functions
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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FunctionsFunctions
CS/APMA 202CS/APMA 202
Rosen section 1.8Rosen section 1.8
Aaron BloomfieldAaron Bloomfield
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Definition of a functionDefinition of a function
A function takes an element from a set A function takes an element from a set and maps it to a UNIQUE element in and maps it to a UNIQUE element in another setanother set
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Function terminologyFunction 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)
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More functionsMore 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
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Even more functionsEven 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
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Function arithmeticFunction arithmetic
Let fLet f11(x) = 2x(x) = 2x
Let fLet f22(x) = x(x) = x22
ff11+f+f22 = (f = (f11+f+f22)(x) = f)(x) = f11(x)+f(x)+f22(x) = 2x+x(x) = 2x+x22
ff11*f*f22 = (f = (f11*f*f22)(x) = f)(x) = f11(x)*f(x)*f22(x) = 2x*x(x) = 2x*x22 = 2x = 2x33
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One-to-one functionsOne-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 A function is one-to-one if each element in the co-domain has a unique pre-imagethe co-domain has a unique pre-image
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More on one-to-oneMore on one-to-one
Injective is synonymous with one-to-oneInjective is synonymous with one-to-one ““A function is injective”A function is injective”
A function is an injection if it is one-to-oneA function is an injection if it is one-to-one
Note that there can Note that there can be un-used elements be un-used elements in the co-domainin the co-domain
1
2
3
4
5
a
e
i
o
A one-to-one function
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Onto functionsOnto 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 A function is onto if each element in the co-domain is an image of some pre-imageco-domain is an image of some pre-image
1
2
3
4
a
e
i
o
u
An onto function
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1
2
3
4
a
e
i
o
u
An onto function
More on ontoMore on onto
Surjective is synonymous with ontoSurjective is synonymous with onto ““A function is surjective”A function is surjective”
A function is an surjection if it is ontoA function is an surjection if it is onto
Note that there can Note that there can be multiply used be multiply used elements in the elements in the co-domainco-domain
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Onto vs. one-to-oneOnto vs. one-to-oneAre the following functions onto, one-to-Are the following functions onto, one-to-one, both, or neither?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
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BijectionsBijections
Consider a function that isConsider a function that isboth one-to-one and onto:both one-to-one and onto:
Such a function is a one-to-one Such a function is a one-to-one correspondence, or a bijectioncorrespondence, or a bijection
1
2
3
4
a
b
c
d
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Identity functionsIdentity functions
A function such that the image and the A function such that the image and the pre-image are ALWAYS equalpre-image are ALWAYS equal
f(x) = 1*xf(x) = 1*x
f(x) = x + 0f(x) = x + 0
The domain and the co-domain must be The domain and the co-domain must be the same setthe same set
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Inverse functionsInverse 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
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More on inverse functionsMore on inverse functions
Can we define the inverse of the following Can we define the inverse of the following functions?functions?
An inverse function can ONLY be done defined An inverse function can ONLY be done defined on a bijectionon 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!
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Compositions of functionsCompositions of functions
Let (f ○ g)(x) = f(g(x))Let (f ○ g)(x) = f(g(x))
Let f(x) = 2x+3Let f(x) = 2x+3 Let g(x) = 3x+2Let g(x) = 3x+2
g(1) = 5, f(5) = 13g(1) = 5, f(5) = 13
Thus, (f ○ g)(1) = f(g(1)) = 13Thus, (f ○ g)(1) = f(g(1)) = 13
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Compositions of functionsCompositions of functions
g f
f ○ g
g(a) f(a)
(f ○ g)(a)
g(a)f(g(a))a
A B C
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Compositions of functionsCompositions 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
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Compositions of functionsCompositions of functions
Does f(g(x)) = g(f(x))?Does f(g(x)) = g(f(x))?
Let f(x) = 2x+3Let f(x) = 2x+3 Let g(x) = 3x+2Let g(x) = 3x+2
f(g(x)) = 2(3x+2)+3 = 6x+7f(g(x)) = 2(3x+2)+3 = 6x+7
g(f(x)) = 3(2x+3)+2 = 6x+11g(f(x)) = 3(2x+3)+2 = 6x+11
Function composition is not commutative!Function composition is not commutative!
Not equal!
