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Functions. CS/APMA 202 Rosen section 1.8 Aaron Bloomfield. Definition of a function. A function takes an element from a set and maps it to a UNIQUE element in another set. Function terminology. f maps R to Z. f. R. Z. Co-domain. Domain. f(4.3). 4. 4.3. Pre-image of 4.

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functions

Functions

CS/APMA 202

Rosen section 1.8

Aaron Bloomfield

definition of a function
Definition of a function
  • A function takes an element from a set and maps it to a UNIQUE element in another set
function terminology
Function terminology

f maps R to Z

f

R

Z

Co-domain

Domain

f(4.3)

4

4.3

Pre-image of 4

Image of 4.3

more functions

“a”

“bb“

“cccc”

“dd”

“e”

1

2

3

4

5

Alice

Bob

Chris

Dave

Emma

A

B

C

D

F

A string length function

A class grade function

More functions

The image

of A

A pre-image

of 1

Domain

Co-domain

even more functions

a

e

i

o

u

1

2

3

4

5

“a”

“bb“

“cccc”

“dd”

“e”

1

2

3

4

5

Some function…

Even more functions

Range

Not a valid function!

Also not a valid function!

function arithmetic
Function arithmetic
  • Let f1(x) = 2x
  • Let f2(x) = x2
  • f1+f2 = (f1+f2)(x) = f1(x)+f2(x) = 2x+x2
  • f1*f2 = (f1*f2)(x) = f1(x)*f2(x) = 2x*x2 = 2x3
one to one functions

a

e

i

o

1

2

3

4

5

a

e

i

o

1

2

3

4

5

A one-to-one function

A function that is

not one-to-one

One-to-one functions
  • A function is one-to-one if each element in the co-domain has a unique pre-image
more on one to one

a

e

i

o

1

2

3

4

5

A one-to-one function

More on one-to-one
  • Injective is synonymous with one-to-one
    • “A function is injective”
  • A function is an injection if it is one-to-one
  • Note that there can be un-used elements in the co-domain
onto functions

a

e

i

o

u

1

2

3

4

a

e

i

o

1

2

3

4

5

An onto function

A function that is not onto

Onto functions
  • A function is onto if each element in the co-domain is an image of some pre-image
more on onto

a

e

i

o

u

1

2

3

4

An onto function

More on onto
  • Surjective is synonymous with onto
    • “A function is surjective”
  • A function is an surjection if it is onto
  • Note that there can be multiply used elements in the co-domain
onto vs one to one

a

b

c

1

2

3

4

a

b

c

d

1

2

3

a

b

c

d

a

b

c

d

1

2

3

4

1

2

3

4

a

b

c

1

2

3

4

Onto vs. one-to-one
  • Are the following functions onto, one-to-one, both, or neither?

1-to-1, not onto

Both 1-to-1 and onto

Not a valid function

Onto, not 1-to-1

Neither 1-to-1 nor onto

bijections

a

b

c

d

1

2

3

4

Bijections
  • Consider a function that isboth one-to-one and onto:
  • Such a function is a one-to-one correspondence, or a bijection
identity functions
Identity functions
  • A function such that the image and the pre-image are ALWAYS equal
  • f(x) = 1*x
  • f(x) = x + 0
  • The domain and the co-domain must be the same set
inverse functions
Inverse functions

Let f(x) = 2*x

f

R

R

f-1

f(4.3)

8.6

4.3

f-1(8.6)

Then f-1(x) = x/2

more on inverse functions

a

b

c

1

2

3

4

a

b

c

d

1

2

3

More on inverse functions
  • Can we define the inverse of the following functions?
  • An inverse function can ONLY be done defined on a bijection

What is f-1(2)?

Not onto!

What is f-1(2)?

Not 1-to-1!

compositions of functions
Compositions of functions
  • Let (f ○ g)(x) = f(g(x))
  • Let f(x) = 2x+3 Let g(x) = 3x+2
  • g(1) = 5, f(5) = 13
  • Thus, (f ○ g)(1) = f(g(1)) = 13
compositions of functions1
Compositions of functions

f ○ g

A

B

C

g

f

g(a)

f(a)

a

f(g(a))

g(a)

(f ○ g)(a)

compositions of functions2
Compositions of functions

Let f(x) = 2x+3 Let g(x) = 3x+2

f ○ g

R

R

R

g

f

g(1)

f(5)

f(g(1))=13

1

g(1)=5

(f ○ g)(1)

f(g(x)) = 2(3x+2)+3 = 6x+7

compositions of functions3
Compositions of functions

Does f(g(x)) = g(f(x))?

