All units combined

Question 1

Mean Value Theorem

Answer 1

If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there exists a point c ∈ (a, b) where

f’(c) = f(b) - f(a) / b - a

Question 2

L’Hospital’s rule (0/0 case)

Answer 2

lim f’(x)/g’(x) = L
implies lim f(x)/g(x)=L

Question 3

Continuous Limit Theorem

Answer 3

Let (fn) be a sequence of functions defined on A ⊆ R that converges uniformly on A to a function f. If each fn is continuous at c ∈ A, then f is continuous at c.

Question 4

Intermediate Value Property

Answer 4

A function f has the intermediate value property on an
interval [a, b] if for all x < y in [a, b] and all L between f(x) and f(y), it is always possible to find a point c ∈ (x, y) where f(c) = L.

Question 5

Differentiability
Let g : A → R be a function defined on an interval A. Given c ∈ A, the derivative of g at c is defined by

Answer 5

g’(c) = lim g(x) - g(c) / x - c

Question 6

Pointwise convergence

Answer 6

For each n ∈ N, let fn be a function defined on a set A ⊆ R.
The sequence (fn) of functions converges pointwise on A to a function f if, for all x ∈ A, the sequence of real numbers fn(x) converges to f(x)

Question 7

Uniform convergence of functions

Answer 7

Let (fn) be a sequence of functions defined on a set A ⊆ R. Then, (fn) converges uniformly on A to a limit
function f defined on A if, for every ϵ > 0, there exists an N ∈ N such that
|fn(x) − f(x)| < ϵ whenever n ≥ N and x ∈ A

Question 8

pointwise and uniform convergence of a series

Answer 8

For each n ∈ N, let fn and f be functions defined on a set A ⊆ R. The infinite series converges pointwise on A to f(x) if the sequence sk(x) of partial sums defined by

sk(x) = f1(x) + f2(x) + ··· + fk(x)

converges pointwise to f(x)

The series converges uniformly on A to f if the sequence sk(x) converges uniformly on A to f(x).

Question 9

power series

Answer 9

f(x) = Σan * x^n = a0 + a1x + a2x^2 + a3x^3
from n = 0 to n = ∞

Question 10

Taylor series (Taylor’s formula)

Answer 10

f(x) = a0 + a1x + a2x^2 + a3x^3 + a4x^4

where an = f^n(0) / n! (the nth derivative)

Question 11

Partition

Answer 11

A partition P of [a, b] is a finite set of points from [a, b] that includes both a and b. The notational convention is to always list the points of a partition P = {x0, x1, x2,…,xn} in increasing order

thus,
a = x0 < x1 < x2 < ··· < xn = b.

Question 12

Refinement

Answer 12

A partition Q is a refinement of a partition P if Q contains all of the points of P; that is, if P ⊆ Q.

Question 13

higher integrals

Answer 13

Let P be the collection of all possible partitions of the
interval [a, b].

The upper integral of f is defined to be
U(f) = inf{U(f,P) : P ∈ P}

Question 14

lower integrals

Answer 14

Let P be the collection of all possible partitions of the
interval [a, b]

In a similar way, define the lower integral of f by
L(f) = sup{L(f,P) : P ∈ P}.

Question 15

Riemann Integrability

Answer 15

A bounded function f defined on the interval [a, b] is Riemann-integrable if U(f) = L(f). In this case, we define ∫f or ∫f(x)dx to be this common value.

∫f = U(f) = L(f)

Question 16

Interior Extremum Theorem PROOF

Answer 16

Theorem:
Let f be differentiable on an open interval (a, b). If f attains a maximum value at some point c ∈ (a, b)
(i.e., f(c) ≥ f(x) for all x ∈ (a, b)), then f’(c)=0. The same is true if f(c) is a minimum value.
.
Proof:
.
Because c is in the open interval (a, b), we can construct two sequences (xn) and (yn), which converge to c and satisfy xn

