] with f(a) ≠ f(b), and k is any number between f(a) and f(b), then there exists a c ∈ [a, b] such that f(c) = k. We proved this as follows. If k = f(a) or k = f(b), then there is nothing to prove. Hence we must consider the case that k is strictly between the two values f(a) and f(b). Suppose, for instance, that f(a) < k < f(b). (The other case f(b) < k < f(a) can be proved similarly.) Consider

S = {x ∈ [a, b] | f(x) ≤ k}

Then S is not empty since for instance a ∈ S. S is also bounded since it is a subset of [a, b]. Hence the LUB axiom implies that c = lub(S) exists in R. We claim that f(c) = k. First note that c < b since f(b) > k. Hence if we let x_{n} → c with c ≤ x_{n} ≤ b, we will have f(x_{n}) > k for all n, and hence (since f is continuous at c, lim_{n→∞} f(x_{n}) = f(c) ≥ k. On the other hand, since c = lub(S), if ε > 0, then there will exist x ∈ S with c − ε < x ≤ c. Apply this for each ε = \frac{1}{n} for n ∈ N. We get a a sequence x′_{n} ∈ S with x′_{n} → c. Hence lim_{n→∞} f(x′_{n}) = f(c) ≤ k since f(x′_{n}) ≤ k for all n. The two inequalities show that f(c) = k.

V. A) (15) Using the limit definition of the derivative, compute f′(c) for f(x) = \frac{1}{(x+3)^{2}} at a

general c ≠ −3.

Solution: We have

f′(c) = lim_{x→c} \frac{f(x) − f(c)}{x − c}

= lim_{x→c} \frac{\frac{1}{(x+3)^{2}} − \frac{1}{(c+3)^{2}}}{x − c}

= lim_{x→c} \frac{(c + 3)^{2} − (x + 3)^{2}}{(x + 3)^{2}(c + 3)^{2}(x − c)}

= lim_{x→c} \frac{c^{2} − x^{2} + 6(c − x)}{(x + 3)^{2}(c + 3)^{2}(x − c)}

= lim_{x→c} \frac{−c − x − 6}{(x + 3)^{2}(c + 3)^{2}}

= \frac{−2}{(c + 3)^{3}}

(Note: It is not permissible to use L’Hopital’s Rule to compute this limit because that requires you to use a derivative formula to derive f′(x) = \frac{−2}{(x+3)^{3}} . Using the derivative rule to compute the derivative by the definition is an example of circular reasoning!)

B) (10) What theorem guarantees that

F (x) = ∫_{1}^{x} \frac{1}{(t + 3)^{2}} dt

is differentiable at x = 2? Exactly why does it apply here? What does it say about F′(2)?

Solution: The theorem is the first part of the Fundamental Theorem of Calculus. It applies because the function f(x) = \frac{1}{(x+3)^{2}} is continuous on the closed interval [1, a] for all a > 2. It implies F′(2) = f(2) = \frac{1}{25}.

C) (10) Show that if f(x) = \frac{e^{x}+e^{−x}}{2} , then for every real k, there exists a solution c of the equation f′(c) = \frac{e^{c}−e^{−c}}{2} = k.

Solution: Note that f′(x) = \frac{e^{x}−e^{−x}}{2} exists and is continuous at all x ∈ R. This function satisfies lim_{x→+∞} f′(x) = +∞ and lim_{x→−∞} f′(x) = −∞. So whatever the value k, we can find an interval [−a, a] such that f′(−a) < k < f′(a). On the interval [−a, a], f′ is continuous, and then the Intermediate Value Theorem implies there is some c ∈ [−a, a] such that f′(c) = k.

Alternatively, it would also be possible to solve the equation \frac{e^{c}−e^{−c}}{2} = k explicitly like this:

\frac{e^{c} − e^{−c}}{2} = k ⇔ e^{2c} − 2ke^{c} − 1 = 0
⇔ e^{c} = \frac{2k + \sqrt{4k^{2} + 4}}{2} (by the quadratic formula)
⇔ e^{c} = k + \sqrt{k^{2} + 1}
⇔ c = ln(k + \sqrt{k^{2} + 1})

The function g(x) = ln(x + \sqrt{x^{2} + 1}) is also known as the inverse hyperbolic sine function, since f′(x) = \frac{e^{x}−e^{−x}}{2} is the hyperbolic sine.

