Mathematical-Reasoning · Unit 21 · Video 3 · Interactive Practice

Derivatives, Averages and Sequences as Functions

IKey Formulas

Formula	Name	Domain and codomain
$D(f) = f'$	Derivative function $D : F \to T$	$F$ = all differentiable real functions, $T$ = all real functions; inputs and outputs are functions
$\text{avg}(A) = \dfrac{a_1 + a_2 + \cdots + a_n}{n}$	Average function $\text{avg} : \mathcal{F}(\mathbb{R}) \to \mathbb{R}$	$A = \{a_1, a_2, \dots, a_n\}$ a nonempty finite set of reals with $n$ distinct elements
$\langle a_n \rangle$ , $a(n) = a_n$	Sequence	A function whose domain is $\mathbb{N}$ or an infinite subset of $\mathbb{N} \cup \{0\}$

Key Insight: A process becomes a function once its domain and codomain are named and each input has exactly one output in the codomain. The domain leaves out the inputs on which the rule fails: non-differentiable functions for $D$ , the empty set and infinite sets for $\text{avg}$ .

IIVisualization 1 — Functions as Inputs and Outputs

The derivative function $D : F \to T$ takes a whole function as its input and returns another whole function.

IIIVisualization 2 — Which Sets Can Be Averaged?

The average function accepts a set only when it is nonempty and finite, and counts each element once.

IVVisualization 3 — A Sequence Is a Function of Its Index

A sequence is a function on $\mathbb{N}$ or an infinite subset of $\mathbb{N} \cup \{0\}$ , so its domain decides which terms exist.

Sequence Input n 6

VQuiz Questions

Problem 1 · Averaging a Set Written with a Repeat

Given: $B = \{2, -4, 2, 6, 5\}$ — find $\text{avg}(B)$ .

✅ Correct! As a set,

B = \{-4, 2, 5, 6\}

has

4

elements with sum

9

, so

\text{avg}(B) = \tfrac{9}{4}

❌ Not quite. Writing

2

twice does not make two elements: you added

2

twice and divided by

5

, but

B

has only

4

elements.

❌ Close. The sum

9

is right, but

n

is the number of elements of the set, which is

4

, not the

5

entries written.

❌ Close. Dividing by

4

is right, but the sum

11

still counts the repeated

2

twice; the sum of the elements is

9

Show solution

A set contains each element once, however it is written, so $B = \{2, -4, 6, 5\}$ , with $n = 4$ elements.

\text{avg}(B) = \frac{2 + (-4) + 6 + 5}{4} = \frac{9}{4} = 2.25

Since $B$ is nonempty and finite, $B \in \mathcal{F}(\mathbb{R})$ , and the average function gives it exactly one output.

Problem 2 · Which Process Is a Function?

Given: four processes, each with a proposed domain and codomain. Here $P$ is the set of all polynomial functions with real coefficients and $\mathcal{F}(\mathbb{R})$ is the set of nonempty finite subsets of $\mathbb{R}$ . Which one is a function from the stated domain to the stated codomain?

✅ Correct! Every polynomial is differentiable and its derivative is again a polynomial, so each

p \in P

has exactly one output

p' \in P

❌ Not quite. The empty set is a finite subset of

\mathbb{R}

, and it has no average:

n = 0

would mean dividing by zero. That is why

\mathcal{F}(\mathbb{R})

contains only nonempty sets.

❌ Not quite. An infinite set such as

\mathbb{N}

is a subset of

\mathbb{R}

, and its number of elements is not an element of

\mathbb{N} \cup \{0\}

. Counting is a function only on finite sets.

❌ Not quite.

\mathbb{N}

is a nonempty subset of

\mathbb{R}

with no largest element, so

\max(\mathbb{N})

does not exist. On

\mathcal{F}(\mathbb{R})

, where every set is finite,

\max

is a function.

Show solution

A process is a function only if every element of the domain has exactly one output in the codomain. Test each with one input:

(a) The input $\emptyset$ has no average, since $\tfrac{\text{sum}}{0}$ is undefined.
(b) The input $\mathbb{N}$ does not have a number of elements in $\mathbb{N} \cup \{0\}$ .
(c) The input $\mathbb{N}$ has no largest element (the input $\mathbb{R}$ fails too).
(d) If $p(x) = c_k x^k + \cdots + c_1 x + c_0$ , then $p'(x) = k c_k x^{k-1} + \cdots + c_1$ , which is again a polynomial (a constant polynomial gives the zero polynomial). So $D_P : P \to P$ is a function.

Like $D : F \to T$ , the function $D_P$ has functions as inputs and as outputs; it simply has a smaller domain and codomain.

Problem 3 · Starting the Index at Zero

Given: the odd natural numbers $1$ , $3$ , $5$ , $7$ , … are listed as a sequence $\langle b_n \rangle$ with domain $\mathbb{N} \cup \{0\}$ , so the first term is $b_0 = 1$ . Find a formula for $b_n$ and the value of $b_{10}$ .