Question 17

Continuous Limit Theorem PROOF

Answer 17

Theorem:
Let (fn) be a sequence of continuous functions defined on A ⊆ R that converges uniformly on A to a function f. If each fn is continuous at c ∈ A, then f is continuous at c.
.
Proof:
.
Fix c ∈ A and let ϵ > 0. Choose N so that
|fN(x) − f(x)| < ϵ/3
for all x ∈ A. Because fN is continuous, there exists a δ > 0 for which
|fN(x) − fN(c)| < ϵ/3
is true whenever |x − c| < δ. But this implies

|f(x) − f(c)|
= |f(x) − fN(x) + fN(x) − fN(c) + fN(c) − f(c)|
≤ |f(x) − fN(x)|+|fN(x) − fN(c)|+|fN(c) − f(c)|
< ϵ/3 + ϵ/3 + ϵ/3 = ϵ

Question 18

Characterization of Compactness in R PROOF

Answer 18

A set K ⊆ R is compact if and only if it is closed and bounded
Let K be compact. We will first prove that K must be bounded, so assume, for contradiction, that K is not a bounded set. The idea is to produce a sequence in K that marches off to infinity in such a way that it cannot have a convergent subsequence as the definition of compact requires. To do this, notice that because K is not bounded there must exist an element x1 ∈ K satisfying |x1| > 1. Likewise, there must exist x2 ∈ K with |x2| > 2, and in general, given any n ∈ N, we can produce xn ∈ K such that |xn| > n.

Now, because K is assumed to be compact, (xn) should have a convergent subsequence (xnk). But the elements of the subsequence must satisfy |xnk| > nk, and consequently (xnk) is unbounded. Because convergent sequences are bounded (Theorem 2.3.2), we have a contradiction. Thus, K must at least be a bounded set.

Next, we will show that K is also closed. To see that K contains its limit points, we let x = lim xn, where (xn) is contained in K and argue that x must be in K as well. By Definition 3.3.1, the sequence (xn) has a convergent subsequence (xnk), and by Theorem 2.5.2, we know (xnk) converges to the same limit x. Finally, Definition 3.3.1 (A set K ⊆ R is compact if every sequence in K has a subsequence that converges to a limit that is also in K) requires that x ∈ K. This proves that K is closed

Question 19

Baire’s Theorem PROOF

Answer 19

The set of real numbers R cannot be
written as the countable union of nowhere-dense sets

To start, assume that E1, E2, E3, . . . are each nowhere-dense and satisfy R = infinite union of En

By the definition of En being nowhere-dense, the closure En contains no nonempty open intervals meaning we can apply Exercise 3.5.5 to conclude that the infinite union of En-closure is not equal to R

Since each En is a subset of En-closure, then we know that the infinite untion of En is also not equal to R, which is a contradiction

Question 20

Uniform Continuity of Compact Sets

Answer 20

A function that is continuous on a compact set K is uniformly continuous on K.

Assume f : K → R is continuous at every point of a compact set K ⊆ R.

To prove that f is uniformly continuous on K we argue by contradiction.

By the criterion in Theorem 4.4.5, if f is not uniformly continuous on K, then there exist two sequences (xn) and (yn) in K such that lim |xn − yn| = 0 while |f(xn) − f(yn)| ≥ ε0 for some particular
ε0 > 0. Because K is compact, the sequence (xn) has a convergent subsequence (xnk) with x = lim xnk also in K.

We could use the compactness of K again to produce a convergent subsequence of (yn), but notice what happens when we consider the particular subsequence (ynk) consisting of those terms in (yn) that correspond to the terms in the convergent subsequence (xnk). By the Algebraic Limit Theorem,

lim(ynk) = lim((ynk − xnk) + xnk) = 0 + x.

The conclusion is that both (xnk) and (ynk) converge to x ∈ K. Because f isassumed to be continuous at x, we have lim f(xnk) = f(x) and lim f(ynk) = f(x), which implies lim(f(xnk) − f(ynk)) = 0.

