Test 3

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 11

Refinement

Answer 11

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 11

higher integrals

Answer 11

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 12

lower integrals

Answer 12

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 12

Riemann Integrability

Answer 12

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 13

Interior Extremum Theorem PROOF

Answer 13

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 <c<yn for all n ∈ N. The fact that f(c) is a maximum implies that f(yn) − f(c) ≤ 0 for all n, and thus

f’(c) = lim f(yn) - f(c) / yn - c ≤ 0

by the Order Limit Theorem (Theorem 2.3.4). In a similar way,

f(xn) − f(c) / xn − c ≥ 0

for each xn because both numerator and denominator are negative. This implies that

f’(c) = lim f(xn) − f(c) / xn − c ≥ 0,
and therefore f’(c) = 0, as desired.

Question 14

Continuous Limit Theorem PROOF

Answer 14

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