Which formula gives $b_n$ ?

What is $b_{10}$ ?

✅ Correct!

b_n = 2n + 1

gives

b_0 = 1

b_1 = 3

b_2 = 5

, …, and

b_{10} = 21

is the eleventh term of the list.

❌ Not quite.

2n - 1

is the formula when the index starts at

n = 1

. With domain

\mathbb{N} \cup \{0\}

it gives

b_0 = -1

, which is not the first term

1

❌ Not quite.

2n

gives

b_0 = 0

and

b_1 = 2

: those are the even numbers.

❌ Not quite.

2n + 3

gives

b_0 = 3

, so the list would skip the first term

1

❌ Not quite. Check your formula at

n = 0

1

2

against the terms

1

3

5

❌ Close.

19

is the tenth odd number, but with the index starting at

0

the tenth term is

b_9

. The index

10

gives

b_{10} = 2 \cdot 10 + 1

❌ Not quite. Every term of this sequence is odd; substitute

n = 10

into

b_n = 2n + 1

❌ Not quite.

23 = 2 \cdot 11 + 1

b_{11}

; the index here is

10

❌ Not quite. Substitute

n = 10

into the formula for

b_n

Show solution

Consecutive odd numbers differ by $2$ , so $b_n = 2n + c$ for some constant $c$ . The first term is $b_0 = 2 \cdot 0 + c = c$ , and it must equal $1$ , so $c = 1$ :

b_n = 2n + 1 \quad (n \in \mathbb{N} \cup \{0\})

Check: $b_1 = 3$ , $b_2 = 5$ , $b_3 = 7$ . Then

b_{10} = 2 \cdot 10 + 1 = 21.

Because the index starts at $0$ , the inputs $0, 1, \dots, 10$ are eleven inputs, so $b_{10}$ is the eleventh term. The same list with domain $\mathbb{N}$ would need the formula $2n - 1$ .

Problem 4 · Ranges and Equal Sequences

Given: the sequences $a, c, d : \mathbb{N} \to \mathbb{R}$ defined by $a_n = (-1)^n \cdot n$ , $c_n = \dfrac{n}{n + 1}$ and $d_n = 1 - \dfrac{1}{n + 1}$ . Find the range of $\langle a_n \rangle$ and decide whether $\langle c_n \rangle = \langle d_n \rangle$ .

What is the range of $\langle a_n \rangle$ ?

Is $\langle c_n \rangle = \langle d_n \rangle$ ?

✅ Correct! The terms of

\langle a_n \rangle

are

-1

2

-3

4

, …, and

1 - \tfrac{1}{n+1} = \tfrac{n}{n+1}

for every

n

, so the two sequences are the same function.

❌ Not quite. Not every nonzero integer is a term:

1

would need

(-1)^n \cdot n = 1

, which forces

n = 1

, but

a_1 = -1

. Likewise

-2

never occurs.

❌ Not quite.

\{-1, 1\}

is the range of

\langle (-1)^n \rangle

. Here the factor

n

makes

|a_n| = n

, so the range is infinite.

❌ Check the signs. Odd

n

give

(-1)^n = -1

, so the odd values are negative:

a_1 = -1

a_3 = -3

; even

n

give positive values.

❌ Not quite. List the first few terms

a_1

a_2

a_3

a_4

and look at their signs.

❌ Not quite. Two sequences are equal when they have the same domain and the same output at every input; different formulas can give the same outputs. Simplify

1 - \tfrac{1}{n+1}

❌ Not quite. They are not merely close:

1 - \tfrac{1}{n+1} = \tfrac{n+1-1}{n+1} = \tfrac{n}{n+1}

exactly, for every

n

❌ Recompute $d_1$ .

d_1 = 1 - \tfrac{1}{1 + 1} = \tfrac{1}{2}

, which equals

c_1

❌ Not quite. Compare

c_n

and

d_n

at a general

n

, not only at a few values.

Show solution

Range. For odd $n$ , $(-1)^n = -1$ , so $a_n = -n$ ; for even $n$ , $(-1)^n = 1$ , so $a_n = n$ . The terms are $-1$ , $2$ , $-3$ , $4$ , $-5$ , $6$ , …, so

\text{range} = \{-1, -3, -5, \dots\} \cup \{2, 4, 6, \dots\}.

The values $1$ , $-2$ , $3$ , … never occur, so the range is not all of $\mathbb{Z} - \{0\}$ .

Equality. Both sequences are functions from $\mathbb{N}$ to $\mathbb{R}$ . For every $n \in \mathbb{N}$ ,

d_n = 1 - \frac{1}{n + 1} = \frac{(n + 1) - 1}{n + 1} = \frac{n}{n + 1} = c_n.

Same domain, same codomain and the same output at every input, so $\langle c_n \rangle = \langle d_n \rangle$ : the first terms are $\tfrac{1}{2}$ , $\tfrac{2}{3}$ , $\tfrac{3}{4}$ , … for both.

Solved: 0 / 4