Math 104: Solutions to sample final problems

1. For the first series

β = lim sup |a_{n}|^{1/n} = lim sup |n|^{3/n} = 1

so the radius of convergence is R = ^{1}/β = 1. At n = 1, the series is

∑_{n=1}^{∞} \frac{1}{n^{3}}

which converges (and can be verified using the integral test). At n = −1, the series is

∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{3}}

which converges by the alternating series test. Hence the exact interval of convergence is [−1, 1]. For the second series, only every third term is non-zero, and thus

β = lim sup |a_{n}|^{1/n} = lim_{n→∞} |a_{3n}|^{1/3n} = lim_{n→∞} | \frac{1}{2n}|^{1/3n} = 1

so the radius of convergence is 1. At x = 1, the series is

∑_{n=1}^{∞} \frac{1}{2n}

which diverges. At x = −1, the series is

∑_{n=1}^{∞} \frac{(−1)^{n}}{2n}

which converges by the alternating series test. Hence the exact interval of convergence is (−1, 1]. For the third series

β = lim sup |a_{n}|^{1/n} = lim_{n→∞} |a_{2n!}|^{1/2n!} = lim_{n→∞} 1 = 1

and hence the radius of convergence is 1. At x = 1

∑_{n=0}^{∞} x^{2n!} = ∑_{n=0}^{∞} 1

which diverges. Since the series is even, it diverges at x = −1 also. Hence the exact interval of convergence is (−1, 1).

2. (a) Suppose that s_{n} → s. Then for all ϵ > 0 there exists an N ∈ such that for all

n > N,

|s_{n} − s| < ϵ.

To show that s_{n}^{2} → s^{2}, consider any ϵ > 0. First, note that

|s_{n}^{2} − s^{2}| = |s_{n} − s| · |s_{n} + s|.

There exists an N_{1} ∈ such that n > N_{1} implies that |s_{n} − s| < 1, and hence that

|s_{n} + s| ≤ |s_{n}| + |s| < (|s| + 1) + |s| = 2|s| + 1.

Similarly, there exists an N_{2} ∈ such that n > N_{2} implies that

|s_{n} − s| < \frac{ϵ}{2|s| + 1} .

Hence, if N = max{N_{1}, N_{2}}, then n > N implies

|s_{n}^{2} − s^{2}| ≤ |s_{n} − s| · |s_{n} + s|
< \frac{ϵ}{2|s| + 1} (2|s| + 1) = ϵ.

(b) A function f is continuous at a point x if for all sequences (x_{n}) such that lim_{n→∞} x_{n} = x, then lim_{n→∞} f(x_{n}) = f(x). By part (a), for all x ∈ , if x_{n} → x, then x_{n}^{2} → x^{2}. Thus f is continuous for all x ∈ . Alternatively, to use the ϵ-δ property, choose ϵ > 0, and fix x_{0} ∈ . Then

|f(x) − f(x_{0})| = |x^{2} − x_{0}^{2}| = |x − x_{0}| · |x + x_{0}|.

If |x − x_{0}| < 1, then |x + x_{0}| ≤ |x| + |x_{0}| < 2|x_{0}| + 1. Pick δ = min\{\frac{ϵ}{2|x_{0}|+1} , 1\}. Then |x − x_{0}| < δ implies

|f(x) − f(x_{0})| < \frac{ϵ}{2|x_{0}| + 1} (2|x_{0}| + 1) = ϵ

and hence f is continuous at x_{0}. Since x_{0} is arbitrary, f is continuous on .

3. By the mean value theorem, there exists a c ∈ (r, s) such that

f′(c) = \frac{f(s) − f(r)}{s − r} = 0

and there exists and d ∈ (s, t) such that

f′(d) = \frac{f(t) − f(s)}{t − s} = 0.

By construction, d > c. Hence there exists an x ∈ (c, d) ⊆ (0, 1) such that

f′′(x) = \frac{f′(d) − f′(c)}{d − c} = 0.

4. (a) If ∑ _{n=0}^{∞} a_{n} converges to a limit a, then for all ϵ > 0, there exists an N ∈ such that k > N implies

|a − ∑_{n=0}^{k} a_{n}| < ϵ.

Now consider any ϵ > 0 and define N as above. If k > N/2, then

|a − ∑_{n=0}^{k} b_{n}| = |a − ∑_{n=0}^{2k+1} a_{n}| < ϵ.

Hence ∑ _{n=0}^{∞} b_{n} converges.

(b) Suppose a_{n} = (−1)^{n}. Then

∑_{n=0}^{N} a_{n} = \{ \begin{matrix}1 & if N is even, \\ 0 & if N is odd; \end{matrix}

this series diverges. However b_{n} = 0 for all n, and thus ∑ _{n=0}^{∞} b_{n} converges.