A contradiction arises when we recall that (xn) and (yn) were chosen to satisfy
|f(xn) − f(yn)| ≥ ε0
for all n ∈ N. We conclude, then, that f is indeed uniformly continuous on K

Question 21

Cantor Set

Answer 21

a set Cn consisting of 2^n closed intervals each having length 1/(3^&n). Finally, we define the Cantor set C (Fig. 3.1) to be the intersection of all the Cns

Properties:
C is uncountable
C contains only the endpoints of each sub interval
C is a perfect set

Question 22

open set

Answer 22

A set O ⊆ R is open if for all points a ∈ O there exists an ε-neighborhood Vε(a) ⊆ O

Question 23

limit point

Answer 23

A point x is a limit point of a set A if every ε-neighborhood Vε(x) of x intersects the set A at some point other than x.

x is quite literally the limit of a sequence in A

Question 24

isolated point

Answer 24

A point a ∈ A is an isolated point of A if it is not a limit point of A

Question 25

closed set

Answer 25

A set F ⊆ R is closed if it contains its limit points

Question 26

closure

Answer 26

Given a set A ⊆ R, let L be the set of all limit points of A. The closure of A is defined to be A-closure = A ∪ L.

Question 27

complement

Answer 27

Ac = {x ∈ R : x /∈ A}.

A set O is open if and only if Oc is closed. Likewise, a set F is closed if and only if Fc is open

Question 28

compact set

Answer 28

A set K ⊆ R is compact if and only if it is closed and bounded.

Question 29

open cover

Answer 29

An open cover for A is a (possibly infinite) collection of open sets {Oλ : λ ∈ Λ} whose union contains the set A; that is, A ⊆ UOλ.

Given an open cover for A, a finite subcover is a finite subcollection of open sets from the original open cover whose union still manages to completely contain A.

A set is compact if every open cover of A contains a finite subcover of A

Question 30

perfect set

Answer 30

A set P ⊆ R is perfect if it is closed and contains no isolated points.

Closed intervals (other than the singleton sets [a, a]) serve as the most
obvious class of perfect sets, but there are more interesting examples.

Question 31

separated sets

Answer 31

Two nonempty sets A, B ⊆ R are separated if A-closure ∩ B and
A ∩ B-closure are both empty

Question 32

connected sets

Answer 32

A set that is not disconnected is called a connected set.

Question 33

disconnected set

Answer 33

A set E ⊆ R is disconnected if it can be written as E = A ∪ B, where A and B are nonempty separated sets.

Question 34

Fσ sets

Answer 34

A set A ⊆ R is called an Fσ set if it can be written as the countable union of closed sets

Question 35

Gd set

Answer 35

A set B ⊆ R is called a Gd set if it can be
written as the countable intersection of open sets.

Question 36

nowhere-dense sets

Answer 36

A set E is nowhere-dense if E-closure contains no nonempty open intervals.

Question 37

limit of a function

Answer 37

Let f : A → R, and let c be a limit
point of the domain A. We say that limx→c f(x) = L provided that, for all
ε > 0, there exists a δ > 0 such that whenever 0 < |x − c| < δ (and x ∈ A) it
follows that |f(x) − L| < ε

Question 38

continuous

Answer 38

A function f : A → R is continuous at a
point c ∈ A if, for all ε > 0, there exists a
δ > 0 such that whenever |x − c| < δ
(and x ∈ A) it follows that
|f(x) − f(c)| < ε

If f is continuous at every point in the domain A, then we say that f is continuous on A

Question 39

uniform continuity

Answer 39

A function f : A → R is uniformly
continuous on A if for every ε > 0 there exists a δ > 0 such that for all x, y ∈ A,
|x − y| < δ implies |f(x) − f(y)| < ε

Question 40

intermediate value property

Answer 40

A function f has the intermediate value property on an interval [a, b] if for all
x < y in [a, b] and all L between f(x) and f(y), it is always possible to find a point c ∈ (x, y) where f(c) = L

Question 41

sets of discontinuity (Df)

Answer 41

Given a function f : R → R, define Df ⊆ R to be the set of points where the function f fails to be continuous

Question 42

monotone, increasing, decreasing functions

Answer 42

A function f : A → R is increasing on A if f(x) ≤ f(y) whenever x < y

decreasing if f(x) ≥ f(y) whenever x < y

A monotone function is one that is either increasing or decreasing.

