Mathematics 242 – Principles of Analysis Solutions for Problem Set 9 – Due: Friday, April 25

‘A’ Section

1. Let f(x) = x^{2} + 4x − 1 on [2, 3].

(a) Show that f is integrable on [2, 3] directly using the definition (that is, do not use any general theorems giving criteria for integrability). Hints: Use regular partitions of [2, 3], and the summation rules

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

Solution: Consider the regular partition P_{n} of the interval [2, 3], so x_{i} = 2 + \frac{i}{n} for i = 0, 1, …, n. We have f′(x) = 2x + 4 > 0 for all x ∈ [2, 3]. Hence f is increasing on the interval and thus m_{i} comes from evaluating at the left endpoints and M_{i} comes from evaluating at the right endpoints:

L(f, P_{n}) = ∑_{i=0}^{n−1} ((2 + \frac{i}{n})^{2} + 4 (2 + \frac{i}{n}) − 1) \frac{1}{n}

= \frac{1}{n^{3}} ∑_{i=0}^{n−1} i^{2} + \frac{8}{n^{2}} ∑_{i=0}^{n−1} i + \frac{11}{n} ∑_{i=0}^{n−1} 1

= \frac{(n − 1)n(2n − 1)}{6n^{3}} + \frac{8(n − 1)n}{2n^{2}} + 11.

Similarly, evaluating at the right endpoints,

U(f, P_{n}) = ∑_{i=1}^{n} ((2 + \frac{i}{n})^{2} + 4 (2 + \frac{i}{n}) − 1) \frac{1}{n}

= \frac{1}{n^{3}} ∑_{i=1}^{n} i^{2} + \frac{8}{n^{2}} ∑_{i=1}^{n} i + \frac{11}{n} ∑_{i=1}^{n} 1

= \frac{n(n + 1)(2n + 1)}{6n^{3}} + \frac{8n(n + 1)}{2n^{2}} + 11.

Hence

U(f, P_{n}) − L(f, P_{n}) = \frac{9}{n}.

We can make this < ε for any ε > 0 by letting n > \frac{9}{ε}. Hence f is integrable by our definition.

(b) Determine the value of ∫_{2}^{3} x^{2} + 4x − 1 dx.

Solution: The value of the integral is

lim_{n→∞} U(f, P_{n}) = lim_{n→∞} \frac{n(n + 1)(2n + 1)}{6n^{3}} + \frac{8n(n + 1)}{2n^{2}} + 11 = \frac{1}{3} + \frac{8}{2} + 11 = \frac{46}{3}.

(This can be checked by the FTC:

∫_{2}^{3} x^{2} + 4x − 1 dx = \frac{x^{3}}{3} + \frac{4x^{2}}{2} − x|_{1}^{2}

= (\frac{27}{3} − \frac{8}{3}) + (18 − 8) + (3 − 2)

= \frac{46}{3}.)

2. Let F (x) = ∫_{0}^{x} f(t) dt where

f(t) = \{ \begin{matrix}1 − t & if 0 ≤ t ≤ 1 \\ 1 + t & if 1 < t ≤ 2 \end{matrix}

(a) Find an explicit formula for F (x) valid for all 0 ≤ x ≤ 2. (Use the FTC in an appropriate way for this.)

Solution: Using the FTC, we have

F (x) = t − \frac{t^{2}}{2} |_{0}^{x} = x − \frac{x^{2}}{2}

if 0 ≤ x ≤ 1. Note that the value at x = 1 is F (1) = \frac{1}{2}. Then for x between 1 and 2, we want an antiderivative that takes the value \frac{1}{2} at x = 1, so

F (x) = x + \frac{x^{2}}{2} − 1

for 1 ≤ x ≤ 2. Therefore, putting these together:

F (x) = \{ \begin{matrix}x − \frac{x^{2}}{2} & if 0 ≤ x ≤ 1 \\ \frac{x^{2}}{2} + x − 1 & if 1 < x ≤ 2 \end{matrix}

(The constant term in the formula on the second half of the interval ensures that F is continuous on [0, 2].)

(b) Is F differentiable at x = 1? Why or why not? Does this contradict the first part of the FTC?

Solution: No, F is not differentiable at x = 1, since

lim_{x→1^{−}} \frac{F (x) − F (1)}{x − 1} = lim_{x→1^{−}} \frac{\frac{x^{2}}{2} − x + \frac{1}{2}}{x − 1} = lim_{x→1^{−}} \frac{1}{2}x − \frac{1}{2} = 0,

but

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

This does not contradict the first part of the FTC since f is not continuous on [0, 2]. It has a jump discontinuity at x = 1 since lim_{t→1^{−}} f(t) = 0, but lim_{t→1^{+}} f(t) = 2.

3. Find the following derivatives using the FTC (and other derivative rules as needed):

(a) \frac{d}{dx} ∫_{1}^{x} \frac{e^{t}}{t} dt (Note that the integrand is not defined at t = 0, so 0 should not be included in the interval of integration, or else the FTC does not apply.)