5. If f(a)f(b) < 0, then it follows that either f(a) < 0 and f(b) > 0, or f(a) > 0 and f(b) < 0. In either case, zero lies between f(a) and f(b), and thus the intermediate value theorem can be applied to show that there exists an x ∈ (a, b) such that f(x) = 0.

6. Let f be a real-valued function defined on an interval [0, b] as

f(x) = \{ \begin{matrix}x & for x ∈ , \\ 0 & for x ∉ . \end{matrix}

Consider a partition P = {0 = t_{0} < t_{1} < … < t_{n} = b}. First note that

m(f, [t_{k−1} , t_{k}]) = 0, M(f, [t_{k−1} , t_{k}]) = t_{k}

for all k = 1, …, n. The first expression follows because any interval must contain an irrational number, whereas the second expression follows because the interval must take rational values arbitrarily close to t_{k}. Hence

U(f, P) = ∑_{k=1}^{n} (t_{k}− t_{k−1} )M(f, [t_{k−1} , t_{k}]) = ∑_{k=1}^{n} (t_{k}− t_{k−1} )t_{k}.

For a general partition this cannot be calculated explicitly, but it will always be strictly positive, since each term in the sum is strictly positive. The lower Darboux sum is given by

L(f, P) = ∑_{k=1}^{n} (t_{k}− t_{k−1} )m(f, [t_{k−1} , t_{k}]) = 0.

To show that f is not integrable on [0, b], consider any partition P. Then

U(f, P) = ∑_{k=1}^{n} (t_{k}− t_{k−1} )t_{k}
≥ ∑_{k=1}^{n} (t_{k}− t_{k−1} )\frac{(t_{k}+ t_{k−1} )}{2}
= \frac{1}{2} ∑_{k=1}^{n} (t_{k}^{2} − t_{k−1}^{2} )
= \frac{t_{n}^{2} − t_{0}^{2}}{2} = \frac{b^{2}}{2}.

Thus for any partition,

U(f, P) − L(f, P) ≥ \frac{b^{2}}{2}

and thus f is not integrable on [0, b].

7. Suppose that ∑ _{n=1}^{∞} f(n) converges, and consider the integral

∫_{1}^{b}f(x) dx.

Let k be the largest integer such that k < b. Hence b − k ≤ 1. Consider the partition P = {1 = t_{0} < t_{1} < … < t_{k−1} < t_{k} = b}, where t_{i} = i + 1 for i = 0, …, k − 1. Then

U(f, P) = ∑_{j=1}^{k} M(f, [t_{j−1} , t_{j}])(t_{j} − t_{j−1} )
= ∑_{j=1}^{k} f(t_{j−1} )(t_{j} − t_{j−1} )
= f(k)(b − k) + ∑_{j=1}^{k−1} f(j)
≤ f(k) + ∑_{j=1}^{k−1} f(j)
= ∑_{j=1}^{k} f(j)

so

∫_{1}^{b}f ≤ U(f, P) ≤ ∑_{j=1}^{k} f(j) ≤ ∑_{j=1}^{∞} f(j).

Since the integral is an increasing function and is bounded above, it follows that it must converge.

Now suppose that ∫_{1}^{x} f converges, and consider ∑ _{n=1}^{N} f(n). Consider the partition P = {1 = t_{0} < t_{1} < … < t_{N} − 1 = N} where t_{i} = i + 1 for all i = 0, …, N − 1. Then

L(f, P) = ∑_{j=1}^{N−1} m(f, [t_{j−1} , t_{j}])(t_{j} − t_{j−1} )

= ∑_{j=1}^{N−1} m(f, [j, j + 1])

= ∑_{j=1}^{N−1} f(j + 1)

= ∑_{j=2}^{N} f(j)

and hence

∑_{j=1}^{N} f(j) = f(1) + L(f, P) ≤ f(1) + ∫_{1}^{N} f ≤ f(1) + ∫_{1}^{∞} f.

Since ∑ _{j=1}^{N} f(j) is increasing and bounded above, then it converges.

8. (a) From the definition, it is clear that d(x, y) = 0 if and only if x = y, and that d(x, y) = d(y, x). To prove the triangle inequality, consider any x, y, z ∈ . If x = z, then d(x, z) = 0, and since d(x, y) + d(y, z) ≥ 0, the triangle inequality is satisfied. If x ≠ z, then d(x, z) = 1. Either y ≠ z or y ≠ x, and hence d(x, y) + d(y, z) ≥ 1, so the triangle inequality is satisfied. Hence d is a metric.

(b) The neighborhood of radius^{1}/2 at 0 is

{x ∈ : d(0, x) < ^{1}/2} = {0}.

It only contains zero, since all other points are a distance of 1 away.