Let f(x) = 2x+3 Let g(x) = 3x+2

f(g(x)) = 2(3x+2)+3 = 6x+7

g(f(x)) = 3(2x+3)+2 = 6x+11

Function composition is not commutative!

Not equal!

graphs of functions
Graphs of functions

f(x)=3

x=1

Let f(x)=2x+1

Plot (x, f(x))

This is a plot

of f(x)

f(x)=5

x=2

useful functions
Useful functions
  • Floor: x means take the greatest integer less than or equal to the number
  • Ceiling: x means take the lowest integer greater than or equal to the number
  • round(x) = floor(x+0.5)
rosen question 8 1 8
Rosen, question 8 (§1.8)
  • Find these values
  • 1.1
  • 1.1
  • -0.1
  • -0.1
  • 2.99
  • -2.99
  • ½+½
  • ½ + ½ + ½

1

2

-1

0

3

-2

½+1 = 3/2 = 1

0 + 1 + ½ = 3/2 = 2

ceiling and floor properties
Ceiling and floor properties

Let n be an integer

(1a) x = n if and only if n ≤ x < n+1

(1b) x = n if and only if n-1 < x ≤ n

(1c) x = n if and only if x-1 < n ≤ x

(1d) x = n if and only if x ≤ n < x+1

(2) x-1 < x ≤ x ≤ = x < x+1

(3a) -x = - x

(3b) -x = - x

(4a) x+n = x+n

(4b) x+n = x+n

ceiling property proof
Ceiling property proof
  • Prove rule 4a: x+n = x+n
    • Where n is an integer
    • Will use rule 1a: x = n if and only if n ≤ x < n+1
  • Direct proof!
    • Let m = x
    • Thus, m ≤ x < m+1 (by rule 1a)
    • Add n to both sides: m+n ≤ x+n < m+n+1
    • By rule 4a, m+n = x+n
    • Since m = x, m+n also equals x+n
    • Thus, x+n = m+n = x+n
factorial
Factorial
  • Factorial is denoted by n!
  • n! = n * (n-1) * (n-2) * … * 2 * 1
  • Thus, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
  • Note that 0! is defined to equal 1
proving function problems
Proving function problems
  • Rosen, question 32, §1.8
  • Let f be a function from A to B, and let S and T be subsets of A. Show that
proving function problems1
Proving function problems
  • Rosen, question 32 (a): f(SUT) = f(S) U f(T)
  • Will show that each side is a subset of the other
  • Two cases!
  • Show that f(SUT)  f(S) U f(T)
    • Let b  f(SUT). Thus, b=f(a) for some aS U T
    • Either aS, in which case bf(S)
    • Or aT, in which case bf(T)
    • Thus, bf(S) U f(T)
  • Show that f(S) U f(T)  f(S U T)
    • Let b  f(S) U f(T)
    • Either b  f(S) or b  f(T) (or both!)
    • Thus, b = f(a) for some a  S or some a  T
    • In either case, b = f(a) for some a  S U T
proving function problems2
Proving function problems
  • Rosen, question 32 (b): f(S∩T)  f(S) ∩ f(T)
  • Let b  f(S∩T). Then b = f(a) for some a  S∩T
  • This implies that a  S and a  T
  • Thus, b  f(S) and b  f(T)
  • Therefore, b  f(S) ∩ f(T)
proving function problems3
Proving function problems
  • Rosen, question 62, §1.8
  • Let f be an invertible function from Y to Z
  • Let g be an invertible function from X to Y
  • Show that the inverse of f○g is:
    • (f○g)-1 = g-1 ○ f-1
proving function problems4
Proving function problems
  • Rosen, question 62, §1.8
  • Thus, we want to show, for all zZ and xX
  • The second equality is similar
quick survey
Quick survey
  • I felt I understood the material in this slide set…
  • Very well
  • With some review, I’ll be good
  • Not really
  • Not at all
quick survey1
Quick survey
  • The pace of the lecture for this slide set was…
  • Fast
  • About right
  • A little slow
  • Too slow
quick survey2
Quick survey
  • How interesting was the material in this slide set? Be honest!
  • Wow! That was SOOOOOO cool!
  • Somewhat interesting
  • Rather borting
  • Zzzzzzzzzzz
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