Question 43

removable discontinuity

Answer 43

If limx→c f(x) exists but has a value different from f(c), the discontinuity
at c is called removable

Question 44

jump discontinuity

Answer 44

If limx→c+ f(x) =/= limx→c− f(x),
then f has a jump discontinuity at c

Question 45

essential discontinuity

Answer 45

If limx→c f(x) does not exist for some other reason, then the discontinuity
at c is called an essential discontinuity.

Question 46

α-continuous

Answer 46

Let f be defined on R, and let α > 0. The function f is α-continuous at x ∈ R if there exists a δ > 0 such that for all y, z ∈ (x−δ, x+δ) it follows that |f(y) − f(z)| < α

Question 47

Df-α

Answer 47

Df-alpha = {x ∈ R : f is not α-continuous at x}.

Question 48

Characterization of Compactness in R

Answer 48

A set K ⊆ R is compact if and only if it is closed and bounded

Question 49

Nested Compact Set Property

Answer 49

If K1 ⊇ K2 ⊇ K3 ⊇ K4 ⊇ …
is a nested sequence of nonempty compact sets, then the infinite intersection Kn is not empty.

Question 50

A nonempty perfect set is

Answer 50

uncountable

Question 51

Sequential Criterion for Functional Limits

Answer 51

Given a function f : A → R and a limit point c of A, the following two statements are
equivalent:
(i) lim (x→c) f(x) = L

(ii) For all sequences (xn) ⊆ A satisfying xn =/= c and (xn) → c, it follows that f(xn) → L.

Question 52

Characterization of Continuity

Answer 52

The function f is continuous at c if and only if any one of the following
three conditions is met:

(i) For all ε > 0, there exists a δ > 0 such that |x−c| < δ (and x ∈ A) implies
|f(x) − f(c)| < ε

(ii) For all V(f(c)), there exists a Vδ(c) with the property that x ∈ Vδ(c) (and x ∈ A) implies f(x) ∈ Vε(f(c))

(iii) For all (xn) → c (with xn ∈ A), it follows that f(xn) → f(c)

If c is a limit point of A, then the above conditions are equivalent to
(iv) lim x→c f(x) = f(c).

Question 53

Algebraic Limit Theorem for Functional Limits

Answer 53

(i) lim x→c kf(x) = kL for all k ∈ R,
(ii) lim x→c [f(x) + g(x)] = L + M,
(iii) limx→c [f(x)g(x)] = LM
(iv) lim x→c f(x)/g(x) = L/M, M =/= 0

Question 54

Divergence criterion for functional Limits

Answer 54

Let f be a function defined on A, and let c be a limit point of A. If there exist two
sequences (xn) and (yn) in A with
xn =/= c and yn =/= c and
lim xn = lim yn = c
but lim f(xn) =/= lim f(yn),

then we can conclude that the functional limit lim x→c f(x) does not exist

Question 55

Criterion for Discontinuity

Answer 55

Let f : A → R, and let c ∈ A be a limit point of A. If there exists a sequence (xn) ⊆ A where (xn) → c but such that f(xn) does not converge to f(c), we may conclude that f is not continuous at c.

Question 56

Algebraic Continuity Theorem

Answer 56

Assume f : A → R and g : A → R are continuous at a point c ∈ A. Then,

(i) kf(x) is continuous at c for all k ∈ R;
(ii) f(x) + g(x) is continuous at c;
(iii) f(x)g(x) is continuous at c; and
(iv) f(x)/g(x) is continuous at c, provided the quotient is defined.

Question 57

Composition of Continuous
Functions

Answer 57

Given f : A→R and g : B → R, assume that the range f(A) = {f(x) : x ∈ A} is contained in the domain B so that the composition g ◦ f(x) = g(f(x)) is defined on A. If f is continuous at c ∈ A, and if g is continuous at f(c) ∈ B, then g ◦ f is
continuous at c.