(c) Consider an arbitrary set S ⊆ . To show that S is open, consider any x ∈ S. Then, by the same argument as in (b), N_{1/2}(x) = {x} ⊆ S so x is an interior point. Since all points are interior, it follows that S is open. Consider an open cover S of S. Suppose S has finitely many points, so that S = {s_{1}, s_{2}, …, s_{n}}. Then there exist sets S_{1}, …, S_{n} ∈ S such that s_{k} ∈ S_{k} for k = 1, …, n, and it follows that {S_{k} : k = 1, …n} is a finite subcover. Hence S is compact.

−1 −0.50 0.5 1 −1 −0.5  0  0.5  1 y  x

Figure 1: A plot of the neighborhood of radius 1 at (0, 0), considered in Problem 9(b).

Suppose S has infinitely many points. Consider the cover S = {{x} : x ∈ S}. By above, the sets {x} are open. Consider a subcover T ⊆ S. For any x ∈ S, {x} must be T , since x is not an element of any other set in S. Thus T = S. Hence S has no finite subcover and S is not compact.

9. Let x = (x_{1}, x_{2}) and y = (y_{1}, y_{2}) be in ^{2}. Consider the function

d(x, y) = |x_{1} − y_{1}| + |x_{2} − y_{2}|.

(a) Consider the three properties of a metric:

M1. Note that

d(x, x) = |x_{1} − x_{1}| + |x_{2} − x_{2}| = 0

and if d(x, y) = 0, then

0 = |x_{1} − y_{1}| + |x_{2} − y_{2}|

from which it follows that x_{1} = y_{1} and x_{2} = y_{2}, so x = y. M2. For all x, y ∈ ^{2},

d(x, y) = |x_{1} − y_{1}| + |x_{2} − y_{2}| = |y_{1} − x_{1}| + |y_{2} − x_{2}| = d(y, x),

and thus d is symmetric.

M3. For all x, y, z ∈ ^{2},

d(x, y) + d(y, z) = |x_{1} − y_{1}| + |x_{2} − y_{2}| + |y_{1} − z_{1}| + |y_{2} − z_{2}|
≤ |x_{1} − z_{1}| + |x_{2} − z_{2}|
= d(x, z),

where the usual triangle inequality on has been applied twice. Hence d satisfies the triangle inequality.

(b) The neighborhood of radius 1 at (0, 0) is given by

N_{1}((0, 0)) = {x ∈ ^{2} : d(x, (0, 0)) < 1} = {x ∈ ^{2} : |x_{1}| + |x_{2}| < 1}

Consider the quadrant where x_{1} ≥ 0 and x_{2} ≥ 0. Then |x_{1}| + |x_{2}| = x_{1} + x_{2}, and thus the boundary of the neighborhood is given by x_{2} = 1 − x_{1}. This is a straight line which intercepts the x_{2} axis at 1, and has slope −1. By symmetry, it follows that the neighborhood is diamond, with corners at (0, 1), (0, 1), (0, −1), and (−1, 0). It is plotted in Fig. 1.

10. Consider a function f defined on which satisfies

|f(x) − f(y)| ≤ (x − y)^{2}

for all x, y ∈ . Prove that f is a constant function. Suppose that f is not constant. Then there exists x < y such that f(x) ≠ f(y). For an arbitrary n ∈ consider the points

t_{k} = x − \frac{(y − x)k}{n}

for k = 0, …, n. Then by the triangle inequality,

|f(x) − f(y)| ≤ ∑_{k=1}^{n} |f(t_{k}) − f(t_{k−1} )|

≤ ∑_{k=1}^{n} (t_{k}− t_{k−1} )^{2}

= ∑_{k=1}^{n} ( \frac{y − x}{n} ) ^{2}

= \frac{(y − x)^{2}}{n} .

Since n is arbitrary, there exists an n such that

n > \frac{(y − x)^{2}}{|f(x) − f(y)|} .

which leads to a contradiction. An alternative approach is to first note that for all x, y ∈

| \frac{f(y) − f(x)}{y − x} | ≤ | \frac{(y − x)^{2}}{y − x} | = |y − x|.

Since |y − x| → 0 as y → x, it follows that

| \frac{f(y) − f(x)}{y − x} |

as y → x, and hence

lim_{y→x} \frac{f(y) − f(x)}{y − x} = 0

for all x ∈ . Thus f is differentiable on and f′(x) = 0 for all x ∈ . For any bounded interval I ⊆ , f must be constant. Hence f is constant on .

11. Suppose that f is differentiable on , and that 2 ≤ f′(x) ≤ 3 for x ∈ . If f(0) = 0, prove that 2x ≤ f(x) ≤ 3x for all x ≥ 0. Suppose x > 0. By the mean value theorem, there exists a y ∈ (0, x) such that

\frac{f(x) − f(0)}{x − 0} = f′(y)

and hence

\frac{f(x)}{x} = f′(y).

