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1 Functions August 18 2013

Functions

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Functions. August 18 2013. 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. Image of 4.3. “a” “bb“ “cccc” “dd” “e”. - PowerPoint PPT Presentation

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Page 1: Functions

1

FunctionsAugust 18 2013

Page 2: Functions

2

Definition of a function

A function takes an element from a set and maps it to a UNIQUE element in another set

Page 3: Functions

3

Function terminology

R Zf

4.3 4

Domain Co-domain

Pre-image of 4 Image of 4.3

f maps R to Z

f(4.3)

Page 4: Functions

4

More functions

1

2

3

4

5

“a”

“bb“

“cccc”

“dd”

“e”

A string length function

A

B

C

D

F

Alice

Bob

Chris

Dave

Emma

A class grade function

Domain Co-domainA pre-image

of 1

The imageof A

Page 5: Functions

5

Even more functions

1

2

3

4

5

“a”

“bb“

“cccc”

“dd”

“e”

Not a valid function!Also not a valid function!

1

2

3

4

5

a

e

i

o

u

Some function…

Range

Page 6: Functions

6

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

Page 7: Functions

7

One-to-one functions

1

2

3

4

5

a

e

i

o

A one-to-one function

1

2

3

4

5

a

e

i

o

A function that is not one-to-one

A function is one-to-one if each element in the co-domain has a unique pre-image

Page 8: Functions

8

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

1

2

3

4

5

a

e

i

o

A one-to-one function

Page 9: Functions

9

Onto functions

1

2

3

4

5

a

e

i

o

A function that is not onto

A function is onto if each element in the co-domain is an image of some pre-image

1

2

3

4

a

e

i

o

u

An onto function

Page 10: Functions

10

1

2

3

4

a

e

i

o

u

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

Page 11: Functions

11

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

1

2

3

4

a

b

c

1

2

3

a

b

c

d

1

2

3

4

a

b

c

d

1

2

3

4

a

b

c

d

1

2

3

4

a

b

c

1-to-1, not onto

Onto, not 1-to-1

Both 1-to-1 and onto Not a valid function

Neither 1-to-1 nor onto

Page 12: Functions

13

Bijections

Consider a function that isboth one-to-one and onto:

Such a function is a one-to-one correspondence, or a bijection

1

2

3

4

a

b

c

d

Page 13: Functions

14

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

Page 14: Functions

15

Inverse functions

R Rf

4.3 8.6

Let f(x) = 2*x

f-1

f(4.3)

f-1(8.6)

Then f-1(x) = x/2

Page 15: Functions

16

More on inverse functions

Can we define the inverse of the following functions?

An inverse function can ONLY be done defined on a bijection

1

2

3

4

a

b

c

1

2

3

a

b

c

d

What is f-1(2)?Not onto!

What is f-1(2)?Not 1-to-1!

Page 16: Functions

17

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

Page 17: Functions

18

Compositions of functions

g f

f ○ g

g(a) f(a)

(f ○ g)(a)

g(a)f(g(a))a

A B C

Page 18: Functions

19

Compositions of functions

g f

f ○ g

g(1) f(5)

(f ○ g)(1)

g(1)=5

f(g(1))=131

R R R

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

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

Page 19: Functions

20

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!

Page 20: Functions

21

Graphs of functions

Let f(x)=2x+1

Plot (x, f(x))

This is a plotof f(x)

x=1

f(x)=3

x=2

f(x)=5

Page 21: Functions

22

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)

Page 22: Functions

23

Rosen, question 8 (§1.8)

Find these values

1.11.1-0.1-0.12.99-2.99½+½½ + ½ + ½

12-103-2½+1 = 3/2 = 10 + 1 + ½ = 3/2 = 2

Page 23: Functions

24

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

Page 24: Functions

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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

Page 25: Functions

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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

Page 26: Functions

27

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

)()()( )

)()()( )

TfSfTSfb

TfSfTSfa

Page 27: Functions

28

Proving function problems

Rosen, question 32 (a): f(SUT) = f(S) U f(T)Will show that each side is a subset of the otherTwo cases!Show that f(SUT) f(S) U f(T)

Let b f(SUT). Thus, b=f(a) for some aS U T Either aS, in which case bf(S) Or aT, in which case bf(T) Thus, bf(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

Page 28: Functions

29

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 TThus, b f(S) and b f(T)

Therefore, b f(S) ∩ f(T)

Page 29: Functions

30

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

Page 30: Functions

31

Proving function problems

Rosen, question 62, §1.8Thus, we want to show, for all zZ and xX

The second equality is similar

xxgffg

zzfggf

)(

)(11

11

z

zff

zfggf

zfggf

zfggfzfggf

)(

)(

)(

)()(

1

11

11

1111

Page 31: Functions

32

Quick survey

I felt I understood the material in this slide set…

a) Very wellb) With some review, I’ll be goodc) Not reallyd) Not at all

Page 32: Functions

33

Quick survey

The pace of the lecture for this slide set was…

a) Fastb) About rightc) A little slowd) Too slow

Page 33: Functions

34

Quick survey

How interesting was the material in this slide set? Be honest!

a) Wow! That was SOOOOOO cool!b) Somewhat interestingc) Rather bortingd) Zzzzzzzzzzz