Question 58

Preservation of Compact Sets

Answer 58

Let f : A → R be continuous on A. If K ⊆ A is compact, then f(K) is compact as well.

Question 59

Extreme Value Theorem

Answer 59

If f : K → R is continuous on
a compact set K ⊆ R, then f attains a maximum and minimum value.

In other words, there exist x0, x1 ∈ K such that f(x0) ≤ f(x) ≤ f(x1) for all x ∈ K.

Question 60

Sequential Criterion for Absence of Uniform Continuity

Answer 60

A function f : A → R fails to be uniformly continuous on A if and only if there exists a particular ε0 > 0 and two sequences (xn) and (yn) in A satisfying

|xn − yn| → 0 but |f(xn) − f(yn)| ≥ ε0.

Question 61

A set containing only isolated points must be closed

Answer 61

FALSE

Question 62

The set Z is a nowhere-dense set

Answer 62

TRUE

Question 63

Let A be a set and c element of A be an isolated point. Then

lim x->c f(x) = f(a)

Answer 63

FALSE

Question 64

The function f(x) = x^2 is uniformly continuous on [0, 1)

Answer 64

FALSE

Question 65

Suppose A is an open set. Then A-closure is perfect

Answer 65

TRUE

Question 66

Let f(x) : R -> R be a continuous function, and suppose for any x except possibly 1, f(x) > 0. Then, f(1) >= 0

Answer 66

TRUE

Question 67

Let f : R ! R be a function. If x ∈ Df , then x ∈ Df-a for every alpha.

Answer 67

FALSE

Question 68

Let (Ai) be a countable sequence of open sets. Then
The closure of the union of all Ai is equal to the union of all Ai closures

Answer 68

FALSE

Question 69

If A is open and unbounded, then Ac is compact

Answer 69

FALSE

Question 70

Nested Interval Property

Answer 70

In order to show that the infinite intersection is not empty, we are going to use the Axiom of Completeness (AoC) to produce a single real number x satisfying x ∈ In for every n ∈ N. Now, AoC is a statement about bounded sets, and theone we want to consider is the set of left-hand endpoints of the intervals:
A = {an : n ∈ N}

Because the intervals are nested, we see that every bn serves as an upper bound for A. Thus, we are justified in setting
x = sup A

Now, consider a particular In = [an, bn]. Because x is an upper bound for A, we have an ≤ x. The fact that each bn is an upper bound for A and that x is the least upper bound implies x ≤ bn

Altogether then, we have an ≤ x ≤ bn, which means x ∈ In for every choice of n ∈ N. Hence, x ∈ the infinite intersection and the intersection is not empty.

Question 71

Monotone Convergence Theorem

Answer 71

Let (an) be monotone and bounded. To prove (an) converges using the definition of convergence, we are going to need a candidate for the limit. Let’s assume the sequence is increasing (the decreasing case is handled similarly), and consider the set of points {an : n ∈ N}. By assumption, this set is bounded, so
we can let
s = sup{an : n ∈ N}.
It seems reasonable to claim that lim an = s

To prove this, let ε > 0. Because s is the least upper bound for {an: n ∈ N}, s − ε is not an upper bound, so there exists a point in the sequence aN such that s − ε < aN .Now, the fact that (an) is increasing implies that if n ≥ N, then aN ≤ an. Hence,
s − ε < aN ≤ an ≤ s < s + ε
which implies |an − s| < ε, as desired

Question 72

Bolzano-Weierstrass Theorem

Answer 72

Let (an) be a bounded sequence so that there exists M > 0 satisfying|an| ≤ M for all n ∈ N. Bisect the closed interval [−M,M] into the two closed intervals [−M, 0] and [0, M]. (The midpoint is included in both halves.) Now, it must be that at least one of these closed intervals contains an infinite number of the terms in the sequence (an). Select a half for which this is the case and label that interval as I1. Then, let an1 be some term in the sequence (an) satisfying an1 ∈ I1.