The constraint on the derivative shows that

2 ≤ \frac{f(x)}{x} ≤ 3

and hence

2x ≤ f(x) ≤ 3x.  (1)

Since f(0) = 0, it follows that Eq. 1 holds for all x ≥ 0.

12. Suppose that f is integrable on [a, b]. Then for all ϵ > 0, there exists a partition P

such that

U(f, P) − L(f, P) < ϵ.

Consider the partition Q composed of the values a, c, d, and b; in general this will have four points, but if c = a or d = b it may have fewer. Define a new partition T = P ∪ Q. Then since T is a refinement,

U(f, T) − L(f, T) < ϵ.

Write T = {a = t_{0} < t_{1} < … < t_{n} = b}. There exists i and j such that t_{i} = c and t_{j} = d. Consider the partition S = {t_{i} < … < t_{j}} of [c, d]. Then

U(f, S) − L(f, S) = ∑_{k=i+1}^{j} (M(f, [t_{k−1} , t_{k}]) − m(f, [t_{k−1} , t_{k}]))

≤ ∑_{k=1}^{n} (M(f, [t_{k−1} , t_{k}]) − m(f, [t_{k−1} , t_{k}]))
= U(f, T) − L(f, T)
< ϵ.

Since this is true for an arbitrary ϵ > 0, it follows that f is integrable on [c, d].

13. (a) Suppose that r^{1/3} is rational. Then

r^{1/3} = \frac{p}{q}

for p, q ∈ , where q ≠ 0. Hence

r = \frac{p^{3}}{q^{3}}

and thus r is rational. Thus is r is irrational, then r^{1/3} is irrational also. Similarly, if r + 1 is rational, then

r + 1 = \frac{p}{q}

for p, q ∈ , where q ≠ 0, and

r = \frac{p}{q} − 1 = \frac{p − q}{q}

so r is rational. Thus if r is irrational, then r + 1 is irrational also.

(b) First, consider the number x = 5 + \sqrt{2}. Then

x − 5 = \sqrt{2}
(x − 5)^{2} = 2
x^{2} − 10x + 25 = 2
x^{2} − 10x + 23 = 0.

By the rational zeroes theorem, if x is rational, then x = ±1 or x = ±23. But x > 5 and x < 5 + 2 = 7, so none of these possibilities are valid, and thus x is irrational. By the results of part (a), (5 + \sqrt{2})^{1/3} is irrational, and (5 + \sqrt{2})^{1/3} + 1 is irrational also.

14. For the first limit, L’Hôpital’s rule can be applied once to show that

lim_{x→0} \frac{x}{1 − e^{−x^{2}−3x}} = lim_{x→0} \frac{1}{(2x + 3)e^{−x^{2}−3x}} = \frac{1}{3}.

The second limit can first be rewritten as

lim_{x→0} ( \frac{1}{sin x} − \frac{1}{x}) = lim_{x→0} \frac{x − sin x}{x sin x}

after which L’Hôpital’s rule can be applied twice to show that

lim_{x→0} \frac{x − sin x}{x sin x} = lim_{x→0} \frac{1 − cos x}{sin x + x cos x} = lim_{x→0} \frac{sin x}{2 cos x + x sin x} = \frac{0}{2} = 0.

For the third limit, L’Hôpital’s rule can be applied three times to show that

lim_{x→0} \frac{x^{3}}{sin x − x} = lim_{x→0} \frac{3x^{2}}{cos x − 1}
= lim_{x→0} \frac{6x}{− sin x}
= lim_{x→0} \frac{6}{− cos x}
= \frac{6}{−1} = −6.

15. Suppose a > 0. Then there exists an N ∈ such that 2^{−2N} < a. Hence for M > 2N,

sup{s_{n} : n > M} = a

and hence lim sup s_{n} = a. Now consider any l > 0. Then there exists an N ∈ such that 2^{−n} < l for n > 2N. Hence l is not a lower bound for the set {s_{n} : n > M} for any M. However, since 0 is lower bound, it follows that it must be the greatest lower bound, and hence lim sup s_{n} = 0.

Now suppose a ≤ 0. Then inf{s_{n} : n > M} = a for all M ∈ , so lim inf s_{n} = a. Also,

sup\{s_{n} : n > M\} = \{ \begin{matrix}2^{−M−1} & if M is odd, \\ 2^{−M−2} & if M is even. \end{matrix}

so lim sup s_{n} = 0.

If a = 0 then lim sup s_{n} = lim inf s_{n} and hence the series converges with limit zero. Otherwise, lim sup s_{n} ≠ lim inf s_{n} and the series does not converge.