Solution: The first part of the FTC implies the derivative is \frac{e^{x}}{x}.

(b) \frac{d}{dx} ∫_{−x^{2}}^{x^{4}} cos(e^{u}) du

Solution: By the first part of the FTC and the chain rule, the derivative is

4x^{3} cos(e^{x^{4}}) + 2x cos(e^{−x^{2}}).

‘B’ Section

1. Let f be integrable on the interval [a, b], and assume ∫_{a}^{b} f(x) dx < 0. Show that there exist k < 0 and an interval [c, d] ⊆ [a, b] such that f(x) < k < 0 for all x ∈ [c, d].

Solution: If the function is integrable, then the definition of integrability implies that for every ε > 0, there exists a partition P such that

(1)  U(f, P) − L(f, P) < ε.

Moreover, it is always true that

(2)  L(f, P) ≤ ∫_{a}^{b} f(x) dx ≤ U(f, P).

Since the integral is strictly negative, this means that if we take 0 < ε < |∫_{a}^{b} f(x) dx|, it follows from (1) and (2) that

U(f, P) < L(f, P) + ε < ∫_{a}^{b} f(x) dx + |∫_{a}^{b} f(x) dx| = 0.

In other words, the lower sum U(f, P) < 0 for some partition. This can only happen if M_{i} = glb{f(x) | x ∈ [x_{i−1}, x_{i}]} < 0 over some interval [x_{i−1}, x_{i}] ⊂ [a, b] from the partition P. Hence f(x) ≤ M_{i} < M_{i}/2 < 0 for all x ∈ [c, d] = [x_{i−1}, x_{i}]. We can take k = m/2 to get the strict inequality desired.

2. Let f be continuous on [a, b]. Show that there exists c ∈ [a, b] such that

∫_{a}^{b} f(x) dx = f(c)(b − a).

(Hint: Look at Theorem 5.2.5 (b).)

Solution: Since f is continous on [a, b], by the Extreme Value Theorem, it attains a maximum M and a minimum m on the interval By the theorem indicated,

m(b − a) ≤ ∫_{a}^{b} f(x) dx ≤ M(b − a),

so

m ≤ \frac{1}{b − a} ∫_{a}^{b} f(x) dx ≤ M.

But then by the Intermediate Value Theorem (in the second version we discussed), it follows that

\frac{1}{b − a} ∫_{a}^{b} f(x) dx = f(c)

for some c ∈ [a, b]. Hence after multiplying through again by b − a, we get:

∫_{a}^{b} f(x) dx = f(c)(b − a)

for some c ∈ [a, b]. (Comment: this result is sometimes called the Mean Value Theorem for Integrals. It is related to the method of computing the average value of a function f on [a, b] that you may have seen as an application of integration in Calculus 2.)

3. Logarithm and Exponential. In this problem, we will construct the natural logarithm and and the exponential function e^{x} “from scratch,” without relying on intuition about exponentials (as you probably did in calculus). We start by considering the function

(1)  L(x) = ∫_{1}^{x} \frac{1}{t} dt,

for x > 0. Note that \frac{1}{t} is continuous on (0, +∞), hence the FTC applies to show that L is a differentiable function. From calculus you probably recognize that L(x) = ln(x). We want to show directly that this makes sense and use this function to construct the inverse function E(x) = L^{−1}(x) which is callled E(x) = e^{x}. Why would we proceed “backwards” like this? The issue is that, while a^{m/n} = (a^{1/n})^{m} makes immediate sense for any positive a ∈ R and any rational exponent, what does a^{x} actually mean if x ∉ Q? Instead of trying to define that directly, we will take an end run around the question.

(a) We claim that the function L “has the right property to be a logarithm” – namely that L(x · x′) = L(x) + L(x′) for all x, x′ > 0. Prove this by showing that for any constant x′ > 0, the function L(xx′) also satisfies \frac{d}{dx}L(xx′) = \frac{1}{x} for all x > 0. Deduce that L(xx′) = L(x) + c for some constant c, then determine c by substituting an appropriate value for x.

Solution: By the first part of the FTC and the chain rule, we have \frac{d}{dx}L(xx′) = \frac{1}{xx′} ·x′ = \frac{1}{x}. Since L(x) also satisfies L′(x) = \frac{1}{x}, this implies by one of our corollaries of the Mean Value Theorem that L(xx′) − L(x) = c is a constant. By definition L(1) = 0, so if we substitute x = 1, we get c = L(x′). Therefore L(xx′) = L(x) + L(x′).

(b) Show that L(x) is strictly increasing for x > 0, hence is a 1-1 function on the domain (0, +∞). Hence L has an inverse function E : R → {x ∈ R | x > 0}.

Solution: Most of this follows since L′(x) = \frac{1}{x} > 0 on (0, +∞), by the first part of the FTC. Hence L is strictly increasing by one of the corollaries of the MVT, hence 1-1.