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Graphs of functionsGraphs 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
2222
Useful functionsUseful functions
Floor: Floor: xx means take the greatest integer means take the greatest integer less than or equal to the numberless than or equal to the number
Ceiling: Ceiling: xx means take the lowest integer means take the lowest integer greater than or equal to the numbergreater than or equal to the number
round(x) = floor(x+0.5)round(x) = floor(x+0.5)
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Rosen, question 8 (§1.8)Rosen, question 8 (§1.8)
Find these valuesFind these values
1.11.11.11.1-0.1-0.1-0.1-0.12.992.99-2.99-2.99½+½+½½½½ + + ½½ + ½ + ½
1122-1-10033-2-2½+1½+1 = = 3/23/2 = 1 = 10 + 1 + ½0 + 1 + ½ = = 3/23/2 = 2 = 2
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Ceiling and floor propertiesCeiling and floor properties
Let n be an integerLet n be an integer(1a)(1a) xx = n if and only if n ≤ x < n+1 = n if and only if n ≤ x < n+1(1b)(1b) xx = n if and only if n-1 < x ≤ n = n if and only if n-1 < x ≤ n(1c)(1c) xx = n if and only if x-1 < n ≤ x = n if and only if x-1 < n ≤ x(1d)(1d) xx = n if and only if x ≤ n < x+1 = n if and only if x ≤ n < x+1(2)(2) x-1 < x-1 < xx ≤ x ≤ = ≤ x ≤ = xx < x+1 < x+1 (3a)(3a) -x-x = - = - xx(3b)(3b) -x-x = - = - xx(4a)(4a) x+nx+n = = xx+n +n (4b)(4b) x+nx+n = = xx+n+n
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Ceiling property proofCeiling property proof
Prove rule 4a: Prove rule 4a: x+nx+n = = xx+n+n Where n is an integerWhere n is an integer Will use rule 1a: Will use rule 1a: xx = n if and only if n ≤ x < = n if and only if n ≤ x <
n+1n+1
Direct proof!Direct proof! Let m = Let m = xx Thus, m ≤ x < m+1 (by rule 1a)Thus, m ≤ x < m+1 (by rule 1a) Add n to both sides: m+n ≤ x+n < m+n+1Add n to both sides: m+n ≤ x+n < m+n+1 By rule 4a, m+n = By rule 4a, m+n = x+nx+n Since m = Since m = xx, m+n also equals , m+n also equals xx+n+n Thus, Thus, xx+n = m+n = +n = m+n = x+nx+n
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FactorialFactorial
Factorial is denoted by n!Factorial is denoted by n!
n! = n * (n-1) * (n-2) * … * 2 * 1n! = n * (n-1) * (n-2) * … * 2 * 1
Thus, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720Thus, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
Note that 0! is defined to equal 1Note that 0! is defined to equal 1
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Proving function problemsProving function problems
Rosen, question 32, §1.8Rosen, question 32, §1.8
Let Let ff be a function from A to B, and let S be a function from A to B, and let S and T be subsets of A. Show thatand T be subsets of A. Show that
)()()( )
)()()( )
TfSfTSfb
TfSfTSfa
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Proving function problemsProving function problems
Rosen, question 32 (a): f(SUT) = f(S) U f(T)Rosen, question 32 (a): f(SUT) = f(S) U f(T)Will show that each side is a subset of the otherWill show that each side is a subset of the otherTwo cases!Two cases!Show that f(SUT) Show that f(SUT) f(S) U f(T) f(S) U f(T)
Let b Let b f(SUT). Thus, b=f(a) for some a f(SUT). Thus, b=f(a) for some aS U TS U T Either aEither aS, in which case bS, in which case bf(S)f(S) Or aOr aT, in which case bT, in which case bf(T)f(T) Thus, bThus, bf(S) U f(T)f(S) U f(T)
Show that f(S) U f(T) Show that f(S) U f(T) f(S U T) f(S U T) Let b Let b f(S) U f(T) f(S) U f(T) Either b Either b f(S) or b f(S) or b f(T) (or both!) f(T) (or both!) Thus, b = f(a) for some a Thus, b = f(a) for some a S or some a S or some a T T In either case, b = f(a) for some a In either case, b = f(a) for some a S U T S U T
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Proving function problemsProving function problems
Rosen, question 32 (b): f(S∩T) Rosen, question 32 (b): f(S∩T) f(S) ∩ f(T) f(S) ∩ f(T)
Let b Let b f(S∩T). Then b = f(a) for some a f(S∩T). Then b = f(a) for some a S∩T S∩T
This implies that a This implies that a S and a S and a T T
Thus, b Thus, b f(S) and b f(S) and b f(T) f(T)
Therefore, b Therefore, b f(S) ∩ f(T) f(S) ∩ f(T)
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Proving function problemsProving function problems
Rosen, question 62, §1.8Rosen, question 62, §1.8
Let Let ff be an invertible function from Y to Z be an invertible function from Y to Z
Let Let gg be an invertible function from X to Y be an invertible function from X to Y
Show that the inverse of f○g is:Show that the inverse of f○g is: (f○g)(f○g)-1-1 = g = g-1 -1 ○ f○ f-1-1
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Proving function problemsProving function problems
Rosen, question 62, §1.8Rosen, question 62, §1.8Thus, we want to show, Thus, we want to show, for all for all zzZ and Z and xxXX
The second equality is similarThe second equality is similar
xxgffg
zzfggf
)(
)(11
11
z
zff
zfggf
zfggf
zfggfzfggf
)(
)(
)(
)()(
1
11
11
1111
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Quick surveyQuick survey
I felt I understood the material in this I felt I understood the material in this slide set…slide set…
a)a) Very wellVery well
b)b) With some review, I’ll be goodWith some review, I’ll be good
c)c) Not reallyNot really
d)d) Not at allNot at all
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Quick surveyQuick survey
The pace of the lecture for this The pace of the lecture for this slide set was…slide set was…
a)a) FastFast
b)b) About rightAbout right
c)c) A little slowA little slow
d)d) Too slowToo slow
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Quick surveyQuick survey
How interesting was the material in How interesting was the material in this slide set? Be honest!this slide set? Be honest!
a)a) Wow! That was SOOOOOO cool!Wow! That was SOOOOOO cool!
b)b) Somewhat interestingSomewhat interesting
c)c) Rather bortingRather borting
d)d) ZzzzzzzzzzzZzzzzzzzzzz