Next, we bisect I1 into closed intervals of equal length, and let I2 be a half that again contains an infinite number of terms of the original sequence. Because there are an infinite number of terms from (an) to choose from, we can select an an2 from the original sequence with n2 > n1 and an2 ∈ I2. In general, we construct the closed interval Ik by taking a half of Ik−1 containing an infinite
number of terms of (an) and then select nk > nk−1 > ··· > n2 > n1 so that ank ∈ Ik.

We want to argue that (ank ) is a convergent subsequence, but we need a candidate for the limit. The sets I1 ⊇ I2 ⊇ I3 ⊇···
form a nested sequence of closed intervals, and by the Nested Interval Property there exists at least one point x ∈ R contained in every Ik. This provides us with the candidate we were looking for. It just remains to show that (ank) → x

Let ε > 0. By construction, the length of Ik is M(1/2)k−1 which converges to zero. (This follows from Example 2.5.3 and the Algebraic Limit Theorem.) Choose N so that k ≥ N implies that the length of Ik is less than ε. Because x and ank are both in Ik, it follows that |ank − x| < ε

Question 73

Let A subset of R be a set and suppose inf A = a exists. Prove there is a sequence (an) where an ∈ A for all n, and lim an = a. Hint: Specifying values of ε equal to 1/n for integers n may help.

Answer 73

Let s = inf A
There is an an ∈ A with an < s + 1/n
Thus s - 1/n < s <= an < s + 1/n
Thus |an - s | < 1/n

Let ε > 0 be given
choose N so that 1/N < ε.
Then if n > N, |an - s| < 1/n < 1/N < ε

Question 74

Let A subset of R be a bounded nonempty set and let c > 0. Define the set cA = {ca : a ∈ A}.
Prove that inf cA = c inf A

Answer 74

Let s = inf A
thus s <= a implying cs <= ca, so cs is a lower bound (which is one half of the lema)

Then there is an a ∈ A with a < s + ε/c
Thus ca < cs + ε
Thus cs <= ca < cs + ε
Thus cs is the greatest lower bound

Question 75

There is an infinite collection of sets Ai (subset of R) where each Ai is
uncountably infinite but for which the infinite intersection of Ai is empty.

Answer 75

TRUE

Question 76

if B is an uncountable set with B ∈ (0, 1), then for some n ∈ N, B ∈ [1/(n + 1), 1/n] must be uncountable.

Answer 76

TRUE

Question 77

Let (an) be a sequence and a ∈ R. If every epsilon neighborhood of a contains infinitely many terms of the sequence, then (an) converges to a

Answer 77

FALSE

Question 78

Let Ai = (ai , bi) be an open interval in R with the property that Ai+1 is a proper subset of Ai
Then the infinite intersection of Ai is the empty set

Answer 78

FALSE

Question 79

If (an) is unbounded, then (an) does not have a converging subsequence

Answer 79

FALSE

Question 80

Suppose B is a subset of R, and A is a subset of B. If A is bounded above, there is a least upper bound of A in B.

Answer 80

FALSE

Question 81

Cantor’s Diagonalization Method is used to show that the rationals are countable.

Answer 81

FALSE

Question 82

If (an) is unbounded, then (an) does not have a converging subsequence.

Answer 82

FALSE

Question 83

If lim(anbn) exists and lim bn = b =/= 0, then lim an exists

Answer 83

TRUE

Question 84

if (an) is monotonic and bounded, then it is Cauchy

Answer 84

TRUE

Question 85

Let A and B be subsets of R with
sup A <= sup B. Then, there is an element b ∈ B that is an (least) upper bound of A.

(This applies to both)

Answer 85

FALSE

Question 86

Let A be a nonempty set, and let f : A -> {0, 1} be a continuous, surjective function. (That is, there are points a ∈ A with f(a) = 0 and there are points b ∈ A
with f(b) = 1) Show that A is a disconnected set.