16. Consider any ϵ > 0. Then

|f(x_{1}, x_{2}) − f(0, 0)| = | \frac{1}{x_{1}^{2} + x_{2}^{2} + 1} − \frac{1}{1}|

= | \frac{x_{1}^{2} + x_{2}^{2}}{x_{1}^{2} + x_{2}^{2} + 1} |
≤ |x_{1}^{2} + x_{2}^{2}|.

Suppose d((x_{1}, x_{2}), (0, 0)) < δ where δ = \sqrt{ϵ}. Then

|f(x_{1}, x_{2}) − f(0, 0)| ≤ (d((x_{1}, x_{2}), (0, 0)))^{2} < δ^{2} = ϵ.

Now consider the point (0, 1):

|f(x_{1}, x_{2}) − f(0, 1)| = | \frac{1}{x_{1}^{2} + x_{2}^{2} + 1} − \frac{1}{2}|

= | \frac{x_{1}^{2} + x_{2}^{2} − 1}{2(x_{1}^{2} + x_{2}^{2} + 1)} |

≤ \frac{|x_{1}^{2} + x_{2}^{2} − 1|}{2}
= \frac{|x_{1}^{2} + (x_{2} − 1)(x_{2} + 1)|}{2}
≤ \frac{|x_{1}|^{2} + |x_{2} − 1| · |x_{2} + 1|}{2} .

Suppose d((x_{1}, x_{2}), (0, 1)) < 1, so that

x_{1}^{2} + (x_{2} − 1)^{2} < 1.

from which it follows that |x_{1}| < 1 and |x_{2} − 1| < 1. Hence |x_{2} + 1| ≤ |x_{2}| + 1 ≤ 3. Define δ = min\{1, \frac{ϵ}{3}\}. Then δ < \frac{ϵ}{3} and δ < \sqrt{ϵ}. Hence |x_{1}| < \sqrt{ϵ} and |x_{2} − 1| < \frac{ϵ}{3}. Then

|f(x_{1}, x_{2}) − f(0, 1)| ≤ \frac{|x_{1}|^{2} + |x_{2} − 1| · |x_{2} + 1|}{2} < \frac{(\sqrt{ϵ})^{2} + 3\frac{ϵ}{3}}{2} = ϵ.

Hence f is continuous at (0, 1).

17. (a) For p ≠ 1, the improper integral can be written as

∫_{0}^{1} x^{−p} dx = lim_{c→0^{+}} ∫_{c}^{1} x^{−p} dx
= lim_{c→0^{+}} [ \frac{x^{−p+1}}{−p + 1} ] _{c}^{1}
= lim_{c→0^{+}} \frac{1 − c^{1−p}}{1 − p} .

Then if 0 < p < 1, the exponent is positive, so

∫_{0}^{1} x^{−p} dx = \frac{1}{1 − p}

whereas if p > 1, the exponent is negative, so

∫_{0}^{1} x^{−p} dx = ∞.

(b) The improper integral can be written as

∫_{0}^{1} x^{−p} dx = lim_{c→0^{+}} ∫_{c}^{1} x^{−p} dx + lim_{d→∞} ∫_{1}^{d}x^{−p} dx.

Since both terms are non-negative, it follows that if one term is ∞, then the integral must ∞ also. By the result above, if p > 1, then the first term is ∞. Now suppose p = 1. Then the second term is

lim_{d→∞} ∫_{1}^{d}x^{−1} dx = lim_{d→∞} [log x]_{1}^{d}
= lim_{d→∞} log d
= ∞.

If 0 < p < 1, then second term is

lim_{d→∞} ∫_{1}^{d}x^{−p} dx = lim_{d→∞} [ \frac{x^{−p+1}}{1 − p} ] _{1}^{d}
= lim_{d→∞} \frac{d^{−p+1} − 1}{1 − p}
= ∞.

Hence in all cases, ∫_{0}^{1} x^{−p} dx = ∞.

18. Since f is integrable on [a, b], it is bounded, so there exists a B > 0 such that f(x) < B for all x ∈ [a, b]. Assume that if f is integrable on [a, b] then it is integrable on any interval [c, d] ⊆ [a, b]; for full details see Problem 12.

To show the above limit, consider any ϵ > 0, and examine c ∈ (b − δ, b) where δ = ϵ/B. Then

|∫_{a}^{c}f − ∫_{a}^{b}f| = |∫_{c}^{b}f|
≤ ∫_{c}^{b}|f|
≤ (b − c)B
< δB = ϵ.

Hence

lim_{d→b^{−}} ∫_{a}^{d}f(x) dx = ∫_{a}^{b}f(x) dx.