Comment: To give every detail here, we should also show that the range of L is R to show that the domain of the inverse function is actually equal to R. To see this, we can “borrow” from the fact that the harmonic series is divergent, like this: We want to show first that {L(x) | x ∈ (0, ∞)} is not bounded above. Let x = n be an integer n > 1. By considering the lower sum of ∫_{1}^{x} \frac{dt}{t} for the partition P = \{1, 2, …, n\} for the interval [1, n], we can see that

L(n) = ∫_{1}^{n} \frac{dt}{t} > \frac{1}{2} + ⋯ + \frac{1}{n} = L (\frac{1}{t}, P) .

As n → ∞, though, the right side increases without any bound, since the harmonic series diverges. Hence the numbers L(n) have no upper bound either. On the other hand by part (a) we have

0 = L(1) = L (n · \frac{1}{n}) = L(n) + L (\frac{1}{n}) .

Hence the values L (\frac{1}{n}) are not bounded below either (they go to −∞ as n → ∞). Since L is continuous, its range must equal (−∞, ∞) = R (by an application of the Intermediate Value Theorem – do you see why?)

(c) Show that the inverse function E satisfies the equation E(x+x′) = E(x)·E(x′), hence E looks like an exponential function.

Solution: By the definition of an inverse function we have

y = E(x) ⇔ x = L(y).

Let y, y′ be any two positive real numbers and let x = L(y), x′ = L(y′), so y = E(x) and y′ = E(x′). By part (a) we know that

L(yy′) = L(y) + L(y′).

Hence

E(x)E(x′) = yy′ = E(L(y) + L(y′)) = E(x + x′).

(d) We define x = e as the unique solution of the equation L(x) = 1. Show that x = e^{m/n} satisfies L(x) = m/n for all rational numbers x = m/n.

Solution: The given information says L(e) = 1. First we claim L(e^{m}) = m for all m ∈ N. This follows by induction since L(e^{1}) = 1 by definition, and if L(e^{k}) = k, then L(e^{k+1}) = L(e^{k} · e) = L(e^{k}) + L(e) = k + 1 by part (a) and the induction hypothesis. Next, since L(e^{m})+L(e^{−m}) = L(e^{m} ·e^{−m}) = L(1) = 0, we have L(e^{−m}) = −m, which shows that L(e^{m}) = m for all integers m. Next, by part (a) again, if n is a positive integer, 1 = L(e) = L((e^{1/n})^{n}) = nL(e^{1/n}). So L(e^{\frac{1}{n}}) = \frac{1}{n} whenever n is a positive integer. Finally, combining all of these steps we can write any rational number m/n as a fraction in which n > 0 and then

L(e^{m/n}) = L((e^{\frac{1}{n}})^{m}) = mL(e^{1/n}) = \frac{m}{n}L(e) = \frac{m}{n}.

(e) Show that the inverse function E(x) of L(x) is differentiable and satisfies E′(x) = E(x). Hint: Look at Section 4.4. What happens if you differentiate on both sides of the equation L(E(x)) = x?

Solution: Theorem 4.4.4 implies that E is differentiable at all real x, since L′(x) = \frac{1}{x} is never equal to zero. Hence if we differentiate the equation L(E(x)) = x on both sides using the chain rule we get L′(E(x))E′(x) = 1. But L′(u) = \frac{1}{u}, so this says \frac{1}{E(x)} · E(x) = 1 for all x, so E′(x) = E(x). This is the familiar differentiation rule for the exponential function: \frac{d}{dx}e^{x} = e^{x}.

(f) Show that

E(x) = e^{x} = lim_{n→∞} (1 + \frac{x}{n})^{n}

for all x ∈ R. Hint: Take the logarithm and use L’Hopital’s Rule.

Solution: We have

L ((1 + \frac{x}{n})^{n}) = nL (1 + \frac{x}{n}) = \frac{L (1 + \frac{x}{n})}{\frac{1}{n}}.

This is the value of a ratio g(u)/h(u) at u = \frac{1}{n}. The top is g(u) = L(1 + ux) and the bottom is h(u) = u. Both of these are “nice functions” differentiable at u = 0. Note that lim_{u→0} g(u) = 0 = lim_{u→0} h(u). So we can apply the version of L’Hopital’s Rule developed on Problem Set 8. We have g′(u) = \frac{x}{1+ux} and h′(u) = 1. The ratio g′(u)/h′(u) does have a limit as u → 0:

lim_{u→0} \frac{g′(u)}{h′(u)} = lim_{u→0} \frac{x}{1 + ux} = x.

Therefore by L’Hopital’s Rule we know

lim_{u→0} \frac{g(u)}{h(u)} = x

as well. Since u = \frac{1}{n} → 0 as n → ∞, this implies

lim_{n→∞} \frac{L (1 + \frac{x}{n})}{\frac{1}{n}} = x.

Since L is differentiable function at all x > 0, it also continuous everywhere. Hence

L ( lim_{n→∞} (1 + \frac{x}{n})^{n}) = lim_{n→∞} L ((1 + \frac{x}{n})^{n}) = x

This implies

lim_{n→∞} (1 + \frac{x}{n})^{n} = E(x) = e^{x}.