Answer 86

Define F as elements of A where f(x) = 0
The same for G where f(x) = 1
Clearly F∩G is empty and FUG = A

Let y be a limit point of F, then there is a sequence with (yn) -> y
Since f is continuous, f(yn) -> f(y), but since f(yn) = 0, f(y) = 0, and y ∈ F, f MUST be closed. Similarly G is closed

Thus F closure = F and the same for G, thus if F∩G is empty then they are separated, thus FUG represents a disconnected set, which is A.

Question 87

Let f : R -> R be continuous and let A = {x ∈ R : f(x)=0} Show that A is a closed
set.

Answer 87

Let y be a limit point of A, then there is a sequence yn with (yn) -> y. Since F is continuous, f(yn) -> f(y). Since both are zero, y must be in A, hence A is closed

Question 88

Let h(x) = 1/x^2
Show that h(x) is not uniformly continuous on (0,1)

Answer 88

Let xn = 1/x and yn = 1/x^2
We can use these because (xn) (yn) -> 0
Thus (xn - yn) -> 0

However, |h(xn) - h(yn)| = n^2|n^2 - 1|
If n >= 2, n^2 >= 4 and n^2 - 1 >= 3
Thus |h(xn) - h(yn)| >= 3n^2 which approaches infinity as n does, thus n is not uniformly continuous

Question 89

The complement of a connected open set is perfect.

Answer 89

TRUE

Question 90

If f : (0, 1) ∈ R is continuous and {xn} ⇢ (0, 1) is Cauchy,
then {f(xn)} is Cauchy.

Answer 90

FALSE

Question 91

Let f: (-1, 1) -> R be given. If f’ exists on (-1,1), then f continuous at 0.

Answer 91

TRUE

Question 92

Let (fn) be a sequence of functions defined on A ⊂ R. If (fn) converges uniformly on A to a function f, then f is continuous

Answer 92

FALSE

Question 93

The radius of convergence of the power series for f (x) = 2x / (1 + 4x) is R = 1/4

Answer 93

TRUE

Question 94

Suppose f : [a, b] -> R is differentiable. If f(a) = f(b) = 0, f’(a) > 0 and f’(b) > 0, then there is a c ∈ (a, b) with f(c) = 0

Answer 94

TRUE

Question 95

The series sqrt(x^2 + 1/n) / n^2 converges uniformly on [0,2]

Answer 95

TRUE

Question 96

Let f : [0, 1] -> R. If f is differentiable on [0,1], then f is integrable on [0,1]

Answer 96

TRUE

Question 97

Suppose f : R -> R is infinitely differentiable. It is possible for the Taylor series of f (centered at 0) to converge only on [0, 1), and diverge elsewhere.

Answer 97

FALSE

Question 98

Let f : [0, 1] ⊂ R, and assume f has countably many discontinuities. Then f is integrable

Answer 98

FALSE

Question 99

Let f : R -> R be differentiable. If f(0) = 0 and f’(x) < 1 for all x ∈ R, then f(1) < 1

Answer 99

TRUE

Question 100

How do you show that a derivative and a second derivative exists at a point a?

Answer 100

Set up:
lim (x->a)
f(x) - f(a) / x - a
Simplify
Find limit

If f’(x) exists, do the same thing but with
f’(x) - f’(a) / x - a
And then solve

Question 101

How do you show that a sequence of functions converges uniformly?

Answer 101

|fn(x) - f(x)| < ϵ
use inequalities so that you get ~ < ϵ
Assume n >= N
Choose N = ~
Work backwards until you get to fn(x)
(you can use x ∈ [a, b] in the inequalities)

Question 102

How do you find the radius of convergence if you already have the series function?

Answer 102

R is where
|Fn+1(x) / Fn(x)| < 1
The center is where the expression containing x is equal to zero
If you want the interval you must check the end points

Question 103

How do you find the power series of a function?