19. If λ = 0, then s_{n} = 0 for all n, so s_{n} is a constant convergent sequence. Similarly if λ = 1, then s_{n} = 1 for all n, so s_{n} is also constant and convergent. In general,

s_{n} = λ^{2^{n−1}} .

If 0 < λ < 1, then since 0 < λ^{2^{n−1}} ≤ λ^{n} for n ∈ and λ^{n} → 0 as n → ∞ it follows that s_{n} → 0. If λ > 1, then since λ^{2^{n−1}} ≥ λ^{n} for n ∈ and λ^{n} → ∞ as n → ∞, it follows that s_{n} → ∞.

Finally, suppose λ < 0. Then for n ≥ 2, the resulting sequence (s_{n}) will be the same as the case for −λ, and will therefore have the same convergence properties. Hence (s_{n}) converges if and only if |λ| ≤ 1.

20. The minimum element of [0, \sqrt{2}] is 0, and since this is also in A it follows that min A = 0. The maximum element of [0, \sqrt{2}] is \sqrt{2}, but this is not in A. Since there are rational numbers arbitrarily close to \sqrt{2}, it follows that A does not have a maximum. The infimum is just inf A = min A = 0. \sqrt{2} is an upper bound for A. For any ϵ > 0, there exist elements in A which are greater than \sqrt{2}− ϵ, and thus \sqrt{2} is the least upper bound. Hence sup A = \sqrt{2}.

For B, note that

x^{2} + x − 1 = (x +^{1}/2)^{2} −^{5}/4.

Since the first term can take any positive value, it follows that B = [−^{5}/4, ∞). Hence min B = inf B = −^{5}/4, the maximum does not exist, and sup B = ∞.

By completing the square, above equation can be written as

x^{2} + x − 1 = ( x + \frac{1 − \sqrt{5}}{2} ) ( x + \frac{1 + \sqrt{5}}{2} ) .

The quadratic will be strictly negative when one of these two factors is strictly negative and the other is strictly positive. Hence C = (−^{1}/2−^{\sqrt{5}}/2, −^{1}/2+^{\sqrt{5}}/2), so the minimum and maximum do not exist, inf C = −^{1}/2−^{\sqrt{5}}/2 and sup C = −^{1}/2+^{\sqrt{5}}/2.

21. (a) Suppose 0 ≤ x < 1. Then

lim_{n→∞} f_{n}(x) = lim_{n→∞} (x − x^{n}) = x

since if |x| < 1, then x^{n} → 0 as n → ∞. For x = 1,

lim_{n→∞} f_{n}(x) = lim_{n→∞} (1 − 1^{n}) = 0.

0 0.2 0.4 0.6 0.8 1 0  0.2  0.4  0.6  0.8  1 y  x y = f_{2}(x) y = f_{4}(x) y = f_{7}(x) y = f_{11}(x) y = f_{16}(x) y = f(x)

Figure 2: Graph for Problem 21, on pointwise and uniform convergence.

Hence f_{n} converges pointwise to a limit f on [0, 1] given by

f(x) = \{ \begin{matrix}x & if 0 ≤ x < 1, \\ 0 & if x = 1. \end{matrix}

Graphs of several of the f_{n} and the limit f are shown in Fig. 2.

(b) As can be seen from Fig. 2, the convergence does not appear to be uniform on [0, 1], since it does not appear that the f_{n} will ever lie within a strip of a fixed width ϵ around f. To see this mathematically, for a given n, consider the point x = (^{1}/2)^{1/n}. Then

|f_{n}(x) − f(x)| = |x − x^{n} − x|
= |x^{n}|
= ^{1}/2

Hence if ϵ = ^{1}/2, there does not exist an N such that n > N implies that |f_{n}(x) − f(x)| < ϵ for all x ∈ [0, 1].

(c) Consider the interval [0,^{1}/2]. Then for any x in this interval

|f_{n}(x) − f(x)| = |x^{n}| ≤ 2^{−n}

Consider any ϵ > 0. Then there exists an N such that n > N implies 2^{−n} < ϵ, and thus |f_{n}(x) − f(x)| < ϵ. Hence f_{n} → f uniformly on [0,^{1}/2].

(d) Since continuous functions are integrable, it follows immediately that f_{n} is integrable for all n ∈ . For a specific n,

∫_{0}^{1} f_{n} = ∫_{0}^{1}(x − x^{n})dx
= [ \frac{x^{2}}{2} − \frac{x^{n+1}}{n + 1} ] _{0}^{1}
= \frac{1}{2} − \frac{1}{n + 1}
= \frac{n − 1}{2(n + 1)} .