VI. (20) In this question, you may use the summation formulas:

∑_{i=1}^{n} 1 = n ∑_{i=1}^{n} i = \frac{n(n + 1)}{2} ∑_{i=1}^{n} i^{2} = \frac{n(n + 1)(2n + 1)}{6}.

Show that f(x) = x^{2} + x is integrable on [a, b] = [0, 2] by considering upper and lower sums for f and determine the value of ∫_{0}^{2} x^{2} + x dx.

Solution: f′(x) = 2x + 1 > 0 for all x ∈ [0, 2]. Therefore, f is increasing on [0, 2] and as in the proof of the integrability of all monotone functions, if P_{n} is the regular partition of [0, 2] into n equal subintervals,

U(f, P_{n}) − L(f, P_{n}) = (f(2) − f(0))\frac{(2 − 0)}{n} = \frac{12}{n}.

If ε > 0 is given and n > \frac{12}{ε}, then

U(f, P_{n}) − L(f, P_{n}) < ε.

The limit of the upper sum for P_{n} as n → ∞ is

∫_{0}^{2} x^{2} + x dx = lim_{n→∞} ∑_{i=1}^{n} ((\frac{2i}{n})^{2} + \frac{2i}{n}) \frac{2}{n}

= lim_{n→∞} (\frac{8}{n^{3}} ∑_{i=1}^{n} i^{2} + \frac{4}{n^{2}} ∑_{i=1}^{n} i)

= lim_{n→∞} (\frac{8}{n^{3}} · \frac{n(n + 1)(2n + 1)}{6} + \frac{4}{n^{2}} · \frac{n(n + 1)}{2})

= \frac{8}{3} + 2 = \frac{14}{3}.

Check: By Part 2 of the FTC:

∫_{0}^{2} x^{2} + x dx = \frac{x^{3}}{3} + \frac{x^{2}}{2} |_{0}^{2} = \frac{8}{3} + 2 = \frac{14}{3}.

VII. A) (5) State the definition of convergence for an infinite series ∑_{n=1}^{∞} a_{n}.

Solution: The infinite series ∑_{n=1}^{∞} a_{n} converges to S ∈ R if the sequence of partial sums s_{k} = ∑_{n=1}^{k} a_{n} converges to S (as a sequence).

B) (15) For which x ∈ R does the power series

∑_{n=1}^{∞} \frac{3^{n}x^{n}}{n7^{n}}

converge? (Apply the Ratio Test and test the series at the endpoints of your interval separately.)

Solution: We have

lim_{n→∞} |\frac{3^{n+1}x^{n+1}}{(n + 1)7^{n+1}} · \frac{n7^{n}}{3^{n}x^{n}} | = lim_{n→∞} \frac{3n}{7(n + 1)}|x| = \frac{3|x|}{7}.

By the Ratio Test, the series converges absolutely whenever \frac{3|x|}{7} < 1, or |x| < \frac{7}{3}, and it diverges when |x| > \frac{7}{3}. At x = ±\frac{7}{3}, we have the following. If x = \frac{7}{3}, we substitute and simplify to obtain ∑_{n=1}^{∞} \frac{1}{n}. This is the harmonic series, which diverges. If x = −\frac{7}{3}, we substitute and simplify to obtain ∑_{n=1}^{∞} \frac{(−1)^{n}}{n}. This series converges by the Alternating Series Test. Therefore the series converges at all x ∈ [−\frac{7}{3}, \frac{7}{3}) (and only for those x).

Extra Credit (20) Prove the result mentioned in part B of question IV above: if f is continuous at c and x_{n} → c is a sequence contained in the domain of f, then f(x_{n}) → f(c).

Solution: Since f is continuous at c, given any ε > 0, there is some δ > 0 such that

(1)  |f(x) − f(c)| < ε

for all x with |x − c| < δ. Then, since x_{n} → c, given that δ, there is an n_{0} such that |x_{n} − c| < δ for all n ≥ n_{0}. But then substituting x = x_{n} from the sequence into the inequality (1) we get |f(x_{n}) − f(c)| < ε for all n ≥ n_{0}. This implies that the sequence f(x_{n}) → f(c).