Answer 103

alter the function until it is of the form
1 / (1 - x)
if there is an x on the outside, put it in front of the normal geometric series Σx^n
if there is a 1 / (1 - x)^2 , you simply take the derivative of the original geometric series function (Σnx^n-1)
plug in - (x) into the geometric series

Question 104

Suppose f : A → R and g : B → R are differentiable on their domains and that f (A) ⊆ B. Then g ◦ f is differentiable on its domain

Answer 104

TRUE

Question 105

Let (fn) be a sequence of functions defined on [a, b] and suppose
(f’n) converges uniformly on [a, b]. Then (fn) converges uniformly on [a, b].

Answer 105

False,

Question 106

Every Taylor series converges on all of R.

Answer 106

FALSE

Question 107

A power series that converges on (R, R] is continuous on(R, R]

Answer 107

TRUE

Question 108

The sequence of functions
1/(1 + x^n)
converges uniformly on
[0, 1]

Answer 108

False, this is only if |x| < 1

Question 109

It is possible for a function to be discontinuous at countably
many points and still be integrable.

Answer 109

True, this is false for Reimann integrability, but true for Lebesgue integrability. It does not mean that every function is, only that it is possible.

Question 110

Suppose f is bounded on [a, b]. If f is integrable on [a, c] for every c ∈ (a, b), then f is integrable on [a, b].

Answer 110

TRUE

Question 111

If f is a bounded function on [a, b], then the upper sum satisfies
U(f + g, P) = U(f, P) + U(g, P) for all partitions P

Answer 111

False on many levels, first of all g is not bounded and thus it can go to infinity, but additionally the supremums across the sub intervals may be in different places for both f and g, thus
U(f+g,P) <= U(f,P) + U(g,P)

Question 112

Suppose g : [a, b] ! R is differentiable, g(a) > 0 and g(y) > g(x) for all x, y ∈ [a, b] with y > x. Then there is a c ∈ (a, b) with

(inverse derivitive = inverse average derivative)

Answer 112

True, g(y) > g(x) just gaurentees that g’(x) > 0, thus the inverse cannot be undefined, and the rest of this just fits the mean value theorem

Question 113

How to show that functions are differentiable on a range?

Answer 113

If it is piecewise and contains polynomials, you only need to check that the derivatives from both sides are equal using the definition of a derivative

Question 114

How to show that sum functions are differentiable and continuous

Answer 114

Use comparison test to other functions to show that the function must converge, thus it must converge uniformly

Let fn be differentiable functions defined on an interval A, and assume ∞
n=1 fn(x) converges uniformly to a limit

g(x) on A. If there exists a point x0 ∈ [a, b] where ∞
n=1 fn(x0) converges, then the series ∞
n=1 fn(x) converges uniformly to a
differentiable function f(x) satisfying f
(x) = g(x) on A

Question 115

Cauchy Criterion for Uniform Convergence of Series

Answer 115

A series Σfn converges uniformly on A ⊆ R if and only if for every ϵ > 0
there exists an N ∈ N such that

Question 116

Weierstrass M-Test

Answer 116

Basically term by term |fn(x)|<= M^n and ΣM^n converging means that the Σfn converges uniformly

Question 117

Generalized Mean Value Theorem

Answer 117

If f and g are continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c ∈ (a, b) where
[f(b) − f(a)]g’(c) = [g(b) − g(a)]f’(c)
Thus if g’(c) =/= 0
f’(c)/g’(c) = f(b) − f(a) / g(b) − g(a)

Question 118

Lagrange’s Remainder Theorem

Answer 118

Basically, if
f^(N+1)(c)/(N + 1)! * x^N+1
converges to zero, then the Taylor series converges across the range

This is used to prove that Taylor series converges uniformly across a range by showing that En (the function above) never gets out of control

Question 119

Darboux’s Theorem

Answer 119

Basically intermediate value theorem but for differentiability

If f is differentiable on an interval
[a, b], and if α satisfies f’(a) < α α > f’(b)), then there exists a point c ∈ (a, b) where f’(c) = α

Question 120

Term-by-term Continuity Theorem

Answer 120

Continuous Limit Theorem but for series

Let fn be continuous functions defined on a set A ⊆ R, and assume Σfn converges uniformly on A to a function f. Then, f is continuous on A.