To show that f is integrable, consider the function

g(x) = \{ \begin{matrix}0 & if 0 ≤ x < 1, \\ 1 & if x = 1. \end{matrix}

Choose any ϵ > 0, and examine the partition P = {0 = t_{0} < t_{1} < t_{2} = 1} where t_{1} = 1 −^{ϵ}/2. Then

L(f, P) = ∑_{k=1}^{2} (t_{k}− t_{k−1} )m(f, [t_{k−1} , t_{k}]) = ∑_{k=1}^{2} 0 = 0

−2.5 −2 −1.5 −1 −0.50 0.5 1 1.5−3  −2  −1  0  1  2  3 y  x  y = f(x) y = F(x)

Figure 3: Graph for Problem 22, on the second Fundamental Theorem of Calculus.

and

U(f, P) = ∑_{k=1}^{2} (t_{k}− t_{k−1} )M(f, [t_{k−1} , t_{k}]) = (t_{1} − t_{0}) · 0 + (t_{2} − t_{1})1 = \frac{ϵ}{2}.

Thus U(f, P) − L(f, P) < ϵ. Since a partition such as this can be constructed for an arbitrary ϵ > 0, it follows that g is integrable and ∫_{0}^{1} g = 0. Since f(x) = x − g(x) and both x and g are integrable, it follows that f is integrable and ∫_{0}^{1} f = ∫_{0}^{1} x − ∫_{0}^{1} g = ^{1}/2. Note that

lim_{n→∞} ∫_{0}^{1} f_{n} = lim_{n→∞} \frac{n − 1}{2(n + 1)} = \frac{1}{2} = ∫_{0}^{1} f.

22. (a) If 0 ≤ x ≤ 1, then

F(x) = ∫_{0}^{x} f = ∫_{0}^{x} 1 dt = x.

If 1 < x ≤ 2, then

F(x) = ∫_{0}^{1} f + ∫_{1}^{x} = 1 − 2(x − 1) = 3 − 2x.

If x > 2, then

F(x) = F(2) + ∫_{2}^{x} 0 = F(2) = −1.

For negative values of x, note that f is an even function, and thus

F(−x) = ∫_{0}^{−x} f(t)dt = ∫_{0}^{x} f(−s)(−ds) = − ∫_{0}^{x} f(s)ds = −F(x)

so F is odd. Hence

F(x) = \{ \begin{matrix}1 & if x < −2, \\ −3 − 2x & if −2 ≤ x < 1, \\ x & if −1 ≤ x ≤ 1, \\ 3 − 2x & if 1 < x ≤ 2, \\ −1 & if x > 2. \end{matrix}

(b) The functions f and F are plotted in Fig. 3.

(c) By the second Fundamental Theorem of Calculus, if f is continuous at x, and then F is differentiable at x and F′(x) = f(x). Thus the only points where F may not be defined are x = ±1, ±2. Since

lim_{x→1^{−}} \frac{F(x) − F(1)}{x − 1} = lim_{x→1^{−}} \frac{x − 1}{x − 1} = 1

but

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

so F is not differentiable at 1. Similarly

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

but

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

so F is not differentiable at 2. Since F is odd, it follows that F is not differentiable at −1 and −2 also. Hence F′is defined on /{−2, −1, 1, 2} and

F(x) = \{ \begin{matrix}0 & if x < −2, \\ −2 & if −2 < x < 1, \\ 1 & if −1 < x < 1, \\ −2 & if 1 < x < 2, \\ 0 & if x > 2. \end{matrix}

23. (a) Let f and g be continuous functions on [a, b] such that ∫_{a}^{b}f = ∫_{a}^{b}g. Prove that there exists an x ∈ [a, b] such that f(x) = g(x). If ∫_{a}^{b}f = ∫_{a}^{b}g, then if h(x) = f(x) − g(x), then ∫_{a}^{b}h = 0. Consider the partition P = {a = t_{0} < t_{1} = b}. Then

0 = ∫_{a}^{b}h ≤ U(h, P) = (b − a)M(h, [a, b])

and

0 = ∫_{a}^{b}h ≥ L(h, P) = (b − a)m(h, [a, b]).

Since a continuous function on a closed interval achieves its bounds, there exist x_{1} and x_{2} such that h(x_{1}) = M(h, [a, b]) and h(x_{2}) = m(h, [a, b]). Either h(x_{1}) = 0 or h(x_{2}) = 0, or otherwise h(x_{1}) > 0 and h(x_{2}) < 0. In the latter case, the intermediate value theorem can be applied to show that there exists an x_{3} between x_{1} and x_{2} such that h(x_{3}) = 0. In all cases there exists an x such that h(x) = 0 and hence f(x) = g(x).

(b) On the interval [−1, 1], define

f(x) = \{ \begin{matrix}−1 & if x < 0, \\ 1 & if x ≥ 0, \end{matrix}

and let g(x) = −f(x). By construction f(x) ≠ g(x) for all x ∈ [−1, 1]. To find the integral of f, choose ϵ > 0 and consider the partition P = {a = −1 < t_{1} < t_{2} < t_{3} = 1} where t_{1} = −^{ϵ}/5 and t_{2} = ^{ϵ}/5. Then

L(f, P) = ∑_{k=1}^{3} (t_{k}− t_{k−1} )m(f, [t_{k−1} , t_{k}])
= ( 1 − \frac{ϵ}{5}) (−1) + \frac{2ϵ(−1)}{5} + ( 1 − \frac{ϵ}{5}) (1)
= −\frac{2ϵ}{5}.

Similarly

U(f, P) = ∑_{k=1}^{3} (t_{k}− t_{k−1} )M(f, [t_{k−1} , t_{k}])
= ( 1 − \frac{ϵ}{5}) (−1) + \frac{2ϵ(1)}{5} + ( 1 − \frac{ϵ}{5}) (1)
= \frac{2ϵ}{5}.

Then U(f, P) − L(f, P) = ^{4ϵ}/5 < ϵ, and since ϵ is arbitrary it follows that f is integrable, and that ∫_{−1}^{1} f = 0. In addition, so ∫_{−1}^{1} g = ∫_{−1}^{1} (−f) = − ∫_{−1}^{1} f = 0.

Thus ∫_{−1}^{1} f = ∫_{−1}^{1} g but f(x) ≠ g(x) for all x ∈ [−1, 1].

0 0.5 1 1.5 2 2.5 3−2  −1.5  −1  −0.5  0  0.5  1  1.5  2 y  x  y = h_{1}(x) y = h_{2}(x) y = h_{3}(x)

Figure 4: Graphs of several functions h_{n}(x) used in Problem 24 on integration limits.

24. (a) The functions h_{1}, h_{2}, and h_{3} are plotted in Fig. 4

(b) Consider any x ≠ 0, and ϵ > 0. Then there exists an N ∈ such that 1/N < 2|x|. Then if n > N, h_{n}(x) = 0. Hence lim_{n→∞} h_{n}(x) = 0. At x = 0,

h_{n}(x) = n

which tends to ∞ as n → ∞.

(c) Consider any ϵ > 0. Then since f is continuous, there exists a δ > 0 such that |x| < δ implies |f(x) − f(0)| < ϵ. Then there exists an N such that 1/2N < δ. For n > N,

lim_{n→∞} ∫_{−∞}^{∞} h_{n}f.

(d) Consider

g(x) = \{ \begin{matrix}0 & if x < 0, \\ 1 & if x ≥ 0. \end{matrix}

Then for any n ∈ ,

∫_{∞}^{∞} h_{n}g = ∫_{−1/2n}^{1/2n} ng = ∫_{−1/2n}^{0} 0 + ∫_{0}^{1/2n} n = \frac{1}{2}

and hence

lim_{n→∞} ∫_{−∞}^{∞} h_{n}g = \frac{1}{2}

which does not equal g(0) = 1.

25. (a) For x > 0,

f(x) = \frac{1}{1 + \frac{1}{x}}

and since 1/x → 0 as x → ∞, it follows that f(x) → 0 as x → ∞. On the interval under consideration 0 ≤ x < 1 + x, so 0 ≤ f(x) < 1.

(b) The function f is shown in Fig. 5.

(c) Since

f(x) = 1 − \frac{1}{x + 1} ,

it follows that the derivatives are

f′(x) = \frac{1}{(x + 1)^{2}}

and

f′′(x) = −\frac{2}{(x + 1)^{3}} .

0 0.2 0.4 0.6 0.8 1 0  0.5  1  1.5  2  2.5  3 y  x  y = f(x) y = f_{T}(x)

Figure 5: A graph of the function f and a Taylor series approximation f_{T} at x = 1, discussed in Problem 25.

Hence f(1) = ^{1}/2, f′(1) = ^{1}/4, f′′(1) = −^{1}/4 and thus

f_{T}(x) = ∑_{n=0}^{2} \frac{(x − 1)^{n}f^{(n)}(1)}{n!}
= f(1) + f′(1)(x − 1) + f′′(1)\frac{(x − 1)^{2}}{2}
= \frac{1}{2} + \frac{x − 1}{4} − \frac{(x − 1)^{2}}{8} .

(d) f_{T} can rewritten as

f_{T}(x) = \frac{1}{2} + \frac{x}{4} − \frac{1}{4} − \frac{x^{2}}{8} + \frac{x}{4} − \frac{1}{8}.
= \frac{1}{8} + \frac{x}{2} − \frac{x^{2}}{8}

which is a quadratic.

(e) The function f_{T} is shown in Fig. 5. By construction, the curves intersect at x = 1, and have the same slope and curvature there.